Problem 2.1-1 (2-1 in text): Buried Tubes Figure P2.1-1 illustrates two tubes that are buried in the ground behind your house that transfer water to and from a wood burner. The left hand tube carries hot water from the burner back to your house at Tw,h = 135°F while the right hand tube carries cold water from your house back to the burner at Tw,c = 70°F. Both tubes have outer diameter Do = 0.75 inch and thickness th = 0.065 inch. The conductivity of the tubing material is kt = 0.22 W/m-K. The heat transfer coefficient between the water and the tube internal surface (in both tubes) is hw = 250 W/m2-K. The center to center distance between the tubes is w = 1.25 inch and the length of the tubes is L = 20 ft (into the page). The tubes are buried in soil that has conductivity ks = 0.30 W/m-K. ks = 0.30 W/m-K th = 0.065 inch
kt = 0.22 W/m-K
Tw,c = 70°F 2 hw = 250 W/m -K
Tw,h = 135°F 2 hw = 250 W/m -K w = 1.25 inch Do = 0.75 inch
Figure P2.1-1: Tubes buried in soil.
a.) Estimate the heat transfer from the hot water to the cold water due to their proximity to one another. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in D_o=0.75 [inch]*convert(inch,m) "outer diameter of tube" th=0.065 [inch]*convert(inch,m) "thickness of tube" T_hw=converttemp(F,K,135) "hot water temperature" T_cw=converttemp(F,K,70) "cold water temperature" L=20 [ft]*convert(ft,m) "length of tubes" w=1.25 [inch]*convert(inch,m) "center to center distance" k_t=0.22 [W/m-K] "conductivity of teflon" k_s=0.30 [W/m-K] "conductivity of sand" h_w=250 [W/m^2-K] "heat transfer coefficient between the water and tube inner surfaces"
The heat transfer is resisted by convection within the tube, conduction through the tube, and conduction in the soil. The convection resistance is calculated according to: Rconv =
1 hw L π ( Do − 2 th )
The conduction resistance associated with the tube is calculated according to:
(1)
Rcond
⎛ Do ⎞ ln ⎜ ⎟ Do − 2 th ⎠ ⎝ = 2 kt L π
R_conv=1/(h_w*L*pi*(D_o-2*th)) R_cond=ln(D_o/(D_o-2*th))/(2*pi*k_t*L)
(2) "convection resistance" "conduction resistance"
The shape factor for parallel, buried tubes (SF) is obtained using EES’ internal library and used to compute the resistance due to conduction through the soil:
Rtubetotube =
1 SF ks
SF=SF_4(D_o,D_o,w,L) R_tubetotube=1/(SF*k_s)
(3) "shape factor" "tube to tube resistance"
The heat transfer rate is therefore: q =
Tw,h − Tw,c 2 Rconv + 2 Rcond + Rtubetotube
q_dot=(T_hw-T_cw)/(2*R_conv+2*R_cond+R_tubetotube)
(4)
"heat transfer rate"
which leads to q = 137.3 W. b.) To do part (a) you should have needed to determine a shape factor; calculate an approximate value of the shape factor and compare it to the accepted value. The shape factor can be thought of as the ratio of the effective area for conduction, Aeff, to the effective length required for conduction, Leff. The effective area for conduction can be estimated according to:
Aeff ≈ w L
(5)
while the length for conduction is approximately:
Leff ≈ w − Do
(6)
and the approximate value of the shape factor should be:
SFapp ≈ A_eff=L*w
Aeff
(7)
Leff "approximate area"
L_eff=(w-D_o) SF_app=A_eff/L_eff
"approximate length" "approximate shape factor"
The exact value of the shape factor returned by the function is SF = 17.4 m while the approximate value is SFapp = 15.2 m; these are sufficiently close for a sanity check. c.) Plot the rate of heat transfer from the hot water to the cold water as a function of the center to center distance between the tubes. The heat transfer rate is shown in Figure 2 as a function of the center to center distance between the tubes.
Figure 2: Heat transfer rate as a function of the center to center distance between the tubes
Problem 2.1-2 Currently, the low-pressure steam exhausted from a steam turbine at the power plant is condensed by heat transfer to cooling water. An alternative that has been proposed is to transport the steam via an underground pipe to a large building complex and use the steam for space heating. You have been asked to evaluate the feasibility of this proposal. The building complex is located 0.2 miles from the power plant. The pipe is made of uninsulated PVC (thermal conductivity of 0.19 W/m-K) with an inner diameter of 8.33 in and wall thickness 0.148 in. The pipe will be buried underground at a depth of 4 ft in soil that has an estimated thermal conductivity of 0.5 W/m-K. The steam leaves the power plant at 6.5 lbm/min, 8 psia with a 95% quality. The outdoor temperature is 5°F. Condensate is returned to the power plant in a separate pipe as, approximately, saturated liquid at 8 psia. You may neglect the resistance due to convection to the air and the steam. a.) Neglecting the inevitable pressure loss, estimate the state of the steam that is provided to the building complex. The inputs are entered in EES: $UnitSytem SI K J Pa $TabStops 0.2 0.4 3.5 in "known information" L=0.2 [mile]*convert(mile,m) D_i=8.33 [in]*convert(in,m) thk=0.148 [in]*convert(in,m) k_soil=0.5 [W/m-K] k_PVC=0.19 [W/m-K] T_air=convertTemp(F,K,5 [F]) P_steam=8 [psia]*convert(psia,Pa) T_steam=T_sat(Steam,P=P_steam) W=4 [ft]*convert(ft,m) m_dot=6.5 [lb_m/min]*convert(lb_m/min,kg/s)
"pipe length" "pipe inner diameter" "pipe wall thickness" "soil thermal conductivity" "PVC thermal conductivity" "air temperature in K" "steam pressure" "steam temperature in K" "distance of pipe below ground level" "mass flow rate of the steam"
The outer diameter of the pipe is calculated: Do = Di + 2 th
(1)
where Di is the inner diameter and th is the pipe thickness. The resistance to conduction through the pipe is:
R pipe
⎛D ⎞ ln ⎜ o ⎟ ⎝ Di ⎠ = 2 π k PVC L
(2)
where kPVC is the conductivity of the PVC pipe and L is the length of the pipe. The resistance between the pipe surface and the surface of the soil is obtained using the shape factor (S) obtained using the function SF_2 in EES. Rsoil =
1 S k soil
(3)
The heat transfer from the pipe to the soil surface is: q =
(Tsteam − Tair ) R pipe + Rsoil
(4)
where Tsteam is the temperature of the steam and Tair is the temperature of the air. D_o=D_i+2*thk R_pipe=ln(D_o/D_i)/(2*pi*k_pvc*L) R_soil=1/(k_soil*S) S=SF_2(D_o,W,L) q_dot_loss=(T_steam-T_air)/(R_pipe+R_soil)
"pipe outer diameter" "resistance to conduction through pipe wall" "resistance of soil" "shape factor for buried pipe" "heat transfer rate through pipe wall"
which leads to q = 107.3 kW. An energy balance on the steam leads to:
q = m ( is ,in − is ,out )
(5)
where is,in and is,out are the inlet and outlet enthalpies of the steam, respectively. The outlet enthalpy is used to determine the outlet quality, xs,out: i_s_in=enthalpy(Steam,P=p_steam,x=0.95) q_dot_loss=m_dot*(i_s_in-i_s_out) x_out=quality(Steam,P=p_steam,h=i_s_out)
"enthalpy of inlet steam" "energy balance on steam" "quality of leaving steam"
which leads to xs,out = 0.673. b.) Are the thermal losses experienced in the underground pipe transport process significant in your opinion? Do you recommend insulating this pipe? Yes, the losses are significant. The quality of the steam is reduced from 95% to 67.3% and therefore you have lost approximately 30% of the heating content of the steam. The pipe should be insulated. c.) Provide a sanity check on the shape factor that you used to solve this problem. The shape factor represents, approximately, the ratio of the cross-sectional area for conduction to the length for conduction. A crude estimate of the shape factor for this problem is:
S app =
⎛ ⎝
π ⎜ Do + W
W⎞ ⎟L 2⎠
(6)
which leads to Sapp = 687.3 m; this is approximately in line with the value provided by the EES function, S = 651.8 m.
Problem 2.1-3 (2-2 in text) A solar electric generation system (SEGS) employs molten salt as both the energy transport and storage fluid. The molten salt is heated to Tsalt = 500°C and stored in a buried semi-spherical tank. The top (flat) surface of the tank is at ground level. The diameter of the tank before insulation is applied Dt = 14 m. The outside surfaces of the tank are insulated with tins = 0.30 m thick fiberglass having a thermal conductivity of kins = 0.035 W/m-K. Sand having a thermal conductivity of ksand = 0.27 W/m-K surrounds the tank, except on its top surface. Estimate the rate of heat loss from this storage unit to the Tair = 25°C surroundings. You may neglect the resistance due to convection. The inputs are entered in EES: $UnitSytem SI K J Pa $TabStops 0.2 0.4 3.5 in "known information" t_ins=0.3 [m] D_t=14 [m] k_sand=0.27 [W/m-K] k_ins=0.035 [W/m-K] T_air=convertTemp(C,K,25 [C]) T_salt=converttemp(C,K,500 [C])
"insulation thickness" "tank diameter" "sand thermal conductivity" "insulation thermal conductivity" "air temperature in K" "salt temperature in K"
The resistance between the bottom (buried, hemispherical) surface of the tank and the air is due to the conduction through the spherical shell of insulation
Rins ,bottom
⎛ 1 ⎞ 1 + ⎜ ⎟ Dt / 2 ( Dt / 2 + tins ) ⎠ ⎝ = 2 π kins
(1)
and the sand:
Rsand =
1 k sand S
(2)
where S is the shape factor, obtained using the EES function SF_22. The resistance from the top, unburied surface of the tank is only due to conduction through the insulation: Rins ,top =
tins D2 kins π t 4
The heat transfer from the bottom and top surfaces are calculated according to:
(3)
qbottom =
(Tsalt − Tair ) Rsoil + Rins ,bottom
(4)
and
qtop =
(Tsalt − Tair ) Rins ,top
(5)
The total heat loss is:
qtotal = qtop + qbottom R_ins_bottom=(1/(D_t/2)-1/(D_t/2+t_ins))/(2*pi*k_ins) "resistance to conduction through insulation on bottom" S=SF_22(D_t+2*t_ins) "shape factor" R_sand=1/(k_sand*S) "resistance of sand" R_ins_top=t_ins/(k_ins*pi*D_t^2/4) "resistance to conduction through insulation on top" q_dot_bottom=(T_salt-T_air)/(R_sand+R_ins_bottom) "heat loss from bottom surface of tank" q_dot_top=(T_salt-T_air)/(R_ins_top) "heat loss from top surface of tank" q_dot_total=q_dot_bottom+q_dot_top "total heat loss"
which leads to qtotal = 12.95 kW.
(6)
Problem 2.1-4 A square extrusion is L = 1 m long and has outer dimension W = 3 cm. There is a D = 1 cm diameter hole aligned with the center of the extrusion. The material has conductivity k = 0.5 W/m-K. The external surface of the extrusion is exposed to air at Ta = 20ºC with heat transfer coefficient ha = 50 W/m2-K. The inner surface of the extrusion is exposed to water at Tw = 80 ºC with heat transfer coefficient hw = 150 W/m2-K. a.) Determine the rate of heat transfer between the water and the air. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in L=1 [m] D=1 [cm]*convert(cm,m) W=3 [cm]*convert(cm,m) k=0.5 [W/m-K] h_bar_a=50 [W/m^2-K] T_a=converttemp(C,K,20 [C]) h_bar_w=150 [W/m^2-K] T_w=converttemp(C,K,80 [C])
"diameter of hole" "width of extrusion" "conductivity" "heat transfer coefficient on air-side" "air temperature" "heat transfer coefficient on water-side" "water temperature"
The resistance to convection to air is:
Rconv ,a =
1 ha 4W L
(1)
Rconv , w =
1 hw π D L
(2)
The resistance to convection to water is:
R_conv_a=1/(h_bar_a*4*W*L) R_conv_w=1/(h_bar_w*pi*D*L)
"air side convection resistance" "water side convection resistance"
The shape factor between the inner and outer surfaces of the extrusion (SF) is obtained from the SF_12 function in EES. The resistance to conduction is: Rcond =
1 SF k
(3)
The total rate of heat transfer is: q =
(Tw − Ta ) Rconv ,a + Rcond + Rconv , w
(4)
SF=SF_12(D,W,L) R_cond=1/(SF*k) q_dot=(T_w-T_a)/(R_conv_a+R_cond+R_conv_w)
"shape factor" "conduction resistance" "heat transfer rate"
which leads to q = 79.7 W. b.) Carry out a sanity check on the value of the shape factor that you used in (a). The shape factor can be thought of as the ratio of the effective area to the effective length for conduction. The area for conduction is smallest at the inner surface and largest at the outer surface. The approximate area can be taken to be the average of these two:
Aapp =
π D L + 4W L
(5)
2
The effective length is largest from the surface of the hole to the corner and smallest from the surface of the hole to the middle of the edge. The average of these values is used: Lapp =
⎡ 1 ⎢ (W − D ) + 2⎢ 2 ⎣
(
)
2W − D ⎤ ⎥ ⎥ 2 ⎦
(6)
The approximate shape factor is computed according to: SFapp = A_app=(pi*D*L+4*W*L)/2 L_app=((W-D)/2+(sqrt(2)*W-D)/2)/2 SF_app=A_app/L_app
Aapp
(7)
Lapp "approximate area for conduction" "approximate length for conduction" "approximate shape factor"
which leads to SFapp = 5.78 m which compares well to the shape factor obtained using the EES library, SF = 5.35 m.
$UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" th=1.5 [mm]*convert(mm,m) k=75 [W/m-K] e=1 [-] q``_s=900 [W/m^2] T_infinity=converttemp(C,K,10[C]) h_bar=5 [W/m^2-K] L=8 [cm]*convert(cm,m) T_w=converttemp(C,K,50 [C]) W=1 [m]
"thickness" "conductivity of collector plate" "emissivity of collector plate" "solar flux" "ambient temperature" "heat transfer coefficient" "half-width between tubes" "water temperature" "per unit length of collector"
Nodes are placed along the length of the collector. Only the region from 0 < x < L is considered due to the symmetry of the system. Therefore, the position of each node is:
xi =
L ( i − 1) for i = 1..N ( N − 1)
(8)
L ( N − 1)
(9)
The distance between adjacent nodes is: Δx = N=21 [-] Dx=L/(N-1) duplicate i=1,N x[i]=L*(i-1)/(N-1) end
"number of nodes"
"location of each node"
Energy balances on each of the internal nodes leads to: q LHS ,i + q RHS ,i + qs ,i + qconv ,i + qrad ,i = 0 for i = 2.. ( N − 1)
(10)
where
q LHS ,i =
k W th (Ti −1 − Ti ) Δx
(11)
q RHS ,i =
k W th (Ti +1 − Ti ) Δx
(12)
qs ,i = W Δx q s′′
(13)
qs ,i = W Δx h (T∞ − Ti )
(14)
qrad ,i = W Δx ε σ (T∞4 − Ti 4 )
(15)
"internal nodes" duplicate i=2,(N-1) q_dot_LHS[i]=k*W*th*(T[i-1]-T[i])/Dx "conduction from left hand side node" q_dot_RHS[i]=k*W*th*(T[i+1]-T[i])/Dx "conduction from right hand side node" q_dot_s[i]=W*Dx*q``_s "absorbed solar radiation" q_dot_conv[i]=W*Dx*h_bar*(T_infinity-T[i]) "convection" q_dot_rad[i]=W*Dx*e*sigma#*(T_infinity^4-T[i]^4) "radiation" q_dot_LHS[i]+q_dot_RHS[i]+q_dot_s[i]+q_dot_conv[i]+q_dot_rad[i]=0 "energy balance" end
The temperature of node 1 is assumed to be equal to the water temperature (neglecting any resistance to convection on the water-side):
T1 = Tw
(16)
q LHS , N + qs , N + qconv , N + qrad , N = 0
(17)
An energy balance on node N leads to:
where
q LHS , N =
k W th (TN −1 − TN ) Δx W Δx q s′′ 2
(19)
W Δx h (T∞ − TN ) 2
(20)
W Δx ε σ (T∞4 − TN4 ) 2
(21)
qs , N = qs , N = qrad , N = "node 1" T[1]=T_w
(18)
"node N" q_dot_LHS[N]=k*W*th*(T[N-1]-T[N])/Dx q_dot_s[N]=W*Dx*q``_s/2 q_dot_conv[N]=W*Dx*h_bar*(T_infinity-T[N])/2 q_dot_rad[N]=W*Dx*e*sigma#*(T_infinity^4-T[N]^4)/2 q_dot_LHS[N]+q_dot_s[N]+q_dot_conv[N]+q_dot_rad[N]=0
"conduction from left hand side node" "absorbed solar radiation" "convection" "radiation" "energy balance"
The temperature distribution within the collector is shown in Figure 2: 333
Temperature (K)
331
329
327
325
323 0
0.02
0.04
0.06
0.08
0.1
Position (m)
Figure 2: Temperature as a function of position in the collector.
An energy balance on node 1 provides the rate of energy transfer to the water:
qwater = q RHS ,1 + qs ,1 + qconv ,1 + qrad ,1
(22)
where
q RHS ,1 =
k W th (T2 − T1 ) Δx W Δx q s′′ 2
(24)
W Δx h (T∞ − T1 ) 2
(25)
qs ,1 = qs ,1 = qrad ,1 =
(23)
W Δx ε σ (T∞4 − T14 ) 2
"node 1" q_dot_RHS[1]=k*W*th*(T[2]-T[1])/Dx q_dot_s[1]=W*Dx*q``_s/2 q_dot_conv[1]=W*Dx*h_bar*(T_infinity-T[1])/2 q_dot_rad[1]=W*Dx*e*sigma#*(T_infinity^4-T[1]^4)/2
"conduction from right hand side node" "absorbed solar radiation" "convection" "radiation"
(26)
q_dot_water=q_dot_RHS[1]+q_dot_s[1]+q_dot_conv[1]+q_dot_rad[1] "energy balance"
which leads to qwater = 28.9 W. b.) Determine the efficiency of the collector; efficiency is defined as the ratio of the energy delivered to the water to the solar energy incident on the collector. The efficiency is calculated according to:
η= eta=q_dot_water/(L*W*q``_s)
qwater LW
(27)
"efficiency"
which leads to η = 0.402. c.) Plot the efficiency as a function of the number of nodes used in the solution. Figure 3 illustrates the efficiency as a function of the number of nodes and shows that at least 20-30 nodes are required for numerical convergence. 0.409 0.408 0.407
Efficiency (-)
0.406 0.405 0.404 0.403 0.402 0.401 2
10
100
225
Number of nodes
Figure 3: Efficiency as a function of the number of nodes.
d.) Plot the efficiency as a function of Tw - T∞. Explain your plot. Figure 4 illustrates the efficiency as a function of the water-to-ambient temperature difference. When the water temperature is low then the losses are low (but not zero, because the temperature of the copper plate is elevated by conduction). As the water temperature increases, the temperature of the plate increases and therefore the losses increase and efficiency drops. The drop in efficiency is dramatic for this type of unglazed collector and therefore the collector may be suitable for providing water heating for swimming pools (at low water temperature) but probably is not suitable for providing domestic hot water (at high water temperature).
1 0.9 0.8
Efficiency
0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0
10
20
30
40
50
60
70
80
Water to ambient temperature difference (K)
Figure 4: Solar collector efficiency as a function of the water-to-ambient temperature difference.
Problem 2.1-5 A pipe carrying water for a ground source heat pump is buried horizontally in soil with conductivity k = 0.4 W/m-K. The center of the pipe is W = 6 ft below the surface of the ground. The pipe has inner diameter Di = 1.5 inch and outer diameter Do = 2 inch. The pipe is made of material with conductivity kp = 1.5 W/m-K. The water flowing through the pipe has temperature Tw = 35ºF with heat transfer coefficient hw = 200 W/m-K. The temperature of the surface of the soil is Ts = 0ºF. a.) Determine the rate of heat transfer between the water and the air per unit length of pipe. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in L=1 [m] D_i=1.5 [inch]*convert(inch,m) D_o=2.0 [inch]*convert(inch,m) k_p=1.5 [W/m-K] k=0.4 [W/m-K] W=6 [ft]*convert(ft,m) T_s=converttemp(F,K,0 [F]) T_w=converttemp(F,K,35 [F]) h_bar_w=200 [W/m^2-K]
"inner diameter of tube" "outer diameter of tube" "conductivity of plastic" "conductivity of soil" "depth of tube beneath soil" "temperature of soil surface" "water temperature" "convection coefficient with water"
The resistance to convection to water is:
Rconv , w =
1 hw π Di L
(1)
The resistance to conduction through the pipe is:
Rcond , p
⎛D ⎞ ln ⎜ o ⎟ D = ⎝ i⎠ 2π k p L
R_conv_w=1/(h_bar_w*pi*D_i*L) R_cond_p=ln(D_o/D_i)/(2*pi*k_p*L)
(2)
"resistance to convection with water" "resistance to conduction through pipe"
The shape factor between the outer surface of the pipe and the surface of the soil (SF) is obtained from the SF_2 function in EES. The resistance to conduction through the soil is: Rcond , s = The total rate of heat transfer is:
1 SF k
(3)
q =
(Tw − Ta )
(4)
Rconv , w + Rcond , p + Rcond , s
SF=SF_2(D_o,W,L) "shape factor between outer surface of tube and surface" R_cond_s=1/(SF*k) "resistance to conduction through soil" q_dot=(T_w-T_s)/(R_conv_w+R_cond_p+R_cond_s) "rate of heat transfer"
which leads to q = 9.49 W/m. b.) Plot the heat transfer as a function of the depth of the pipe. Figure 1 illustrates the rate of heat transfer as a function of the depth of the pipe. 17
Rate of heat transfer (W)
16 15 14 13 12 11 10 9 8 7 0
1
2
3
4
5
6
7
8
9
10
11
Depth of pipe (m) Figure 1: Rate of heat transfer as a function of the depth of the pipe (per unit length).
c.) Carry out a sanity check on the value of the shape factor that you used in (a). The shape factor can be thought of as the ratio of the effective area to the effective length for conduction. Assuming that the heat spreads at an angle of approximately 45º as it moves from the pipe to the surface, the approximate area for conduction is:
Aapp = 2 W L
(5)
The effective length is the average of the length from the pipe to the surface perpendicularly and at 45º.
Lapp =
2W +W 2
The approximate shape factor is computed according to:
(6)
SFapp = A_app=sqrt(2)*W*L L_app=(sqrt(2)*W+W)/2 SF_app=A_app/L_app
Aapp Lapp
(7)
"approximate area" "approximate length" "approximate shape factor"
which leads to SFapp = 1.17 m which compares well to the shape factor obtained using the EES library, SF = 1.26 m.
Problem 2.2-1: Model of Welding Process You are evaluating a technique for controlling the properties of welded joints by using aggressive liquid cooling. Figure P2.2-1 illustrates a cut-away view of two plates that are being welded together. Both edges of the plate are clamped and effectively held at temperatures Ts = 25°C. The top of the plate is exposed to a heat flux that varies with position x, measured from joint, according to: qm′′ ( x ) = q ′′j exp ( − x / L j ) where q ′′j =1x106 W/m2 is the maximum heat flux
(at the joint, x = 0) and Lj = 2.0 cm is a measure of the extent of the heat flux. The back side of the plates are exposed to aggressive liquid cooling by a jet of fluid at Tf = -35°C with h = 5000 W/m2-K. A half-symmetry model of the problem is shown in Figure P2.2-1. The thickness of the plate is b = 3.5 cm and the width of a single plate is W = 8.5 cm. You may assume that the welding process is steady-state and 2-D. You may neglect convection from the top of the plate. The conductivity of the plate material is k = 38 W/m-K. both edges are clamped and held at fixed temperature
heat flux joint
impingement cooling with liquid jets qm′′
k = 38 W/m-K
W = 8.5 cm b = 3.5 cm
y
Ts = 25°C
x T f = −35°C 2 h = 500 W/m -K
Figure P2-2-1: Welding process and half-symmetry model of the welding process.
a.) Develop a separation of variables solution to the problem. Implement the solution in EES and prepare a plot of the temperature as a function of x at y = 0, 1.0, 2.0, 3.0, and 3.5 cm. The problem shown in Figure P2.2-1 is governed by the partial differential equation: ∂ 2T ∂ 2T + =0 ∂x 2 ∂y 2
(1)
and has the boundary conditions: ∂T ∂x
=0 x =0
(2)
Tx =W = Ts k
∂T ∂y
= h (Ty =0 − T f
(3)
)
(4)
y =0
k
∂T ∂y
= qm′′
(5)
y =b
Notice that the problem has 3 non-homogeneous boundary conditions and cannot, as it is written, be solved using separation of variables. However, the simple transformation:
θ = T − Ts
(6)
transforms the boundary condition at x = W to a homogeneous boundary condition and provides a homogeneous direction (x). The transformed differential equation and boundary conditions are: ∂ 2θ ∂ 2θ + =0 ∂x 2 ∂y 2
(7)
and has the boundary conditions: dθ dx
=0
(8)
x =0
θ x =W = 0 k
∂θ ∂y
= h (θ y =0 − θ f
(9)
)
(10)
y =0
k
∂θ ∂y
= qm′′
(11)
y =b
where
θ f = T f − Ts
(12)
T = TX TY
(13)
The separated solution is assumed:
and substituted into Eq. (7); the process leads to two ordinary differential equations:
d 2TX + λ 2 TX = 0 2 dx
(14)
d 2TY − λ 2 TY = 0 2 dy
(15)
TX = C1 sin ( λ x ) + C2 cos ( λ x )
(16)
TY = C3 sinh ( λ y ) + C4 cosh ( λ y )
(17)
The solutions to Eqs. (14) and (15) are:
We will solve the eigenproblem first (i.e., the problem in the homogeneous direction). The boundary condition at x = 0, Eq. (8), leads to: dTX dx
x =0
= ⎡⎣C1 λ cos ( λ x ) − C2 λ sin ( λ x ) ⎤⎦ x =0 = C1 λ = 0
(18)
which indicates that C1 = 0 and therefore: TX = C2 cos ( λ x )
(19)
The boundary condition at x = W, Eq. (9), leads to: TX x =W = C2 cos ( λ W ) = 0
(20)
Equation (20) provides our eigenvalues:
λi =
(1 + 2 i ) π 2W
for i = 1..∞
(21)
and the eigenfunctions are: TX i = C2,i cos ( λi x )
(22)
The solution for TX is the sum of the eigenfunctions: ∞
TX = ∑ C2,i cos ( λi x ) i =0
(23)
and the solution for the temperature is: ∞
TX = ∑ C2,i cos ( λi x ) ⎡⎣C3,i sinh ( λi y ) + C4,i cosh ( λi y ) ⎤⎦
(24)
i =0
Note that C2,i can be absorbed in the other constants: ∞
θ = ∑ cos ( λi x ) ⎡⎣C3,i sinh ( λi y ) + C4,i cosh ( λi y ) ⎤⎦
(25)
i =0
Equation (25) is substituted into the boundary condition at y = 0, Eq. (10): ∞
k ∑ cos ( λi x ) ⎣⎡C3,i λi cosh ( λi 0 ) + C4,i λi sinh ( λi 0 ) ⎦⎤ = i =0
(26)
⎧∞ ⎫ h ⎨∑ cos ( λi x ) ⎡⎣C3,i sinh ( λi 0 ) + C4,i cosh ( λi 0 ) ⎤⎦ − θ f ⎬ ⎩ i =0 ⎭ or ∞ ⎡∞ ⎤ k ∑ cos ( λi x ) C3,i λi = h ⎢ ∑ cos ( λi x ) C4,i − θ f ⎥ i =0 ⎣ i =0 ⎦
(27)
We take advantage of the orthogonality of the eigenfunctions by multiplying Eq. (27) by an arbitrary eigenfunction and integrating from x = 0 to x = L in order to achieve: W W ⎡ k C3,i λi ∫ cos 2 ( λi x ) dx = h ⎢C4,i ∫ cos 2 ( λi x ) dx − θ f 0 0 ⎣
The integrals in Eq. (28) are evaluated using Maple: > restart; > assume(i,integer); > lambda:=(1+2*i)*Pi/(2*W);
λ :=
( 1 + 2 i~ ) π 2W
> int((cos(lambda*x))^2,x=0..W);
W 2 > int(cos(lambda*x),x=0..W);
2 ( -1 ) i~ W ( 1 + 2 i~ ) π
⎤
W
∫ cos ( λ x ) dx ⎥⎦ i
0
(28)
which leads to: k C3,i λi W 2
i ⎡ C4,i W 2 ( −1) W ⎤ = h⎢ −θ f ⎥ (1 + 2 i ) π ⎥⎦ ⎢⎣ 2
(29)
or
4 ( −1) kλ θf = C4,i − i C3,i h (1 + 2 i ) π i
(30)
which is one equation for each of the two pairs of unknown coefficients. Equations (25) and Error! Reference source not found. are substituted into the final boundary condition at y = b, Eq. (11): ∞ ⎛ x k ∑ cos ( λi x ) ⎡⎣C3,i λi cosh ( λi b ) + C4,i λi sinh ( λi b ) ⎤⎦ = q ′′j exp ⎜ − ⎜ Lj i =0 ⎝
⎞ ⎟⎟ ⎠
(31)
Again, the orthogonality of the eigenfunctions is used to obtain: W W ⎛ x k ⎡⎣C3,i λi cosh ( λi b ) + C4,i λi sinh ( λi b ) ⎤⎦ ∫ cos 2 ( λi x ) dx = q ′′j ∫ exp ⎜ − ⎜ Lj 0 0 ⎝
⎞ ⎟⎟ cos ( λi x ) dx ⎠
(32)
or C3,i λi cosh ( λi b ) + C4,i λi sinh ( λi b ) =
2 q ′′ Wkj
W
⎛
x ⎞ ⎟⎟ cos ( λi x ) dx j ⎠
∫ exp ⎜⎜ − L 0
⎝
(33)
which is the 2nd relationship between the coefficient pairs. The integral on the right hand side of Eq. (32) is accomplished using Maple: > int(exp(-x/L_j)*cos(lambda*x),x=0..W); ⎛ ⎞ ⎜⎜ ⎟⎟ ⎛ ⎞ ⎛⎜⎜ − L_j ⎞⎟⎟ ⎜⎜ ⎟ ⎝ ⎝ L_j ⎠ ⎠ i~ i~ 2 L_j W ⎝ 2 W e + π L_j ( -1 ) + 2 π L_j ( -1 ) i~ ⎟⎠ e 4 W2 + π 2 L_j2 + 4 π 2 L_j2 i~ + 4 π 2 L_j2 i~ 2 W
and pasted directly into EES in order to accomplish the implementation. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J
W
$TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" b=3.5 [cm]*convert(cm,m) W=8.5 [cm]*convert(cm,m) k=38 [W/m-K] T_s=converttemp(C,K,25) T_f=converttemp(C,K,-35) h=5000 [W/m^2-K] q``_j=1e6 [W/m^2] L_j=2.0 [cm]*convert(cm,m)
"plate thickness" "plate width" "plate conductivity" "edge temperature" "fluid temperature" "heat transfer coefficient" "heat flux at x=0" "extent of the heat flux"
and a position is specified: "Position" x_bar=0.5 x=x_bar*W y_bar=0.5 y=y_bar*b
The solution is implemented:: N=10 [-] "number of terms" theta_f=(T_f-T_s) "transformed fluid temperature" duplicate i=0,N lambda[i]=(1+2*i)*pi/(2*W) theta_f*4*(-1)^i/((1+2*i)*pi)=C_4[i]-k*lambda[i]*C_3[i]/h C_3[i]*lambda[i]*cosh(lambda[i]*b)+C_4[i]*lambda[i]*sinh(lambda[i]*b)=(2*q``_j/(W*k))* (2*W*L_j*(2*W*exp(W/L_j)+Pi*L_j*(-1)^i+2*Pi*L_j*(-1)^i*i)*exp(-W/L_j)/(4*W^2+Pi^2*L_j^2 +4*Pi^2*L_j^2*i+4*Pi^2*L_j^2*i^2)) theta[i]=cos(lambda[i]*x)*(C_3[i]*sinh(lambda[i]*y)+C_4[i]*cosh(lambda[i]*y)) end T=T_s+sum(theta[0..N]) T_C=converttemp(K,C,T)
Note that the bold portion of the code was pasted from Maple. Figure 2 illustrates the temperature as a function of dimensionless axial position (x/W) for various values of y.
Figure 2: Temperature as a function of x/W for various values of y.
b.) Prepare a contour plot of the temperature distribution. Figure 3 illustrates a contour plot of the temperature distribution in the plate.
Figure 3: Contour plot of the temperature distribution in the plate.
Problem 2.2-2: Derive a Partial Differential Equation Figure P2.2-2 illustrates a thin plate that is exposed to air on upper and lower surfaces. The heat transfer coefficient between the top and bottom surfaces is h and the air temperature is Tf. The thickness of the plate is th and its width and height are a and b, respectively. The conductivity of the plate is k. The top edge is fixed at a uniform temperature, T1. The right edge is fixed at a different, uniform temperature, T2. The left edge of the plate is insulated. The bottom edge of the plate is exposed to a heat flux, q ′′ . This problem should be done on paper. T1
a T2 k y
b x q ′′
th
h ,Tf
h ,T f
Figure P.2.2-2: Thin plate exposed to air.
a.) The temperature distribution within the plate can be considered 2-D (i.e., temperature variations in the z-direction can be neglected) if the plate is thin and conductive. How would you determine if this approximation is valid? The Biot number characterizes the ratio of conduction in the z direction (which I want to neglect) to convection from the plate surface (which I will consider): Bi =
Rcond , z Rconv
=
th h a b th h = 2k ab 1 2k
(1)
b.) Derive the partial differential equation and boundary conditions that would need to be solved in order to obtain an analytical solution to this problem. A differential control volume has extend dx x dy and extends through the thickness of the plate, as shown in Figure 2.
Figure 2: Differential control volume used to derive the governing partial differential equation.
The first law balance suggested by Figure 2 is:
q x + q y = q x + dx + q y + dy + qconv
(2)
or 0=
∂q y ∂q x dx + dy + qconv ∂x ∂y
(3)
Substituting the rate equations into the energy balance leads to: 0=
∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ dx + ⎢ − k th dx ⎥ dy + 2 dx dy h (T − T f ) −k th dy ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂y ⎣ ∂y ⎦
(4)
or
∂ 2T ∂ 2T 2 h + − (T − T f ) = 0 ∂x 2 ∂y 2 k th
(5)
The boundary conditions at the upper and right edges are:
Tx = a = T2
(6)
Ty =b = T1
(7)
The boundary condition at the left edge is obtained from an interface balance which requires that the conduction heat flux in the x-direction at this insulated edge be zero:
q ′′x , x =0 = 0
(8)
or −k
∂T ∂x
=0
(9)
x =0
which leads to:
∂T ∂x
=0
(10)
x =0
The boundary condition at the bottom edge is also obtained from an interface energy balance: q ′′ = −k
∂T ∂y
(11) y =0
or ∂T ∂y
=− y =0
q ′′ k
(12)
Problem 2.2-3 (2-3 in text): Heat Transfer Coefficient Measurement Device You are the engineer responsible for a simple device that is used to measure heat transfer coefficient as a function of position within a tank of liquid (Figure P2.2-3). The heat transfer coefficient can be correlated against vapor quality, fluid composition, and other useful quantities. The measurement device is composed of many thin plates of low conductivity material that are interspersed with large, copper interconnects. Heater bars run along both edges of the thin plates. The heater bars are insulated and can only transfer energy to the plate; the heater bars are conductive and can therefore be assumed to come to a uniform temperature as a current is applied. This uniform temperature is assumed to be applied to the top and bottom edges of the plates. The copper interconnects are thermally well-connected to the fluid; therefore, the temperature of the left and right edges of each plate are equal to the fluid temperature. This is convenient because it isolates the effect of adjacent plates from one another which allows each plate to measure the local heat transfer coefficient. Both surfaces of the plate are exposed to the fluid temperature via a heat transfer coefficient. It is possible to infer the heat transfer coefficient by measuring heat transfer required to elevate the heater bar temperature a specified temperature above the fluid temperature. top and bottom surfaces exposed to fluid 2 T∞ = 20°C, h = 50 W/m -K copper interconnet, T∞ = 20°C
a = 20 mm b = 15 mm
plate: k = 20 W/m-K th = 0.5 mm
heater bar, Th = 40°C
Figure P2.2-3: Device to measure heat transfer coefficient as a function of position.
The nominal design of an individual heater plate utilizes metal with k = 20 W/m-K, th = 0.5 mm, a = 20 mm, and b = 15 mm (note that a and b are defined as the half-width and half-height of the heater plate, respectively, and th is the thickness as shown in Figure P2-3). The heater bar temperature is maintained at Th = 40ºC and the fluid temperature is T∞ = 20ºC. The nominal value of the average heat transfer coefficient is h = 50 W/m2-K. a.) Develop an analytical model that can predict the temperature distribution in the plate under these nominal conditions. The problem can be simplified and tackled using the quarter-symmetry model of a single plate, shown in Figure P2.2-3, is specified mathematically in Figure 2(a).
(a)
(b) Figure 2: Quarter symmetry model of a plate.
The lines of symmetry (x =0 and y = 0) are adiabatic, the top edge (y = b) is held at the heater temperature (Th) and the right edge (x = a) is held at the fluid temperature (T∞). These boundary conditions are expressed below: ∂T ∂x
=0 x =0
Tx =a = T∞
∂T ∂y
=0 y =0
Ty =b = Th The differential control volume and associated first law balance are shown in Figure 2(a):
q x + q y = q x + dx + q y + dy + qconv or, after expanding the x + dx and y + dy terms: 0=
∂q y ∂q x dx + dy + qconv ∂x ∂y
Substituting the rate equations into the energy balance leads to: 0=
∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ −k th dy dx + ⎢ − k th dx dy + 2 dx dy h ( T − T∞ ) ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂y ⎣ ∂y ⎥⎦
or
∂ 2T ∂ 2T 2 h + − T − T∞ = 0 ∂x 2 ∂y 2 k th
(
)
(1)
Equation (1) is not homogeneous; however, a transformation variable can be identified by inspection:
θ = T − T∞
(2)
Substituting Eq. (2) into Eq. (1) transforms the governing partial differential equation into a homogeneous equation; i.e., any multiple of θ will satisfy transformed partial differential equation: ∂ 2θ ∂ 2θ + 2 − m2 θ = 0 2 ∂x ∂y
where
m=
2h k th
(3)
Substituting Eq. (2) into the boundary conditions leads to: ∂θ ∂x
=0
θ x =a = 0 ∂θ ∂y
(4)
x =0
=0
(5) (6)
y =0
θ y =b = θ h
(7)
where
θ h = Th − T∞ The transformed problem is shown in Figure 2(b); note that the transformed problem is linear and consists of a homogeneous partial differential equation and homogeneous boundary conditions in the x-direction. Therefore, we can apply a separation of variables solution to the problem and x is our homogeneous direction. The solution is assumed to be the product of θX which is a function of x and θY which is a function of y:
θ ( x, y ) = θ X ( x ) θ Y ( y ) Substituting the product solution into the differential equation leads to:
θY
d 2θ X d 2θ Y θ + X − m2 θ X θY = 0 2 2 dx dy
Dividing by the product θX θY: 2 d 2θ X d θ Y 2 dx 2 + dy − m 2 = 0 θ Y
X θ
−λ 2
λ2
Recall that x is our homogeneous direction; therefore, we need the θX group in the equation to equal a negative constant (-λ2); the remaining part of the equation must equal the positive value
of the same constant (λ2). Therefore, the two ordinary differential equations that result from the separation process are:
d 2θ X + λ2 θ X = 0 2 dx
(8)
d 2θ Y − ( λ 2 + m2 )θY = 0 dy 2
(9)
and
The next step is to solve the eigenproblem; the solution to the ordinary differential equation for θX is:
θ X = C1 cos ( λ x ) + C2 sin ( λ x )
(10)
The 1st boundary conditions in the homogeneous direction, Eq. (4), leads to: dθ X dx
=0
(11)
x =0
Substituting Eq. (10) into Eq. (11) leads to:
dθ X dx
x =0
= −C1 λ sin ( λ 0 ) + C2 λ cos ( λ 0 ) = 0
=0
=1
which can only be true if C2 = 0. the 2nd boundary condition in the x-direction, Eq. (5), leads to:
θ x =a = C1 cos ( λ a ) = 0 which can only be true if the argument of the cosine is 0. Therefore, the eigenfunctions and eigenvalues for the problem are:
θ X i = C1,i cos ( λi x ) where λi =
( 2 i − 1) π 2
a
The solution to the ordinary differential equation in the non-homogeneous direction, Eq. (9) can be obtained using Maple: > restart; > assume((lambda^2+m^2),positive); > ODEy:=diff(diff(thetaY(y),y),y)-(lambda^2+m^2)*thetaY(y)=0;
2 ⎞ ⎛d ODEy := ⎜ 2 thetaY( y ) ⎟⎟ − ( λ∼ 2 + m~ 2 ) thetaY( y ) = 0 ⎜ dy ⎝ ⎠
> qYs:=dsolve(ODEy);
qYs := thetaY( y ) = _C1 e
λ∼ 2 + m~2 y )
(
+ _C2 e
( − λ∼2 + m~2 y )
Note that Maple can convert this exponential form to an equivalent form involving hyperbolic sines and cosines with the convert command. > convert(qYs,'trigh');
thetaY( y ) = ( _C1 + _C2 ) cosh( λ∼ 2 + m~ 2 y ) + ( _C1 − _C2 ) sinh( λ∼ 2 + m~ 2 y )
θ Yi = C3,i cosh
(
)
λi2 + m 2 y + C4,i sinh
(
λi2 + m 2 y
)
The solution associated with the ith eigenvalue is therefore:
θi = θ X i θ Yi = cos ( λi x ) ⎡⎢C3,i cosh ⎣
(
)
λi2 + m 2 y + C4,i sinh
(
)
λi2 + m 2 y ⎤⎥ ⎦
(12)
where the constant C1,i is absorbed into the constants C3,i and C4,i. It is good practice to verify that Eq. (12) satisfies both boundary conditions in the x-direction and the partial differential equation for any value of i: > restart; > assume(i,integer); > lambda:=(2*i-1)*Pi/(2*a);
λ :=
( 2 i~ − 1 ) π 2a
> theta:=(x,y)->cos(lambda*x)*(C_3*cosh(sqrt(lambda^2+m^2)*y)+C_4*sinh(sqrt(lambda^2+m^2)*y));
θ := ( x, y ) → cos( λ x ) ( C_3 cosh( λ 2 + m 2 y ) + C_4 sinh( λ 2 + m 2 y ) ) > eval(diff(theta(x,y),x),x=0);
0 > theta(a,y);
0 > diff(diff(theta(x,y),x),x)+diff(diff(theta(x,y),y),y)-m^2*theta(x,y);
1 ( 2 i~ − 1 ) π x ⎞ 2 2 − cos⎛⎜⎜ ⎟⎟ ( 2 i~ − 1 ) π 2a 4 ⎝ ⎠ ⎛ ⎛ ⎛ ⎞ ( 2 i~ − 1 ) 2 π 2 ⎜ ⎜ ⎜ + 4 m 2 y ⎟⎟ ⎜ ⎜ ⎜ 2 a ⎜ ⎜ ⎟ ⎜ ⎜⎜ C_3 cosh⎜⎜ ⎟⎟ + C_4 sinh⎜⎜ 2 ⎝ ⎝ ⎠ ⎝
⎞⎞ ( 2 i~ − 1 ) 2 π 2 + 4 m 2 y ⎟⎟ ⎟⎟ 2 a ⎟⎟ ⎟⎟ ⎟⎟ 2 ⎠⎠
⎛ ⎜ ⎜ ( 2 i~ − 1 ) π x ⎞ ⎜⎜ a 2 + cos⎛⎜⎜ ⎟⎟ ⎜ 2a ⎝ ⎠⎝ 2 2 ⎛ ⎞ ( 2 i~ − 1 ) π ⎜ + 4 m 2 y ⎟⎟ ⎜ 2 1 a ⎜ ⎟ ⎛ ( 2 i~ − 1 ) 2 π 2 ⎞ ⎜ ⎟⎟ ⎜ + 4 m 2 ⎟⎟ C_3 cosh⎜ 2 ⎜ 2 4 a ⎝ ⎠⎝ ⎠ 2 2 ⎞ ⎛ ⎞ ( 2 i~ − 1 ) π 2 ⎟ ⎜ ⎟ + 4 m y ⎟ ⎜ ⎟ 1 a2 ⎟ ⎜ ⎟ ⎛ ( 2 i~ − 1 ) 2 π 2 ⎞ 2 ⎟⎟ ⎜ + C_4 sinh⎜⎜ + 4 m ⎟⎟ ⎟⎟ − m 2 2 ⎜ 2 4 a ⎝ ⎠⎝ ⎠⎠ ( 2 i~ − 1 ) π x ⎞ cos⎛⎜⎜ ⎟⎟ 2 a ⎝ ⎠ ⎛ ⎞ ⎞⎞ ⎛ ⎛ ( 2 i~ − 1 )2 π2 ( 2 i~ − 1 ) 2 π 2 2 ⎜ ⎜ ⎟ ⎜ + 4 m y + 4 m 2 y ⎟⎟ ⎟⎟ ⎜ ⎜ ⎟ ⎜ 2 2 a a ⎜ ⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ C_3 cosh⎜⎜ ⎟⎟ + C_4 sinh⎜⎜ ⎟⎟ ⎟⎟ 2 2 ⎝ ⎝ ⎠ ⎝ ⎠⎠ > simplify(%);
0
The general solution is the sum of the solution for each eigenvalue: ∞
∞
θ = ∑θi = ∑ cos ( λi x ) ⎡⎢C3,i cosh i =1
⎣
i =1
(
)
λi2 + m 2 y + C4,i sinh
(
)
λi2 + m 2 y ⎤⎥ ⎦
(13)
The general solution must satisfy the boundary conditions in the non-homogeneous direction. Substituting Eq. (13) into Eq. (6) leads to: ∂θ ∂y
⎡ ⎤ 2 2 2 2 2 2 2 2 ⎢ = ∑ cos ( λi x ) C3,i λi + m sinh λi + m 0 + C4,i λi + m cosh λi + m 0 ⎥ = 0 ⎢ i =1
⎥ =0 =1 ⎣⎢ ⎦⎥
y =0
)
(
∞
(
)
or ∞
∑C i =1
4,i
λi2 + m 2 cos ( λi x ) = 0
which can only be true if C4,i = 0: ∞
θ = ∑ Ci cosh i =1
(
)
λi2 + m 2 y cos ( λi x )
(14)
where the subscript 3 has been removed from C3,i because it is the only remaining undetermined constant. Substituting Eq. (14) into Eq. (7) leads to: ∞
θ y =b = ∑ Ci cosh i =1
(
)
λi2 + m 2 b cos ( λi x ) = θ h
This equation is multiplied by cos(λj x) and integrated from 0 to a: ∞
∑ Ci cosh i =1
(
)
a
a
0
0
λi2 + m 2 b ∫ cos ( λi x ) cos ( λ j x ) dx = θ h ∫ cos ( λ j x ) dx
Orthogonality causes all of the terms in the summation to integrate to zero except for the one in which i = j: Ci cosh
(
λ +m b 2 i
2
) ∫ cos (λ x ) dx = θ ∫ cos (λ x ) dx a
a
2
i
h
0
i
0
Maple is used to carry out the integrations: > restart; > assume(i,integer); > lambda:=(2*i-1)*Pi/(2*a);
λ := > int((cos(lambda*x))^2,x=0..a);
( 2 i~ − 1 ) π 2a a 2
> int(cos(lambda*x),x=0..a);
( 1 + i~ )
2 ( -1 ) a ( 2 i~ − 1 ) π
which leads to an equation for each of the constants.
θ h 4 ( −1)
1+ i
Ci =
The inputs are entered in EES: $UnitSystem SI MASS RAD PA C J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs"
cosh
(
)
λi2 + m 2 b ( 2 i − 1) π
th_mm=0.5 [mm] th=th_mm * convert(mm,m) k=20 [W/m-K] a_mm=20 [mm] a= a_mm * convert(mm,m) b= 15 [mm] *convert(mm,m) T_h = converttemp(C,K,40 [C]) T_infinity=converttemp(C,K,20 [C]) h_bar=50 [W/m^2-K]
"thickness of plate in mm" "thickness of plate" "conductivity of plate" "half-width of plate in mm" "half-width of plate" "half-height of plate" "heater temperature" "fluid temperature" "heat transfer coefficient"
The position is specified using dimensionless variables x_bar and y_bar: "position" x_bar=0.5 y_bar=0.5 x=x_bar*a y=y_bar*b
The solution is evaluated: theta_h=T_h-T_infinity "heater temperature difference" m=sqrt(2*h_bar/(k*th)) "solution constant" N=100 "number of terms" duplicate i=1,N lambda[i]=(2*i-1)*pi/(2*a) "eigenvalues" C[i]=theta_h*(4*(-1)^(1+i)/(2*i-1)/Pi)/(cosh(sqrt(lambda[i]^2+m^2)*b)) theta[i]=C[i]*cos(lambda[i]*x)*cosh(sqrt(lambda[i]^2+m^2)*y) end T=sum(theta[1..N])+T_infinity T_C=converttemp(K,C,T)
A parametric table is created and used to generate the contour plot shown in Figure 3.
Figure 3: Contour plot of temperature on the plate.
b.) The measured quantity is the rate of heat transfer to the plate from the heater ( qh ) and therefore the relationship between qh and h (the quantity that is inferred from the heater power) determines how useful the instrument is. Determine the heater power. The heater power can be computed by integrating the conduction heat transfer along the top surface according to: a
qh = 4 ∫ k 0
∂θ ∂y
th dx
(15)
y =b
where the factor of 4 comes from the quarter symmetry of the model. substituted into Eq. (15): ∞
qh = 4 k th ∑ Ci λi2 + m 2 sinh i =1
(
λi2 + m 2 b
Equation (14) is
) ∫ cos (λ x ) dx a
i
0
which leads to: ∞
qh = 2 k th a ∑ Ci λi2 + m 2 sinh i =1
Equation (16) is evaluated in EES: "heater power"
(
λi2 + m 2 b
)
(16)
duplicate i=1,N q_dot_h[i]=2*k*th*a*C[i]*sqrt(lambda[i]^2+m^2)*sinh(sqrt(lambda[i]^2+m^2)*b) end q_dot_h=sum(q_dot_h[1..N])
c.) If the uncertainty in the measurement of the heater power is δ qh = 0.01 W, estimate the uncertainty in the measured heat transfer coefficient ( δ h ). A parametric table is used to explore the relationship between h and qhtr ; the two columns in the table are the variables h_bar and q_dot_h. Figure 4 illustrates the heat transfer coefficient as a function of the heater power, all else held constant.
Figure 4: Heat transfer coefficient as a function of the heater power.
Note that a good measuring device would show a strong relationship between the physical quantity being measured ( h ) and the measurement ( qh ); this would be indicated by a large partial derivative of heat transfer coefficient with respect to heater power. Given the uncertainty in the heater power measurement, δ qh (related to, for example, the resolution of the data acquisition system, noise, etc.), it is possible to estimate the uncertainty in the measurement of the heat transfer coefficient, δ h , according to:
δh =
∂h δ qh ∂qh
For example, if the uncertainty in the heater power is 0.01 W then Fig. 3 suggests that the partial derivative of heat transfer coefficient with respect to heater power is approximately 125 W/m2-
K-W and the uncertainty in the heat transfer coefficient measurement would be δ h = 1.8 W/m2K. As an engineer designing this measurement device, you would like to calculate not the heater power but rather the partial derivative in the heat transfer coefficient with respect to heater power in order to automate the process of computing δ h and therefore evaluate how design changes affect the instrument. This would be difficult to accomplish analytically; note that both m and the constants Ci in Eq. (16) are functions of h . It is more convenient to determine the partial derivative numerically. That is, set the heat transfer coefficient at its nominal value plus a small amount ( h + Δh ) and evaluate the heater power ( qh , h +Δh ); note that Δh should be small relative to the nominal value of the heat transfer coefficient. Then set the heat transfer coefficient at its nominal value less a small amount ( h − Δh ) and evaluate the heater power ( qh , h −Δh ). The partial derivative is approximately: ∂h 2 Δh ≈ ∂qh qh ,h +Δh − qh ,h −Δh
Since qh needs to be evaluated at several values of the heat transfer coefficient, it is convenient to have the computation of the heater power occur within a function that is called twice within the equation window. Because the solution is in the form of an EES code, this is an ideal problem to use a MODULE. A MODULE is a stand-alone EES program that can be called from the main EES equation window. The MODULE is provided with inputs and it calculates outputs. The protocol of a call to a MODULE involves the name of the MODULE followed by a series of inputs separated by a colon from a series of outputs. For example, we will create a MODULE Heaterpower that calculates the value of the variable q_dot_h. It is important to note that all of the variables from the main equation window that are required by the MODULE must be passed to the MODULE; the MODULE can only access variables that are passed to it as parameters or using the $COMMON directive. The MODULE Heaterpower must be placed at the top of the EES code and is a small selfcontained EES program; we create the MODULE by copying those lines of the main EES code that are required to calculate the variable q_dot_h. MODULE Heaterpower(th,k,a,b,T_h,T_infinity,h_bar:q_dot_h) theta_h=T_h-T_infinity "heater temperature difference" m=sqrt(2*h_bar/(k*th)) "solution constant" N=100 "number of terms" duplicate i=1,N lambda[i]=(2*i-1)*pi/(2*a) "eigenvalues" C[i]=theta_h*(4*(-1)^(1+i)/(2*i-1)/Pi)/(cosh(sqrt(lambda[i]^2+m^2)*b)) q_dot_h[i]=2*k*th*a*C[i]*sqrt(lambda[i]^2+m^2)*sinh(sqrt(lambda[i]^2+m^2)*b) end q_dot_h=sum(q_dot_h[1..N]) end
The calling protocol for the MODULE consists of a series of inputs (the variables d, k, a, b, T_htr, T_f, and h) that are separated by a series outputs (in this case only the variable q_dot_htr) by a colon. MODULES are most useful where a certain sequence of code must be executed multiple times; in this case, the MODULE Heaterpower enables the partial derivative to be easily computed. delta_q_dot_htr=0.01 [W] dh=1 [W/m^2-K]
"heater power resolution" "perturbation of heat transfer coefficient"
CALL Q_dot_heater(d,k,a,b,T_htr,T_f,h-dh:q_dot_htr_minus) CALL Q_dot_heater(d,k,a,b,T_htr,T_f,h+dh:q_dot_htr_plus) dhdqdot=2*dh/(q_dot_htr_plus-q_dot_htr_minus) delta_h=dhdqdot*delta_q_dot_htr
The modifications to the EES code verifies that the uncertainty in heat transfer coefficient, delta_h, is 1.26 W/m2-K and provides a convenient tool for assessing the impact of the various design parameters on the performance of the measurement system. For example, Figure 5 illustrates the uncertainty in the heat transfer coefficient as a function of the plate thickness (th) for various values of its half-width (a).
Figure 5: Uncertainty in the measured heat transfer coefficient as a function of the plate thickness for various values of the plate half-width.
Problem 2.2-4: Local Heat Transfer Coefficient Measurement Figure P2.2-4(a) illustrates a proposed device to measure the local heat transfer coefficient from a surface undergoing a boiling heat transfer process. Micro-scale heaters and temperature sensors are embedded in a substrate in a regularly spaced array, as shown. evaporating fluid
heaters computational temperature domain sensors Figure P2.2-4(a): An array of micro-scale heaters and temperature sensors embedded in a substrate.
The heaters are activated, producing a heat flux that is removed primarily from the surface of the substrate exposed to evaporating fluid. The temperature sensors are embedded in sets of two located very near the surface. Each set of thermocouples are used to infer the local heat flux to the surface ( qs′′ ) and the surface temperature (Ts); these quantities are sufficient to measure the heat transfer coefficient. A half-symmetry model of the region of the substrate between two adjacent heaters (see Figure 2.2-4(a)) is shown in Figure P2.2-4(b). T∞ = 20°C 2 h = 2500 W/m -K
y1 = 1.9 mm
y
5 2 qh′′ = 1x10 W/m
b = 2 mm W = 5 mm x
y2 = 1.8 mm
k = 3.5 W/m-K Figure P2.2-4(b): A half-symmetry model of the region of the substrate between two adjacent heaters.
The thickness of the substrate is b = 2 mm and the half-width between adjacent heaters is W = 5 mm. The substrate has conductivity k = 3.5 W/m-K and you may assume that the presence of the temperature sensors does not affect the temperature distribution. The heat flux exposed to the computational domain at x = W is qh′′ = 1x105 W/m2. The heat transfer coefficient between the evaporating fluid at T∞ = 20ºC and the surface is h = 2500 W/m2-K. a.) Develop a separation of variables solution based on the computational domain shown in Figure P2.2-4(b). Implement your solution in EES. The known information is entered in EES: $UnitSystem SI MASS RAD PA C J
$Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" k=3.5 [W/m-K] T_infinity=converttemp(C,K,20 [C]) h=2500 [W/m^2-K] W=5 [mm]*convert(mm,m) b=2 [mm]*convert(mm,m) y_1=1.9 [mm]*convert(mm,m) y_2=1.8 [mm]*convert(mm,m) q_h=1e5 [W/m^2]
"conductivity" "fluid temperature" "heat transfer coefficient" "half-width" "substrate thickness" "temperature sensor #1 position" "temperature sensor #2 position" "edge heat flux"
The problem shown in Figure P2.2-4(b) is governed by the partial differential equation: ∂ 2T ∂ 2T + =0 ∂x 2 ∂y 2
(1)
and has the boundary conditions:
k
∂T ∂x
x =0
∂T ∂x
x =W
∂T ∂y
−k
∂T ∂y
=0
(2)
= qh′′
(3)
=0
(4)
y =0
= h (Ty =b − T∞ )
(5)
y =b
Neither the x- or y-directions are homogeneous. However, by defining the temperature difference relative to T∞, both boundary conditions in the y-direction can be made homogeneous:
θ = T − T∞
(6)
The transformed differential equation and boundary conditions are: ∂ 2θ ∂ 2θ + =0 ∂x 2 ∂y 2
and has the boundary conditions:
(7)
k
∂θ ∂x
x =0
∂θ ∂x
x =W
∂θ ∂y −k
∂θ ∂y
=0
(8)
= qh′′
(9)
=0
(10)
= h θ y =b
(11)
y =0
y =b
The separated solution is assumed:
θ = θ X θY
(12)
and substituted into Eq. (7); the process leads to two ordinary differential equations: d 2θ X − λ2 θ X = 0 2 dx
(13)
d 2θ Y + λ 2 θY = 0 dy 2
(14)
Notice that the sign of the constant is selected so that the solution in the homogeneous direction (y) is sines and cosines. The solution to Eq. (14) is:
θ Y = C1 sin ( λ y ) + C2 cos ( λ y )
(15)
The boundary condition at y = 0, Eq. (10), leads to: dθ Y dy
= C1 λ cos ( λ 0 ) − C2 λ sin ( λ 0 ) = 0
(16)
y =0
which indicates that C1 = 0 and therefore:
θ Y = C2 cos ( λ y )
(17)
The boundary condition at y = b, Eq. (11), leads to: k C2 λ sin ( λ b ) = h C2 cos ( λ b )
(18)
Equation (18) provides the eigencondition for the problem: tan ( λi b ) =
hb k ( λi b )
(19)
The roots of Eq. (19) occur in regular intervals of the argument of the tangent function. A residual is defined by rearranging Eq. (19): Res = tan ( λi b ) −
hb k ( λi b )
(20)
Figure P2.2-4-2 illustrates the residual defined in Eq. (20); the eigenvalues correspond to the points where the residual is zero. 1 0.75 0.5
Residual
0.25 0 -0.25 -0.5 -0.75 -1 0
eigenvalues 1.57 3.14 4.71 6.28 7.85 9.42 10.99 12.56 14.13 15.7
λb
Figure P2.2-4-2: Residual as a function of λb.
The eigenvalues lie in regular intervals of λb, as shown in Figure 2.2-4-2 and are identified automatically in EES by defining arrays and using them as the lower limit, upper limit, and guess values for the entries in the eigenvalue array in the Variable Information window (Figure P2.2-43).
Figure P2.2-4-3: Variable Information window. {Res=tan(lambdab)-h*b/(k*lambdab)}
"residual of the eigencondition"
N=20 [-] duplicate i=1,N upperlimit[i]=(i-1/2)*pi lowerlimit[i]=(i-1)*pi guess[i]=(i-3/4)*pi tan(lambdab[i])-h*b/(k*lambdab[i])=0 lambda[i]=lambdab[i]/b end
The eigenfunctions are:
θ Yi = C2,i cos ( λi y )
(21)
The solution to Eq. (13) for each eigenvalue is:
θ X i = C3,i sinh ( λi x ) + C4,i cosh ( λi x )
(22)
The general solution for for each eigenvalue is:
θi = θ X i θ Yi = C2,i cos ( λi y ) ⎡⎣C3,i sinh ( λi x ) + C4,i cosh ( λi x ) ⎤⎦
(23)
The sum of these solutions is, itself, a solution (note that the constant C2,i is absorbed into the other constants): ∞
∞
i =1
i =1
θ = ∑ θi = ∑ cos ( λi y ) ⎡⎣C3,i sinh ( λi x ) + C4,i cosh ( λi x ) ⎤⎦
(24)
Equation (24) is substituted into the boundary condition at x= 0, Eq. (8): ∂θ ∂x
∞
x =0
= ∑ cos ( λi y ) ⎡⎣C3,i λi cosh ( λi 0 ) + C4,i λi sinh ( λi 0 ) ⎤⎦
(25)
i =1
or ∞
∑ cos ( λ y ) C i =1
i
3,i
λi = 0
(26)
which can only be true if C3,i = 0; therefore: ∞
θ = ∑ Ci cos ( λi y ) cosh ( λi x ) i =1
(27)
where the constant C4,i is referred to simply as Ci since it is the only remaining undetermined constant associated with each eigenvalue. Equation (27) is substituted into the boundary condition at x = W, Eq. (9): ∞
k ∑ Ci cos ( λi y ) λi cosh ( λi W ) = qh′′
(28)
i =1
The eigenfunctions are orthogonal; therefore, Eq. (28) is multiplied by an arbitary eigenfunction and integrated from y = 0 to y = b. The result is: Ci k λi cosh ( λi W )
b
b
∫ cos ( λ y ) dy = q′′ ∫ cos ( λ y ) dy 2
i
y =0
i
(29)
y =0
The two integrals in Eq. (29) are evaluated in Maple: > restart; > int((cos(lambda*y))^2,y=0..b);
1 cos( λ b ) sin( λ b ) + λ b 2 λ
> int(cos(lambda*y),y=0..b);
sin( λ b ) λ
Therefore, the constants can be evaluated according to: Ci = q ′′
sin ( λi b ) k λ cosh ( λi W ) Integrali 2 i
(30)
where ⎡cos ( λi b ) sin ( λi b ) + λi b ⎤⎦ Integrali = ⎣ 2 λi duplicate i=1,N Integral[i]=(cos(lambdab[i])*sin(lambdab[i])+lambdab[i])/(2*lambda[i]) C[i]=q_h*sin(lambdab[i])/(k*lambda[i]^2*cosh(lambdab[i])*Integral[i]) end
The solution is obtained according to Eq. (27) at a particular position: x_bar=0.5 [-] y_bar=0.5 [-] x=x_bar*W y=y_bar*b
(31)
duplicate i=1,N theta[i]=C[i]*cos(lambda[i]*y)*cosh(lambda[i]*x) end T=T_infinity+sum(theta[i],i=1,N) T_C=converttemp(K,C,T)
b.) Prepare a contour plot of the temperature distribution in the substrate. Figure P2.2-4-4 illustrates a contour plot of temperature in the substrate. 1 0.9
contours of constant temperature (°C)
0.8 0.7 55.72
y/b
0.6 49.9
0.5
61.54
44.08
0.4
32.45 38.27
0.3
67.36
0.2
73.17
0.1 0 0
78.99
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x/W
Figure P2.2-4-4: Contour plot of temperature.
The position of temperature sensors #1 and #2 at a particular value of x are y1 = 1.9 mm and y2 = 1.8 mm, respectively (see Figure P2.2-4(b)). The surface temperature measurement extracted from these measured temperatures is associated with a linear extrapolation to the surface at y = 0: Ts ,m = T2 + (T1 − T2 )
( b − y2 ) ( y1 − y2 )
(32)
The heat flux measurement extracted from these measured temperatures is obtained from Fourier's law according to:
qs′′,m = k
(T2 − T1 ) ( y1 − y2 )
(33)
c.) What is the heat transfer coefficient measured by the device at x/W = 0.5? That is, based on the temperatures T1 and T2 predicted by your model at x/W = 0.5, calculate the measured heat transfer coefficient according to:
hm =
(T
qs′′,m
s ,m
− T∞ )
(34)
and determine the discrepancy of your measurement relative to the actual heat transfer coefficient.
The temperatures measured by the sensor set at x/W = 0.5 are evaluated: duplicate i=1,N theta_1[i]=C[i]*cos(lambda[i]*y_1)*cosh(lambda[i]*x) theta_2[i]=C[i]*cos(lambda[i]*y_2)*cosh(lambda[i]*x) end T_1=T_infinity+sum(theta_1[i],i=1,N) T_2=T_infinity+sum(theta_2[i],i=1,N)
The surface temperature, heat flux, and heat transfer coefficient are evaluated according to Eqs. (32) through (34): T_s_m=T_2+(T_1-T_2)*(b-y_2)/(y_1-y_2) q_s_m=k*(T_2-T_1)/(y_1-y_2) h_m=q_s_m/(T_s_m-T_infinity)
which leads to hm = 2358 W/m2-K. This is 5.6% in error relative to the actual heat transfer coefficient. d.) Plot the % error associated with the device configuration (i.e., the discrepancy between the measured and actual heat transfer coefficient from part (c)) as a function of axial position, x. The % error is calculated according to: deltah=100*abs(h-h_m)/h
and shown in Figure P2.2-4-5 as a function of x/W. Error in heat transfer coefficient (%)
20 18 16 14 12 10 8 6 4 2 0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x/W
Figure P2.2-4-5: Error as a function of axial location.
Problem 2.2-5 Three heater blocks provide heat to the back-side of a fin array that must be tested, as shown in Figure P2.2-5. 4 2 q1′′ = 3x10 W/m 4 2 q2′′ = 3x10 W/m
k = 30 W/m-K c = 1 cm
y th = 1 cm
L = 7 cm
d = 4 cm
x 2 h f = 35 W/m -K T f = 35°C N fins = 150 2 As, fin = 8 cm 2 Ab, fin = 0.8 cm η fin = 0.85
2 h = 20 W/m -K T∞ = 20°C
Figure P2.2-5: An array of fins installed on a base plate energized by three heater blocks.
The fins are installed on a base plate that has half-width of L= 7 cm, thickness of th = 1 cm, and width W = 20 cm (into the page). The base plate material has conductivity k = 30 W/m-K. The edge of the base plate (at x = L) is exposed to air at T∞ = 20°C with heat transfer coefficient h = 20 W/m2-K. The middle of the plate (at x = 0) is a line of symmetry and can be modeled as being adiabatic. The bottom of the plate (at y = 0) has an array of Nfin = 150 fins installed. Each fin has surface area As,fin= 8 cm2, base area Ab,fin = 0.8 cm2, and efficiency ηfin = 0.85. The fin and the base material are exposed to fluid at Tf = 35°C with heat transfer coefficient h f = 20 W/m2-K. The top of the plate (at y = th) is exposed to the heat flux from the heater blocks. The heat flux is distributed according to:
q ′′y =th
⎧q1′′ if x < c ⎪ = ⎨0 if c < x < ( d + c ) ⎪q ′′ if x > ( d + c ) ⎩ 2
where q1′′ = 3x104 W/m2, q2′′ = 3x104 W/m2, c = 1 cm and d = 4 cm. a.) Determine an effective heat transfer coefficient that can be applied to the surface at y = th in order to capture the combined effect of the fins and the unfinned base area. The known information is entered in EES: "PROBLEM 2.2-5" $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" k=30 [W/m-K] h_bar=20 [W/m^2-K]
"thermal conductivity" "heat transfer coefficient to ambient air"
W=20 [cm]*convert(cm,m) L=7 [cm]*convert(cm,m) th=1 [cm]*convert(cm,m) h_bar_f=35 [W/m^2-K] A_s_fin=8 [cm^2]*convert(cm^2,m^2) A_b_fin=0.8 [cm^2]*convert(cm^2,m^2) N_fin=150 [-] eta_fin=0.85 [-] T_infinity=converttemp(C,K,20[C]) T_f=converttemp(C,K,35 [C]) c=1 [cm]*convert(cm,m) d=4 [cm]*convert(cm,m) q``_1=3e4 [W/m^2] q``_2=3e4 [W/m^2]
"width of base" "half-length of base" "thickness of base" "heat transfer coefficient to fluid" "surface area of each fin" "base area of each fin" "number of fins" "fin efficiency" "ambient air temperature" "fluid temperature" "width of first heating zone" "space between heating zones" "heat flux in 1st heating zone" "heat flux in 2nd heating zone"
The effective heat transfer coefficient is defined so that it provides the same rate of heat transfer as the finned surface:
W L heff = (W L − N fins Ab , fin ) h f + N fins As , fin η fin h f
(1)
W*L*h_bar_eff=(W*L-N_fin*A_b_fin)*h_bar_f+eta_fin*N_fin*A_s_fin*h_bar_f "effective heat transfer coefficient"
which leads to heff = 260 W/m2-K. b.) Develop a separation of variables solution for the temperature distribution within the fin base material. The problem shown in Figure P2.2-5 is governed by the transformed partial differential equation: ∂ 2θ ∂ 2θ + =0 ∂x 2 ∂y 2
(2)
and has the boundary conditions: ∂θ ∂x
=0
(3)
= h θ x=L
(4)
⎧q1′′ if x < c ⎪⎪ = ⎨0 if c < x < ( d + c ) ⎪ ⎪⎩q2′′ if x > ( d + c )
(5)
−k
∂θ k ∂y
y =th
∂θ ∂y
x =0
x=L
∂θ heff ⎡⎣θ f − θ y =0 ⎤⎦ = − k ∂y
(6) y =0
where
θ = T − T∞
(7)
θ f = T f − T∞
(8)
θ = θ X θY
(9)
and
The separated solution is assumed:
and substituted into Eq. (2); the process leads to two ordinary differential equations:
d 2θ X + λ2 θ X = 0 dx 2
(10)
d 2θ Y − λ 2 θY = 0 2 dy
(11)
Notice that the sign of the constant is selected so that the solution in the homogeneous direction (x) is sines and cosines. The solution to Eq. (10) is:
θ X = C1 sin ( λ x ) + C2 cos ( λ x )
(12)
The boundary condition at x = 0, Eq. (3) requires that C1 = 0 and therefore:
θ X = C2 cos ( λ x )
(13)
The boundary condition at x = L, Eq. (4), leads to the eigencondition for the problem:
tan ( λi L ) =
hL k ( λi L )
(14)
The eigenvalues lie in regular intervals of λL and are identified automatically in EES by defining arrays and using them as the lower limit, upper limit, and guess values for the entries in the eigenvalue array in the Variable Information window. Nterm=11 [-] "number of terms to use in the solution" "Setup guess values and lower and upper bounds for eigenvalues"
duplicate i=1,Nterm lowerlimit[i]=(i-1)*pi upperlimit[i]=lowerlimit[i]+pi/2 guess[i]=lowerlimit[i]+pi/4 end "Identify eigenvalues" duplicate i=1,Nterm tan(lambdaL[i])=h_bar*L/(k*lambdaL[i]) lambda[i]=lambdaL[i]/L end
"eigencondition" "eigenvalue"
The eigenfunctions are:
θ X i = C2,i cos ( λi x )
(15)
The solution to Eq. (11) for each eigenvalue is:
θ Yi = C3,i sinh ( λi y ) + C4,i cosh ( λi y )
(16)
The general solution for for each eigenvalue is:
θi = θ X i θ Yi = C2,i cos ( λi x ) ⎡⎣C3,i sinh ( λi y ) + C4,i cosh ( λi y ) ⎤⎦
(17)
The sum of these solutions is, itself, a solution (note that the constant C2,i is absorbed into the other constants): ∞
∞
i =1
i =1
θ = ∑ θi = ∑ cos ( λi x ) ⎡⎣C3,i sinh ( λi y ) + C4,i cosh ( λi y ) ⎤⎦
(18)
Equation (18) is substituted into the boundary condition at y = 0, Eq. (6): ∞ ∞ ⎡ ⎤ heff ⎢θ f − ∑ C4,i cos ( λi x ) ⎥ = − k ∑ C3,i λi cos ( λi x ) i =1 i =1 ⎣ ⎦
(19)
Equation (19) is multiplied by an eigenfunction and integrated from x = 0 to x = L: L
L
L
0
0
0
heff θ f ∫ cos ( λi x ) dx − heff C4,i ∫ cos 2 ( λi x ) dx = − k C3,i λi ∫ cos 2 ( λi x ) dx
(20)
or L
L
heff θ f ∫ cos ( λi x ) dx = ⎡⎣ heff C4,i − k C3,i λi ⎤⎦ ∫ cos 2 ( λi x ) dx 0 0
Integral1
(21)
The interals in Eq. (21) are evaluated in Maple: > restart; > int(cos(lambda[i]*x),x=0..L);
sin( λ i L ) λi
> int((cos(lambda[i]*x))^2,x=0..L);
1 cos( λ i L ) sin( λ i L ) + λ i L λi 2
The integrals are put into EES, leading to a relationship between C3,i and C4,i : duplicate i=1,Nterm Int1[i]=1/2*(cos(lambda[i]*L)*sin(lambda[i]*L)+lambda[i]*L)/lambda[i] h_bar_eff*(T_f-T_infinity)*sin(lambda[i]*L)/lambda[i]=(h_bar_eff*C4[i]-k*C3[i]*lambda[i])*Int1[i] end
Equation (18) is substituted into the boundary condition at y = th, Eq. (5): ⎧q1′′ if x < c ⎪⎪ λi k ∑ cos ( λi x ) ⎣⎡C3,i cosh ( λi th ) + C4,i sinh ( λi th ) ⎦⎤ = ⎨0 if c < x < ( d + c ) i =1 ⎪ ⎪⎩q2′′ if x > ( d + c ) ∞
(22)
Again, the orthogonality of the eigenfunctions are used to obtain: L
c
λi k ⎡⎣C3,i cosh ( λi th ) + C4,i sinh ( λi th ) ⎤⎦ ∫ cos 2 ( λi x ) dx = q1′′ ∫ cos 2 ( λi x ) dx + q2′′ 0
0
The integrals are carried out in Maple: > int(cos(lambda[i]*x),x=0..c);
sin( λ i c ) λi
> int(cos(lambda[i]*x),x=(d+e)..L);
−sin( λ i ( d + e ) ) + sin( λ i L ) λi
and used to complete the computation of the constants in EES: duplicate i=1,Nterm
L
∫ cos ( λ x ) dx (23) 2
i
d +c
Int2[i]=(-sin(lambda[i]*(d+c))+sin(lambda[i]*L))/lambda[i] k*(C3[i]*lambda[i]*cosh(lambda[i]*L)+C4[i]*lambda[i]*sinh(lambda[i]*L))*Int1[i]=& q``_1*sin(lambda[i]*c)/lambda[i]+q``_2*Int2[i] end
The solution is evaluated at an arbitary position: x_bar=0.5 [-] y_bar=0 [-] x=x_bar*L y=y_bar*th duplicate i=1,Nterm theta[i]=cos(lambda[i]*x)*(C3[i]*sinh(lambda[i]*y)+C4[i]*cosh(lambda[i]*y)) end T=sum(theta[1..Nterm])+T_infinity T_C=converttemp(K,C,T)
c.) Prepare a plot showing the temperature as a function of x at various values of y. Figure 2 illlustrates the temperature as a function of x/L at various values of y/th. 83 y/th = 1
Temperature (°C)
82 y/th = 0.75 81 y/th = 0.5 80
y/th = 0.25
79
78 0
y/th = 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Dimensionless axial position, x/L Figure 2: Temperature as a function of x/L for various values of y/th.
d.) The goal of the base plate is to provide an uniform heat flow to each fin. Assess the performance of the base plate by plotting the rate of heat flux transferred to the fluid as a function of x at y = 0. The heat flux to the fluid is computed according to:
q ′′f = heff (Ty =0 − T f
)
(24)
q``_f=h_bar_eff*(T-T_f)
Figure 3 shows the heat flux to the fluid as a function of axial location. 19000 3 W/m-K
2
Heat flux to fluid (W/m )
18000 17000 16000 15000 14000
10 W/m-K
13000 30 W/m-K 100 W/m-K
12000 11000 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Dimensionless axial location, x/L Figure 3: Heat flux to fluid as a function of x/L.
e.) Overlay on your plot for (d) the rate of heat flux transferred to the fluid for various values of the base plate conductivity. Figure 3 includes various values of conductivity and shows that as conductivity increases, more energy is lost to the edge and the heat flux along the bottom is more uniform.
Problem 2.2-6 (2-4 in text) A laminated composite structure is shown in Figure P2.2-6. H = 3 cm
2 q ′′ = 10000 W/m
Tset = 20°C W = 6 cm
Tset = 20°C
kx = 50 W/m-K ky = 4 W/m-K Figure P2.2-6: Composite structure exposed to a heat flux.
The structure is anisotropic. The effective conductivity of the composite in the x-direction is kx = 50 W/m-K and in the y-direction it is ky = 4 W/m-K. The top of the structure is exposed to a heat flux of q ′′ = 10,000 W/m2. The other edges are maintained at Tset = 20°C. The height of the structure is H = 3 cm and the half-width is W = 6 cm. a.) Develop a separation of variables solution for the 2-D steady-state temperature distribution in the composite. The known information is entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" W=6 [cm]*convert(cm,m) H=3 [cm]*convert(cm,m) q``=10000 [W/m^2] k_x=50 [W/m-K] k_y=4 [W/m-K] T_set=converttemp(C,K,20[C])
"width of laminate" "height of laminate" "heat flux" "conductivity in the x-direction" "conductivity in the y-direction" "specified edge temperatures"
A half-symmetry model of the problem shown in Figure P2.2-6 is governed by the transformed partial differential equation: kx
∂ 2θ ∂ 2θ + k =0 y ∂x 2 ∂y 2
(1)
with the boundary conditions: ∂θ ∂x
=0
(2)
x =0
θ x =W = 0
(3)
ky
∂θ ∂y
= q ′′
(4)
y=H
θ y =0 = 0
(5)
θ = T − Tset
(6)
θ = θ X θY
(7)
where
The separated solution is assumed:
and substituted into Eq. (1):
kx θY
d 2θ X d 2θ Y k θ X + y =0 dx 2 dx 2
(8)
Dividing through by θY θX leads to:
d 2θ X d 2θ Y dx 2 + k y dx 2 = 0 k θY θX
x
−λ2
(9)
λ2
The process leads to two ordinary differential equations:
d 2θ X + λ2 θ X = 0 2 dx
(10)
d 2θ Y − β 2 λ 2 θY = 0 dy 2
(11)
where
β2 =
kx ky
(12)
beta=sqrt(k_x/k_y)
Notice that the sign of the constant is selected so that the solution in the homogeneous direction (x) is sines and cosines. The solution to Eq. (10) is:
θ X = C1 sin ( λ x ) + C2 cos ( λ x )
(13)
The boundary condition at x = 0, Eq. (2) requires that C1 = 0 and therefore:
θ X = C2 cos ( λ x )
(14)
The boundary condition at x = L, Eq. (3), leads to the eigencondition for the problem: cos ( λi W ) = 0
(15)
which requires that:
λi =
(1 + 2 i ) π 2W
for i = 0,1,...∞
N_term=11 [-] duplicate i=0,N_term lambda[i]=(1+2*i)*pi/(2*W) end
(16)
"number of terms" "eigenvalues"
The eigenfunctions are:
θ X i = C2,i cos ( λi x )
(17)
The solution to Eq. (11) for each eigenvalue is:
θ Yi = C3,i sinh ( β λi y ) + C4,i cosh ( β λi y )
(18)
The general solution for for each eigenvalue is:
θi = θ X i θ Yi = C2,i cos ( λi x ) ⎡⎣C3,i sinh ( β λi y ) + C4,i cosh ( β λi y ) ⎤⎦
(19)
The sum of these solutions is, itself, a solution (note that the constant C2,i is absorbed into the other constants): ∞
∞
i =1
i =1
θ = ∑ θi = ∑ cos ( λi x ) ⎡⎣C3,i sinh ( β λi y ) + C4,i cosh ( β λi y ) ⎤⎦
(20)
The boundary condition at y = 0, Eq. (5), leads to C4,i = 0: ∞
∞
i =1
i =1
θ = ∑ θi = ∑ Ci cos ( λi x ) sinh ( β λi y )
(21)
Equation (21) is substituted into the boundary condition at y = H, Eq. (4): ∞
β λi k y ∑ Ci cos ( λi x ) cosh ( β λi H ) = q ′′
(22)
i =1
Equation (22) is multiplied by an eigenfunction and integrated from x = 0 to x = W: L
L
0
0
β λi k y cosh ( β λi H ) Ci ∫ cos 2 ( λi x ) dx = q ′′∫ cos ( λi x ) dx The integrals are carried out in Maple: > restart; > assume(i,integer); > lambda:=(1+2*i)*Pi/(2*W);
λ :=
( 1 + 2 i~ ) π 2W
> int((cos(lambda*x))^2,x=0..W);
W 2 > int(cos(lambda*x),x=0..W);
2 ( -1 )i~ W ( 1 + 2 i~ ) π
and used to complete the computation of the constants in EES: duplicate i=0,N_term k_y*C[i]*beta*lambda[i]*cosh(beta*lambda[i]*H)*W/2=q``*2*(-1)^i*W/((1+2*i)*Pi) end
The solution is evaluated at an arbitary position: x_bar=0.5 [-] y_bar=0.5 [-] x=x_bar*W y=y_bar*H duplicate i=0,N_term theta[i]=C[i]*cos(lambda[i]*x)*sinh(beta*lambda[i]*y) end theta=sum(theta[0..N_term]) T=theta+T_set T_C=converttemp(K,C,T)
b.) Prepare a contour plot of the temperature distribution.
"dimensionless x-position" "dimensionless y-position" "x-position" "y-position"
(23)
Figure 2 illlustrates a contour plot of the temperature distribution. 1
Dimensionless y position, y/H
20 23.12
0.8
26.24 29.36 32.48
0.6
35.6 38.72
0.4
41.84 44.96 48.08
0.2
51.2 0 0
0.2
0.4
0.6
0.8
Dimensionless x position, x/W Figure 2: Contour plot of the temperature distribution.
1
Problem 2.3-1 (2-5 in text): Cryogen Transfer Pipe Figure P2.3-1 illustrates a pipe that connects two tanks of liquid oxygen on a spacecraft. The pipe is subjected to a heat flux, q′′ = 8,000 W/m2, which can be assumed to be uniformly applied to the outer surface of the pipe and entirely absorbed. Neglect radiation from the surface of the pipe to space. The inner radius of the pipe is rin = 6 cm, the outer radius of the pipe is rout = 10 cm, and the half-length of the pipe is L = 10 cm. The ends of the pipe are attached to the liquid oxygen tanks and therefore are at a uniform temperature of TLOx = 125 K. The pipe is made of a material with a conductivity of k = 10 W/m-K. The pipe is empty and therefore the internal surface can be assumed to be adiabatic. rout = 10 cm rin = 6 cm
L = 10 cm TLOx = 125 K
2 k = 10 W/m-K q′′s = 8,000 W/m Figure P2.3-1: Cryogen transfer pipe connecting two liquid oxygen tanks.
a.) Develop an analytical model that can predict the temperature distribution within the pipe. Prepare a contour plot of the temperature distribution within the pipe. A differential control volume leads to the energy balance: qx + qr = qx + dx + qr + dr or
0=
∂qx ∂q dx + r dr ∂x ∂r
Substituting the rate equations: qx = − k 2 π r dr
∂T ∂x
qr = − k 2 π r dx
∂T ∂r
and
into the differential energy balance leads to:
0=
∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ dx + ⎢ −k 2 π r dx ⎥ dr − k 2 π r dr ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂r ⎣ ∂r ⎦
or r
∂ 2T ∂ ⎡ ∂T ⎤ + r =0 ∂x 2 ∂r ⎢⎣ ∂r ⎥⎦
A half-symmetry model of the pipe will be generated; the boundary conditions are therefore: ∂T ∂x
=0 x =0
Tx = L = TLOx
k
∂T ∂r
r = rin
∂T ∂r
r = rout
=0
= q′′
As stated, there are two non-homogeneous boundary conditions; however, the boundary condition at x = L can be made homogeneous by defining the temperature difference:
θ = T − TLOx The partial differential equation and boundary conditions are written in terms of θ: r
∂ 2θ ∂ ⎡ ∂θ ⎤ + r =0 ∂x 2 ∂r ⎢⎣ ∂r ⎥⎦ ∂θ ∂x
=0
(2)
x =0
θ x= L = 0 ∂θ ∂r
(1)
=0 r = rin
(3) (4)
k
∂θ ∂r
= q′′
(5)
r = rout
Note that the two homogeneous boundary conditions are in the x-direction and so the eigenfunctions of the problem will be in this direction. We assume that the solution is separable; that is, the solution is the product of a function only of x (θX) and r (θR):
θ ( x, y ) = θ X ( x ) θ R ( r ) Substituting the product solution into the governing partial differential equation, Eq. (1), leads to: rθ R
d 2θ X d ⎡ dθ R ⎤ +θ X r =0 2 dx dr ⎢⎣ dr ⎥⎦
Dividing by the product r θX θR leads to: d ⎡ dθ R ⎤ d 2θ X ⎢r ⎥ 2 dx + dr ⎣ dr ⎦ = 0 rθ R θX ±λ2
∓λ2
Note that the 1st term is a function only of x while the 2nd term is a function only of r; these two quantities must be equal and opposite constants (±λ2). The choice of the sign is again important; the eigenfunctions must be in x and therefore the two ordinary differential equations must be: d 2θ X + λ2 θ X = 0 dx 2
(6)
d ⎡ dθ R ⎤ r − λ2 rθ R = 0 ⎢ ⎥ dr ⎣ dr ⎦
(7)
The eigenproblem will be solved first; the solution to Eq. (6) is:
θ X = C1 cos ( λ x ) + C2 sin ( λ x ) The boundary condition at x = 0, Eq. (2), eliminates the sine term. The boundary condition at x = L, Eq. (3), leads to: C1 cos ( λ L ) = 0
which provides the eigenvalues and the eigenfunctions:
θ X i = C1,i cos ( λi x ) where λi =
( 2 i − 1) π 2L
for i = 1, 2,..∞
The ordinary differential equation for θR, Eq. (7), is solved by Bessel functions, as discussed in Section 1.8. Equation (7) is a form of Bessel's equation: d ⎛ p dθ ⎞ 2 s ⎜x ⎟±c x θ =0 dx ⎝ dx ⎠ where s = 1 and p = 1; the quantity s – p +2 is therefore not equal to zero and we are directed toward the left-side of the Bessel function flow chart presented in Section 1.8.4 where the parameters n = 0, a = 1, and n/a = 0 are computed. The last term is negative and therefore the general solution to Eq. (7) is:
θ Ri = C3,i BesselI ( 0,λi r ) + C4,i BesselK ( 0,λi r ) This general solution could have been obtained using Maple: > restart; > ODEr:=diff(r*diff(thetar(r),r),r)-lambda^2*r*thetar(r)=0; 2 d ⎛d ⎞ ⎛ ⎞ ⎜ ODEr := ⎜⎜ thetar( r ) ⎟⎟ + r ⎜ 2 thetar( r ) ⎟⎟ − λ 2 r thetar( r ) = 0 ⎝ dr ⎠ ⎝ dr ⎠
> thetars:=dsolve(ODEr);
thetars := thetar( r ) = _C1 BesselI( 0, λ r ) + _C2 BesselK( 0, λ r )
The general solution for each eigenvalue is therefore:
θi = θ X i θ Ri = cos ( λi x ) ⎡⎣C3,i BesselI ( 0,λi r ) + C4,i BesselK ( 0,λi r ) ⎤⎦ The general solution is entered as a function of x and y in Maple: > restart; > assume(i,integer); > lambda:=(2*i-1)*Pi/(2*L);
λ :=
( 2 i~ − 1 ) π 2L
> theta:=(x,r)->cos(lambda*x)*(C3*BesselI(0,lambda*r)+C4*BesselK(0,lambda*r));
θ := ( x, r ) → cos( λ x ) ( C3 BesselI( 0, λ r ) + C4 BesselK( 0, λ r ) )
Verify that it solves both boundary conditions in the x-directions, Eqs. (2) and (3): > eval(diff(theta(x,r),x),x=0);
0
> theta(L,r);
0
and the partial differential equation, Eq. (1): > r*diff(diff(theta(x,r),x),x)+diff(r*diff(theta(x,r),r),r);
1 ( 2 i~ − 1 ) π x ⎞ 2 2 − r cos⎛⎜⎜ ⎟⎟ ( 2 i~ − 1 ) π 2 L 4 ⎝ ⎠ ⎛ ( 2 i~ − 1 ) π r ⎞ + C4 BesselK⎛ 0, ( 2 i~ − 1 ) π r ⎞ ⎞ L 2 + ⎛ ⎜⎜ C3 BesselI⎜⎜ 0, ⎟⎟ ⎜⎜ ⎟⎟ ⎟⎟ 2L 2L ⎝ ⎝ ⎠ ⎝ ⎠⎠ ( 2 i~ − 1 ) π r ⎛ ⎛ ⎞ ( 2 i~ − 1 ) π ⎜ C3 BesselI⎜⎜ 1, ⎟⎟ ⎜ ( 2 i~ − 1 ) π x 1 2 L ⎝ ⎛ ⎞ ⎠ cos⎜⎜ ⎟⎟ ⎜⎜ 2 L 2 L ⎝ ⎠⎝
⎛ ⎜ ⎜ ( 2 i~ − 1 ) π r ⎞ ⎞ ⎛ ⎜ C4 BesselK⎜⎜ 1, ⎟⎟ ( 2 i~ − 1 ) π ⎟ ⎜ ⎟ 2L 1 ( 2 i~ − 1 ) π x ⎜ ⎛ ⎞ ⎝ ⎠ ⎟ + r cos⎜ − ⎟ ⎟ ⎜ ⎟ ⎜⎜ L 2 2 L ⎠ ⎝ ⎠⎝ ( 2 i~ − 1 ) π r ⎞ ⎞ ⎛ 2 L BesselI⎛⎜⎜ 1, ⎜ ⎟⎟ ⎟ ⎜ ( 2 i~ − 1 ) π r 2L ⎞ ⎛ ⎝ ⎠ ⎟⎟ ( 2 i~ − 1 ) 2 π 2 ⎜ C3 ⎜ BesselI⎜⎜ 0, ⎟⎟ − ⎟ 1 2L ( 2 i~ − 1 ) π r ⎠ ⎝ ⎠ ⎝ − 2 4 L ( 2 i~ − 1 ) π r ⎞ ⎞ ⎛ 2 L BesselK⎛⎜⎜ 1, ⎜ ⎟⎟ ⎟ ⎜ ( 2 i~ − 1 ) π r ⎞ 2L ⎝ ⎠ ⎟⎟ ( 2 i~ − 1 ) 2 π 2 − C4 ⎜⎜ −BesselK⎛⎜⎜ 0, ⎟⎟ ⎟ 1 2L ( 2 i~ − 1 ) π r ⎝ ⎠ ⎠ ⎝ 2 4 L
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎟ ⎠
> simplify(%);
0
The general solution for θ is the series: ∞
∞
i =1
i =1
θ = ∑ θi = ∑ cos ( λi x ) ⎡⎣C3,i BesselI ( 0,λi r ) + C4,i BesselK ( 0,λi r ) ⎤⎦
(8)
The solution must satisfy the boundary conditions in the non-homogeneous direction; the boundary condition at r = rin, Eq. (4), leads to:
∂θ ∂r
∞
r = rin
= ∑ cos ( λi x ) i =1
d ⎡C3,i BesselI ( 0,λi r ) + C4,i BesselK ( 0,λi r ) ⎤⎦ r = r = 0 in dr ⎣
(9)
The derivatives of the Bessel functions may either be evaluated using the equations provided in Section 1.8.4 or, more conveniently, using Maple:
> restart; > diff(BesselI(0,lambda*r),r);
BesselI( 1, λ r ) λ
> diff(BesselK(0,lambda*r),r);
−BesselK( 1, λ r ) λ
Therefore Eq. (9) can be written as: ∞
∑ cos ( λ x ) ⎡⎣C i
i =1
3,i
λi BesselI (1,λi rin ) − C4,i λi BesselK (1,λi rin ) ⎤⎦ = 0
which can only be true if: C3,i λi BesselI (1,λi rin ) − C4,i λi BesselK (1,λi rin ) = 0
Unlike most of the problems we have previously encountered, neither of the constants is eliminated; instead we see that there is a relationship between the two constants: C4,i = C3,i
BesselI (1,λi rin ) BesselK (1,λi rin )
(10)
Substituting Eq. (10) into Eq. (8) leads to: ∞
∞
⎡
i =1
i =1
⎣
θ = ∑ θi = ∑ Ci cos ( λi x ) ⎢ BesselI ( 0,λi r ) +
⎤ BesselI (1,λi rin ) BesselK ( 0,λi r ) ⎥ BesselK (1,λi rin ) ⎦
(11)
Equation (11) is substituted into the remaining non-homogeneous boundary condition at r = rout , Eq. (5), in order to obtain: k
∂θ ∂r
r = rout
∞ ⎡ ⎤ BesselI (1,λi rin ) BesselK (1,λi rout ) ⎥ = q′′ = k ∑ Ci λi cos ( λi x ) ⎢ BesselI (1,λi rout ) − BesselK (1,λi rin ) i =1 ⎣ ⎦
This equation is multiplied by an arbitrary eigenfunction, cos(λj x), and integrated between the homogeneous boundary conditions (from x = 0 to x= L); using the orthogonality property of the eigenfunctions we obtain: L ⎡ ⎤L BesselI (1,λi rin ) k Ci λi ⎢ BesselI (1,λi rout ) − BesselK (1,λi rout ) ⎥ ∫ cos 2 ( λi x ) dx = q′′∫ cos ( λi x ) dx BesselK (1,λi rin ) 0 ⎣ ⎦0
which leads to an expression for each constant: ⎤ BesselI (1,λi rin ) sin ( λi L ) k Ci λi L ⎡ BesselK (1,λi rout ) ⎥ = q′′ ⎢ BesselI (1,λi rout ) − 2 BesselK (1,λi rin ) λi ⎣ ⎦ Solving for Ci: Ci =
2 q′′ sin ( λi L ) ⎡
λi2 k L ⎢ BesselI (1,λi rout ) − ⎣
⎤ BesselI (1,λi rin ) BesselK (1,λi rout ) ⎥ BesselK (1,λi rin ) ⎦
The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" q``_dot=8000 [W/m^2] r_in=6.0 [cm]*convert(cm,m) r_out=10.0 [cm]*convert(cm,m) L = 10.0 [cm]*convert(cm,m) T_LOx=125 [K] k=10 [W/m-K]
"Heat flux on pipe surface" "Pipe inner radius" "Pipe outer radius" "Pipe half-length" "Pipe end temperature" "Pipe conductivity"
A position is specified in terms of dimensionless coordinates: "dimensionless position" r_bar=0.5 x_bar=0.5 r=r_in+(r_out-r_in)*r_bar x=x_bar*L
The solution is implemented for the 1st N terms of the series: N=100 "Number of terms" duplicate i=1,N lambda[i]=(2*i-1)*pi/(2*L) C[i]=2*q``_dot*sin(lambda[i]*L)/(lambda[i]^2*k*L*(BesselI(1,lambda[i]*r_out)-& BesselI(1,lambda[i]*r_in)*BesselK(1,lambda[i]*r_out)/BesselK(1,lambda[i]*r_in))) theta[i]=C[i]*cos(lambda[i]*x)*(BesselI(0,lambda[i]*r)+BesselI(1,lambda[i]*r_in)*& BesselK(0,lambda[i]*r)/BesselK(1,lambda[i]*r_in)) end T=sum(theta[1..N])+T_LOx
The temperature distribution is computed over a range of positions and the results are used to generate the contour plot shown in Figure 2.
Figure 2: Contour plot of temperature.
Problem 2.3-2 (2-6 in text) Figure P2.3-2 illustrates a cylinder that is exposed to a concentrated heat flux at one end. extends to infinity k = 168 W/m-K rout = 200 μm Ts = 20°C rexp = 21 μm q ′′ = 1500 W/cm
2
adiabatic
Figure P2.3-2: Cylinder exposed to a concentrated heat flux at one end.
The cylinder extends infinitely in the x-direction. The surface at x = 0 experiences a uniform heat flux of q′′ = 1500 W/cm2 for r < rexp = 21 μm and is adiabatic for rexp < r < rout where rout = 200 μm is the outer radius of the cylinder. The outer surface of the cylinder is maintained at a uniform temperature of Ts = 20ºC. The conductivity of the cylinder material is k = 168 W/m-K. a.) Develop a separation of variables solution for the temperature distribution within the cylinder. Plot the temperature as a function of radius for various values of x. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" r_out=200 [micron]*convert(micron,m) q``_dot=1500 [W/cm^2]*convert(W/cm^2,W/m^2) k=168 [W/m-K] r_exp=21 [micron]*convert(micron,m)
A differential control volume leads to the energy balance: qx + qr = qx + dx + qr + dr or
0= Substituting the rate equations:
∂qx ∂q dx + r dr ∂x ∂r
"outer radius of domain" "exposure flux" "conductivity of work piece" "radius of exposure zone"
qx = − k 2 π r dr
∂T ∂x
qr = − k 2 π r dx
∂T ∂r
and
into the differential energy balance leads to: 0=
∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ dx + ⎢ −k 2 π r dx ⎥ dr − k 2 π r dr ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂r ⎣ ∂r ⎦
or r
∂ 2T ∂ ⎡ ∂T ⎤ + r =0 ∂x 2 ∂r ⎢⎣ ∂r ⎥⎦
The boundary conditions are: −k
∂T ∂x
x =0
⎧⎪q′′ for r < rexp =⎨ ⎪⎩0 for r > rexp
Tx →∞ = Ts Tr =0 must be finite
Tr = rout = Ts As stated, there are two non-homogeneous boundary conditions; however, the boundary condition at r = rout can be made homogeneous by defining the temperature difference:
θ = T − Ts The partial differential equation and boundary conditions are written in terms of θ: r
−k
∂ 2θ ∂ ⎡ ∂θ ⎤ + r =0 ∂x 2 ∂r ⎢⎣ ∂r ⎥⎦
∂θ ∂x
x =0
⎧⎪q′′ for r < rexp =⎨ ⎪⎩0 for r > rexp
(1)
(2)
θ x →∞ = 0
(3)
θ r =0 must be finite
(4)
θ r =r = 0
(5)
out
Note that the two homogeneous boundary conditions are in the x-direction and so the eigenfunctions of the problem will be in this direction. We assume that the solution is separable; that is, the solution is the product of a function only of x (θX) and r (θR):
θ ( x, y ) = θ X ( x ) θ R ( r ) Substituting the product solution into the governing partial differential equation, Eq. (1), leads to: d 2θ X d ⎡ dθ R ⎤ rθ R +θ X r =0 2 dx dr ⎢⎣ dr ⎥⎦
Dividing by the product r θX θR leads to: d ⎡ dθ R ⎤ d 2θ X ⎢r ⎥ 2 dx + dr ⎣ dr ⎦ = 0 rθ R θX ±λ2
∓λ2
Note that the 1st term is a function only of x while the 2nd term is a function only of r; these two quantities must be equal and opposite constants (±λ2). The choice of the sign is again important; the eigenfunctions must be in r and therefore the two ordinary differential equations must be: d 2θ X − λ2 θ X = 0 dx 2
(6)
d ⎡ dθ R ⎤ r + λ2 rθ R = 0 ⎢ ⎥ dr ⎣ dr ⎦
(7)
The eigenproblem will be solved first; the solution to Eq. (7) is:
θ R = C1 BesselJ ( 0,λ r ) + C2 BesselY ( 0,λ r ) The boundary condition at r = 0, Eq. (4), requires that C2 = 0.
θ R = C1 BesselJ ( 0,λ r )
The boundary condition at r = rout, Eq. (5), leads to: C1 BesselJ ( 0,λ rout ) = 0
(8)
The 0th order Bessel function of the 1st kind (i.e., Bessel_J(0,x)) oscillates about zero every time the argument changes by 2π in the same way that sine and cosine do; therefore, there are an infinite number of eigenvalues λi that will satisfy Eq. (8) associated with an infinite number of eigenfunctions. The eigencondition for this problem cannot be used to explicitly solve for the eigenvalues; rather, an implicit equation for the eigenvalues results from Eq. (8): BesselJ ( 0,λi rout ) = 0 where i = 1, 2,...∞
(9)
The eigenfunctions for this problem are:
θ Ri = C1,i BesselJ ( 0,λi r ) where BesselJ ( 0,λi b ) = 0 for i = 1, 2,...∞ N=51 duplicate i=1,N lowerlimit[i]=(i-1)*pi/r_out upperlimit[i]=i*pi/r_out guess[i]=lowerlimit[i]+pi/(2*r_out) end
(11)
"number of terms in solution" "lower limit of eigenvalue" "upper limit of eigenvalue" "guess value for eigenvalue"
duplicate i=1,N BesselJ(0,lambda[i]*r_out)=0 end
"solve for eigenvalues"
The arrays lowerlimit[], upperlimit[], and guess[] are used to constrain the solution for the array lambda[] in the Variable Information window. The solution to the ordinary differential equation for θX, Eq. (6), is:
θ X i = C3,i exp ( λi x ) + C4,i exp ( −λi x ) The boundary condition at x → ∞, Eq. (3), requires that C3,i = 0:
θ X i = C4,i exp ( −λi x ) and so the general solution for each eigenvalue is:
θ i = θ Ri θ X i = Ci BesselJ ( 0,λi r ) exp ( −λi x ) The series solution for θ is:
∞
∞
i =1
i =1
θ = ∑ θi = ∑ Ci BesselJ ( 0,λi r ) exp ( −λi x )
(10)
Substituting Eq. (10) into the boundary condition at x = 0, Eq. (2), leads to: ∞ ⎧⎪q′′ for r < rexp k ∑ Ci λi BesselJ ( 0,λi r ) = ⎨ i =1 ⎪⎩0 for r > rexp
(11)
Equation (11) is multiplied by r BesselJ ( 0,λi r ) and integrated from r = 0 to r = rout: k Ci λi
rout
∫
r BesselJ ( 0,λi r ) dr = q′′ 2
0
rexp
∫ r BesselJ ( 0,λ r ) dr
(12)
i
0
Integral 1
Integral 2
The integrals in Eq. (12) are carried out in Maple: > restart; > Integral1:=int(r*(BesselJ(0,lambda[i]*r))^2,r=0..r_out);
Integral1 :=
2
2
1 r_out ( π λ i r_out BesselJ( 0, λ i r_out ) + π λ i r_out BesselJ( 1, λ i r_out ) ) 2 πλ i
> Integral2:=int(r*BesselJ(0,lambda[i]*r),r=0..r_exp);
Integral2 :=
r_exp BesselJ( 1, r_exp λ i ) λi
and used to complete the solution in EES: duplicate i=1,N Integral1[i]=1/2*r_out/Pi^(1/2)/lambda[i]*(Pi^(1/2)*lambda[i]*r_out*BesselJ(0,lambda[i]*r_out)^2+& Pi^(1/2)*lambda[i]*r_out*BesselJ(1,lambda[i]*r_out)^2) Integral2[i]= 1/lambda[i]*r_exp*BesselJ(1,r_exp*lambda[i]) k*lambda[i]*C[i]*Integral1[i]=q``_dot*Integral2[i] end
The solution is obtained at an arbitrary position according to: x=0 [micron]*convert(micron,m) r=0 [micron]*convert(micron,m) duplicate i=1,N theta[i]=C[i]*BesselJ(0,lambda[i]*r)*exp(-lambda[i]*x) end theta=sum(theta[1..N])
"x-position" "r-position" "i'th term"
Temperature difference relative to Ts (K)
Figure 2 shows the temperature elevation relative to Ts as a function of radius for various values of x. 1.2 x=0 x = 5 μm x = 10 μ m x = 25 μ m x = 50 μ m
1 0.8 0.6 0.4 0.2 0 0
20
40
60
80
100
120
140
160
180
200
Radius (μm) Figure 2: Temperature difference as a function of r for various values of x.
b.) Determine the average temperature of the cylinder at the surface exposed to the heat flux. The average temperature of the cylinder over the region x = 0 and 0 < r < rexp is: 1 T = 2 π rexp
rexp
∫ 2π r T
x =0
(13)
dr
0
or: rexp
1 T = Ts + 2 π rexp
∫ 2π r θ
x =0
dr
(14)
0
Substitutinge Eq. (10) into Eq. (14) leads to: T = Ts +
1 2 π rexp
rexp
∫
∞
2 π r ∑ Ci BesselJ ( 0,λi r ) dr
(15)
i =1
0
which can be rearranged: 2 T = Ts + 2 rexp
rexp
∞
∑ C ∫ r BesselJ ( 0,λ r ) dr i =1
i
i
0
Integral 2
(16)
"compute average temperature in flux region" duplicate i=1,N theta_bar[i]=C[i]*2*Integral2[i]/r_exp^2 end theta_bar=sum(theta_bar[1..N])
c.) Define a dimensionless thermal resistance between the surface exposed to the heat flux and Ts. Plot the dimensionless thermal resistance as a function of rout/rin. A dimensionless thermal resistance is defined by normalizing the actual resistance against the reistance to axial conduction through a cylinder that is rout in radius and rout long: 2 2 R π rout k (T − Ts ) π rout k (T − Ts ) ⎛ rout R= k⎜ = = 2 rout q′′ π rexp rout q′′ rout ⎜⎝ rexp
⎞ ⎟⎟ ⎠
2
(17)
actual resistance
R_bar=theta_bar*k/(q``_dot*r_exp)
"dimensionless thermal resistance"
Figure 3 illustrates the dimensionless thermal resistance as a function of rout/rexp. Dimensionless thermal resistance
80
10
1
0.1 1
10
100
Ratio of cylinder to exposure radii, rout/rexp Figure 3: Dimensionless thermal resistance as a function of rout/rexp.
d.) Show that your plot from (c) does not change if the problem parameters (e.g., Ts, k, etc.) are changed. The values of various parameters were changed and did not affect Figure 3 (the plots are overlaid onto Figure 3).
Problem 2.3-3 A disk-shaped window in an experiment is shown in Figure P2.3-3.
th = 1 cm x
2 h = 50 W/m -K T∞ = 20°C k = 1.2 W/m-K
Rw = 3 cm r
Tedge = 25°C ′′ = 1000 W/m2 qrad Figure P2.3-3: Window.
The inside of the window (the surface at x = 0) is exposed to vacuum and therefore does not ′′ = 1000 experience any convection. However, this surface is exposed to a radiation heat flux qrad 2 W/m . The window is assumed to be completely opaque to this radiation and therefore it is absorbed at x = 0. The edge of the window at Rw = 3 cm is maintained at a constant temperature Tedge = 25°C. The outside of the window (the surface at x = th) is cooled by air at T∞ = 20°C with heat transfer coefficient h = 50 W/m2-K. The conductivity of the window material is k = 1.2 W/m-K. a.) Is the extended surface approximation appropriate for this problem? That is, can the temperature in the window be approximated as being 1-D in the radial direction? Justify your answer. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" th=1 [cm]*convert(cm,m) R_w=3 [cm]*convert(cm,m) q``_rad=1000 [W/m^2] k=1.2 [W/m-K] T_edge=converttemp(C,K,25 [C]) T_infinity=converttemp(C,K,20 [C]) h_bar=50 [W/m^2-K]
"thickness of window" "outer radius of window" "radiation heat flux" "conductivity" "edge temperature" "ambient temperature" "heat transfer coefficient"
A Biot number that characterizes the resistance to conduction in the axial direction to convection is:
Bi = Bi=h_bar*th/k
"Biot number"
h th k
(1)
which leads to Bi = 0.42; this is not sufficiently small relative to 1 to justify an extended surface solution. b.) Assume that your answer to (a) is no. Develop a 2-D separation of variables solution to this problem. A differential control volume leads to the energy balance:
qx + qr = qx + dx + qr + dr or
0=
∂qx ∂q dx + r dr ∂x ∂r
Substituting the rate equations:
qx = − k 2 π r dr
∂T ∂x
qr = − k 2 π r dx
∂T ∂r
and
into the differential energy balance leads to: 0=
∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ dx + ⎢ −k 2 π r dx ⎥ dr − k 2 π r dr ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂r ⎣ ∂r ⎦
or r
∂ 2T ∂ ⎡ ∂T ⎤ + r =0 ∂x 2 ∂r ⎢⎣ ∂r ⎥⎦
The boundary conditions are: −k
−k
∂T ∂x
∂T ∂x
′′ = qrad x =0
= h (Tx =th − T∞ ) x = th
Tr =0 must be finite
Tr = Rw = Tedge As stated, there are two non-homogeneous boundary conditions; however, the boundary condition at r = Rw can be made homogeneous by defining the temperature difference:
θ = T − Tedge The partial differential equation and boundary conditions are written in terms of θ: ∂ 2θ ∂ ⎡ ∂θ ⎤ r 2 + ⎢r =0 ∂x ∂r ⎣ ∂r ⎥⎦ −k
−k
∂θ ∂x
∂θ ∂x
(2)
′′ = qrad
(3)
= h (θ x =th − θ ∞ )
(4)
x =0
x = th
θ r =0 must be finite
(5)
θr=R = 0
(6)
θ = T − Tedge
(7)
θ ∞ = T∞ − Tedge
(8)
w
where
and
theta_infinity=T_infinity-T_edge
Note that the two homogeneous boundary conditions are in the x-direction and so the eigenfunctions of the problem will be in this direction. We assume that the solution is separable; that is, the solution is the product of a function only of x (θX) and r (θR):
θ ( x, y ) = θ X ( x ) θ R ( r ) Substituting the product solution into the governing partial differential equation, Eq. (2), leads to:
rθ R
d 2θ X d ⎡ dθ R ⎤ +θ X r =0 2 dx dr ⎢⎣ dr ⎥⎦
Dividing by the product r θX θR leads to:
d ⎡ dθ R ⎤ d 2θ X ⎢r ⎥ 2 dx + dr ⎣ dr ⎦ = 0 rθ R θX ±λ2
∓λ2
Note that the 1st term is a function only of x while the 2nd term is a function only of r; these two quantities must be equal and opposite constants (±λ2). The choice of the sign is again important; the eigenfunctions must be in r and therefore the two ordinary differential equations must be:
d 2θ X − λ2 θ X = 0 dx 2
(9)
d ⎡ dθ R ⎤ r + λ2 rθ R = 0 ⎢ ⎥ dr ⎣ dr ⎦
(10)
The eigenproblem will be solved first; the solution to Eq. (10) is:
θ R = C1 BesselJ ( 0,λ r ) + C2 BesselY ( 0,λ r ) The boundary condition at r = 0, Eq. (5), requires that C2 = 0.
θ R = C1 BesselJ ( 0,λ r ) The boundary condition at r = Rw, Eq. (6), leads to: C1 BesselJ ( 0,λ rout ) = 0
(11)
The 0th order Bessel function of the 1st kind (i.e., Bessel_J(0,x)) oscillates about zero every time the argument changes by 2π in the same way that sine and cosine do; therefore, there are an infinite number of eigenvalues λi that will satisfy Eq. (11) associated with an infinite number of eigenfunctions. The eigencondition for this problem cannot be used to explicitly solve for the eigenvalues; rather, an implicit equation for the eigenvalues results from Eq. (11): BesselJ ( 0,λi rout ) = 0 where i = 1, 2,...∞
The eigenfunctions for this problem are:
(12)
θ Ri = C1,i BesselJ ( 0,λi r ) where BesselJ ( 0,λi b ) = 0 for i = 1, 2,...∞ N=11 duplicate i=1,N lowerlimit[i]=(i-1)*pi/R_w upperlimit[i]=i*pi/R_w guess[i]=lowerlimit[i]+pi/(2*R_w) end
(11)
"number of terms in solution" "lower limit of eigenvalue" "upper limit of eigenvalue" "guess value for eigenvalue"
duplicate i=1,N BesselJ(0,lambda[i]*R_w)=0 end
"solve for eigenvalues"
The arrays lowerlimit[], upperlimit[], and guess[] are used to constrain the solution for the array lambda[] in the Variable Information window. The solution to the ordinary differential equation for θX, Eq. (9), is:
θ X i = C3,i sinh ( λi x ) + C4,i cosh ( λi x ) The series solution for θ is: ∞
∞
i =1
i =1
θ = ∑ θ Ri θ X i = ∑ BesselJ ( 0,λi r ) ⎡⎣C3,i sinh ( λi x ) + C4,i cosh ( λi x ) ⎤⎦ The boundary condition at x = 0, Eq. (3), leads to: ∞
′′ −k ∑ BesselJ ( 0,λi r ) C3,i λi = qrad
(13)
i =1
Equation (13) is multiplied by r BesselJ ( 0,λi r ) and integrated from r = 0 to r = Rw
−k C3,i λi
Rw
Rw
∫ r BesselJ ( 0,λ r ) dr = q′′ ∫ r BesselJ ( 0,λ r ) dr 2
i
rad
0
(14)
i
0
Integral 1
Integral 2
The integrals in Eq. (14) are evaluated using Maple: > restart; > Int1[i]:=int(r*(BesselJ(0,lambda[i]*r))^2,r=0..R_w); 2
2
1 R_w ( π λ i R_w BesselJ( 0, λ i R_w ) + π λ i R_w BesselJ( 1, λ i R_w ) ) Int1i := 2 πλ i
> Int2[i]=int(r*BesselJ(0,lambda[i]*r),r=0..R_w);
Int2i =
R_w BesselJ( 1, λ i R_w ) λi
and pasted into EES in order to obtain the constant C3,i for each eigenvalue: duplicate i=1,N Int1[i]=(R_w/Pi^(1/2)/lambda[i]*(Pi^(1/2)*lambda[i]*R_w*BesselJ(0,lambda[i]*R_w)^2+& Pi^(1/2)*lambda[i]*R_w*BesselJ(1,lambda[i]*R_w)^2))/2 Int2[i] = 1/lambda[i]*R_w*BesselJ(1,lambda[i]*R_w)& -k*C3[i]*lambda[i]*Int1[i]=q``_rad*Int2[i] end
The boundary condition at x = th, Eq. (4), leads to: ∞
− k ∑ BesselJ ( 0,λi r ) λi ⎡⎣C3,i cosh ( λi th ) + C4,i sinh ( λi th ) ⎤⎦ = i =1
⎛ ∞ ⎞ h ⎜ ∑ BesselJ ( 0,λi r ) ⎡⎣C3,i sinh ( λi th ) + C4,i cosh ( λi th ) ⎤⎦ − θ ∞ ⎟ ⎝ i =1 ⎠
(15)
Equation (15) is multiplied by r BesselJ ( 0,λi r ) and integrated from r = 0 to r = Rw h θ∞
Rw
∫ r BesselJ ( 0,λ r ) dr = i
0
Integral 2
C4,i ⎡⎣ k λi sinh ( λi th ) + h cosh ( λi th ) ⎤⎦
Rw
∫ r BesselJ ( 0,λ r ) dr 2
i
0
Integral 1
+C3,i ⎡⎣ k λi cosh ( λi th ) + h sinh ( λi th ) ⎤⎦
Rw
∫ r BesselJ ( 0,λ r ) dr 2
i
0
Integral 1
where the integrals were previously accomplished in Maple: duplicate i=1,N h_bar*theta_infinity*Int2[i]=Int1[i]*(C4[i]*(k*lambda[i]*sinh(lambda[i]*th)+& h_bar*cosh(lambda[i]*th))+C3[i]*(k*lambda[i]*cosh(lambda[i]*th)+& h_bar*sinh(lambda[i]*th))) end
The solution is evaluated at an arbitrary position: x_bar=0 [-] r_bar=0.5 [-] x=x_bar*th
(16)
r=r_bar*R_w duplicate i=1,N theta[i]=BesselJ(0,lambda[i]*r)*(C3[i]*sinh(lambda[i]*x)+C4[i]*cosh(lambda[i]*x)) end theta=sum(theta[1..N]) T=theta+T_edge T_C=converttemp(K,C,T)
c.) Plot the temperature as a function of r for various values of x. Figure 2 illustrates the temperature as a function or r for various values of x.
Temperature (°C)
40 38
x/th = 0
36
x/th = 0.25
34
x/th = 0.5 x/th = 0.75
32
x/th = 1
30 28 26 24 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Dimensionless radius, r/Rw Figure 2: Temperature as a function of r for various values of x.
d.) Prepare a contour plot of the temperature in the window. The requested contour plot is shown in Figure 3.
Dimensionless x position, x/th
1 26.33
0.9
27.66
0.8
28.98
0.7
30.31 0.6
31.64
0.5
32.97
0.4
34.3
0.3
35.62
0.2
36.95 38.28
0.1 0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Dimensionless radial position, r/Rw
Figure 3: Contour plot of the temperature in the window.
Problem 2.3-4 Reconsider Problem 2.3-3. The window is not opaque to the radiation but does absorb some of it. The radiation that is absorbed is transformed to thermal energy. The volumetric rate of ′′ α exp ( −α x ) where α = 100 m-1 is the thermal energy generation is given by: g ′′′ = qrad absorption coefficient. The radiation that is not absorbed is transmitted. Otherwise the problem remains the same. a.) Develop a separation of variables solution to this problem using the techniques discussed in Section 2.3. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" th=1 [cm]*convert(cm,m) R_w=3 [cm]*convert(cm,m) q``_rad=1000 [W/m^2] k=1.2 [W/m-K] T_edge=converttemp(C,K,25 [C]) T_infinity=converttemp(C,K,20 [C]) h_bar=50 [W/m^2-K] alpha=100 [1/m]
"thickness of window" "outer radius of window" "radiation heat flux" "conductivity" "edge temperature" "ambient temperature" "heat transfer coefficient" "absorption coefficient"
A differential control volume leads to the energy balance: q x + qr + g = q x + dx + qr + dr
(1)
where g is the rate of generation of thermal energy. Equation (1) can be expanded:
g =
∂q x ∂q dx + r dr ∂x ∂r
Substituting the rate equations:
q x = − k 2 π r dr
∂T ∂x
qr = − k 2 π r dx
∂T ∂r
and ′′ α exp ( −α x ) g = 2 π r dr dx qrad
(2)
into the differential energy balance leads to: ′′ α exp ( −α x ) = 2 π r dr dx qrad
∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ dx + ⎢ − k 2 π r dx ⎥ dr −k 2 π r dr ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂r ⎣ ∂r ⎦
or r
′′ α ∂ 2T ∂ ⎡ ∂T ⎤ qrad + ⎢r + exp ( −α x ) r = 0 2 ⎥ ∂x ∂r ⎣ ∂r ⎦ k
(3)
The boundary conditions are: −k
−k
∂T ∂x
∂T ∂x
=0
(4)
= h (Tx =th − T∞ )
(5)
x =0
x = th
Tr =0 must be finite
(6)
Tr = Rw = Tedge
(7)
Note that the radiant heat flux is no longer absorbed at x = 0 but rather is absorbed volumetrically throughout the domain; it shows up as the volumetric generation in the partial differential equation rather than as a heat flux at a boundary. With the simple transformation:
θ = T − T∞
r
′′ α ∂ 2θ ∂ ⎡ ∂θ ⎤ qrad exp ( −α x ) r = 0 + ⎢r + 2 ⎥ k ∂x ∂r ⎣ ∂r ⎦
(8)
(9)
The boundary conditions are: −k
−k
∂θ ∂x
∂θ ∂x
=0
(10)
= h θ x =th
(11)
x =0
x = th
θ r =0 must be finite
(12)
θ r = R = θ edge
(13)
θ edge = Tedge − T∞
(14)
w
where
theta_edge=T_edge-T_infinity
The partial differential equation is non-homogeneous and there is no simple transformation that can rectify this. Therefore, it is necessary to use the solution technique provided in Section 2.3. The first step is to split the solution into a homogeneous, 2-D solution that can be solved using separation of variables and one or more 1-D particular solution:
θ = θh + X + R
(15)
Substituting Eq. (15) into Eq. (9) leads to: r
′′ α ∂ 2θ h d 2 X ∂ ⎡ ∂θ h ⎤ d ⎡ dR ⎤ qrad + + ⎢r + + exp ( −α x ) r = 0 r r 2 2 ∂x ∂r ⎣ ∂r ⎥⎦ dr ⎢⎣ ∂r ⎥⎦ dx k
(16)
In order to solve the problem for θh using separation of variables it is necessary that: ∂ 2θ h ∂ ⎡ ∂θ h ⎤ r 2 + ⎢r =0 ∂x ∂r ⎣ ∂r ⎥⎦
(17)
Therefore, the ODEs for X and R are:
′′ α d 2 X qrad + exp ( −α x ) = 0 2 dx k
(18)
d ⎡ dR ⎤ r =0 dr ⎢⎣ ∂r ⎥⎦
(19)
′′ dX qrad = exp ( −α x ) + C1 dx k
(20)
and
Solving Eq. (18) leads to:
X =−
′′ qrad exp ( −α x ) + C1 x + C2 kα
(21)
Solving Eq. (19) leads to: R = C3 ln ( r ) + C4
(22)
where C1 through C4 are undetermined constants. In order to be able to solve θh using separation of variables it is necessary that one direction have homogeneous boundary conditions. This problem only works if x is the homogeneous direction. Substituting Eq. (15) into Eq. (10) leads to: −k
∂θ h ∂x
−k x =0
dX dx
=0
(23)
x =0
Therefore: −k
∂θ h ∂x
x =0
dX dx
x =0
=0
(24)
=0
(25)
′′ qrad + C1 = 0 k
(26)
and −k
Substituting Eq. (25) into Eq. (20) leads to:
or: C1 = −
′′ qrad k
(27)
C_1=-q``_rad/k
Therefore: ′′ q ′′ dX qrad exp ( −α x ) − rad = dx k k
(28)
X =−
′′ qrad q ′′ exp ( −α x ) − rad x + C2 kα k
(29)
Substituting Eq. (15) into Eq. (11) leads to: −k
∂θ h ∂x
−k x =th
dX dx
x = th
= h (θ h , x =th + X x =th + R )
(30)
Equation (30) requires that: −k
∂θ h ∂x
x = th
dX dx
x =th
= h θ h , x =th
(31)
= h X x =th
(32)
as well as: −k
and R=0
(33)
Therefore C3 and C4 must both be zero. Substituting Eqs. (28) and (29) into Eq. (32) leads to: ⎡ q ′′ ⎤ q ′′ ⎤ q ′′ ⎡ q ′′ − k ⎢ rad exp ( −α th ) − rad ⎥ = h ⎢ − rad exp ( −α th ) − rad th + C2 ⎥ k ⎦ k ⎣ k ⎣ kα ⎦
(34)
which can be solved for C2: -k*(q``_rad*exp(-alpha*th)/k-q``_rad/k)=h_bar*(-q``_rad*th/k+C_2-q``_rad*exp(-alpha*th)/(k*alpha))
Finally, Eq. (15) is substituted into Eqs. (12) and (13) in order to determine the nonhomogeneous boundary conditions for θh:
θ h,r =0 + X must be finite
(35)
θ h,r = R + X = θ edge
(36)
θ h,r =0 must be finite
(37)
w
Therefore:
and
θ h ,r = R = θ edge − w
′′ qrad q ′′ exp ( −α x ) + rad x − C2 kα k
(38)
The problem for θh is solved using the typical steps for separation of variables. eigencondition is: tan ( λi th ) =
Bi
The
(39)
λi
where Bi =
h th k
(40)
The eigenvalues occur in regular intervals of λ th and are automatically identified using EES: N=11 duplicate i=1,N lowerlimit[i]=(i-1)*pi upperlimit[i]=lowerlimit[i]+pi/2 guess[i]=lowerlimit[i]+pi/4 end duplicate i=1,N tan(lambdath[i])=Bi/lambdath[i] lambda[i]=lambdath[i]/th end
"number of terms in solution" "lower limit of eigenvalue" "upper limit of eigenvalue" "guess value for eigenvalue"
"solve for eigenvalues" "eigenvalue"
The series solution for θh is: ∞
θ = ∑ cos ( λi x ) ⎡⎣C3,i BesselI ( 0,λi r ) + C4,i BesselK ( 0,λi r ) ⎤⎦
(41)
i =1
The boundary condition at r = 0, Eq. (37), leads to C4,i = 0: ∞
θ = ∑ Ci cos ( λi x ) BesselI ( 0,λi r )
(42)
i =1
Substituting Eq. (42) into the boundary condition at r = Rw, Eq. (38), leads to: ∞
∑ C cos ( λ x ) BesselI ( 0,λ R ) = θ i =1
i
i
i
w
edge
−
′′ qrad q ′′ exp ( −α x ) + rad x − C2 kα k
Multiplying Eq. (43) by cos(λj x) and integrating from x = 0 to x = th leads to:
(43)
th
th
Ci BesselI ( 0,λi Rw ) ∫ cos ( λi x ) dx = (θ edge − C2 ) ∫ cos ( λi x ) dx 0 0
2
Integral 1
q ′′ + rad k
Integral 2
q ′′ ∫0 x cos ( λi x ) dx − kradα ∫0 exp ( −α x ) cos ( λi x ) dx
th
th
Integral 3
Integral 4
The integrals in Eq. (44) are carried out using Maple: > restart; > Int1[i]:=int((cos(lambda[i]*x))^2,x=0..th);
Int1i :=
1 cos ( λ i th ) sin( λ i th ) + λ i th λi 2
> Int2[i]:=int(cos(lambda[i]*x),x=0..th);
Int2i :=
sin( λ i th ) λi
> Int3[i]=int(x*cos(lambda[i]*x),x=0..th);
Int3i =
−1 + cos ( λ i th ) + th sin( λ i th ) λ i λi
2
> Int4[i]=int(exp(-alpha*x)*cos(lambda[i]*x),x=0..th);
Int4i =
α−αe
( −α th )
cos ( λ i th ) + λ i e α 2 + λi
( −α th )
sin( λ i th )
2
and copied into EES in order to evaluate the constants: duplicate i=1,N Int1[i] = 1/2*(cos(lambda[i]*th)*sin(lambda[i]*th)+lambda[i]*th)/lambda[i] Int2[i] = 1/lambda[i]*sin(lambda[i]*th) Int3[i] = (-1+cos(lambda[i]*th)+th*sin(lambda[i]*th)*lambda[i])/lambda[i]^2 Int4[i] = (alpha-alpha*exp(-alpha*th)*cos(lambda[i]*th)+& lambda[i]*exp(-alpha*th)*sin(lambda[i]*th))/(alpha^2+lambda[i]^2) C[i]*BesselI(0,lambda[i]*R_w)*Int1[i]=(theta_edge-C_2)*Int2[i]-C_1*Int3[i]+q``_rad*Int4[i]/(k*alpha) end
The solution is obtained at an arbitrary position: x_bar=0.5 [-] r_bar=0.5 [-] x=x_bar*th r=r_bar*R_w Xp=C_1*x+C_2-q``_rad*exp(-alpha*x)/(k*alpha) duplicate i=1,N
(44)
theta_h[i]=C[i]*cos(lambda[i]*x)*BesselI(0,lambda[i]*r) end theta_h=sum(theta_h[1..N]) theta=theta_h+Xp T=theta+T_infinity T_C=converttemp(K,C,T)
b.) Plot the temperature as a function of r for various values of x. Figure 1 illustrates the requested plot. 31 x/th = 0 x/th = 0.25 x/th = 0.5 x/th = 0.75 x/th = 1.0
Temperature (°C)
30 29 28 27 26 25 24 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Dimensionless radius r/Rw Figure 2: Temperature as a function of dimensionless radius for various values of x/th.
c.) Show that your solution limits to the solution from Problem 2.3-3 in the limit that α → ∞. Figure 3 illustrates the temperature at x = 0 predicted by Problem 2.3-3 and by the numerical model developed here with α → ∞.
35 P2.3-4 with α→∞ P2.3-3
34
Temperature (°C)
33 32 31 30 29 28 27 26 25 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Dimensionless radius r/Rw Figure 3: Temperature at x = 0 as a function of dimensionless radius predicted by the solution developed in Problem 2.3-3 overlaid onto the solution developed here with α approaching infinity.
Problem 2.4-1 (2-7 in text) The plate shown in Figure P2.4-1 is exposed to a uniform heat flux q ′′ = 1x105 W/m2 along its top surface and is adiabatic at its bottom surface. The left side of the plate is kept at TL = 300 K and the right side is at TR = 500 K. The height and width of the plate are H = 1 cm and W = 5 cm, respectively. The conductivity of the plate is k = 10 W/m-K. 5 2 q ′′ = 1x10 W/m
W = 5 cm
TL = 300 K
H = 1 cm
y
TR = 500 K
x k = 10 W/m-K Figure P2.4-1: Plate.
a.) Derive an analytical solution for the temperature distribution in the plate. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" H=1 [cm]*convert(cm,m) W=5 [cm]*convert(cm,m) k=10 [W/m-K] q``=100000 [W/m^2] T_R=500 [K] T_L=300 [K]
"height of plate" "thickness of plate" "conductivity" "heat flux" "temperature of right hand surface" "temperature of left hand surface"
A mathematical statement of the transformed problem is: ∂ 2θ ∂ 2θ + =0 ∂x 2 ∂y 2
(1)
θ x =0 = 0
(2)
θ x =W = θ R
(3)
∂θ ∂y
(4)
with boundary conditions:
=0 y =0
k
∂θ ∂y
= q ′′
(5)
y=H
where
θ = T − TL
(6)
θ R = TR − TL
(7)
and
The problem has two, non-homogeneous boundary condition and therefore must be solved using superposition.
θ = θ A + θB
(8)
Problem θA retains the non-homogeneous boundary condition in the x-direction: ∂ 2θ A ∂ 2θ A + 2 =0 ∂x 2 ∂y
(9)
θ A, x = 0 = 0
(10)
θ A, x =W = θ R
(11)
∂θ A ∂y
=0
(12)
=0
(13)
with boundary conditions:
∂θ A ∂y
y =0
y=H
By inspection, the solution for θA is 1-D in x and given by:
θ A = θR x_bar=0.5 [-] x=x_bar*W y_bar=0.5 [-] y=y_bar*H
x L
(14) "dimensionless x location" "x location" "dimensionless y location" "y location"
"sub-problem A solution" theta_R=T_R-T_L surface" theta_A=theta_R*x/W
"temperature difference of right hand "solution for sub-problem A"
Problem θB retains the non-homogeneous boundary condition in the y-direction: ∂ 2θ B ∂ 2θ B + 2 =0 ∂x 2 ∂y
(15)
θ B , x =0 = 0
(16)
θ B , x =W = 0
(17)
∂θ B ∂y
=0
(18)
= q ′′
(19)
with boundary conditions:
k
∂θ B ∂y
y =0
y=H
The solution for θB is 2-D and can be obtained using separation of variables. The eigenfunctions are:
θ X B ,i = sin ( λB ,i x )
(20)
where the eigenvalues are:
λB , i =
iW
π
for i = 1, 2,..∞
"sub-problem B solution" N_term=11 [-] duplicate i=1,N_term lambda_B[i]=i*pi/W end
(21)
"number of terms" "i'th eigenvalue"
The solution in the non-homogeneous direction is:
θ YB ,i = C2,i cosh ( λB ,i y ) The series solution for θB is:
(22)
∞
θ B = ∑ Ci sin ( λB ,i x ) cosh ( λB ,i y )
(23)
i =1
Subsituting Eq. (23) into Eq. (19) leads to: ∞
k ∑ λB ,i Ci sin ( λB ,i x ) sinh ( λB ,i H ) = q ′′
(24)
i =1
Equation (24) is multiplied by an eigenfunction and integrated from x = 0 to x = W: W
W
k λB ,i Ci sinh ( λB ,i H ) ∫ sin ( λB ,i x ) dx = q ′′ ∫ sin ( λB ,i x ) dx 2
0
0
The integrals in Eq. (25) are evaluated in Maple: > restart; > assume(i,integer); > lambda_B:=i*Pi/W;
lambda_B :=
i~ π W
> int((sin(lambda_B*x))^2,x=0..W);
W 2 > int(sin(lambda_B*x),x=0..W);
W ( −1 + ( -1 )i~ ) − i~ π
and used to evaluate the constants: duplicate i=1,N_term C[i]*k*lambda_B[i]*sinh(lambda_B[i]*H)*W/2=q``*(-W*(-1+(-1)^i)/i/Pi) "i'th constant" end
The solution for θB is obtained: duplicate i=1,N_term theta_B[i]=C[i]*sin(lambda_B[i]*x)*cosh(lambda_B[i]*y) end theta_B=sum(theta_B[1..N_term])
"i'th term" "solution to sub-problem B"
and used to obtain the solution for T: theta=theta_A+theta_B T=theta+T_L
"superposition of solutions" "temperature"
(25)
b.) Prepare a contour plot of the temperature. Figure 2 illustrates a contour plot of the temperature distribution in the plate. 1
Dimensionless y-position, y/H
300 345.5
0.8
390.9 436.4 481.8
0.6
527.3 572.7
0.4
618.2 663.6 709.1
0.2
754.5 0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Dimensionless x-position, x/W Figure 2: Contour plot of the temperature distribution.
Problem 2.5-1 (2-8 in text): A Heating Element Figure P2.5-1 illustrates an electrical heating element that is affixed to the wall of a chemical reactor. The element is rectangular in cross-section and very long (into the page). The temperature distribution within the element is therefore two-dimensional, T(x, y). The width of the element is a = 5.0 cm and the height is b = 10.0 cm. The three edges of the element that are exposed to the chemical (at x = 0, y = 0, and x = a) are maintained at a temperature Tc = 200°C while the upper edge (at y = b) is affixed to the well-insulated wall of the reactor and can therefore be considered adiabatic. The element experiences a uniform volumetric rate of thermal energy generation, g ′′′ = 1x106 W/m3. The conductivity of the material is k = 0.8 W/m-K. reactor wall k = 0.8 W/m-K 6 3 g ′′′ = 1x10 W/m
Tc = 200°C a = 5 cm y x Tc = 200°C
Tc = 200°C b = 10 cm
Figure P2.5-1: Electrical heating element.
a.) Develop a 2-D numerical model of the element using EES. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" a=5.0 [cm]*convert(cm,m) b=10.0 [cm]*convert(cm,m) k=0.8 [W/m-K] T_c=converttemp(C,K,200 [C]) g```_dot=1e6 [W/m^3] L=1 [m]
"width of element" "height of element" "conductivity" "chemical temperature" "rate of volumetric generation" "unit length of element into the page"
The computational domain of the element with the regularly spaced grid of nodes is shown in Figure 2.
Figure 2: The regularly spaced grid used to obtain a numerical solution.
The first step in obtaining a numerical solution is to position the nodes throughout the computational domain. We will use grid with nodes placed on the edges and distributed uniformly throughout. The x and y distance between adjacent nodes (Δx and Δy) are: Δx =
L ( m − 1)
(1)
Δy =
b ( n − 1)
(2)
and the x and y positions of any node i,j are given by:
xi =
( i − 1) a ( m − 1)
(3)
yj =
( j − 1) b ( n − 1)
(4)
where m and n are the number of nodes used in the x and y directions. "Setup nodes" m=11 [-] "number of nodes in the x-direction"
n=11 [-] "number of nodes in the y-direction" Dx=a/(m-1) "distance between nodes in the x-direction" Dy=b/(n-1) "distance between nodes in the y-direction" duplicate i=1,m x[i]=(i-1)*Dx end duplicate j=1,n y[j]=(j-1)*Dy end
"x-position of each node"
"y-position of each node"
The next step in the solution is to write an energy balance for each node. Figure 3 illustrates a control volume and the associated energy transfers for an internal node (see Figure 2) which include conduction from each side ( q RHS and q LHS ), the top ( qtop ), and the bottom ( qbottom ). Note that the direction associated with these energy transfers is arbitrary (i.e., they could have been taken as positive if energy leaves the control volume), but it is important to write the equation in a manner consistent with the chosen directions.
Figure 3: Energy balance for an internal node
The energy balance suggested by Figure 3 is: q RHS [i, j ] + q LHS [i, j ] + qtop [i, j ] + qbottom [i, j ] + g [i, j ] = 0
(5)
The next step is to approximate each of the terms in the energy balance; the material separating the nodes is assumed to behave as a plane wall resistance and therefore: q RHS [i, j ] =
k Δy L (T [i + 1, j ] − T [i, j ]) Δx
(6)
where L is the length of the element (assumed to be 1 m in order to do the problem on a unit length basis); therefore, Δy L is the area for conduction and Δx is the distance over which the conduction heat transfer occurs. Note that the temperature difference is consistent with the direction of the arrow in Figure 3; if Ti+1,j is greater than Ti,j then energy is leaving the node and q RHS is positive. The other heat transfers are approximated using a similar model:
k Δy L (T [i − 1, j ] − T [i, j ]) Δx
(7)
k ΔxW (T [i, j + 1] − T [i, j ]) Δy
(8)
q LHS [i, j ] = qtop [i, j ] =
qbottom [i, j ] =
k ΔxW (T [i, j − 1] − T [i, j ]) Δy
(9)
The generation is the product of the volume of the control volume and the volumetric rate of generation: g [i, j ] = g ′′′ Δx Δy L
(10)
These equations are entered in EES using a nested duplicate statement: "Internal node control volumes" duplicate i=2,(m-1) duplicate j=2,(n-1) q_dot_LHS[i,j]+q_dot_RHS[i,j]+q_dot_top[i,j]+q_dot_bottom[i,j]+gen[i,j]=0 q_dot_LHS[i,j]=k*L*Dy*(T[i-1,j]-T[i,j])/Dx q_dot_RHS[i,j]=k*L*Dy*(T[i+1,j]-T[i,j])/Dx q_dot_top[i,j]=k*L*Dx*(T[i,j+1]-T[i,j])/Dy q_dot_bottom[i,j]=k*L*Dx*(T[i,j-1]-T[i,j])/Dy gen[i,j]=Dx*Dy*L*g```_dot end end
Note that each time the outer duplicate statement iterates once (i.e., i is increased by 1), the inner duplicate statement iterates (n-1) times (i.e., j runs from 2 to n-1). Therefore, all of the internal nodes are considered with these two nested duplicate loops. Also note that the unknowns are placed in an array rather than a vector. The entries in the array T is accessed using two indices contained in square brackets. The boundary nodes have to be treated separately. The left, right, and bottom boundaries are easy as the temperature is specified:
These equations are entered in EES:
T [1, j ] = Tc for j = 1...n
(11)
T [ m, j ] = Tc for j = 1...n
(12)
T [i,1] = Tc for i = 1...m
(13)
"Temperature along right edge" duplicate j=1,n T[1,j]=T_c end "Temperature along left edge" duplicate j=1,n T[m,j]=T_c end "Temperature along bottom edge" duplicate i=2,(m-1) T[i,1]=T_c end
The upper boundary nodes must be considered using energy balances. Figure 4 illustrates an energy balance associated with a node that is located on the top, insulated boundary (see Figure 2).
Figure 4: Energy balance for a node on the top boundary
The energy balance suggested by Figure 4 is: q RHS [i, n ] + q LHS [i, n ] + qbottom [i, n ] + g [i, n ] = 0
(14)
The conduction terms in the x direction must be approximated slightly differently:
q RHS [i, n ] =
k Δy L (T [i + 1, n] − T [i, n]) 2Δx
(15)
q LHS [i, n ] =
k Δy L (T [i − 1, n] − T [i, n]) 2 Δx
(16)
The factor of 2 in the denominator appears because there is half the available area for conduction through the sides of the control volume on the top boundary. The other conduction term is approximated as before: qbottom [i, n ] =
k ΔxW (T [i, n − 1] − T [i, n]) Δy
(17)
The generation term is: g [i, n ] =
Δx Δy L g ′′′ 2
(18)
These equations are entered in EES using a single duplicate statement: "Upper edge" duplicate i=2,(m-1) q_dot_LHS[i,n]+q_dot_RHS[i,n]+q_dot_bottom[i,n]+gen[i,n]=0 q_dot_LHS[i,n]=k*(Dy/2)*L*(T[i-1,n]-T[i,n])/Dx q_dot_RHS[i,n]=k*(Dy/2)*L*(T[i+1,n]-T[i,n])/Dx q_dot_bottom[i,n]=k*Dx*L*(T[i,n-1]-T[i,n])/Dy gen[i,n]=Dx*Dy*L*g```_dot/2 end
Notice that the control volumes at the top left corner (i.e., i =1, j =n) and the top right corner (i.e., i =m, j =n) have already been taken addressed by the equations for the left and right boundaries, Eqs. (11) and (12). Therefore, we have to make sure not to write additional equations related to this node or the problem will be over-specified and so the equations for the top boundary can only be written for i = 2...(m-1). We have derived a total of m x n equations in the m x n unknown temperatures; these equations completely specify the problem and they have now all been entered in EES. Therefore, a solution can be obtained by solving the EES code. The solution is contained in the Arrays Window; each column of the table corresponds to the temperatures associated with one value of i and all of the value of j (i.e., the temperatures in a column are at a constant value of y and varying values of x). b.) Plot the temperature as a function of x at various values of y. What is the maximum temperature within the element and where is it located? The solution is obtained for n=11 and the columns Ti,1 (corresponding to y =0) through Ti,11 (corresponding to y = 10 cm) are plotted in Figure 5 as a function of x.
Figure 5: Temperature as a function of x for various values of y.
The hottest spot in the element is at the adiabatic wall (y = 10 cm) and the center (x = 2.5 cm); the hottest temperature is about 860 K. c.) Prepare a reality check to show that your solution behaves according to your physical intuition. That is, change some aspect of your program and show that the results behave as you would expect (clearly describe the change that you made and show the result). There are several possible answers to this; I increased the conductivity by a factor of 10 and examined the temperature distribution. Figure 6 illustrates the temperature as a function of x for the original conductivity (k = 0.8 W/m-K) and the increased conductivity (k = 8.0 W/m-K); notice that the increased conductivity has had the expected effect of reducing the temperature rise.
Figure 6: Temperature as a function of x for y = 10 cm and two values of conductivity.
Problem 2.5-2 Figure P2.5-2 illustrates a composite material that is being machined on a lathe. composite thins = 100 μm kins = 1.5 W/m-K thm = 200 μm km = 35 W/m-K q′′ l
h depends on RS T∞ = 20°C
Tchuck = 20°C y x W = 12 cm H = 3 cm Figure P2.5-2: Composite material being machined on a lathe.
The composite is composed of alternating layers of insulating material and metal. The insulating layers have thickness thins= 100 μm and conductivity kins = 1.5 W/m-K. The metal layers have thickness thm = 200 μm and conductivity km = 35 W/m-K. The workpiece is actually cylindrical and rotating. However, because the radius is large relative to it thickness and there are no circumferential variations we can model the workpiece as a 2-D problem in Cartesian coordinates, x and y, as shown in Figure 2.5-2. The width of the workpiece is W = 12 cm and the thickness is H = 3 cm. The left surface of the workpiece at x = 0 is attached to the chuck and therefore maintained at Tchuck = 20°C. The inner surface at y = 0 is insulated. The outer surface (at y = H) and right surface (at x = W) are exposed to air at T∞ = 20°C with heat transfer coefficient h that depends on the rotational speed of the chuck, RS in rev/min, according to: ⎡ W min 2 ⎤ 2 h = 2⎢ 2 RS 2⎥ ⎣ m K rev ⎦
In order to extend the life of the tool used for the machining process, the workpiece is preheated by applying laser power to the outer surface. The heat flux applied by the laser depends on the rotational speed and position according to: ⎡ ⎛ x − x ⎞2 ⎤ c ql′′ = a RS exp ⎢ - ⎜ ⎟ ⎥ ⎢⎣ ⎝ pw ⎠ ⎥⎦
where a = 5000 W-min/m2-rev, xc = 8 cm, and pw = 1 cm. a.) What is the effective thermal conductivity of the composite in the x- and y-directions? The inputs are entered in EES and a function is defined in order to provide the heat flux:
$UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in function q_f(RS,x) "Inputs RS - rotational speed (rpm) x - position (m) Output q_f - laser heat flux (W/m^2)" a=5000 [W-min/m^2-rev] x_c=8 [cm]*convert(cm,m) pw=1 [cm]*convert(cm,m) q_f=a*RS*exp(-((x-x_c)/pw)^2) end "Inputs" th_ins=100 [micron]*convert(micron,m) k_ins=1.5 [W/m-K] th_m=200 [micron]*convert(micron,m) k_m=35 [W/m-K] W=12 [cm]*convert(cm,m) H=3 [cm]*convert(cm,m) T_chuck=20 [C] T_infinity=20 [C] RS=60 [rev/min]
"insulation layer thickness" "insulation conductivity" "metal thickness" "metal conductivity" "width of work piece" "thickness of work piece" "chuck temperature" "ambient air temperature" "rotational speed"
The heat transfer coefficient ( h ) is evaluated using the equation provided in the problem statement. h_bar=2 [W-min^2/m^2-K-rev^2]*RS^2 "heat transfer coefficient"
The effective conductivities of a lamination stack was derived in Section 2.9.1. The effective conductivity in the x-direction is:
keff , x =
( km thm + kins thins ) ( thm + thins )
(1)
and the conductivity in the y-direction is:
keff , y =
( thm + thins ) thm thins + km kins
"Determine effective conductivities" k_x=(k_m*th_m+k_ins*th_ins)/(th_m+th_ins) "effective conductivity in the x-direction" k_y=(th_m+th_ins)/(th_m/k_m+th_ins/k_ins) "effective conductivity in the y-direction"
which leads to keff,x = 23.83 W/m-K and keff,y = 4.15 W/m-K.
(2)
b.) Develop a 2-D numerical model of the workpiece in EES. Plot the temperature as a function of x at various values of y, including at least y = 0, H/2, and H. A regularly spaced grid of nodes is uniformly distributed with the first and last nodes in each dimension placed on the boundaries of the domain; the x- and y-positions of any node (i,j) are given by: xi = ( i − 1) Δx for i = 1..M
(3)
y j = ( j − 1) Δy for j = 1..N
(4)
where M and N are the number of nodes used in the x- and y-directions, respectively. The x- and y-distance between adjacent nodes (Δx and Δy, respectively) are: Δx =
W ( M − 1)
(5)
Δy =
H ( N − 1)
(6)
This information is entered in EES: "Setup nodes" M=21 [-] N=11 [-] Dx=W/(m-1) Dy=H/(n-1) duplicate i=1,m x[i]=(i-1)*Dx end duplicate j=1,n y[j]=(j-1)*Dy end
"number of nodes in the x-direction" "number of nodes in the y-direction" "distance between nodes in the x-direction" "distance between nodes in the y-direction" "x-position of each node"
"y-position of each node"
The next step in the solution is to write an energy balance for each node. The energy transfers for an internal node include conduction from each side ( q RHS and q LHS ), the top ( qtop ), and the bottom ( qbottom ). The direction associated with these energy transfers is arbitrary but they are all taken as being into the control volume, as was done in Section 2.5.2.
q RHS + q LHS + qtop + qbottom = 0 The next step is to approximate each of the terms in the energy balance:
(7)
keff , x
Δy Δy Δx Δx Ti +1, j − Ti , j ) + keff , x Ti −1, j − Ti , j ) + keff , y Ti , j +1 − Ti , j ) + keff , y ( ( ( (Ti, j −1 − Ti, j ) = 0 Δx Δx Δy Δy for i = 2... ( M − 1) and j = 2... ( N − 1)
(8)
These equations are entered in EES using nested duplicate statements: "Internal node energy balances" duplicate i=2,(M-1) duplicate j=2,(N-1) Dy*k_x*(T[i+1,j]-T[i,j])/Dx+Dy*k_x*(T[i-1,j]-T[i,j])/Dx+Dx*k_y*(T[i,j+1]-T[i,j])/Dy+& Dx*k_y*(T[i,j-1]-T[i,j])/Dy=0 end end
The left boundary (x= 0) has a specified temperature: T1, j = Tchuck for j = 1...N
(2-9)
where Tchuck is the chuck temperature. These equations are entered in EES: "left boundary" duplicate j=1,N T[1,j]=T_chuck end
The remaining boundary nodes do not have specified temperatures and therefore must be determined using energy balances. An energy balance on a node that is located on the top boundary (at y= H) is: q RHS + q LHS + qbottom + qconv + ql = 0
(2-10)
which leads to: keff , x Δy 2Δx
(T
i +1, N
− Ti , N ) + +
keff , x Δy 2 Δx
(T
i −1, N
− Ti , N ) +
keff , y Δx Δy
(T
i , N −1
Δx h (T∞ − Ti, N ) + ql′′Δx=0 k i = 2... ( M − 1)
These equations are entered in EES using a single duplicate statement: "top boundary" duplicate i=2,(M-1) Dy*k_x*(T[i+1,N]-T[i,N])/(2*Dx)+Dy*k_x*(T[i-1,N]-T[i,N])/(2*Dx)+& Dx*k_y*(T[i,N-1]-T[i,N])/Dy+Dx*h_bar*(T_infinity-T[i,N])+q_f(RS,x[i])*Dx=0 end
− Ti , N )
(2-11)
A similar procedure for the nodes on the lower boundary leads to: keff , x Δy 2Δx
(T
i +1,1
− Ti ,1 ) +
keff , x Δy
(T
i −1,1
2 Δx
− Ti ,1 ) +
keff , y Δx Δy
(T
i ,2
− Ti ,1 ) =0 for i = 2... ( M − 1) (2-12)
"bottom boundary" duplicate i=2,(M-1) Dy*k_x*(T[i+1,1]-T[i,1])/(2*Dx)+Dy*k_x*(T[i-1,1]-T[i,1])/(2*Dx)+Dx*k_y*(T[i,2]-T[i,1])/Dy=0 end
Energy balances for the nodes on the right-hand boundary (x = L) lead to: keff , y Δx 2Δy
(T
M , j +1
− TM , j ) +
keff , y Δx 2 Δy
(T
M , j −1
− TM , j ) +
keff , x Δy Δx
(T
M −1, j
− TM , j )
+Δy h (T∞ − TM , j ) =0 for j = 2... ( N − 1)
(2-13)
"right boundary" duplicate j=2,(N-1) Dx*k_y*(T[M,j+1]-T[M,j])/(2*Dy)+Dx*k_y*(T[M,j-1]-T[M,j])/(2*Dy)+& Dy*k_x*(T[M-1,j]-T[M,j])/Dx+Dy*h_bar*(T_infinity-T[M,j])=0 end
The two corners (right upper and right lower) have to be considered separately. A control volume and energy balance for node (M,N) leads to: keff , y Δx 2 Δy
(T
M , N −1
− TM , N ) +
keff , x Δy 2 Δx
(T
M −1, N
− TM , N )
Δx Δy T∞ − TM , N ) +h +h ( (T∞ − TM , N ) =0 2 2
(2-14)
"upper right corner" Dx*k_y*(T[M,N-1]-T[M,N])/(2*Dy)+Dy*k_x*(T[M-1,N]-T[M,N])/(2*Dx)+& h_bar*Dx*(T_infinity-T[M,N])/2+h_bar*Dy*(T_infinity-T[M,N])/2=0
The energy balance for the right lower boundary, node (M,1), leads to: Δx Δy Δy TM ,2 − TM ,1 ) + TM −1,1 − TM ,1 ) + h ( ( (T∞ − TM ,1 ) = 0 2 Δy 2 Δx 2
(2-15)
"lower right corner" Dx*k_y*(T[M,2]-T[M,1])/(2*Dy)+Dy*k_x*(T[M-1,1]-T[M,1])/(2*Dx)+h_bar*Dy*(T_infinity-T[M,1])/2=0
We have derived a total of M x N equations in the M x N unknown temperatures; these equations completely specify the problem and they have now all been entered in EES. Therefore, a solution can be obtained by solving the EES code. The solution is contained in the Arrays
window; each column of the table corresponds to the temperatures associated with one value of i and all of the value of j (i.e., the temperatures in a column are at a constant value of y and varying values of x). Figure 2 illustrates the temperature as a function of x at various values of y and shows the heating caused by the laser energy at the top surface. 250 y = 30 mm
Temperature (°C)
200
150
y = 24 mm y = 18 mm
100
50
0 0
y = 0 mm y = 6 mm y = 12 mm 0.02
0.04
0.06
0.08
0.1
0.12
Axial position (m)
Figure 2: Temperature as a function of x for various values of y.
c.) Plot the maximum temperature in the workpiece as a function of the rotational speed, RS. If the objective is to preheat the material to its maximum possible temperature, then what is the optimal rotational speed? The maximum temperature (Tmax) occurs on the upper boundary (j = N) and is identified according to: T_max=max(T[1..M,N])
Figure 3 illustrates the maximum temperature as a function of rotational speed and shows that the maximum preheating occurs when the rotational speed is approximately 12 rpm.
Maximum temperature (°C)
90 80 70 60 50 40 30 0
10
20
30
40
50
60
Rotational speed (rev/min)
Figure 3: Maximum temperature as a function of the rotational speed.
Problem 2.5-3 In Section 1.6 the constant cross-section, straight fin shown in Figure P2.5-3 was analyzed under the assumption that it could be treated as an extended surface (i.e., temperature gradients in the y direction can be neglected). In this example, the 2-D temperature distribution within the fin will be determined using separation of variables.
T∞ , h
W
insulated tip
y Tb
x L
g ′′′
th
Figure P2.5-3: Straight, constant cross-sectional area fin.
Assume that the tip of the fin is insulated and that the width (W) is much larger than the thickness (th) so that convection from the edges can be neglected. The length of the fin is L = 5.0 cm. The fin base temperature is Tb = 20ºC and the fin experiences convection with fluid at T∞= 100ºC with average heat transfer coefficient, h = 100 W/m2-K. The fin is th = 2.0 cm thick and has conductivity k = 1.5 W/m-K. a.) Develop a numerical solution for the temperature distribution in the fin using a finite difference technique. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in L=5 [cm]*convert(cm,m) T_b=20 [C] T_infinity=100 [C] h_bar=100 [W/m^2-K] th=3 [cm]*convert(cm,m) k=1.5 [W/m-K] g```_dot=5e5 [W/m^3] W=1 [m]
"length of fin" "base temperature" "fluid temperature" "average heat transfer coefficient" "fin thickness" "fin conductivity" "volumetric generation" "per unit width into page"
A regularly spaced grid of nodes is uniformly distributed with the first and last nodes in each dimension placed on the boundaries of the domain; the x- and y-positions of any node (i,j) are given by:
xi =
( i − 1) L ( M − 1)
(1)
yj =
( j − 1) th 2 ( N − 1)
(2)
where M and N are the number of nodes used in the x- and y-directions, respectively. The x- and y-distance between adjacent nodes (Δx and Δy, respectively) are: Δx =
L ( M − 1)
(3)
Δy =
th 2 ( N − 1)
(4)
N=7 [-] M=11 [-] Dx=L/(M-1) Dy=(th/2)/(N-1) duplicate i=1,M x[i]=L*(i-1)/(M-1) end duplicate j=1,N y[j]=(th/2)*(j-1)/(N-1) end
"number of nodes in the y-direction" "number of nodes in the x-direction" "y-distance between adjacent nodes" "x-position of each node"
"y-position of each node"
An energy balance on an internal node is: q RHS ,i , j + q LHS ,i , j + qtop ,i , j + qbottom ,i , j + g i , j = 0 for i = 2.. ( M − 1) and j = 2.. ( N − 1)
(5)
where
q RHS ,i , j =
k ΔyW (Ti+1, j − Ti, j ) Δx
(6)
q LHS ,i , j =
k ΔyW (Ti −1, j − Ti, j ) Δx
(7)
qtop ,i , j =
k ΔxW (Ti, j +1 − Ti, j ) Δy
(8)
qbottom ,i , j =
k ΔxW (Ti, j −1 − Ti, j ) Δy
g i , j = Δx ΔyW g ′′′
(9) (10)
"Internal nodes" duplicate i=2,(M-1) duplicate j=2,(N-1) q_dot_LHS[i,j]+q_dot_RHS[i,j]+q_dot_top[i,j]+q_dot_bottom[i,j]+g_dot[i,j]=0 q_dot_LHS[i,j]=k*W*Dy*(T[i-1,j]-T[i,j])/Dx q_dot_RHS[i,j]=k*W*Dy*(T[i+1,j]-T[i,j])/Dx q_dot_top[i,j]=k*W*Dx*(T[i,j+1]-T[i,j])/Dy q_dot_bottom[i,j]=k*W*Dx*(T[i,j-1]-T[i,j])/Dy g_dot[i,j]=W*Dx*Dy*g```_dot end end
The left boundary (x= 0) has a specified temperature:
T1, j = Tb for j = 1...N
(11)
"nodes on left edge" duplicate j=1,N T[1,j]=T_b end
An energy balance on the nodes on the upper edge leads to: q RHS ,i , N + q LHS ,i , N + qconv ,i , N + qbottom ,i , N + g i , N = 0 for i = 2.. ( M − 1)
(12)
where q RHS ,i , N =
k ΔyW (Ti+1, N − Ti, N ) 2 Δx
(13)
q LHS ,i , N =
k ΔyW (Ti−1, N − Ti, N ) 2 Δx
(14)
qconv ,i , N = h W Δx (T∞ − Ti , N ) qbottom ,i , N =
(15)
k ΔxW (Ti, N −1 − Ti, N ) Δy
(16)
Δy W g ′′′ 2
(17)
g i , N = Δx
"nodes on upper edge" duplicate i=2,(M-1) q_dot_LHS[i,N]+q_dot_RHS[i,N]+q_dot_conv[i,N]+q_dot_bottom[i,N]+g_dot[i,N]=0
q_dot_LHS[i,N]=k*W*(Dy/2)*(T[i-1,N]-T[i,N])/Dx q_dot_RHS[i,N]=k*W*(Dy/2)*(T[i+1,N]-T[i,N])/Dx q_dot_conv[i,N]=h_bar*W*Dx*(T_infinity-T[i,N]) q_dot_bottom[i,N]=k*W*Dx*(T[i,N-1]-T[i,N])/Dy g_dot[i,N]=Dx*(Dy/2)*W*g```_dot end
This process is continued for all of the edge and corner nodes: "upper right corner node" q_dot_LHS[M,N]+q_dot_conv[M,N]+q_dot_bottom[M,N]+g_dot[M,N]=0 q_dot_LHS[M,N]=k*W*(Dy/2)*(T[M-1,N]-T[M,N])/Dx q_dot_conv[M,N]=h_bar*W*(Dx/2)*(T_infinity-T[M,N]) q_dot_bottom[M,N]=k*W*(Dx/2)*(T[M,N-1]-T[M,N])/Dy g_dot[M,N]=(Dx/2)*(Dy/2)*W*g```_dot "nodes on right edge" duplicate j=2,(N-1) q_dot_LHS[M,j]+q_dot_top[M,j]+q_dot_bottom[M,j]+g_dot[M,j]=0 q_dot_LHS[M,j]=k*W*Dy*(T[M-1,j]-T[M,j])/Dx q_dot_top[M,j]=k*W*(Dx/2)*(T[M,j+1]-T[M,j])/Dy q_dot_bottom[M,j]=k*W*(Dx/2)*(T[M,j-1]-T[M,j])/Dy g_dot[M,j]=(Dx/2)*Dy*W*g```_dot end "lower right corner node" q_dot_LHS[M,1]+q_dot_top[M,1]+g_dot[M,1]=0 q_dot_LHS[M,1]=k*W*(Dy/2)*(T[M-1,1]-T[M,1])/Dx q_dot_top[M,1]=k*W*(Dx/2)*(T[M,2]-T[M,1])/Dy g_dot[M,1]=(Dx/2)*(Dy/2)*W*g```_dot "nodes on bottom edge" duplicate i=2,(M-1) q_dot_LHS[i,1]+q_dot_RHS[i,1]+q_dot_top[i,1]+g_dot[i,1]=0 q_dot_LHS[i,1]=k*W*(Dy/2)*(T[i-1,1]-T[i,1])/Dx q_dot_RHS[i,1]=k*W*(Dy/2)*(T[i+1,1]-T[i,1])/Dx q_dot_top[i,1]=k*W*Dx*(T[i,2]-T[i,1])/Dy g_dot[i,1]=Dx*(Dy/2)*W*g```_dot end
b.) Use the numerical solution to predict and plot the temperature distribution in the heater. The temperature as a function of x for various values of y is shown in Figure 2.
200 180
y = 0 (center)
Temperature (°C)
160 140 y = th/2 (edge)
120 100 80 60 40 20 0
0.01
0.02
0.03
0.04
0.05
Axial position (m) Figure 2: Temperature as a function of x for various values of y.
c.) Use the numerical solution to predict the heater efficiency; the heater efficiency is defined as the ratio of the rate of heat transfer to the fluid to the total rate of thermal energy generation in the fin. The convection heat transfer from node 1,N is computed:
qconv ,1, N = h W
Δx (T∞ − T1, N ) 2
(18)
and the total convective heat transfer from the heater is: M
qhtr = −2 ∑ qconv ,i , N
(19)
i =1
The heater efficiency is:
ηhtr =
qhtr g ′′′W L th
q_dot_conv[1,N]=h_bar*W*(Dx/2)*(T_infinity-T[1,N]) q_dot_htr=-2*sum(q_dot_conv[1..M,N]) eta_htr=q_dot_htr/(W*L*th*g```_dot)
(20) "convection to upper left corner node" "fin heat transfer rate" "heater efficiency"
which leads to ηhtr = 0.357. d.) Plot the heater efficiency as a function of the fin length for various values of the fin thickness. Explain your plot. Figure 3 illustrates the heater efficiency as a function of length for various values of thickness. Note that at small lengths the fin efficiency can become negative because the heater thermally
communicates with the wall and is unable to elevate the temperature above the fluid temperature. Therefore, the fluid is actually cooled by the heater. 1 0.8
Heater efficiency
0.6
th = 3 cm th = 5 cm
0.4 0.2
th = 7 cm
0
th = 9 cm
-0.2 -0.4 -0.6 -0.8 0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
Heater length (m) Figure 3: Heater efficiency as a function of length for various values of thickness.
Problem 2.6-1 (2-9 in text): Model of Welding Process (revisited) Figure P2.6-1 illustrates a cut-away view of two plates that are being welded together. Both edges of the plate are clamped and effectively held at temperatures Ts = 25°C. The top of the plate is exposed to a heat flux that varies with position x, measured from joint, according to: qm′′ ( x ) = q ′′j exp ( − x / L j ) where q ′′j =1x106 W/m2 is the maximum heat flux (at the joint, x = 0)
and Lj = 2.0 cm is a measure of the extent of the heat flux. The back side of the plates are exposed to liquid cooling by a jet of fluid at Tf = -35°C with h = 5000 W/m2-K. A halfsymmetry model of the problem is shown in Figure P2.6-1. The thickness of the plate is b = 3.5 cm and the width of a single plate is W = 8.5 cm. You may assume that the welding process is steady-state and 2-D. You may neglect convection from the top of the plate. The conductivity of the plate material is k = 38 W/m-K. both edges are held at fixed temperature heat flux joint
qm′′
impingement cooling
k = 38 W/m-K W = 8.5 cm
Ts = 25°C
b = 3.5 cm y x 2 h = 5000 W/m -K, T f = −35°C Figure P2.6-1: Welding process and half-symmetry model of the welding process.
a.) Develop a separation of variables solution to the problem (note, this was done previously in Problem 2.2-1). Implement the solution in EES and prepare a plot of the temperature as a function of x at y = 0, 1.0, 2.0, 3.0, and 3.5 cm. b.) Prepare a contour plot of the temperature distribution. See the solution for Problem 2.2-1 for parts (a) and (b). c.) Develop a numerical model of the problem. Implement the solution in MATLAB and prepare a contour or surface plot of the temperature in the plate. The input parameters are entered in the MATLAB function P2p2_1; the input arguments are m and n, the number of nodes in the x and y coordinates while the output arguments are the x and y positions of each node and the predicted temperature at each node. function[xm,ym,T]=P2p6_1(m,n) W=0.085;
%width of plate (m)
b=0.035; k=38; T_s=298.1; T_f=238.2; h=5000; L=1;
%thickness of plate (m) %conductivity of plate material (W/m-K) %side temperature (K) %fluid temperature (K) %heat transfer coefficient (W/m^2-K) %per unit length (m)
end
A sub-function is defined to provide the heat flux on the upper surface: function[qflux]=qf(x) L_j=0.02; qf_j=1e6;
%length scale (m) %heat flux at center (W/m^2)
qflux=qf_j*exp(-x/L_j); end
A 2-D numerical model will be generated using a grid in which the x and y coordinates of each node are:
xi =
( i − 1)W ( m − 1)
for i = 1..m
(1)
yi =
( j − 1)W ( m − 1)
for j = 1..n
(2)
The distance between adjacent nodes is:
%Setup grid for i=1:m x(i,1)=(i-1)*W/(m-1); end Dx=W/(m-1); for j=1:n y(j,1)=(j-1)*W/(n-1); end Dy=b/(n-1);
Δx =
W ( m − 1)
(3)
Δy =
W ( n − 1)
(4)
The problem will be solved by setting up and inverting a matrix containing the algebraic equations that enforce the conservation of energy for each control volume. A control volume for an internal node includes conduction from the left and right sides ( q LHS and q RHS ) and top and bottom ( qtop and qbottom ). The energy balance is:
q RHS + q LHS + qtop + qbottom = 0 The conduction terms are approximated according to: q RHS =
k Δy L (Ti−1, j − Ti, j ) Δx
q LHS =
k Δy L (Ti+1, j − Ti, j ) Δx
qbottom =
k Δx L (Ti, j −1 − Ti, j ) Δy
qtop =
k Δx L (Ti, j +1 − Ti, j ) Δy
where L is the depth of the plate (into the page); L is set to 1.0 m which is consistent with doing the problem on a per unit length basis. Combining these equations leads to: k Δy L k Δy L k Δx L k Δx L Ti −1, j − Ti , j ) + Ti +1, j − Ti , j ) + Ti , j −1 − Ti , j ) + ( ( ( (Ti, j +1 − Ti, j ) = 0 Δx Δx Δy Δy
for i = 2.. ( m − 1) and j = 2.. ( n − 1)
(5)
The equation is rearranged to make it clear what the coefficient is for each unknown temperature: ⎡ k Δy L ⎡ k Δx L ⎤ ⎡ k Δx L ⎤ k Δx L ⎤ ⎡ k Δy L ⎤ ⎡ k Δy L ⎤ Ti , j ⎢ −2 −2 + Ti −1, j ⎢ + Ti +1, j ⎢ + Ti , j −1 ⎢ + Ti , j +1 ⎢ ⎥ ⎥ ⎥=0 ⎥ ⎥ Δx Δy ⎦ ⎣ Δx ⎦ ⎣ Δx ⎦ ⎣ ⎣ Δy ⎦ ⎣ Δy ⎦ for i = 2.. ( m − 1) and j = 2.. ( n − 1) (6) The control volume equations must be placed into the matrix equation: AX =b
where the equation for control volume i,j is placed into row m(j-1)+i of A and Ti,j corresponds to element Xm (j-1)+i in the vector X . Therefore, each coefficient in Eq. (6) (i.e., each term multiplying an unknown temperature on the left side of the equation) must be placed in the row of A corresponding to the control volume being examined and the column of A corresponding to the unknown in X . The matrix assignments consistent with Eq. (6) are: k Δy L k Δx L −2 for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δx Δy
(7)
Am( j −1)+i ,m( j −1)+i −1 =
k Δy L for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δx
(8)
Am( j −1)+i ,m( j −1)+i +1 =
k Δy L for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δx
(9)
Am( j −1)+i ,m( j −1−1)+i =
k Δx L for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δy
(10)
Am( j −1)+i ,m( j +1−1)+i =
k Δx L for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δy
(11)
Am( j −1)+i ,m( j −1)+i = −2
A sparse matrix is allocated in MATLAB for A and the equations derived above are implemented using a nested for loop. The spalloc command requires three arguments, which are the number of rows and columns and the maximum number of elements in the matrix. Note that there are at most 5 non-zero entries in each row of A , corresponding to Eqs. (7) through (11); thus the last argument in the spalloc command which corresponds to the maximum number of non-zero entries in the sparse matrix. A=spalloc(m*n,m*n,5*m*n); %allocate a sparse matrix for A bm=zeros(m*n,1); %allocate a matrix for b %energy balances for internal nodes for i=2:(m-1) for j=2:(n-1) A(m*(j-1)+i,m*(j-1)+i)=-2*k*Dy*L/Dx-2*k*Dx*L/Dy; A(m*(j-1)+i,m*(j-1)+i-1)=k*Dy*L/Dx; A(m*(j-1)+i,m*(j-1)+i+1)=k*Dy*L/Dx; A(m*(j-1)+i,m*(j-1-1)+i)=k*Dx*L/Dy; A(m*(j-1)+i,m*(j+1-1)+i)=k*Dx*L/Dy; end end
The nodes on the right side have a specified temperature: Tm, j = Ts for j = 1..n
The matrix assignments suggested by these equations are: Am( j −1)+ m, m( j −1)+ m = 1 for j = 1..n bm( j −1)+ m = Tst for j = 1..n
These assignments are implemented in MATLAB: %right side temperature is specified for j=1:n A(m*(j-1)+m,m*(j-1)+m)=1; bm(m*(j-1)+m,1)=T_s; end
The nodes along the upper edge must be considered separately, leading to: ⎡ k Δy L ⎤ ⎡ k Δy L ⎤ ⎡ k Δy L k Δx L ⎤ ⎡ k Δx L ⎤ Ti ,n ⎢ − − + Ti −1, n ⎢ + Ti +1, n ⎢ + Ti ,n −1 ⎢ ⎥ ⎥ ⎥ ⎥ = − qm′′ Δx L Δx Δy ⎦ ⎣ ⎣ Δy ⎦ ⎣ 2 Δx ⎦ ⎣ 2 Δx ⎦ for i = 2.. ( m − 1)
(12)
which is expressed in matrix form as: k Δy L k Δx L − for i = 2.. ( m − 1) Δx Δy
(13)
Am( n −1)+i ,m( n −1)+i −1 =
k Δy L for i = 2.. ( m − 1) 2 Δx
(14)
Am( n −1)+i ,m( n −1)+i +1 =
k Δy L for i = 2.. ( m − 1) 2 Δx
(15)
Am( n −1)+i , m( n −1−1)+i =
k Δx L for i = 2.. ( m − 1) Δy
(16)
Am( n −1)+i ,m( n −1)+i = −
bm( n −1)+i ,1 = −qm′′ Δx L for i = 2.. ( m − 1) %upper edge for i=2:(m-1) A(m*(n-1)+i,m*(n-1)+i)=-k*Dy*L/Dx-k*Dx*L/Dy; A(m*(n-1)+i,m*(n-1)+i-1)=k*Dy*L/(2*Dx); A(m*(n-1)+i,m*(n-1)+i+1)=k*Dy*L/(2*Dx); A(m*(n-1)+i,m*(n-1-1)+i)=k*Dx*L/Dy; bm(m*(n-1)+i,1)=-qf(x(i))*Dx*L; end
(17)
The node at the upper left corner must be considered separately, leading to: ⎡ k Δy L k Δx L ⎤ ⎡ k Δy L ⎤ ⎡ k Δx L ⎤ Δx L − + T2,n ⎢ + T1,n −1 ⎢ = − qm′′ T1,n ⎢ − ⎥ ⎥ ⎥ 2 Δy ⎦ 2 ⎣ 2 Δx ⎣ 2 Δx ⎦ ⎣ 2 Δy ⎦
(18)
which is expressed in matrix form as: k Δy L k Δx L − 2 Δx 2 Δy
(19)
Am( n −1) +1,m( n −1)+1+1 =
k Δy L 2 Δx
(20)
Am( n −1)+1, m( n −1−1)+1 =
k Δx L 2 Δy
(21)
Δx L 2
(22)
Am( n −1)+1,m( n −1)+1 = −
bm( n −1)+1,1 = −qm′′
%upper left corner A(m*(n-1)+1,m*(n-1)+1)=-k*Dy*L/(2*Dx)-k*Dx*L/(2*Dy); A(m*(n-1)+1,m*(n-1)+1+1)=k*Dy*L/(2*Dx); A(m*(n-1)+1,m*(n-1-1)+1)=k*Dx*L/(2*Dy); bm(m*(n-1)+1,1)=-qf(x(1))*Dx*L/2;
The node along the left edge must be considered separately, leading to: ⎡ k Δy L ⎤ ⎡ k Δx L ⎤ ⎡ k Δx L ⎤ ⎡ k Δy L k Δx L ⎤ T1, j ⎢ − − + T2 +1, j ⎢ + T1, j −1 ⎢ + T1, j +1 ⎢ ⎥ ⎥ ⎥=0 ⎥ Δx Δy ⎦ ⎣ ⎣ 2 Δx ⎦ ⎣ 2 Δy ⎦ ⎣ 2 Δy ⎦ for j = 2.. ( n − 1)
(23)
which is expressed in matrix form as: k Δy L k Δx L − for j = 2.. ( n − 1) Δx Δy
(24)
Am( j −1)+1,m( j −1)+1+1 =
k Δy L for j = 2.. ( n − 1) Δx
(25)
Am( j −1) +1,m( j −1−1)+1 =
k Δx L for j = 2.. ( n − 1) 2 Δy
(26)
Am( j −1)+1,m( j −1)+1 = −
Am( j −1)+1,m( j +1−1)+1 =
k Δx L for j = 2.. ( n − 1) 2 Δy
(27)
%left edge for j=2:(n-1) A(m*(j-1)+1,m*(j-1)+1)=-k*Dy*L/Dx-k*Dx*L/Dy; A(m*(j-1)+1,m*(j-1)+1+1)=k*Dy*L/Dx; A(m*(j-1)+1,m*(j-1-1)+1)=k*Dx*L/(2*Dy); A(m*(j-1)+1,m*(j+1-1)+1)=k*Dx*L/(2*Dy); end
The node at the lower left corner must be considered separately, leading to: ⎡ k Δy L k Δ x L ⎡ k Δy L ⎤ ⎡ k Δx L ⎤ Δx L ⎤ Δx L T1,1 ⎢ − Tf − −h + T2,1 ⎢ + T1,2 ⎢ = −h ⎥ ⎥ ⎥ 2 Δy 2 ⎦ 2 ⎣ 2 Δx ⎣ 2 Δx ⎦ ⎣ 2 Δy ⎦
(28)
which is expressed in matrix form as: k Δy L k Δx L Δx L − −h 2 Δx 2 Δy 2
(29)
Am(1−1)+1, m(1−1)+1+1 =
k Δy L 2 Δx
(30)
Am(1−1)+1, m(1+1−1)+1 =
k Δx L 2 Δy
(31)
Am(1−1) +1,m(1−1)+1 = −
bm(1−1)+1,1 = −h
Δx L Tf 2
(32)
%lower left corner A(m*(1-1)+1,m*(1-1)+1)=-k*Dy*L/(2*Dx)-k*Dx*L/(2*Dy)-h*L*Dx/2; A(m*(1-1)+1,m*(1-1)+1+1)=k*Dy*L/(2*Dx); A(m*(1-1)+1,m*(1+1-1)+1)=k*Dx*L/(2*Dy); bm(m*(1-1)+1,1)=-h*L*Dx*T_f/2;
The nodes along the bottom edge must be considered separately, leading to: ⎡ k Δy L ⎤ ⎡ k Δy L ⎤ ⎡ k Δy L k Δx L ⎤ ⎡ k Δx L ⎤ Ti ,1 ⎢ − − − h Δx L ⎥ + Ti −1,1 ⎢ + Ti +1,1 ⎢ + Ti ,1+1 ⎢ ⎥ ⎥ ⎥ = − h Δx LT f Δx Δy (33) ⎣ ⎦ ⎣ Δy ⎦ ⎣ 2 Δx ⎦ ⎣ 2 Δx ⎦ for i = 2.. ( m − 1) which is expressed in matrix form as:
k Δy L k Δx L − - h Δx L for i = 2.. ( m − 1) Δx Δy
(34)
Am(1−1) +i , m(1−1)+i −1 =
k Δy L for i = 2.. ( m − 1) 2 Δx
(35)
Am(1−1) +i , m(1−1)+i +1 =
k Δy L for i = 2.. ( m − 1) 2 Δx
(36)
Am(1−1) +i , m(1+1−1)+i =
k Δx L for i = 2.. ( m − 1) Δy
(37)
Am(1−1)+i ,m(1−1)+i = −
bm(1−1)+i ,1 = − h Δx LT f for i = 2.. ( m − 1)
(38)
%lower edge for i=2:(m-1) A(m*(1-1)+i,m*(1-1)+i)=-k*Dy*L/Dx-k*Dx*L/Dy-h*Dx*L; A(m*(1-1)+i,m*(1-1)+i-1)=k*Dy*L/(2*Dx); A(m*(1-1)+i,m*(1-1)+i+1)=k*Dy*L/(2*Dx); A(m*(1-1)+i,m*(1+1-1)+i)=k*Dx*L/Dy; bm(m*(1-1)+i,1)=-h*Dx*L*T_f; end
The vector of unknowns X is obtained through matrix manipulation and then placed into matrix format. Matrices for the x and y positions of each node are also created. X=A\bm; for i=1:m for j=1:n xm(i,j)=x(i); ym(i,j)=y(j); T(i,j)=X(m*(j-1)+i); end end
A surface plot of the result is obtained by typing: >> [x,y,T]=P2p6_1(20,40); >> surf(x,y,T); The plot is shown in Figure 3.
Figure 3: Surface plot of temperature in the plate.
b.) Plot the temperature as a function of x at y = 0, b/2, and b and overlay on this plot the separation of variables solution obtained in part (a) evaluated at the same locations. The comparison is shown in Figure 4.
Figure 4: Temperature as a function of axial position for y = 0, b/2, and b predicted using the separation of variables solution and the numerical solution.
Problem 2.6-2 Prepare a solution to Problem 2.3-3 using a finite difference technique. a.) Plot the temperature as a function of r for various values of x. The input parameters are entered in the MATLAB script P2p2_2. clear all; th=0.01; h_bar=50; T_infinity=20; T_edge=25; k=1.2; R_w=0.03; qf_rad=1000;
% % % % % % %
thickness of window (m) heat transfer coefficient (W/m^2-K) ambient temperature (C) edge temperature (C) conductivity (W/m-K) window radius (m) radiation heat flux (W/m^2)
A 2-D numerical model will be generated using a grid in which the r and x coordinates of each node are:
ri =
( i − 1) RW ( M − 1)
xj =
( j − 1) th ( N − 1)
for i = 1..M
(1)
for j = 1..N
(2)
The distance between adjacent nodes is: Δr =
RW ( M − 1)
(3)
Δx =
th ( N − 1)
(4)
% Setup grid M=51; % number of r-nodes for i=1:M r(i,1)=(i-1)*R_w/(M-1); end Dr=R_w/(M-1); % distance between r-nodes N=101; % number of x-nodes for j=1:N x(j,1)=(j-1)*th/(N-1); end Dx=th/(N-1); % distance between x-nodes
The problem will be solved by setting up and inverting a matrix containing the algebraic equations that enforce the conservation of energy for each control volume. A control volume for
an internal node includes conduction from the left and right sides ( q LHS and q RHS ) and top and bottom ( qtop and qbottom ). The energy balance is:
q RHS + q LHS + qtop + qbottom = 0 The conduction terms are approximated according to:
q RHS
q LHS
Δr ⎞ ⎛ k 2 π ⎜ ri + ⎟ Δx 2 ⎠ ⎝ = (Ti +1, j − Ti, j ) Δr Δr ⎞ ⎛ k 2 π ⎜ ri − ⎟ Δx 2 ⎠ ⎝ = (Ti−1, j − Ti, j ) Δr
qbottom = qtop =
k 2 π ri Δr (Ti, j −1 − Ti, j ) Δx
k 2 π ri Δr (Ti, j +1 − Ti, j ) Δx
Combining these equations leads to: Δr ⎞ Δr ⎞ ⎛ ⎛ k 2 π ⎜ ri + ⎟ Δx k 2 π ⎜ ri − ⎟ Δx 2 ⎠ 2 ⎠ ⎝ ⎝ Ti +1, j − Ti , j ) + ( (Ti−1, j − Ti, j ) Δr Δr k 2 π ri Δr k 2 π ri Δr + Ti , j −1 − Ti , j ) + ( (Ti, j +1 − Ti, j ) = 0 Δx Δx for i = 2.. ( M − 1) and j = 2.. ( N − 1)
(5)
The equation is rearranged to make it clear what the coefficient is for each unknown temperature:
⎡ Δr ⎞ Δr ⎞ ⎤ ⎛ ⎛ ⎢ k 2 π ⎜ ri − 2 ⎟ Δx k 2 π ⎜ ri + 2 ⎟ Δx k 4 π r Δr ⎥ ⎝ ⎠ − ⎝ ⎠ − i ⎥+ Ti , j ⎢ − Δr Δr Δx ⎥ ⎢ ⎢⎣ ⎥⎦ ⎡ Δr ⎞ ⎤ ⎡ Δr ⎞ ⎤ ⎛ ⎛ ⎢ k 2 π ⎜ ri − 2 ⎟ Δx ⎥ ⎢ k 2 π ⎜ ri + 2 ⎟ Δx ⎥ ⎝ ⎠ ⎥ +T ⎢ ⎝ ⎠ ⎥+ Ti −1, j ⎢ i +1, j Δ Δ r r ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎡ k 2 π ri Δr ⎤ ⎡ k 2 π ri Δr ⎤ + Ti , j +1 ⎢ Ti , j −1 ⎢ ⎥ ⎥=0 ⎣ Δx ⎦ ⎣ Δx ⎦ for i = 2.. ( M − 1) and j = 2.. ( N − 1)
(6)
The control volume equations must be placed into the matrix equation: AX =b where the equation for control volume i,j is placed into row M (j-1) + i of A and Ti,j corresponds to element XM(j-1)+i in the vector X . Therefore, each coefficient in Eq. (6) (i.e., each term multiplying an unknown temperature on the left side of the equation) must be placed in the row of A corresponding to the control volume being examined and the column of A corresponding to the unknown in X . The matrix assignments consistent with Eq. (6) are:
AM ( j −1)+i , M ( j −1)+i
Δr ⎞ Δr ⎞ ⎛ ⎛ k 2 π ⎜ ri − ⎟ Δx k 2 π ⎜ ri + ⎟ Δx k 4 π r Δr 2 ⎠ 2 ⎠ ⎝ ⎝ i =− − − Δr Δr Δx for i = 2.. ( M − 1) and j = 2.. ( N − 1)
Δr ⎞ ⎛ k 2 π ⎜ ri − ⎟ Δx 2 ⎠ ⎝ AM ( j −1)+i , M ( j −1)+i −1 = for i = 2.. ( M − 1) and j = 2.. ( N − 1) Δr Δr ⎞ ⎛ k 2 π ⎜ ri + ⎟ Δx 2 ⎠ ⎝ AM ( j −1)+i , M ( j −1)+i +1 = for i = 2.. ( M − 1) and j = 2.. ( N − 1) Δr
(7)
(8)
(9)
AM ( j −1)+i , M ( j −1−1)+i =
k 2 π ri Δr for i = 2.. ( M − 1) and j = 2.. ( N − 1) Δx
(10)
AM ( j −1)+i , M ( j +1−1)+i =
k 2 π ri Δr for i = 2.. ( M − 1) and j = 2.. ( N − 1) Δx
(11)
A sparse matrix is allocated in MATLAB for A and the equations derived above are implemented using a nested for loop. The spalloc command requires three arguments, which are the number of rows and columns and the maximum number of elements in the matrix. Note that there are at most 5 non-zero entries in each row of A , corresponding to Eqs. (7) through (11); thus the last argument in the spalloc command which corresponds to the maximum number of non-zero entries in the sparse matrix. A=spalloc(M*N,M*N,5*M*N); b=zeros(M*N,1); for i=2:(M-1) for j=2:(N-1) A(M*(j-1)+i,M*(j-1)+i)=-k*2*pi*Dx*(r(i)-Dr/2)/Dr-... k*2*pi*Dx*(r(i)+Dr/2)/Dr-k*4*pi*r(i)*Dr/Dx; A(M*(j-1)+i,M*(j-1)+i-1)=k*2*pi*Dx*(r(i)-Dr/2)/Dr; A(M*(j-1)+i,M*(j-1)+i+1)=k*2*pi*Dx*(r(i)+Dr/2)/Dr; A(M*(j-1)+i,M*(j+1-1)+i)=k*2*pi*r(i)*Dr/Dx; A(M*(j-1)+i,M*(j-1-1)+i)=k*2*pi*r(i)*Dr/Dx; end end
The nodes on the right side have a specified temperature: TM , j = Tedge for j = 1..N
The matrix assignments suggested by these equations are: AM ( j −1)+ M , M ( j −1)+ M = 1 for j = 1..N bM ( j −1)+ M = Tedge for j = 1..N
These assignments are implemented in MATLAB: for j=1:N A(M*(j-1)+M,M*(j-1)+M)=1; b(M*(j-1)+M,1)=T_edge; end
The nodes along the bottom edge must be considered separately, leading to: ⎡ k π Δx ⎛ Δr ⎞ k π Δ x ⎛ Δr ⎞ k 2 π ri Δr ⎤ Ti ,1 ⎢ − ⎜ ri − ⎟ − ⎜ ri + ⎟ − Δr ⎝ Δr ⎝ Δx ⎥⎦ 2 ⎠ 2 ⎠ ⎣ ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k 2 π ri Δr ⎤ ′′ 2 π ri Δr +Ti −1,1 ⎢ ⎜ ri − ⎟ ⎥ + Ti +1,1 ⎢ ⎜ ri + ⎟ ⎥ + Ti ,2 ⎢ ⎥ = − qrad 2 ⎠⎦ 2 ⎠⎦ ⎣ Δx ⎦ ⎣ Δr ⎝ ⎣ Δr ⎝ for i = 2.. ( M − 1)
(12)
for i=2:(M-1) A(M*(1-1)+i,M*(1-1)+i)=-k*pi*Dx*(r(i)-Dr/2)/Dr-... k*pi*Dx*(r(i)+Dr/2)/Dr-k*2*pi*r(i)*Dr/Dx; A(M*(1-1)+i,M*(1-1)+i-1)=k*pi*Dx*(r(i)-Dr/2)/Dr; A(M*(1-1)+i,M*(1-1)+i+1)=k*pi*Dx*(r(i)+Dr/2)/Dr; A(M*(1-1)+i,M*(2-1)+i)=k*2*pi*r(i)*Dr/Dx; b(M*(1-1)+i,1)=-qf_rad*2*pi*r(i)*Dr; end
The nodes along the top edge must be considered separately, leading to:
+Ti −1, N
⎡ k π Δx ⎛ Δr ⎞ k π Δ x ⎛ Δr ⎞ k 2 π ri Δr ⎤ Ti , N ⎢ − ⎜ ri − ⎟ − ⎜ ri + ⎟ − Δr ⎝ Δr ⎝ Δx ⎥⎦ 2 ⎠ 2 ⎠ ⎣ ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k 2 π ri Δr ⎤ ⎢ Δr ⎜ ri − 2 ⎟ ⎥ + Ti +1, N ⎢ Δr ⎜ ri + 2 ⎟ ⎥ + Ti , N −1 ⎢ Δx ⎥ = − h 2 π ri Δr T∞ (13) ⎝ ⎠⎦ ⎝ ⎠⎦ ⎣ ⎦ ⎣ ⎣ for i = 2.. ( M − 1)
for i=2:(M-1) A(M*(N-1)+i,M*(N-1)+i)=-k*pi*Dx*(r(i)-Dr/2)/Dr-k*pi*Dx*(r(i)+Dr/2)/Drk*2*pi*r(i)*Dr/Dx-h_bar*2*pi*r(i)*Dr; A(M*(N-1)+i,M*(N-1)+i-1)=k*pi*Dx*(r(i)-Dr/2)/Dr; A(M*(N-1)+i,M*(N-1)+i+1)=k*pi*Dx*(r(i)+Dr/2)/Dr; A(M*(N-1)+i,M*(N-1-1)+i)=k*2*pi*r(i)*Dr/Dx; b(M*(N-1)+i,1)=-h_bar*2*pi*r(i)*Dr*T_infinity; end
The nodes along the left side leads to: ⎡ k 2 π Δx ⎛ Δr ⎞ ⎤ ⎡ k 2 π Δx ⎛ Δr ⎞ ⎤ T1, j ⎢ − ⎜ r1 + ⎟ ⎥ + T2, j ⎢ ⎜ r1 + ⎟ =0 Δr ⎝ 2 ⎠⎦ 2 ⎠ ⎦⎥ ⎣ ⎣ Δr ⎝ for j = 2.. ( N − 1)
(14)
for j=2:(N-1) A(M*(j-1)+1,M*(j-1)+1)=-k*2*pi*(r(1)+Dr/2)*Dx/Dr; A(M*(j-1)+1,M*(j-1)+2)=k*2*pi*(r(1)+Dr/2)*Dx/Dr; end
The node at the lower and upper left corners lead to: ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k π Δx ⎛ Δr ⎞ ⎤ T1,1 ⎢ − ⎜ r1 + ⎟ ⎥ + T2,1 ⎢ ⎜ r1 + ⎟ =0 Δr ⎝ 2 ⎠⎦ 2 ⎠ ⎥⎦ ⎣ ⎣ Δr ⎝
(15)
⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k π Δx ⎛ Δr ⎞ ⎤ T1, N ⎢ − ⎜ r1 + ⎟ ⎥ + T2, N ⎢ ⎜ r1 + ⎟ =0 Δr ⎝ 2 ⎠⎦ 2 ⎠ ⎥⎦ ⎣ ⎣ Δr ⎝
(16)
A(M*(1-1)+1,M*(1-1)+1)=-k*pi*(r(1)+Dr/2)*Dx/Dr;
A(M*(1-1)+1,M*(1-1)+2)=k*pi*(r(1)+Dr/2)*Dx/Dr; A(M*(N-1)+1,M*(N-1)+1)=-k*pi*(r(1)+Dr/2)*Dx/Dr; A(M*(N-1)+1,M*(N-1)+2)=k*pi*(r(1)+Dr/2)*Dx/Dr;
The vector of unknowns X is obtained through matrix manipulation and then placed into matrix format. X=A\b; for i=1:M for j=1:N T(i,j)=X(M*(j-1)+i); end end
The temperature as a function of radius at various values of x is shown in Figure 1. 40 x/th = 0
Temperature (°C)
38
x/th = 0.25
36
x/th = 0.5
34 32 30 x/th = 0.75 x/th = 1
28 26
Problem 2.6-2 Problem 2.3-3
24 0
0.005
0.01
0.015
0.02
0.025
0.03
Radius (m) Figure 1: Temperature as a function of radius at various values of axial position.
b.) Prepare a contour plot of the temperature in the window. A contour plot of the temperature in the window is shown in Figure 2.
0.03
36 0.025
34
Axial position (m)
0.02
32 0.015
30 0.01
28 0.005
26 0
0
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0.008
0.009
0.01
Radius (m) Figure 2: Contour plot of the temperature.
c.) Verify that your solution agrees with the analytical solution from Problem 2.3-3. Figure 1 illustrates the solution obtained from Problem 2.3-3 and shows that the agreement is nearly exact.
Problem 2.6-3 Prepare a solution to Problem 2.3-4 using a finite difference technique. a.) Plot the temperature as a function of r for various values of x. The input parameters are entered in the MATLAB script P2p6d3. clear all; th=0.01; h_bar=50; T_infinity=20; T_edge=25; k=1.2; R_w=0.03; qf_rad=1000; alpha=100;
% % % % % % % %
thickness of window (m) heat transfer coefficient (W/m^2-K) ambient temperature (C) edge temperature (C) conductivity (W/m-K) window radius (m) radiation heat flux (W/m^2) absorption coefficient (1/m)
A 2-D numerical model will be generated using a grid in which the r and x coordinates of each node are:
ri =
( i − 1) RW ( M − 1)
xj =
( j − 1) th ( N − 1)
for i = 1..M
(1)
for j = 1..N
(2)
The distance between adjacent nodes is: Δr =
RW ( M − 1)
(3)
Δx =
th ( N − 1)
(4)
% Setup grid M=51; % number of r-nodes for i=1:M r(i,1)=(i-1)*R_w/(M-1); end Dr=R_w/(M-1); % distance between r-nodes N=101; % number of x-nodes for j=1:N x(j,1)=(j-1)*th/(N-1); end Dx=th/(N-1); % distance between x-nodes
The problem will be solved by setting up and inverting a matrix containing the algebraic equations that enforce the conservation of energy for each control volume. A control volume for
an internal node includes conduction from the left and right sides ( q LHS and q RHS ) and top and bottom ( qtop and qbottom ). The energy balance is:
q RHS + q LHS + qtop + qbottom + g = 0 The conduction terms are approximated according to:
q RHS
q LHS
Δr ⎞ ⎛ k 2 π ⎜ ri + ⎟ Δx 2 ⎠ ⎝ = (Ti +1, j − Ti, j ) Δr Δr ⎞ ⎛ k 2 π ⎜ ri − ⎟ Δx 2 ⎠ ⎝ = (Ti−1, j − Ti, j ) Δr
qbottom = qtop =
k 2 π ri Δr (Ti, j −1 − Ti, j ) Δx
k 2 π ri Δr (Ti, j +1 − Ti, j ) Δx
′′ α exp ( −α x j ) g = 2 π ri Δr Δx qrad Combining these equations leads to: Δr ⎞ Δr ⎞ ⎛ ⎛ k 2 π ⎜ ri + k 2 π ⎜ ri − ⎟ Δx ⎟ Δx 2 ⎠ 2 ⎠ ⎝ ⎝ Ti +1, j − Ti , j ) + ( (Ti−1, j − Ti, j ) Δr Δr k 2 π ri Δr k 2 π ri Δr ′′ α exp ( −α x j ) Ti , j −1 − Ti , j ) + + ( (Ti, j +1 − Ti, j ) = −2 π ri Δr Δx qrad Δx Δx for i = 2.. ( M − 1) and j = 2.. ( N − 1)
(5)
The equation is rearranged to make it clear what the coefficient is for each unknown temperature:
⎡ Δr ⎞ Δr ⎞ ⎤ ⎛ ⎛ ⎢ k 2 π ⎜ ri − 2 ⎟ Δx k 2 π ⎜ ri + 2 ⎟ Δx k 4 π r Δr ⎥ ⎝ ⎠ − ⎝ ⎠ − i ⎥+ Ti , j ⎢ − Δr Δr Δx ⎢ ⎥ ⎢⎣ ⎥⎦ ⎡ Δr ⎞ ⎤ ⎡ Δr ⎞ ⎤ ⎛ ⎛ ⎢ k 2 π ⎜ ri − 2 ⎟ Δx ⎥ ⎢ k 2 π ⎜ ri + 2 ⎟ Δx ⎥ ⎝ ⎠ ⎥ +T ⎢ ⎝ ⎠ ⎥+ Ti −1, j ⎢ i +1, j Δ r Δ r ⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎡ k 2 π ri Δr ⎤ ⎡ k 2 π ri Δr ⎤ ′′ α exp ( −α x j ) Ti , j −1 ⎢ + Ti , j +1 ⎢ ⎥ ⎥ = −2 π ri Δr Δx qrad ⎣ Δx ⎦ ⎣ Δx ⎦ for i = 2.. ( M − 1) and j = 2.. ( N − 1)
(6)
The control volume equations must be placed into the matrix equation: AX =b where the equation for control volume i,j is placed into row M (j-1) + i of A and Ti,j corresponds to element XM(j-1)+i in the vector X . Therefore, each coefficient in Eq. (6) (i.e., each term multiplying an unknown temperature on the left side of the equation) must be placed in the row of A corresponding to the control volume being examined and the column of A corresponding to the unknown in X . The matrix assignments consistent with Eq. (6) are:
AM ( j −1)+i , M ( j −1)+i
Δr ⎞ Δr ⎞ ⎛ ⎛ k 2 π ⎜ ri − ⎟ Δx k 2 π ⎜ ri + ⎟ Δx k 4 π r Δr 2 ⎠ 2 ⎠ ⎝ ⎝ i =− − − Δr Δr Δx for i = 2.. ( M − 1) and j = 2.. ( N − 1)
Δr ⎞ ⎛ k 2 π ⎜ ri − ⎟ Δx 2 ⎠ ⎝ AM ( j −1)+i , M ( j −1)+i −1 = for i = 2.. ( M − 1) and j = 2.. ( N − 1) Δr Δr ⎞ ⎛ k 2 π ⎜ ri + ⎟ Δx 2 ⎠ ⎝ AM ( j −1)+i , M ( j −1)+i +1 = for i = 2.. ( M − 1) and j = 2.. ( N − 1) Δr
(7)
(8)
(9)
AM ( j −1)+i , M ( j −1−1)+i =
k 2 π ri Δr for i = 2.. ( M − 1) and j = 2.. ( N − 1) Δx
(10)
AM ( j −1)+i , M ( j +1−1)+i =
k 2 π ri Δr for i = 2.. ( M − 1) and j = 2.. ( N − 1) Δx
(11)
′′ α exp ( −α x j ) for i = 2.. ( M − 1) and j = 2.. ( N − 1) bM ( j −1)+i = −2 π ri Δr Δx qrad
(12)
A sparse matrix is allocated in MATLAB for A and the equations derived above are implemented using a nested for loop. The spalloc command requires three arguments, which are the number of rows and columns and the maximum number of elements in the matrix. Note that there are at most 5 non-zero entries in each row of A , corresponding to Eqs. (7) through (11); thus the last argument in the spalloc command which corresponds to the maximum number of non-zero entries in the sparse matrix. A=spalloc(M*N,M*N,5*M*N); b=zeros(M*N,1); for i=2:(M-1) for j=2:(N-1) A(M*(j-1)+i,M*(j-1)+i)=-k*2*pi*Dx*(r(i)-Dr/2)/Drk*2*pi*Dx*(r(i)+Dr/2)/Dr-k*4*pi*r(i)*Dr/Dx; A(M*(j-1)+i,M*(j-1)+i-1)=k*2*pi*Dx*(r(i)-Dr/2)/Dr; A(M*(j-1)+i,M*(j-1)+i+1)=k*2*pi*Dx*(r(i)+Dr/2)/Dr; A(M*(j-1)+i,M*(j+1-1)+i)=k*2*pi*r(i)*Dr/Dx; A(M*(j-1)+i,M*(j-1-1)+i)=k*2*pi*r(i)*Dr/Dx; b(M*(j-1)+i,1)=-2*pi*r(i)*Dr*Dx*qf_rad*alpha*exp(-alpha*x(j)); end end
The nodes on the right side have a specified temperature: TM , j = Tedge for j = 1..N
The matrix assignments suggested by these equations are: AM ( j −1)+ M , M ( j −1)+ M = 1 for j = 1..N bM ( j −1)+ M = Tedge for j = 1..N
These assignments are implemented in MATLAB: for j=1:N A(M*(j-1)+M,M*(j-1)+M)=1; b(M*(j-1)+M,1)=T_edge; end
The nodes along the bottom edge must be considered separately, leading to:
⎡ k π Δx ⎛ Δr ⎞ k π Δ x ⎛ Δr ⎞ k 2 π ri Δr ⎤ Ti ,1 ⎢ − ⎜ ri − ⎟ − ⎜ ri + ⎟− 2 ⎠ 2 ⎠ Δr ⎝ Δr ⎝ Δx ⎥⎦ ⎣ ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k 2 π ri Δr ⎤ +Ti −1,1 ⎢ ⎜ ri − ⎟ ⎥ + Ti +1,1 ⎢ ⎜ ri + ⎟ ⎥ + Ti ,2 ⎢ ⎥= 2 ⎠⎦ 2 ⎠⎦ ⎣ Δx ⎦ ⎣ Δr ⎝ ⎣ Δr ⎝
(13)
′′ α exp ( −α x1 ) −π ri Δr Δx qrad
for i = 2.. ( M − 1) for i=2:(M-1) A(M*(1-1)+i,M*(1-1)+i)=-k*pi*Dx*(r(i)-Dr/2)/Dr-k*pi*Dx*(r(i)+Dr/2)/Dr-... k*2*pi*r(i)*Dr/Dx; A(M*(1-1)+i,M*(1-1)+i-1)=k*pi*Dx*(r(i)-Dr/2)/Dr; A(M*(1-1)+i,M*(1-1)+i+1)=k*pi*Dx*(r(i)+Dr/2)/Dr; A(M*(1-1)+i,M*(2-1)+i)=k*2*pi*r(i)*Dr/Dx; b(M*(1-1)+i,1)=-pi*r(i)*Dr*Dx*qf_rad*alpha*exp(-alpha*x(1)); end
The nodes along the top edge must be considered separately, leading to:
+Ti −1, N
⎡ k π Δx ⎛ Δr ⎞ k π Δ x ⎛ Δr ⎞ k 2 π ri Δr ⎤ Ti , N ⎢ − ⎜ ri − ⎟ − ⎜ ri + ⎟− 2 ⎠ 2 ⎠ Δr ⎝ Δr ⎝ Δx ⎥⎦ ⎣ ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k 2 π ri Δr ⎤ ⎢ Δr ⎜ ri − 2 ⎟ ⎥ + Ti +1, N ⎢ Δr ⎜ ri + 2 ⎟ ⎥ + Ti , N −1 ⎢ Δx ⎥ = ⎝ ⎠⎦ ⎝ ⎠⎦ ⎣ ⎦ ⎣ ⎣
(14)
′′ α exp ( −α xN ) −h 2 π ri Δr T∞ − π ri Δr Δx qrad
for i = 2.. ( M − 1) for i=2:(M-1) A(M*(N-1)+i,M*(N-1)+i)=-k*pi*Dx*(r(i)-Dr/2)/Dr-k*pi*Dx*(r(i)+Dr/2)/Dr-... k*2*pi*r(i)*Dr/Dx-h_bar*2*pi*r(i)*Dr; A(M*(N-1)+i,M*(N-1)+i-1)=k*pi*Dx*(r(i)-Dr/2)/Dr; A(M*(N-1)+i,M*(N-1)+i+1)=k*pi*Dx*(r(i)+Dr/2)/Dr; A(M*(N-1)+i,M*(N-1-1)+i)=k*2*pi*r(i)*Dr/Dx; b(M*(N-1)+i,1)=-h_bar*2*pi*r(i)*Dr*T_infinity-... pi*r(i)*Dr*Dx*qf_rad*alpha*exp(-alpha*x(N)); end
The nodes along the left side leads to: ⎡ k 2 π Δx ⎛ Δr ⎞ ⎤ ⎡ k 2 π Δx ⎛ Δr ⎞ ⎤ T1, j ⎢ − ⎜ r1 + ⎟ ⎥ + T2, j ⎢ ⎜ r1 + ⎟ =0 2 ⎠⎦ 2 ⎠ ⎥⎦ Δr ⎝ ⎣ ⎣ Δr ⎝ for j = 2.. ( N − 1) for j=2:(N-1) A(M*(j-1)+1,M*(j-1)+1)=-k*2*pi*(r(1)+Dr/2)*Dx/Dr; A(M*(j-1)+1,M*(j-1)+2)=k*2*pi*(r(1)+Dr/2)*Dx/Dr; end
(15)
The node at the lower and upper left corners lead to: ⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k π Δx ⎛ Δr ⎞ ⎤ T1,1 ⎢ − ⎜ r1 + ⎟ ⎥ + T2,1 ⎢ ⎜ r1 + ⎟ =0 Δr ⎝ 2 ⎠⎦ 2 ⎠ ⎥⎦ ⎣ ⎣ Δr ⎝
(16)
⎡ k π Δx ⎛ Δr ⎞ ⎤ ⎡ k π Δx ⎛ Δr ⎞ ⎤ T1, N ⎢ − ⎜ r1 + ⎟ ⎥ + T2, N ⎢ ⎜ r1 + ⎟ =0 Δr ⎝ 2 ⎠⎦ 2 ⎠ ⎥⎦ ⎣ ⎣ Δr ⎝
(17)
A(M*(1-1)+1,M*(1-1)+1)=-k*pi*(r(1)+Dr/2)*Dx/Dr; A(M*(1-1)+1,M*(1-1)+2)=k*pi*(r(1)+Dr/2)*Dx/Dr; A(M*(N-1)+1,M*(N-1)+1)=-k*pi*(r(1)+Dr/2)*Dx/Dr; A(M*(N-1)+1,M*(N-1)+2)=k*pi*(r(1)+Dr/2)*Dx/Dr;
The vector of unknowns X is obtained through matrix manipulation and then placed into matrix format. X=A\b; for i=1:M for j=1:N T(i,j)=X(M*(j-1)+i); end end
The temperature as a function of radius at various values of x is shown in Figure 1. 31 x/th = 0 x/th = 0.25 x/th = 0.5 x/th = 0.75 x/th = 1.0
Temperature (°C)
30 29 28 27 26 25 24 0
Problem 2.3-4 Problem 2.6-3
0.005
0.01
0.015
0.02
0.025
0.03
Radius (m) Figure 1: Temperature as a function of radius at various values of axial position.
b.) Prepare a contour plot of the temperature in the window. A contour plot of the temperature in the window is shown in Figure 2.
0.03 30
29.5
0.025
29
Axial position (m)
0.02
28.5
28 0.015 27.5
27 0.01 26.5
26
0.005
25.5
0 0
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0.008
0.009
0.01
25
Radius (m) Figure 2: Contour plot of the temperature.
c.) Verify that your solution agrees with the analytical solution from Problem 2.3-4. Figure 1 illustrates the solution obtained from Problem 2.3-4 and shows that the agreement is nearly exact.
Problem 2.7-1 (2-10 in text): A Double Paned Window Figure P2.7-1(a) illustrates a double paned window. The window consists of two panes of glass each of which is tg = 0.95 cm thick and W = 4 ft wide by H = 5 ft high. The glass panes are separated by an air gap of g = 1.9 cm. You may assume that the air is stagnant with ka = 0.025 W/m-K. The glass has conductivity kg = 1.4 W/m-K. The heat transfer coefficient between the inner surface of the inner pane and the indoor air is hin = 10 W/m2-K and the heat transfer coefficient between the outer surface of the outer pane and the outdoor air is hout = 25 W/m2-K. You keep your house heated to Tin = 70°F. width of window, W = 4 ft tg = 0.95cm tg = 0.95 cm g = 1.9 cm
Tin = 70°F 2 hin = 10 W/m -K H = 5 ft
Tout = 23°F 2 hout = 25 W/m -K ka = 0.025 W/m-K kg = 1.4 W/m-K
casing shown in P2.10(b)
Figure P2.7-1(a): Double paned window.
The average heating season in Madison lasts about time = 130 days and the average outdoor temperature during this time is Tout = 23°F. You heat with natural gas and pay, on average, ec = 1.415 $/therm (a therm is an energy unit =1.055x108 J). a.) Calculate the average rate of heat transfer through the double paned window during the heating season. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" tg=0.375*convert(inch,m) g=0.75*convert(inch,m) k_g=1.4 [W/m-K] k_a=0.025 [W/m-K] H=5 [ft]*convert(ft,m) W=4[ft]*convert(ft,m) T_in=converttemp(F,K,70 [F]) h_in=10 [W/m^2-K] T_out=converttemp(F,K,23 [F]) h_out=25 [W/m^2-K] time=130 [day]*convert(day,s) ec=1.415 [$/therm]*convert($/therm,$/J)
"glass thickness" "air gap" "glass conductivity" "air conductivity" "height of window" "width of window" "indoor air temperature" "heat transfer coefficient on inside of window" "outdoor air temperature" "heat transfer coefficient on outside of window" "heating season duration" "cost of energy"
The heat transfer is resisted by convection on the inner and outer surfaces:
Rconv ,in =
1 hin W H
(1)
Rconv ,out =
1 hout W H
(2)
and conduction through the glass panes and the air:
R_conv_in=1/(h_in*W*H) R_cond_g=tg/(k_g*W*H) R_cond_a=g/(k_a*W*H) R_conv_out=1/(h_out*W*H)
Rcond , g =
tg kg W H
(3)
Rcond ,a =
g ka W H
(4)
"convection resistance on inside of window" "conduction resistance of glass pane" "conduction resistance of air gap" "convection resistance on outside of window"
The heat transfer rate through the window is: q =
Tin − Tout Rconv ,in + 2 Rcond , g + Rcond ,a + Rconv ,out
(5)
q_dot=(T_in-T_out)/(R_conv_in+2*R_cond_g+R_cond_a+R_conv_out) "rate of heat transfer through the window"
which leads to q = 53.0 W. b.) How much does the energy lost through the window cost during a single heating season? The total amount of energy lost over the course of a heating season is: Q = q time
(6)
cost = ec Q
(7)
and the associated cost is:
Q=q_dot*time cost=ec*Q
which leads to cost = $7.98/heating season.
"total energy loss" "cost to heat house per window"
There is a metal casing that holds the panes of glass and connects them to the surrounding wall, as shown in Figure P2.7-1(b). Because the metal casing is high conductivity, it seems likely that you could lose a substantial amount of heat by conduction through the casing (potentially negating the advantage of using a double paned window). The geometry of the casing is shown in Figure P2.7-1(b); note that the casing is symmetric about the center of the window. glass panes 1.9 cm Tin = 70°F 2 hin = 10 W/m -K
0.95 cm air
Tout = 23°F 2 hout = 25 W/m -K
2 cm 4 cm
metal casing km = 25 W/m-K
0.5 cm
3 cm 0.4 cm
wood
Figure P2-10(b) Metal casing.
All surfaces of the casing that are adjacent to glass, wood, or the air between the glass panes can be assumed to be adiabatic. The other surfaces are exposed to either the indoor or outdoor air. c.) Prepare a 2-D thermal analysis of the casing using FEHT. Turn in a print out of your geometry as well as a contour plot of the temperature distribution. What is the rate of energy lost via conduction through the casing per unit length (W/m)? The geometry from Figure 3 is entered approximately, as shown in Figure 4.
Figure 4: Approximate geometry.
Each of the points are selected individually and their exact coordinates are entered, which leads to the more precise geometry shown in Figure 4.
Figure 5: Geometry.
The material properties are specified by selecting the outline and selecting Material Properties from the Specify menu. The boundary conditions are specified by selecting each type of boundary and then selecting Boundary Conditions from the Specify menu. A crude mesh is generated, as shown in Figure 6.
Figure 6: Crude Mesh.
The temperature contours are shown in Figure 7.
Figure 7: Temperature Contours.
The heat transfer per unit length is obtained by selecting Heat Flow from the View menu and selecting all of the boundaries that are exposed to either the indoor or outdoor air. The heat flow is 10.0 W/m. d.) Show that your numerical model has converged by recording the rate of heat transfer per length for several values of the number of nodes. The mesh is refined several times and each time the heat transfer rate per unit length of casing is recorded; the results are shown in Figure 8.
Figure 8: Heat transfer through the casing per unit length as a function of the number of nodes.
e.) How much does the casing add to the cost of heating your house?
′ ) is entered in EES: The result from FEHT, the heat transfer per unit length ( qcasing q`_dot_casing=10.0 [W/m]
"casing heat transfer per unit meter"
The total heat transfer through the casing is: ′ qcasing = qcasing 2 (W + H )
(8)
The total heat lost through the casing is: Qcasing = qcasing time
and the associated cost is:
(9)
costcasing = ec Qcasing q_dot_casing=q`_dot_casing*2*(W+H) Q_casing=q_dot_casing*time cost_casing=ec*Q_casing "cost to heat house per window due to casing"
(10) "heat transfer rate through casing" "total energy flow through casing"
which leads to an additional cost of $8.27 per heating season.
Problem 2.7-2: Energy recovery system Relatively hot gas flows out of the stack of a power plant. The energy associated with these combustion products is useful for providing hot water or other low grade energy. The system shown in Figure P2.7-2(a) has been developed to recover some of this energy. water flows through holes
hot gas
stack structure
Figure P2.7-2(a): Energy recovery system for hot exhaust gas.
The system is fabricated from a high temperature material in the form of a ring; 16 fluid channels are integrated with the stack in a circular array. The water to be heated flows through these channels. The inner surface of the liner is finned to increase its surface area and is exposed to the hot gas while the outer surface is cooled externally by ambient air. A cut-away view of the stack is shown in Figure P2.7-2(b). unit cell Tair = 20°C 2 hair = 20 W/m -K Tw = 30°C 2 hw = 500 W/m -K Tgas = 800°C 2 k = 25 W/m-K hgas = 50 W/m -K Figure P2.7-2(b): Problem specification for the energy recovery problem.
At a particular section, the water can be modeled as being at a uniform temperature of Tw = 30°C with a heat transfer coefficient, hw = 500 W/m2-K. The hot gas is at Tgas = 800°C with a heat transfer coefficient, hgas = 50 W/m2-K. The ambient air external to the stack is at Tair = 20°C and hair = 20 W/m2-K. The stack material has conductivity k = 25 W/m-K. The liner geometry is relatively complex and includes curved segments as well as straight sections. Only a single unit cell of the structure (see Figure P2.7-2(b)) needs to be simulated. FEHT does not allow curved sections to be simulated; rather, a curved section must be approximated as a polygon. Rather than attempting to draw the geometry manually, it is preferable to import a drawing (e.g., from a computer aided drawing package) and trace the drawing in FEHT. More advanced finite element tools will have automated processes for importing geometry from various sources. A drawing of the unit cell with a scale can be copied onto the clipboard (from the website for this text, using Microsoft Powerpoint) and pasted into FEHT. a.) Use FEHT to develop a finite element model of the stack and determine the heat transfer to the water per unit length of stack.
Open FEHT and specify that you are doing a steady state heat transfer problem in cartesian coordinates using the C temperature scale using the Setup menu (Figure 3).
Figure 3: Specify menu.
Open the program that contains the unit cell drawing and copy it to the clip board. In FEHT, select Paste from Clipboard from the File menu; the drawing should appear in the drawing space (Figure 4).
Figure 4: Drawing pasted onto drawing area.
Move the template to the center of the screen and select Size/Move Template from the draw menu; this will allow the grid to show through the template (Figure 5).
Figure 5: Template and drawing grid.
Finally, resize your drawing grid so that 1 cm on your template corresponds to 1 cm on your model. First, identify the size of the square that is contained on the template for this purpose. Select Scale and Size from the Setup menu and then adjust the Grid Spacing option until the size of grid matches the size of the square (Figures 6 and 7).
Figure 6: Adjust Grid Spacing until a grid matches the scale on the template.
Figure 7: Grid spacing adjusted to match the template scale.
In Figure 7, the grid spacing is 1.45 cm x 1.45 cm; therefore, the 1.45 cm of screen is equivalent to 1 cm of drawing distance. Select Scale and Size from the Setup menu and set the scale such that 1 cm of screen is equal to 1/1.45 cm (or 0.69 cm) of drawing distance (Figure 8).
Figure 8: Adjust scale to match template.
Now it is possible to draw the geometry by tracing the template. You need to specify the outer region and then “punch” holes in it (rather than drawing the flow passage and then overwriting it with the outer region). Select Outline from the Draw menu and “trace” the outside edge of the drawing. It is important to recognize that the curved surfaces must be represented by a discrete number of straight lines that are defined by the nodes. The more nodes you use, the better you can simulate the curve but the more difficult it will be to generate a mesh. Start with just a few nodes, as shown in Figure 9.
Figure 9: Adjust scale to match template.
Select Hide Patterns from the Display menu so that the material is transparent (Figure 10).
Figure 10: Pattern removed to make the template visible.
Select Outline from the Draw menu and trace the circle (use an octagon to represent the curve), as shown in Figure 11.
Figure 11: Circular flow passage defined in FEHT.
Select Hide Template from the Display menu and the geometry will remain. Save your file. Having defined the geometry it is necessary to define the material properties, boundary conditions, and mesh in order to solve the problem. The liquid within the flow passage can be represented conveniently as a single temperature. In this case it is possible to mesh the liquid and specify a very high conductivity; however, this is a waste of computational resources given that we know the answer. Instead, FEHT allows you to specify that a region is a Lumped/Fluid element which has a single temperature. This is a nice feature for many problems that reduces the complexity of the model without affecting the accuracy of the predictions. Select the outline of the fluid passage and then select Material Properties from the Specify menu (Figure 12). Select not specified from the list and then give the element the name Lumped Water, click on Type until Fluid/Lump appears.
Figure 12: Specify Properties dialog.
Select Show Patterns from the Display menu to see the water. Next select the outline of the stack material and specify the conductivity.
The edges that define the unit cell are adiabatic; select these edges and then select Boundary Conditions from the Specify menu. Set the heat flux to 0 W/m2 (Figure 13).
Figure 13: Specify Boundary Conditions dialog.
Set the boundary conditions on the internal surface (exposed to combustion gas) and external surface (exposed to ambient air) in the normal way. You also need to specify a heat transfer coefficient between the water and the fluid passage; select the octagon that defines this boundary and then select Boundary Conditions from the Specify menu. Set the heat transfer coefficient to hwater (Figure 14).
Figure 14: Specify Boundary Conditions for the boundary of the lumped element.
You will need to specify the temperature of the lumped element; click on the lumped element and select Lumped Information from the Specify menu (Figure 15). Set the temperature to be Twater
Figure 15: Specify Lumped Information dialog.
It is necessary to specify a crude mesh; you do not need to mesh the lumped element because the governing equations will not be solved within this element (it is just assigned a temperature). The mesh is specified by selecting Element Lines from the Draw menu; all of the mesh elements must be triangular. A reasonable initial mesh is shown in Figure 16.
Figure 16: Mesh for the stack material.
Select Calculate from the Run menu and then select Temperature Contours from the View menu (Figure 17).
Figure 17: Solution with crude mesh.
It is possible to determine the total heat transferred to the water by selecting Heat Flows from the View menu and then selecting the 8 boundaries that define the water passage (Figure 18). Notice that the sum of the total heat flow is displayed on the toolbar.
Figure 18: Heat flows.
The total heat flow to a single tube is 3076 W/m; therefore, the total heat transfer to the 16 passages filled with water will be 49.2 kW. b.) Verify that your solution has converged numerically.
In order to evaluate how many nodes are required, refine the mesh (select Input from the View menu and then Reduce Mesh from the Draw menu) and solve the problem again. Record the number of nodes or elements (the number of elements is displayed by selecting Check from the Run menu) and the heat flow to the water as the mesh is progressively refined. The result is shown in Figure 19 and suggests that at least 400 nodes should be used.
Figure 19: Predicted heat flows per passage per unit length as a function of the number of elements.
c.) Sanity check your results against a simple model. The geometry is too irregular to allow an analytical solution. Therefore, it is possible to compare the answer to a very rough resistance calculation. The known information is entered in EES: $UnitSystem SI MASS RAD PA K J "Inputs" T_gas=converttemp(C,K,800) h_gas=50 [W/m^2-K] T_water=converttemp(C,K,30) h_water=500 [W/m^2-K] k=25 [W/m-K] L=1 [m]
"hot gas" "hot gas heat transfer coefficient" "water temperature" "water heat transfer coefficient" "material conductivity" "per unit length of stack"
There are, approximately, three resistances between the hot gas and the water. The resistance to convection with the flue gas, Rconv,gas, is: Rconv , gas =
1 hgas L pgas
where pgas is the perimeter of the stack material that is exposed to gas - approximately 7.0 cm.
(1)
p_gas=7.0 [cm]*convert(cm,m) R_conv_gas=1/(h_gas*L*p_gas)
"perimeter of stack exposed to gas" "resistance due to convection to the gas"
The resistance to convection with the water, Rconv,water, is:
Rconv , water =
hwater
1 L π Dwater
(2)
where Dwater is the diameter of the water passage, approximately 1.5 cm. D_water=1.5 [cm]*convert(cm,m) R_conv_water=1/(h_water*L*pi*D_water)
"diameter of water passage" "resistance due to convection to water"
Finally, the resistance due to conduction within the stack material, Rcond, is approximately: Rcond =
Lc k L wc
(3)
where Lc is the distance heat must be conducted and L wc is the area for conduction; Lc is approximately 3.0 cm and wc is approximately 2.5 cm. L_c=3.0 [cm]*convert(cm,m) w_c=2.5 [cm]*convert(cm,m) R_cond=L_c/(k*w_c*L)
"distance for conduction" "width for conduction" "resistance for conduction"
The total heat transfer rate ( q ) should be approximately:
q ≈
(T
gas
− Twater )
Rconv , gas + Rconv , water + Rcond
(4)
q_dot=(T_gas-T_water)/(R_conv_gas+R_conv_water+R_cond) "heat transfer rate"
The approximate model predicts that Rconv,gas= 0.29 K/W, Rconv,water = 0.04 K/W, Rcond = 0.05 K/W which leads to q =2050 W; this is 30% less than the FEHT model prediction. The agreement is not expected to be substantially better than this given the extremely crude approximations involved in the resistance model. However, the calculations suggest that no substantial errors were made in the finite element model. Furthermore, the magnitude of the three resistances provides some insight into what thermal resistance controls the problem (the convection with the air) and therefore how the design could be most effectively improved.
P2.7-3 (2-11 in text): A Spacecraft Radiator A radiator panel extends from a spacecraft; both surfaces of the radiator are exposed to space (for the purposes of this problem it is acceptable to assume that space is at 0 K); the emittance of the surface is ε = 1.0. The plate is made of aluminum (k = 200 W/m-K and ρ = 2700 kg/m3) and has a fluid line attached to it, as shown in Figure 2.7-3(a). The half-width of the plate is a=0.5 m wide while the height of the plate is b=0.75m. The thickness of the plate is a design variable and will be varied in this analysis; begin by assuming that the thickness is th = 1.0 cm. The fluid lines carry coolant at Tc = 320 K. Assume that the fluid temperature is constant although the fluid temperature will actually decrease as it transfers heat to the radiator. The combination of convection and conduction through the panel-to-fluid line mounting leads to an effective heat transfer coefficient of h = 1,000 W/m2-K over the 3.0 cm strip occupied by the fluid line. k = 200 W/m-K ρ = 2700 kg/m3 ε = 1.0
space at 0 K
a = 0.5 m
3 cm th = 1 cm
b = 0.75 m
fluid at Tc = 320 K
half-symmetry model of panel, Figure P2-11(b)
Figure 2.7-3(a): Radiator panel
The radiator panel is symmetric about its half-width and the critical dimensions that are required to develop a half-symmetry model of the radiator are shown in Figure 2.7-3(b). There are three regions associated with the problem that must be defined separately so that the surface conditions can be set differently. Regions 1 and 3 are exposed to space on both sides while Region 2 is exposed to the coolant fluid one side and space on the other; for the purposes of this problem, the effect of radiation to space on the back side of Region 2 is neglected. Region 1 (both sides exposed to space) Region 2 (exposed to fluid - neglect radiation to space) Region 3 (both sides exposed to space) (0.50,0.75) (0.50,0.55) (0.50,0.52)
y x (0.50,0)
(0,0) (0.22,0)
(0.25,0)
line of symmetry
Figure 2.7-3(b): Half-symmetry model.
a.) Prepare a FEHT model that can predict the temperature distribution over the radiator panel.
The radiator panel can be modeled as 2-D problem because it is thin and has high conductivity. The Biot number compares the ratio of the internal conduction resistance to the external resistance; in this case due to radiation.
Rcond Rrad
(1)
th 2 k As
(2)
1 2 As σ ε s (T + Tsur )(Ts + Tsur )
(3)
Bi = where
Rcond = Using the concept of a radiation resistance: Rrad =
2 s
The Biot number is therefore: Bi =
2 th σ ε s (Ts2 + Tsur )(Ts + Tsur )
2k
(4)
The surrounding temperature in Eq. (4), Tsur corresponds to space which is essentially 0 K and therefore Eq. (4) reduces to: Bi =
th σ ε s Ts3 2k
(5)
We don’t know the value of the panel surface temperature, Ts in Eq. (5), but we can assume that it will be near the fluid temperature for any well-designed radiator. Using Ts = Tf in Eq. (5) results in a Biot number of 4.6x10-5 which is very small; clearly for any reasonable value of Ts the problem will be 2-D. FEHT can simulate 2-D problems of this type using the Extended Surface mode. Start FEHT and select Extended Surface from the Subject menu. Each of the three regions must be drawn separately using the Outline option from the Draw menu. It is easiest to set an appropriate grid using the Scale and Size selection from the Setup menu (Figure 2).
Figure 2: Scale and Size dialog window
Generate outlines for the 3 regions that are close to the correct scale and then double-click on each node and position it exactly (Figure 3).
Figure 3: Geometry definition.
The material properties for all three regions can be set by selecting each while holding down the shift key and then selecting Specify Properties from the Specify menu. Select Aluminum from the list of materials and then modify the conductivity to be 200 W/m-K and the thickness to be 0.01 m to match the problem statement (Figure 4).
Figure 4: Specify material properties.
The surface conditions for Regions 1 and 3 can be set by selecting these regions with the shift key held down and then selecting Surface Conditions from the Specify menu. These regions are radiating to space but there is no option listed in the Extended Surface Conditions for a radiative surface condition. The radiation heat flux is:
′′ = ε σ T 4 qrad
(6)
Equation (6) can be rewritten according to: ′′ = ε σ T 3 (T − 0 ) qrad
(7)
Equation (7) is similar to a convection equation:
′′ = hrad (T − T f ) qrad
(8)
where fluid temperature (Tf) is 0 K and the convection coefficient (hrad) is a function of temperature:
hrad = ε σ T 3
(9)
FEHT allows the specification of the surface conditions in terms of position (X and Y) as well as temperature (T); therefore, the radiative surface condition can be modeled as shown in Figure 5.
Figure 5: Extended Surface Conditions Dialog window for Regions 1 and 3.
Region 2 experiences a convection boundary condition with the coolant, but only from one side (as opposed to the double sided condition assumed by FEHT). The heat transfer from a differential element in Region 2 is given by: q = h dA (T − Tc )
(10)
whereas the convection surface condition in FEHT assumes convection from both sides of the plate and therefore: q = hF 2 dA (T − Tc )
(11)
where hF is the heat transfer coefficient that should be set in FEHT in order to simulate the single-sided convection coefficient represented by Eq. (10). Comparing Eqs. (10) and (11) leads to:
hF =
h 2
(12)
Click on Region 2 and select Surface Condition from the Specify menu; specify the surface conditions as shown in Figure 6.
Figure 6: Extended Surface dialog window for Region 2.
Finally, the boundary conditions along each edge of the computational domain must be specified; the line of symmetry is adiabatic and the remaining edges are also assumed to be adiabatic. Generate a reasonable but crude mesh, as shown in Figure 7(a) and then refine it. Note that the mesh within Region 2 will start relatively refined due to its small width and need not be refined as much as the mesh in Regions 1 and 3. However, it is possible to refine the mesh in a single
region by selecting that region and then selecting Reduce Mesh from the Draw menu. The result should be similar to Figure 7(b).
(a)
(b) Figure 7: (a) coarse and (b) refined mesh.
Solve the problem by selecting Calculate from the Run menu; the problem is non-linear due to the temperature dependent heat transfer coefficient and therefore the solution process will be iterative. Plot the temperature distribution by selecting Temperature Contours from the View menu (Figure 8).
Figure 8: Temperature distribution for a 1 cm thick plate.
b.) Export the solution to EES and calculate the total heat transferred from the radiator and the radiator efficiency (defined as the ratio of the radiator heat transfer to the heat transfer from the radiator if it were isothermal and at the coolant temperature). Select Tabular Output from the View menu and then Select All and Save As. Save the data as a file called ‘1 cm’. Open EES and select Open Lookup Table from the Tables menu; navigate the file ‘1 cm’ and open it; the solution (the temperature at each node together with the locations of the nodes) is contained in the lookup table. Using the technique discussed in EXAMPLE 16-1 it is possible to use the Interpolate2D function to obtain the temperature at an arbitrary x and y location on the radiator plate. $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "INPUTS" e=1.0 [-] a=0.5 [m] b=0.75 [m] T_f=320 [K] th=0.01 [m] rho=2700 [kg/m^3] k=200 [W/m-K] "location on plate" x=0.1 [m] y=0.5 [m] "2-D interpolation from table of nodal data"
"emittance of panel surface" "half-width of panel" "height of panel" "fluid temperature" "thickness" "density" "conductivity"
T=Interpolate2D('1 cm',X,Y,T,X=x,Y=y)
"temperature interpolated from data"
In order to obtain the total heat transferred from the panel it is necessary to integrate the heat flux over the entire area of the plate. This process can be accomplished manually by dividing the plate into many small integration areas, calculating the heat flux within each area, and summing the result. The plate is divided into Nx segments in the x direction, each with width: Δx =
a Nx
(13)
The x position of each segment is: xi = ( i − 0.5 ) Δx
(14)
The plate is divided into Ny segments in the y direction, each with height: Δy =
b Ny
(15)
The y position of each segment is: y j = ( j − 0.5 ) Δy
(16)
The corresponding EES code is: "Double integration manually" Nx=25 [-] "number of integration areas in x" Ny=10 [-] "number of integration areas in y" "size of an integration area" Dx=a/Nx Dy=b/Ny "setup position of x and y areas" duplicate i=1,Nx x[i]=(i-0.5)*Dx end duplicate j=1,Ny y[j]=(j-0.5)*Dy end
The total heat transferred from the plate ( q ) is calculated using the double integral: b a
q = ∫ ∫ q ′′x , y dx dy 0 0
where the heat flux is:
(17)
q ′′x , y = ε σ Tx4, y
(18)
The numerical summation that approximates the integration in Eq. (17) is: N y Nx
q = ∑∑ q ′′xi , y j Δx Δy
(19)
q ′′xi , y j = ε σ Tx4i , y j
(20)
j =1 i =1
where
The corresponding EES code is: duplicate i=1,Nx duplicate j=1,Ny "2-D interpolation from table of nodal data" T[i,j]=Interpolate2D('1 cm',X,Y,T,X=x[i],Y=y[j]) "temperature interpolated from data" q``_dot[i,j]=2*e*sigma#*T[i,j]^4 "heat flux" end end q_dot=sum(q``_dot[1..Nx,1..Ny])*Dx*Dy
Note that the same process can be accomplished using EES’ native Integral command. The Integral command carries out numerical integration using a more sophisticated algorithm than simply assuming a constant value over each step. EES’ Integral command requires 4 arguments and allows a 5th, optional argument; the protocol for calling the function is: F=Integral(Integrand,VarName,LowerLimit,UpperLimit,StepSize) where Integrand is the EES variable or expression that must be integrated, VarName is the integration variable, and LowerLimit and UpperLimit define the limits of integration. StepSize is optional and defines the numerical step that is used to accomplish the integration; a small value of StepSize will lead to more accurate results but take longer to calculate. The double integral in Eq. (17) can be accomplished by calling the Integral function twice; Eq. (17) is rewritten as: b
q = ∫ q′y dy 0
where
(21)
a
q ′y = ∫ q ′′x , y dx
(22)
0
Equation (22) is evaluated using the Integral command as shown in the EES code below: "Double integration using EES' Integral command" "2-D interpolation from table of nodal data" T=Interpolate2D('1 cm',X,Y,T,X=x,Y=y) q``_dot=2*e*sigma#*T^4 y=0 q`_dot=INTEGRAL(q``_dot,x,0,a,0.02)
"temperature interpolated from data" "heat flux" "heat transfer per unit length"
Note that the value of y was set as q ′y is a function of y but the value of x is not set as x is the integration variable. To evaluate Eq. (21), integrate q ′y from y=0 to y=b: "Double integration using EES' Integral command" "2-D interpolation from table of nodal data" T=Interpolate2D('1 cm',X,Y,T,X=x,Y=y) q``_dot=2*e*sigma#*T^4 q`_dot=INTEGRAL(q``_dot,x,0,a,0.02) q_dot=INTEGRAL(q`_dot,y,0,b,0.02)
"temperature interpolated from data" "heat flux" "heat transfer per unit length" "total heat transfer rate"
Note that the value of y is no longer set as y is the integration variable in the 2nd integral. The results of the manual integration should agree with the use of the Integral function; either should yield 339 W. The radiator efficiency (η) and mass (M) are computed according to:
η=
q 2 a b ε σ T f4
(23)
and
M = a b th ρ
(24)
c.) Explore the effect of thickness on the radiator efficiency and mass. The solution procedure described above is repeated for several values of the radiator thickness. Figure 10 shows the predicted temperature distribution for several values of the thickness. Note that the same contour levels were selected for each plot by selecting User in the Temperature Contour Information dialog window and specifying the range from 200 K to 320 K (Figure 9); this allows the different cases to be compared directly and shows clearly that the reduced thickness increases the resistance to conduction along the plate and therefore leads to progressively larger temperature gradients through the plate.
Figure 9: Temperature Contour Information dialog window.
Figure 10: Temperature distribution in the panel for thicknesses of (a) 1.0 cm, (b) 0.75 cm, (c) 0.5 cm, (d) 0.3 cm, (e) 0.2 cm, and (f) 0.1 cm.
Figure 11 illustrates the radiator efficiency and mass as a function of the thickness. As the thickness is reduced, the efficiency drops but so does the mass; clearly there must be a trade-off between these effects.
Figure 11: Efficiency and mass as a function of thickness.
Figure 12 shows the efficiency as a function of mass and makes the trade-off clearer; above a panel with a mass of nominally 2 kg there is a region of diminishing return where additional mass provides only a small gain in efficiency.
Figure 12: Efficiency as a function of mass.
A complete analysis would require more information then is given; specifically, how should the radiator performance be compared with its mass to determine a true optimal thickness. Barring additional constraints related to, for example, structural stability, Figure 12 suggests that any optimization process will result in a panel that is approximately 0.2 cm thick with a mass of 2 kg and an efficiency of 60%.
Problem 2.7-4 Gas turbine power cycles are used for the generation of power; the size of these systems can range from 10’s of kWs for the microturbines that are being installed on-site at some commercial and industrial locations to 100’s of MWs for natural gas fired power plants. The efficiency of a gas turbine power plant increases with the temperature of the gas entering from the combustion chamber; this temperature is constrained by the material limitations of the turbine blades which tend to creep (i.e., slowly grow over time) in the high temperature environment in their high centrifugal stress state. One technique for achieving high gas temperatures is to cool the blades internally; often the air is bled through the blade surface using a technique called transpiration. A simplified version of a turbine blade that will be analyzed in this problem is shown in Figure P2.7-4. combustion gas Tg = 1800 K
k = 15 W/m-K
hg= 850 W/m2-K
coordinates (in cm): point 1 (0, 0), point 2 (2, 0), point 3 (3.5, 0.25), point 4 (4, 0), point 5 (5, 0.75), point 6 (5.5, 0), point 7 (6.25, 1.25), point 8 (7, 0), point 9 (7, 2)
cooling air Tc = 500 K hc = 250 W/m2-K
Figure 2.7-4: A simplified schematic of an air cooled blade.
The high temperature combustion gas is at Tg = 1800 K and the heat transfer coefficient between the gas and blade external surface is hg = 850 W/m2-K. The blades are cooled by three internal air passages. The cooling air in the passages is at Tc = 500 K and the air-to-blade heat transfer coefficient is ha= 250 W/m2-K. The blade material has conductivity k = 15 W/m-K. The coordinates of the points required to define the geometry are indicated in Figure P2.7-4. a.) Generate a ½ symmetry model of the blade in FEHT. Generate a figure showing the temperature distribution in the blade predicted using a very crude mesh. The half-symmetry model is generated by selecting Scale and Size from the Setup menu and setting up a convenient grid to approximately place the outer nodes; the scale was set to 1 cm = 0.5 cm with grid spacing 0.5 cm x 0.5 cm (Figure 1). The temperatures are set to K and the problem set to a steady-state problem in cartesian coordinates by making the appropriate selections in the Setup menu.
Figure 2: Scale and Size window.
The nodes required to define the shape are approximately positioned using the grid by selecting Outline from the Draw menu (Figure 3(a)); each node is subsequently double-clicked on and the exact coordinates are entered (Figure 3(b)).
(a)
(b) Figure 3: FEHT model with (a) nodes approximately placed on the grid and (b) nodes exactly positioned.
The material properties are specified by clicking on the outline and selecting Material Properties from the Specify menu (Figure 4). It is best to set a lighter color for the material so that the grid can be seen.
Figure 4: Specifying material properties.
The boundary conditions along the external surfaces are specified by clicking on these two lines and selecting Boundary Conditions from the Specify menu (Figure 5).
Figure 5: External surface boundary condition specification.
The remaining boundary conditions are specified using the same process (notice that the boundaries along the line of symmetry are adiabatic); the result is shown in Figure 6.
Figure 6: Boundary conditions A very crude grid is generated (Figure 7); recall that each element has to be a triangle.
Figure 7: Crude mesh.
The mesh can be checked by selecting Check from the Run menu. If there are no errors then select Calculate from the Run menu and view the temperature distribution by selecting Temperature Contours from the View menu; the result is shown in Figure 8.
Figure 8: Temperature distribution with crude mesh.
b.) Refine your mesh and keep track of the temperature experienced at the trailing edge of the blade (i.e., at position 9 in Figure P2.7-4) as a function of the number of nodes in your mesh. Prepare a plot of this data that can be used to establish that your model has converged to the correct solution. The nodal data can be viewed by selecting either Tabular Output or Temperatures from the View menu. The number of entries in the Tabular Output table leads to the number of nodes and the temperature at the trailing edge can be read from the model. The number of unknown temperatures is also displayed each time the model is calculated. The mesh is refined over an over again (by selecting Reduce Mesh from the Draw menu) in order to obtain the results listed in Table 1 and plotted in Figure 9. Table 1: Mesh convergence data Number of nodes Temperature at trailing edge of blade 20 1689 K 57 1686 K 185 1685 K 657 1685 K
Table 1 and Figure 9 suggest that the solution has converged at 657 nodes. The temperature distribution for 657 nodes is shown in Figure 10.
Figure 9: Trailing edge temperature as a function of the number of nodes.
Figure 10: Temperature distribution in the turbine blade with 657 nodes.
c.) Do your results make sense? Use a very simple, order-of-magnitude analysis based on thermal resistances to decide whether your predicted blade surface temperature is reasonable (hint – there are three thermal resistances that govern the behavior of the blade, estimate each one and show that your results are approximately correct given these thermal resistances). From a very approximate standpoint, the problem is governed by the thermal resistance to convection on the internal and external surfaces (Rconv,in and Rconv,ext) and conduction through the blade (Rcond). These resistances are estimated here:
Rconv ,c =
1 hc pc L
(1)
where pc is the perimeter of the cooling passages (7.75 cm, from Figure 2.7-4) and L is a unit length of the turbine blade.
Rconv , g =
where (
pg
is
the
( 7 cm ) + ( 2 cm ) 2
2
perimeter
of
the
1 hg pg L
blade
(2) exposed
to
the
combustion
gas
+ 2 cm ) from Figure 2.7-4.
Rcond =
Lcond k Acond
(3)
where Lcond is the approximate length that must be conducted through (about 0.5 cm from Figure 2.7-4) and Acond is the approximate area for conduction (about 5.0 cm x L from Figure 2.7-4). The calculations are carried out in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" T_c=500 [K] h_c=250 [W/m^2-K] T_g=1800 [K] h_g=850 [W/m^2-K] k=15 [W/m-K] L=1 [m]
"cooling air temperature" "cooling air heat transfer coefficient" "combustion gas temperature" "combustion gas heat transfer coefficient" "conductivity" "calculations are per meter"
p_c=7.75 [cm]*convert(cm,m) p_g=(sqrt((7 [cm])^2+(2 [cm])^2)+2[cm])*convert(cm,m) L_cond=0.5 [cm]*convert(cm,m) A_cond=5.0 [cm]*convert(cm,m)*L
"perimeter of cooling channels" "perimeter of blade" "conduction length" "conduction area"
The heat transfer per unit length of blade is: q =
Tg − Tc Rconv , g + Rcond + Rconv ,c
(4)
and can be used to estimate the surface temperature of the blade (Ts,g): Ts , g = Tg − q Rconv , g q_dot=(T_g-T_c)/(R_conv_g+R_cond+R_conv_c) T_s_g=T_g-q_dot*R_conv_g
"heat transfer per unit length per half-blade" "estimated surface temperature"
The result is 1570 K which is approximately consistent with the results shown in Figure 10.
(5)
Problem 2.7-5 a.) Show how the construction of the finite element problem changes with the addition of volumetric generation. The governing differential equation, including volumetric generation is: ∂ ⎛ ∂T ⎜ −k ∂x ⎝ ∂x
⎞ ∂ ⎛ ∂T ⎟ + ⎜ −k ∂y ⎠ ∂y ⎝
⎞ ⎟ − g ′′′ = 0 ⎠
(1)
The average weighted residual equation is: ∂ ⎛
∂Tˆ ⎞
∂ ⎛
∂Tˆ ⎞
∫∫ f ∂x ⎜⎝ −k ∂x ⎟⎠ dx dy + ∫∫ f ∂y ⎜⎝ −k ∂y ⎟⎠ dx dy − ∫∫ f g ′′′ dx dy = weighted average residual
A A
Integral 1
(2)
A
Integral 2
and the weak form of the equation is:
⎡ ∂f ∂Tˆ ∂f ∂Tˆ ⎤ k k + ∫∫A ⎢⎣ ∂x ∂x ∂y ∂y ⎥⎦ dA − ∫∫A f g ′′′ dx dy = ∫B f qn′′ ds
(3)
Substituting the approximate temperature distribution: Tˆ ( x, y ) = wT Tˆ
(4)
into Eq. (3) leads to: ⎡ ∂f ∂
∂f ∂
⎤
∫∫ ⎢⎣k ∂x ∂x ( w Tˆ ) + k ∂y ∂y ( w Tˆ )⎥⎦ dA − ∫∫ f g ′′′ dx dy = ∫ f A
T
T
A
qn′′ ds
B
Using the Galerkin technique for the weighting functions leads to: ⎡ ⎛ ∂w1 ∂w1 ⎡ ⎛ ∂w ∂w ∂w ∂w ⎞ ⎤ ∂w ∂w ⎞ ⎤ + k 1 1 ⎟ dA⎥ Tˆ1 + ⎢ ∫∫ ⎜ k 1 2 + k 1 2 ⎟ dA⎥ Tˆ2 + ... ⎢ ∫∫ ⎜ k ∂y ∂y ⎠ ⎦ ∂y ∂y ⎠ ⎦ ⎣ A ⎝ ∂x ∂x ⎣ A ⎝ ∂x ∂x ⎡ ⎛ ∂w ∂w ∂w ∂w ⎞ ⎤ + ⎢ ∫∫ ⎜ k 1 N + k 1 N ⎟ dA⎥ TˆN − ∫∫ w1 g ′′′ dx dy = ∫ w1 qn′′ ds ∂y ∂y ⎠ ⎦ A B ⎣ A ⎝ ∂x ∂x
(5)
⎡ ⎛ ∂w2 ∂w1 ⎡ ⎛ ∂w ∂w ∂w ∂w ⎞ ⎤ ∂w ∂w ⎞ ⎤ + k 2 1 ⎟ dA⎥ Tˆ1 + ⎢ ∫∫ ⎜ k 2 2 + k 2 2 ⎟ dA⎥ Tˆ2 + ... ⎢ ∫∫ ⎜ k ∂y ∂y ⎠ ⎦ ∂y ∂y ⎠ ⎦ ⎣ A ⎝ ∂x ∂x ⎣ A ⎝ ∂x ∂x ⎡ ⎛ ∂w ∂wN ∂w ∂wN ⎞ ⎤ ˆ + ⎢ ∫∫ ⎜ k 2 +k 2 ⎟ dA⎥ TN − ∫∫ w2 g ′′′ dx dy = ∫ w2 qn′′ ds ∂y ∂y ⎠ ⎦ A B ⎣ A ⎝ ∂x ∂x ...
(6)
⎡ ⎛ ∂wN ∂w1 ⎡ ⎛ ∂w ∂w2 ∂w ∂w ⎞ ⎤ ∂w ∂w2 ⎞ ⎤ ˆ + k N 1 ⎟ dA⎥ Tˆ1 + ⎢ ∫∫ ⎜ k N +k N ⎢ ∫∫ ⎜ k ⎟ dA⎥ T2 + ... x x y y x x y y ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎝ ⎠ ⎝ ⎠ ⎦ ⎣A ⎦ ⎣A ⎡ ⎛ ∂w ∂wN ∂w ∂wN ⎞ ⎤ ˆ + ⎢ ∫∫ ⎜ k N +k N ⎟ dA⎥ TN − ∫∫ wN g ′′′ dx dy = ∫ wN qn′′ ds x x y y ∂ ∂ ∂ ∂ ⎝ ⎠ ⎦ A B ⎣A Equation (6) can be written more concisely in vector notation: ⎡ ⎛ ∂ w ∂ wT ∂ w ∂ wT ⎞ ⎤ ˆ + ⎢ ∫∫ k ⎜ ⎟ dA⎥ T − ∫∫ w g ′′′ dx dy = ∫ w qn′′ ds x x y y ∂ ∂ ∂ ∂ A B ⎠ ⎦⎥ ⎣⎢ A ⎝
g
K
(7)
q
The construction of the matrix K and the vector q are discussed in Section 2.7.2. This problem focuses on the construction of the generation matrix g . The generation vector is assembled by the summation of element generation vectors according to: Ne
g = ∑ ge
(8)
i =1
where the element generation vector is given by:
g e = ∫∫ w g e′′′dx dy
(9)
Ae
where g e′′′ is the volumetric generation within element e. Within element e, only the weighting functions wi, wj, and wk are nonzero (where i, j, and k are the indices of the nodes that define the element): T w = ⎡⎣ 0 ... 0 wi 0 ... 0 w j 0 ... 0 wk 0 ... 0 ⎤⎦
Substituting Eq. (10) into Eq. (9) leads to:
(10)
⎡0 ⎤ ⎢ ⎥ ⎢... ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢ wi ⎥ ⎢0 ⎥ ⎢ ⎥ ⎢... ⎥ ⎢0 ⎥ ⎢ ⎥ g e = g e′′′∫∫ ⎢ w j ⎥ dx dy ⎥ Ae ⎢ ⎢0 ⎥ ⎢... ⎥ ⎢ ⎥ ⎢0 ⎥ ⎢w ⎥ ⎢ k⎥ ⎢0 ⎥ ⎢ ⎥ ⎢... ⎥ ⎢⎣0 ⎥⎦
(11)
The integral of a weighting function over the element is obtained from Eq. (S2.7.2-55):
∫∫ w
a i
wbj wkc dA =
Ae
a !b !c ! 2A ( a + b + c + 2 )! e
(S2.7.2-55)
or
∫∫ w
i
Ae
dA = ∫∫ wi1 w0j wk0 dA = Ae
1!0!0! 1 2 Ae = Ae 3 (1 + 0 + 0 + 2 )!
(12)
Therefore, Eq. (11) can be written as:
⎡. ⎤ ⎢⎥ ⎢1⎥ ⎢. ⎥ e′′′ Ae ⎢ ⎥ g ge = 1 3 ⎢⎢ ⎥⎥ . ⎢⎥ ⎢1⎥ ⎢⎣. ⎥⎦ The overall matrix equation becomes:
row i row j row k (13)
( K + H ) Tˆ = Q + g
(14)
b.) Re-solve the problem discussed in Section 2.7.2 assuming that the material experiences a volumetric generation rate of g ′′′ = 1x103 W/m3, as shown in Figure P2.7-5.
(0,0.75)
2 ht = 100 W/m -K T∞,t = 300 K
(0.5,0.75)
k = 0.25 W/m-K 3 g ′′′ = 1500 W/m
2 q ′′ = 100 W/m
(0.5,0.25) 2 hb = 50 W/m -K T∞,b = 350 K Figure P2.7-5: Two-dimensional conduction problem used to illustrate the finite element solution with generation. The coordinates of points are shown in m.
(0,0)
(0.25,0)
The solution proceeds as discussed in Section 2.7.2. The information about the sub-domains is contained in a matrix sd . The number of columns in the matrix sd is equal to the number of sub-domains in the computational domain. The information about the sub-domains is contained in the column. The first row of each column contains the conductivity of the sub-domain and the second row the generation. The matrix sd is setup in the MATLAB script S2p7d5 according to: clear all;
sdm=[0.25 1500]'; bcm=[0,0,0; 50,350,0; 0,0,0; 100,300,0; 0,0,100]';
% setup details of computational domain % conductivity and generation of each sub-domain % boundary condition for each boundary
The boundary condition matrix is unchanged: bcm=[0,0,0; 50,350,0; 0,0,0; 100,300,0; 0,0,100]';
The same mesh is used, shown in Figure 2:
% boundary condition for each boundary
n11 b7 n6 b8 n5 b9 n1
b6 n10 b5 n9 e11 e8 b4 e9 e10 n8 n7 e6 e5 b3 e7 e4 n3 n4 e1 n = node# e3 e2 b2 e = element # b = boundary segment # b1 n2 Figure 2: A crude mesh.
The point, edge, and triangle matrices remain unchanged: % setup details of the grid N_n=11; % number of nodes % setup point matrix - coordinates of each node pm=[0,0; 0.25,0; 0.5,0.25; 0.25,0.25; 0,0.25; 0,0.5; 0.25,0.5; 0.5,0.5; 0.5,0.75; 0.25,0.75; 0,0.75]'; N_b=9; % number of boundary segments % setup edge matrix - nodes that define each segment and boundary em=[1,2,1; 2,3,2; 3,8,3; 8,9,3; 9,10,4; 10,11,4; 11,6,5; 6,5,5; 5,1,5]'; N_e=11; % number of elements % setup triangle matrix - nodes that define each element and subdomain tm=[5,1,4,1; 1,2,4,1; 4,2,3,1; 4,3,8,1; 7,4,8,1; 6,4,7,1; 6,5,4,1; 11,6,10,1; 6,7,10,1; 10,7,8,1; 10,8,9,1]';
The global conduction matrix and generation vector are setup together, element-by-element. Both are initialized: K=zeros(N_n,N_n); g=zeros(N_n,1);
% global conduction matrix initialization % global generation matrix initialization
These for loop cycles through all of the elements: for e=1:N_e
The indices of the three nodes that define the element and the subdomain that the element belongs to are obtained from the triangle matrix: i=tm(1,e); j=tm(2,e); k=tm(3,e); d=tm(4,e);
% index of node i % index of node j % index of node k % subdomain of element
The x- and y-distances separating the nodes (xij, xik, etc.) are computed; the coordinates of the nodes are obtained from the point matrix: xij=pm(1,j)-pm(1,i);
% x-distance between nodes j and i in element e
xik=pm(1,k)-pm(1,i); xjk=pm(1,k)-pm(1,j); yij=pm(2,j)-pm(2,i); yik=pm(2,k)-pm(2,i); yjk=pm(2,k)-pm(2,j);
% x-distance between nodes k and i in element e % x-distance between nodes k and j in element e % y-distance between nodes j and i in element e % y-distance between nodes k and i in element e % y-distance between nodes k and j in element e
The parameter bijk and the area of the element are computed: bijk=xij*yjk-xjk*yij; Ae=abs(bijk)/2;
% area of element e
The conductivity of the element is obtained from the matrix sd and the elements of the element conduction matrix are added to the global conduction matrix. % add element conduction matrix to global conduction matrix ke=sdm(1,d); % conductivity of element K(i,i)=K(i,i)+Ae*ke*(xjk^2+yjk^2)/bijk^2; K(i,j)=K(i,j)-Ae*ke*(xjk*xik+yjk*yik)/bijk^2; K(i,k)=K(i,k)+Ae*ke*(xjk*xij+yjk*yij)/bijk^2; K(j,i)=K(j,i)-Ae*ke*(xik*xjk+yik*yjk)/bijk^2; K(j,j)=K(j,j)+Ae*ke*(xik^2+yik^2)/bijk^2; K(j,k)=K(j,k)-Ae*ke*(xik*xij+yik*yij)/bijk^2; K(k,i)=K(k,i)+Ae*ke*(xij*xjk+yij*yjk)/bijk^2; K(k,j)=K(k,j)-Ae*ke*(xij*xik+yij*yik)/bijk^2; K(k,k)=K(k,k)+Ae*ke*(xij^2+yij^2)/bijk^2;
The generation within the element is obtained from the matrix sd and the elements of the element conduction matrix are added to the global conduction matrix according to Eq. (13). % add element generation matrix to global generation matrix ge=sdm(2,d); % generation of element g(i,1)=g(i,1)+Ae*ge/3; g(j,1)=g(j,1)+Ae*ge/3; g(k,1)=g(k,1)+Ae*ge/3; end
The vector Q and matrix H are unchanged: Q=zeros(N_n,1); H=zeros(N_n,N_n); for b=1:N_b i=em(1,b); j=em(2,b); bndry=em(3,b); xij=pm(1,j)-pm(1,i); yij=pm(2,j)-pm(2,i); sij=sqrt(xij^2+yij^2); qfsb=bcm(3,bndry); hb=bcm(1,bndry);
% global boundary vector initialization % global convection matrix initialization % index of node i % index of node j % boundary of boundary segment % x-distance between nodes j and i on boundary segment b % y-distance between nodes j and i on boundary segment b % length of boundary segment b % specified heat flux (W/m^2) % heat transfer coefficient (W/m^2-K)
Tinfb=bcm(2,bndry);
% ambient temperature (K)
% add segment boundary vector to global boundary vector Q(i,1)=Q(i,1)+qfsb*sij/2+hb*Tinfb*sij/2; Q(j,1)=Q(j,1)+qfsb*sij/2+hb*Tinfb*sij/2; % add segment convection matrix to global convective matrix H(i,i)=H(i,i)+hb*sij/3; H(i,j)=H(i,j)+hb*sij/6; H(j,i)=H(j,i)+hb*sij/6; H(j,j)=H(j,j)+hb*sij/3; end
Equation (14) is solved and the result is plotted: T=(K+H)\(Q+g); % obtain temperatures at each node pdeplot(pm,em,tm,'xydata',T,'contour','on');
The solution is shown in Figure 3.
Figure 3: Solution to the problem shown in Figure 2.7-5 with the mesh shown in Figure 2.
Note that the PDE toolbox can be used to generate a more refined mesh, as shown in Figure 4.
Figure 4: Refined mesh generated by the PDE toolbox.
The mesh is exported to the workspace and the script is altered as shown below (changes are highlighted in bold): %clear all; % setup details of computational domain sdm=[0.25 1500]'; % conductivity and generation of each sub-domain bcm=[0,0,0; 50,350,0; 0,0,0; 100,300,0; 0,0,100]'; % boundary condition for each boundary % % setup details of the grid % N_n=11; % number of nodes % % setup point matrix - coordinates of each node % pm=[0,0; 0.25,0; 0.5,0.25; 0.25,0.25; 0,0.25; 0,0.5; 0.25,0.5; 0.5,0.5; 0.5,0.75; 0.25,0.75; 0,0.75]'; % % N_b=9; % number of boundary segments % % setup edge matrix - nodes that define each segment and boundary % em=[1,2,1; 2,3,2; 3,8,3; 8,9,3; 9,10,4; 10,11,4; 11,6,5; 6,5,5; 5,1,5]'; % % N_e=11; % number of elements % % setup triangle matrix - nodes that define each element and subdomain % tm=[5,1,4,1; 1,2,4,1; 4,2,3,1; 4,3,8,1; 7,4,8,1; 6,4,7,1; 6,5,4,1; 11,6,10,1; 6,7,10,1; 10,7,8,1; 10,8,9,1]'; % % setup details of the grid % N_n=11; % number of nodes % % setup point matrix - coordinates of each node % pm=[0,0; 0.25,0; 0.5,0.25; 0.25,0.25; 0,0.25; 0,0.5; 0.25,0.5; 0.5,0.5; 0.5,0.75; 0.25,0.75; 0,0.75]'; % % N_b=9; % number of boundary segments % % setup edge matrix - nodes that define each segment and boundary % em=[1,2,1; 2,3,2; 3,8,3; 8,9,3; 9,10,4; 10,11,4; 11,6,5; 6,5,5; 5,1,5]'; % % N_e=11; % number of elements % % setup triangle matrix - nodes that define each element and subdomain
% tm=[5,1,4,1; 1,2,4,1; 4,2,3,1; 4,3,8,1; 7,4,8,1; 6,4,7,1; 6,5,4,1,; 11,6,10,1; 6,7,10,1; 10,7,8,1; 10,8,9,1]'; [g,N_n]=size(p);% number of nodes pm=p; % point matrix is equal to the one produced by pdetool [g,N_e]=size(t);% number of elements tm=t; % triangle matrix is equal to the one produced by pdetool [g,N_b]=size(e);% number of boundary segments em=zeros(3,N_b);% initialize edge matrix em(1,:)=e(1,:); % index of node i em(2,:)=e(2,:); % index of node j em(3,:)=e(5,:); % boundary K=zeros(N_n,N_n); % global conduction matrix initialization g=zeros(N_n,1); % global generation matrix initialization for e=1:N_e i=tm(1,e); % index of node i j=tm(2,e); % index of node j k=tm(3,e); % index of node k d=tm(4,e); % subdomain of element xij=pm(1,j)-pm(1,i); % x-distance between nodes j and i in element e xik=pm(1,k)-pm(1,i); % x-distance between nodes k and i in element e xjk=pm(1,k)-pm(1,j); % x-distance between nodes k and j in element e yij=pm(2,j)-pm(2,i); % y-distance between nodes j and i in element e yik=pm(2,k)-pm(2,i); % y-distance between nodes k and i in element e yjk=pm(2,k)-pm(2,j); % y-distance between nodes k and j in element e bijk=xij*yjk-xjk*yij; Ae=abs(bijk)/2; % area of element e % add element conduction matrix to global conduction matrix ke=sdm(1,d); % conductivity of element K(i,i)=K(i,i)+Ae*ke*(xjk^2+yjk^2)/bijk^2; K(i,j)=K(i,j)-Ae*ke*(xjk*xik+yjk*yik)/bijk^2; K(i,k)=K(i,k)+Ae*ke*(xjk*xij+yjk*yij)/bijk^2; K(j,i)=K(j,i)-Ae*ke*(xik*xjk+yik*yjk)/bijk^2; K(j,j)=K(j,j)+Ae*ke*(xik^2+yik^2)/bijk^2; K(j,k)=K(j,k)-Ae*ke*(xik*xij+yik*yij)/bijk^2; K(k,i)=K(k,i)+Ae*ke*(xij*xjk+yij*yjk)/bijk^2; K(k,j)=K(k,j)-Ae*ke*(xij*xik+yij*yik)/bijk^2; K(k,k)=K(k,k)+Ae*ke*(xij^2+yij^2)/bijk^2; % add element generation matrix to global generation matrix ge=sdm(2,d); % generation of element g(i,1)=g(i,1)+Ae*ge/3; g(j,1)=g(j,1)+Ae*ge/3; g(k,1)=g(k,1)+Ae*ge/3; end Q=zeros(N_n,1); % global boundary vector initialization H=zeros(N_n,N_n); % global convection matrix initialization for b=1:N_b i=em(1,b); % index of node i j=em(2,b); % index of node j bndry=em(3,b); % boundary of boundary segment xij=pm(1,j)-pm(1,i); % x-distance between nodes j and i on boundary segment b
yij=pm(2,j)-pm(2,i); % y-distance between nodes j and i on boundary segment b sij=sqrt(xij^2+yij^2); % length of boundary segment b qfsb=bcm(3,bndry); hb=bcm(1,bndry); Tinfb=bcm(2,bndry);
% specified heat flux (W/m^2) % heat transfer coefficient (W/m^2-K) % ambient temperature (K)
% add segment boundary vector to global boundary vector Q(i,1)=Q(i,1)+qfsb*sij/2+hb*Tinfb*sij/2; Q(j,1)=Q(j,1)+qfsb*sij/2+hb*Tinfb*sij/2; % add segment convection matrix to global convective matrix H(i,i)=H(i,i)+hb*sij/3; H(i,j)=H(i,j)+hb*sij/6; H(j,i)=H(j,i)+hb*sij/6; H(j,j)=H(j,j)+hb*sij/3; end T=(K+H)\(Q+g); % obtain temperatures at each node pdeplot(pm,em,tm,'xydata',T,'contour','on');
The solution with the refined mesh is shown in Figure 5.
Figure 5: Solution to the problem shown in Figure 2.7-5 with the mesh shown in Figure 4.
Problem 2.7-6 Figure P2.7-6 illustrates a tube in a water-to-air heat exchanger with a layer of polymer coating that can be easily etched away in order to form an array of fin-like structures that increase the surface area exposed to air. ha = 10 W/m -K Ta = 290 K 2
polymer, kp = 1.5 W/m-K
hw = 250 W/m -K Tw = 320 K tube, kt = 15 W/m-K unit cell of fin structure Figure P2.7-6: Tube coated with polymer and a unit cell showing fin-like structures etched into polymer. 2
The water flowing through the tube has temperature Tw = 320 K and heat transfer coefficient hw = 250 W/m2-K. The air has temperature Ta = 290 K and heat transfer coefficient ha = 10 W/m2-K. The thermal conductivity of the polymer and tube material is kp = 1.5 W/m-K and kt = 15 W/mK, respectively. a.) Generate a finite element solution for the temperature distribution within the unit cell shown in Figure 2.7-6 using the mesh shown in Figure 2.7-6(b). The coordinates of the nodes are listed in Table P2.7-6. 14
13 12
11 10 7 6
5
8
9 4
3 1 2 Figure P2.7-6(b): Mesh for finite element solution.
Node 1 2 3 4 5 6 7
Table 2.7-6: Coordinates of nodes in Figure 2.7-6(b). x-coord. (m) y-coord. (m) Node x-coord. (m) 0 0 8 0.01 0.01 0 9 0.015 0.015 0 10 0 0.015 0.008 11 0.007 0.005 0.008 12 0.005 0 0.008 13 0 0.006 0.01 14 0.009
y-coord. (m) 0.01 0.01 0.015 0.02 0.025 0.03 0.03
The first step is to define the sub-domains and boundaries that specify the problem. The information about the sub-domains is contained in a matrix sd . The number of columns in the matrix sd is equal to the number of sub-domains in the computational domain. The information about the sub-domains is contained in the column. For this problem, only the conductivity is required. There are two sub-domains, as shown in Figure 3, and therefore the matrix sd is setup in the MATLAB function S2p7p6 according to: function[]=P2p7d6(h_a) % Input: % h_a - heat transfer coefficient on the air-side (W/m^2-K) % % setup details of computational domain sdm=[15; 1.5]'; % conductivity and generation of each sub-domain
subdomain 2
boundary 3
boundary 4
subdomain 1
boundary 2
boundary 1 Figure 3:Sub-domains and boundaries.
The information about the boundary conditions is contained in a matrix bc . The number of columns in the matrix bc is equal to the number of boundaries in the computational domain. The information about the boundaries is contained in the column: the first row holds the heat transfer coefficient, the second row holds the adjacent fluid temperature, and the third row holds the specified heat flux into the computational domain. There are four boundaries for the problem, as shown in Figure 3, and therefore the matrix bc is setup according to: bcm=[250,320,0; 0,0,0; h_a,290,0; 0,0,0]'; % boundary condition for each boundary
The mesh shown in Figure P2.7-6(b) is defined by Nn = 14 nodes. Information about the nodes is stored in the point matrix, p . There are Nn columns in the point matrix. The first row of each column contains the x-coordinate of the corresponding node and the second row contains the ycoordinate. The point matrix is setup: % setup details of the grid
N_n=14; % number of nodes % setup point matrix - coordinates of each node pm=[0,0; 0.01,0; 0.015,0; 0.015,0.008; 0.005,0.008; 0,0.008; 0.006,0.01; 0.01,0.01; 0.015,0.01;... 0,0.015; 0.007,0.02; 0.005,0.025; 0,0.03; 0.009,0.03]';
The mesh is defined by Nb = 12 boundary segments. Information about the boundary segments is stored in the edge matrix, e . There are Nb columns in the edge matrix. The first two rows of each column contain the indices of the nodes that define the boundary segment. The third row contains the index of the boundary that the boundary segment lies on. N_b=12; % number of boundary segments % setup edge matrix - nodes that define each segment and boundary em=[1,2,1; 2,3,1; 3,4,2; 4,9,2; 9,8,3; 8,7,3; 7,11,3; 11,14,3; 14,13,3; 13,10,4; 10,6,4; 6,1,4]';
The mesh is defined by Ne = 14 elements. Information about the edge is stored in the triangle matrix, t . There are Ne columns in the triangle matrix. The first three rows contain the indices of the three nodes that define the elements, given in clockwise order. The fourth row contains the number of the subdomain that the element lies in. N_e=14; % number of elements % setup triangle matrix - nodes that define each element and subdomain tm=[6,1,5,1; 1,2,5,1; 5,2,4,1; 2,3,4,1; 8,4,9,2; 5,4,8,2; 7,5,8,2;... 7,6,5,2; 10,6,7,2; 10,7,11,2; 12,10,11,2; 13,10,12,2; 12,11,14,2;... 13,12,14,2]';
The element conduction matrix was derived in Eq. (S2.7.2-66), which is repeated below:
⎡. . ⎢ 2 2 ⎢. (x jk + y jk ) ⎢ . ⎢. Ae ke K e = 2 ⎢. − (xik x jk + yik y jk ) bijk ⎢ ⎢. . ⎢ ⎢. (xij x jk + yij y jk ) ⎢ . ⎣. column i
.
.
.
. − (x jk xik + y jk yik ) . . . . . .
.
(x
2 ik
(x
.
jk
xij + y jk yij )
.
.
+ yik2 )
. − (xik xij + yik yijk )
.
.
.
− (xij xik + yij yik ) . . column j
.
(x
2 ij
+ yij2 ) .
.⎤ ⎥ .⎥ row i ⎥ .⎥ .⎥ row j ⎥ .⎥ ⎥ .⎥ row k ⎥ .⎦
column k (S2.7.2-66)
The elements associated with the element conduction matrix are added, element by element, to the global conduction matrix according to: Ne
K = ∑ Ke i =1
(S2.7.2-1)
The global conduction matrix is initialized: K=zeros(N_n,N_n); % global conduction matrix initialization
The global conduction matrix is constructed element-by-element within a for loop: for e=1:N_e
The indices of the three nodes that define the element and the subdomain that the element belongs to are obtained from the triangle matrix: i=tm(1,e); j=tm(2,e); k=tm(3,e); d=tm(4,e);
% % % %
index of node i index of node j index of node k subdomain of element
The x- and y-distances separating the nodes (xij, xik, etc.) are computed; the coordinates of the nodes are obtained from the point matrix: xij=pm(1,j)-pm(1,i); xik=pm(1,k)-pm(1,i); xjk=pm(1,k)-pm(1,j); yij=pm(2,j)-pm(2,i); yik=pm(2,k)-pm(2,i); yjk=pm(2,k)-pm(2,j);
% x-distance between nodes j and i in element e % x-distance between nodes k and i in element e % x-distance between nodes k and j in element e % y-distance between nodes j and i in element e % y-distance between nodes k and i in element e % y-distance between nodes k and j in element e
The parameter bijk and the area of the element are computed: bijk=xij*yjk-xjk*yij; Ae=abs(bijk)/2;
% area of element e
The conductivity of the element is obtained from the matrix sd and the elements of the element conduction matrix are added to the global conduction matrix according to Eq. (S2.2-66).. % add element conduction matrix to global conduction matrix ke=sdm(1,d); % conductivity of element K(i,i)=K(i,i)+Ae*ke*(xjk^2+yjk^2)/bijk^2; K(i,j)=K(i,j)-Ae*ke*(xjk*xik+yjk*yik)/bijk^2; K(i,k)=K(i,k)+Ae*ke*(xjk*xij+yjk*yij)/bijk^2; K(j,i)=K(j,i)-Ae*ke*(xik*xjk+yik*yjk)/bijk^2; K(j,j)=K(j,j)+Ae*ke*(xik^2+yik^2)/bijk^2; K(j,k)=K(j,k)-Ae*ke*(xik*xij+yik*yij)/bijk^2; K(k,i)=K(k,i)+Ae*ke*(xij*xjk+yij*yjk)/bijk^2; K(k,j)=K(k,j)-Ae*ke*(xij*xik+yij*yik)/bijk^2; K(k,k)=K(k,k)+Ae*ke*(xij^2+yij^2)/bijk^2; end
The vector Qb and matrix H b were derived in Eq. (S2.7.2-84), which is repeated below:
⎡. ⎢ ⎡.⎤ ⎡.⎤ ⎢. ⎢1⎥ ⎢1⎥ ⎢ qs′′,b sij ⎢ ⎥ hb T∞,b sij ⎢ ⎥ ⎢.⎥ + ⎢ . ⎥ − hb sij ⎢. qb = ⎢ 2 ⎢⎥ 2 ⎢⎥ 1 1 ⎢. ⎢⎥ ⎢⎥ ⎢ ⎢⎣ . ⎥⎦ ⎢⎣ . ⎥⎦ ⎢ row j ⎣. row i
Qb
Hb
.
.
.
1 1 . 3 6 . . . 1 1 . 6 3 . . .
.⎤ ⎥ .⎥ row i ⎥ .⎥ Tˆ ⎥ .⎥ row j ⎥ ⎥ .⎦
column i column j
(S2.7.2-84)
These variables are initialized: Q=zeros(N_n,1); H=zeros(N_n,N_n);
% global boundary vector initialization % global convection matrix initialization
and constructed boundary segment-by-boundary segment using a for loop: for b=1:N_b
The indices of the two nodes that define the boundary segment and the boundary that the segment belongs to are obtained from the edge matrix: i=em(1,b); j=em(2,b); bndry=em(3,b);
% index of node i % index of node j % boundary of boundary segment
The x- and y-distance separating the nodes and the linear distance between the nodes, sij, are computed: xij=pm(1,j)-pm(1,i); yij=pm(2,j)-pm(2,i); sij=sqrt(xij^2+yij^2);
% x-distance between nodes j and i on boundary segment b % y-distance between nodes j and i on boundary segment b % length of boundary segment b
The specified heat flux, heat transfer coefficient, and ambient temperature associated with the boundary segment is obtained from the matrix bc and the elements of Qb and H b are added to the global vector Q and matrix H according to Eq. (S2.7.2-84). qfsb=bcm(3,bndry); hb=bcm(1,bndry); Tinfb=bcm(2,bndry);
% specified heat flux (W/m^2) % heat transfer coefficient (W/m^2-K) % ambient temperature (K)
% add segment boundary vector to global boundary vector Q(i,1)=Q(i,1)+qfsb*sij/2+hb*Tinfb*sij/2;
Q(j,1)=Q(j,1)+qfsb*sij/2+hb*Tinfb*sij/2; % add segment convection matrix to global convective matrix H(i,i)=H(i,i)+hb*sij/3; H(i,j)=H(i,j)+hb*sij/6; H(j,i)=H(j,i)+hb*sij/6; H(j,j)=H(j,j)+hb*sij/3; end
The temperatures at each node are obtained: T=(K+H)\Q;
% obtain temperatures at each node
and plotted according to: pdeplot(pm,em,tm,'xydata',T,'contour','on');
which leads to Figure 4.
Figure 4: Contour plot of the finite element solution.
b.) Plot the average heat flux at the internal surface of the tube as a function of the air-side heat transfer coefficient with and without the polymer coating. You should see a cross-over point where it becomes disadvantageous to use the polymer coating; explain this. The heat transfer into the different boundaries can be evaluated by summing the heat transfer to the boundary segments that lie on that boundary. The heat transfer from the water and the air is set to 0 and then each boundary segment is considered one by one: q_dot_w=0; q_dot_a=0;
for b=1:N_b
The indices of the nodes that define the boundary segment and the boundary that the segment lies on are obtained from the edge matrix: i=em(1,b); j=em(2,b); bndry=em(3,b);
% index of node i % index of node j % boundary of boundary segment
The length of the segment is computed: xij=pm(1,j)-pm(1,i); % x-distance between nodes j and i on boundary segment b yij=pm(2,j)-pm(2,i); % y-distance between nodes j and i on boundary segment b sij=sqrt(xij^2+yij^2); % length of boundary segment b
If the boundary segment lies on boundary 1 then it is exposed to water and the heat transfer into that boundary is added to the total heat transfer to the water. The heat transfer is the product of the length of the segment, the heat transfer coefficient and the difference between the adjacent water temperature and the average temperature on the boundary. if(bndry==1) hb=bcm(1,bndry); % heat transfer coefficient (W/m^2-K) Tinfb=bcm(2,bndry); % ambient temperature (K) T_b=(T(i)+T(j))/2; % average temperature on boundary (K) q_dot_w=q_dot_w+sij*hb*(Tinfb-T_b); % rate of heat transfer per unit length (W/m) end
If the boundary segment lies on boundary 3 then it is exposed to air and the heat transfer into that boundary is added to the total heat transfer to the air. if(bndry==3) hb=bcm(1,bndry); % heat transfer coefficient (W/m^2-K) Tinfb=bcm(2,bndry); % ambient temperature (K) T_b=(T(i)+T(j))/2; % average temperature on boundary (K) q_dot_a=q_dot_a+sij*hb*(Tinfb-T_b); % rate of heat transfer per unit length (W/m) end end
The heat flux from the water is the heat transfer over the area: q_dot_flux=q_dot_w/0.015;
% heat flux
end
which leads to 570.6 W/m2. The function P2p7d6_makefig runs the finite element solution over a range of air-side heat transfer coefficients: h_a=1; for i=1:101 h_av(i)=h_a; [q_dot_fluxv(i)]=P2p7d6(h_a); h_a=h_a*1.08;
end
Figure 5 illustrates the heat flux as a function of air-side heat transfer coefficient predicted by the finite element solution. 7000
2
Heat flux (W/m )
with polymer fins 1000
without polymer fins 100
10 1
10
100
1000 2250 2
Air-side heat transfer coefficient (W/m -K) Figure 5: Heat flux as a function of air-side heat transfer coefficient with the polymer fins and without the polymer (bare tube).
If the polymer is removed, then the problem is a 1-D conduction problem that can be solved in EES. The inputs are entered: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" {h_bar_a=10 [W/m^2-K]} s=0.015 [m] h_bar_w=250 [W/m^2-K] k_t=15 [W/m-K] th_t=0.008 [m] T_w=320 [K] T_a=290 [K]
"heat transfer coefficient on air-side" "width of tube" "heat transfer coefficient on water-side" "conductivity of tube" "tube thickness" "water temperature" "air temperature"
The resistance to conduction through the tube, per unit area, is:
Rtube =
tht kt
The resistance to convection on the air- and water-sides, per unit area, are computed according to:
Rconv ,a =
1 ha
Rconv , w =
1 hw
The total resistance per unit area is: Rtotal = Rconv ,a + Rtube + Rconv , w
and the heat flux associated with the bare tube is: q ′′ =
(Tw − Ta ) Rtotal
The heat flux associated with the bare tube (without the polymer fins) as a function of the airside heat transfer coefficient is shown in Figure 5. The addition of the polymer fins adds a resistance to conduction through the polymer but reduces the resistance to air-side convection. If the resistance to air-side convection is sufficiently small, then the reduction of the air-side convetion resistance is not sufficient to make up for the additional resistance due to conduction through the polymer. Therefore, there is a cross-over point in Figure 5.
Problem 2.7-7 Figure P2.7-7 illustrates a power electronics chip that is used to control the current to a winding of a motor. 2 silicon chip ha = 10 W/m -K ks = 80 W/m-K Ta = 20°C spreader k = 45 W/m-K 0.5 cm sp 0.2 cm 0.5 cm
dielectric layer kd = 1 W/m-K
0.1 cm
1.4 cm hw = 1000 W/m -K Tw = 10°C 2
Figure P2.7-7: Power electronics chip.
The silicon chip has dimensions 0.2 cm by 0.5 cm and conductivity ks = 80 W/m-K. A generation of thermal energy occurs due to losses in the chip; the thermal energy generation can be modeled as being uniformly distributed with a value of g ′′′ = 1x108 W/m3 in the upper 50% of the silicon. The chip is thermally isolated from the spreader by a dielectric layer with thickness 0.1 cm and conductivity kd = 1 W/m-K. The spreader has dimension 0.5 cm by 1.4 cm and conductivity ksp = 45 W/m-K. The external surfaces are all air cooled with ha = 10 W/m 2 -K and Ta = 20ºC except for the bottom surface of the spreader which is water cooled with hw = 1000 W/m 2 -K and Tw = 10ºC. a.) Develop a numerical model of the system using FEHT.
FEHT is opened and Heat Transfer is selected from the Subject menu. Steady-state and Temperatures in C are selected from the setup menu and a grid of dimension 0.2 cm x 0.2 cm is specified. A rough geometry is obtained using the Outline selection from the Draw menu, as shown in Figure 2.
Figure 2: Rough geometry.
The precise coordinates of each point are specified by double-clicking on each one in turn. The result is shown in Figure 3.
Figure 3: Geometry completely specified.
The material properties are specified by selecting an outline and then selecting Material Properties from the Specify menu. The boundary conditions and generation are specified in the same way. A very crude mesh is obtained by selecting Element Lines from the Draw menu. The result is shown in Figure 4.
Figure 4: Crude mesh.
The problem is solved by selecting Calculate from the Run menu. The temperatures are shown at each node by selecting Temperatures from the View menu, as shown in Figure 5. The maximum temperature is 153ºC.
Figure 5: Temperature predicted at each node.
b.) Plot the maximum temperature in the system as a function of the number of nodes. The mesh is successively refined (by selecting Reduce Mesh from the Draw menu. Figure 6 illustrates a highly reduced mesh and Figure 7 illustrates the maximum temperature in the silicon as a function of the number of nodes in the mesh.
Figure 6: Highly reduced mesh.
154.4
Maximum temperature (°C)
154.2 154 153.8 153.6 153.4 153.2 153 0
100
200
300
400
500
600
Number of nodes Figure 7: Maximum temperature as a function of the number of nodes.
c.) Develop a simple sanity check of your results using a resistance network. Figure 8 illustrates the temperature contours predicted by the FEHT model.
Figure 8: Temperature contours predicted by the FEHT model.
Figure 8 shows that the temperature is reduced primarily by conduction across the dielectric and convection from the surface of the spreader to the water. The temperature gradients caused by conduction within the spreader and silicon are small. These observations are consistent with a simple resistance network in which the heat transfer related to generation: q =
1x108 W 1 m 0.001 m 0.005 m =500 W m3
(1)
must pass through a resistance due to conduction through the silicon: Rcond , s =
0.002 m
mK K = 0.005 0.005 m 1 m 80 W W
(2)
a resistance due to conduction through the dielectric layer:
Rcond ,d =
0.001 m
mK K = 0.2 0.005 m 1 m 1 W W
(3)
a resistance due to conduction through the spreader: Rcond , sp =
0.005 m
mK K = 0.008 0.014 m 1 m 45 W W
(4)
and a resistance due to convection to the water: Rconv , w =
m2 K K = 0.071 1000 W 0.014 m 1 m W
(5)
The resistance to conduction through the dielectric and convection to the water are important, the others are not. The temperature difference between the silicon and the spreader is approximately: ΔTd = Rcond , d q =
0.2 K 500 W = 100 K W
(6)
and the temperature difference between the water and the spreader is approximately: ΔTconv ,v = Rconv , w q =
0.071 K 500 W = 35.7 K W
(7)
These values are consistent with the FEHT predictions, providing some confidence in the results.
Problem 2.7-8 Figure P2.7-8(a) illustrates a heat exchanger in which hot fluid and cold fluid flows through alternating rows of square channels that are installed in a piece of material. C
C H
C
C H
C H
C H
C H
C H
C H
C H
C H
H unit cell of heat exchanger
C H
H
H = channels carrying hot fluid C = channels carrying hot fluid Figure P2.7-8(a): Heat exchanger.
You are analyzing this heat exchanger and will develop a model of the unit cell shown in Figure P2.7-8(a) and illustrated in more detail in Figure 2.7-8(b). L = 6 mm
h = 150 W/m -K Th = 80°C 2
th/2 =0.4 mm
p/2 = 4 mm thb = 0.8 mm h = 150 W/m -K Tc = 25°C 2
k = 12 W/m-K
Figure 2.7-8(b): Details of unit cell shown in Figure 2.7-8(a).
The metal struts separating the square channels form fins. The length of the fin (the half-width of the channel) is L = 6 mm and the fin thickness is th = 0.8 mm. The thickness of the material separating the channels is thb = 0.8 mm. The distance between adjacent fins is p = 8 mm. The channel structure for both sides (hot and cold) are identical. The conductivity of the metal is k = 12 W/m-K. The hot fluid has temperature Th = 80ºC and heat transfer coefficient h = 150 W/m2-K. The cold fluid has temperature Tc = 25ºC and the same heat transfer coefficient. a.) Prepare a numerical model of the unit cell shown in Figure 2.7-8(b) using FEHT. The geometry is entered in FEHT using the Outline selection from the draw menu. Each point is precisely positioned by double-clicking on it and entering its coordinates. The result is shown in Figure 3.
Figure 3: Geometry entered in FEHT.
The material properties and boundary conditions are entered. Note that the lines of symmetry (i.e., the lines at the ends and centers of the fins) are adiabatic. A crude mesh is drawn. The result is shown in Figure 4.
Figure 4: FEHT model with crude mesh.
The problem is solved, leading to the temperature contours shown in Figure 5.
Figure 5: Temperature distribution.
b.) Plot the rate of heat transfer from the hot fluid to the cold fluid within the unit cell as a function of the number of nodes. The boundary adjacent to the cold fluid is selected and Heat Flows is selected from the View menu. The mesh is successively refined by selecting Reduce Mesh from the Draw menu and the heat flow as a function of the number of nodes is shown in Figure 6.
32.2
Heat transfer rate (W/m)
32 31.8 31.6 31.4 31.2 31 0
100
200
300
400
500
600
700
Number of nodes Figure 6: Heat transfer rate (per length of heat exchanger) as a function of the number of nodes.
c.) Develop a simple model of the unit cell using a resistance network and show that your result from (b) makes sense. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in L=6 [mm]*convert(mm,m) th=0.8 [mm]*convert(mm,m) p=8 [mm]*convert(mm,m) th_b=0.8 [mm]*convert(mm,m) k=12 [W/m-K] h_bar=150 [W/m^2-K] T_h=80 [C] T_c=25 [C] W=1 [m]
"length of hot-side fins" "thickness of hot-side fins" "fin-to-fin distance" "base thickness" "conductivity of material" "heat transfer coefficient" "temperature of hot fluid" "temperature of cold fluid" "per unit length of heat exchanger"
The fin efficiency (ηf) for an adiabatic tipped constant cross-section fin is obtained using the function eta_fin_straight_rect in EES. The resistance of each of the fins is: Rf =
1 h LW η f
(1)
where W is a unit length of heat exchanger and only the area of the fin within the unit cell is used. The resistance of the unfinned region of the base is: Ruf =
The total resistance is:
2 h ( p − th ) W
(2)
Rtotal
⎛ 1 1 ⎞ = 2⎜ + ⎟⎟ ⎜R ⎝ f Ruf ⎠
−1
(3)
The rate of heat transfer is:
q = eta_f=eta_fin_straight_rect(th,L,h_bar,k) R_f=1/(h_bar*W*L*eta_f) R_uf=1/(h_bar*W*(p-th)/2) R_total=2*(1/R_f+1/R_uf)^(-1) q_dot=(T_h-T_c)/R_total
(Th − Tc ) Rtotal
(4) "fin efficiency" "fin resistance" "unfinned surface resistance" "total resistance between the streams" "total heat transfer rate"
which leads to q = 32.6. Note that this is somewhat larger than the prediction from (b) because the conduction resistance of the base (both across the base and along the base) is not considered.
Problem 2.8-1 (2-12 in text): Cryogenic Thermal Switch There are several cryogenic systems that require a “thermal switch”, a device that can be used to control the thermal resistance between two objects. One class of thermal switch is activated mechanically and an attractive method of providing mechanical actuation at cryogenic temperatures is with a piezoelectric stack; unfortunately, the displacement provided by a piezoelectric stack is very small, typically on the order of 10 microns. A company has proposed an innovative design for a thermal switch, shown in Figure P2.81(a). Two blocks are composed of th = 10 μm laminations that are alternately copper (kCu = 400 W/m-K) and plastic (kp = 0.5 W/m-K). The thickness of each block is L = 2.0 cm in the direction of the heat flow. One edge of each block is carefully polished and these edges are pressed together; the contact resistance associated with this joint is Rc′′ = 5x10-4 K-m2/W. th = 10 μm plastic laminations kp = 0.5 W/m-K L = 2 cm L = 2 cm TH
direction of actuation
TC Figure P2.8-1(b) “on” position
“off” position -4 2 contact resistance, Rc′′ = 5x10 m -K/W
th = 10 μm copper laminations kCu = 400 W/m-K
Figure P2.8-1(a): Thermal switch in the “on” and “off” positions.
Figure P2.8-1(a) shows the orientation of the two blocks when the switch is in the “on” position; notice that the copper laminations are aligned with one another in this configuration which provides a continuous path for heat through high conductivity copper (with the exception of the contact resistance at the interface). The vertical location of the right-hand block is shifted by 10 μm to turn the switch "off". In the “off” position, the copper laminations are aligned with the plastic laminations; therefore, the heat transfer is inhibited by low conductivity plastic. Figure P2.8-1(b) illustrates a closer view of half (in the vertical direction) of two adjacent laminations in the “on” and “off” configurations. Note that the repeating nature of the geometry means that it is sufficient to analyze a single lamination set and assume that the upper and lower boundaries are adiabatic.
L = 2 cm
L = 2 cm TC
TH th/2 = 5 μm th/2 = 5 μm
kp = 0.5 W/m-K kCu = 400 W/m-K
Rc′′ = 5x10 m -K/W -4
2
“on” position TC
TH “off” position
Figure P2.8-1(b): A single set consisting of half of two adjacent laminations in the “on” and "off” positions.
The key parameter that characterizes a thermal switch is the resistance ratio (RR) which is defined as the ratio of the resistance of the switch in the “off” position to its resistance in the “on” position. The company claims that they can achieve a resistance ratio of more than 100 for this switch. a) Estimate upper and lower bounds for the resistance ratio for the proposed thermal switch using 1-D conduction network approximations. Be sure to draw and clearly label the resistance networks that are used to provide the estimates. Use your results to assess the company’s claim of a resistance ratio of 100. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" th = 10 [micron]*convert(micron,m) k_Cu=400 [W/m-K] k_p=0.5 [W/m-K] L = 2.0 [cm]*convert(cm,m) R``_c=5e-4 [K-m^2/W] W=1 [m]
"thickness of laminations" "conductivity of copper laminations" "conductivity of plastic laminations" "lengt of laminations" "area specific contact resistance of interface" "unit depth into page"
The isothermal and adiabatic models provide two limiting cases. Both result in a 1-D problem that can be represented using a resistance network. The “adiabatic” approximation does not allow heat transfer in the y-direction (i.e., perpendicular to the laminations). The resistance network for the adiabatic approximation is shown in Figure 3(a) for the “on” position and in Figure 3(b) for the “off” position.
(a) (b) Figure 3: Resistance network for the “adiabatic” approximation in the (a) “on” and (b) “off” positions.
Note that the parameter Alam in Figure 3 is the cross-sectional area associated a single lamination: Alam = W th where W is the depth into the page (assumed to be 1 m). The resistance associated with the adiabatic approximation in the “on” state, Rad,on from Figure 3(a) is:
Rad ,on
⎡ ⎤ ⎢ ⎥ 2 ⎢ 1 1 ⎥ = + Alam ⎢ 2 L + R′′ 2 L + R′′ ⎥ c c ⎥ ⎢k kCu ⎣ p ⎦
−1
A_lam=W*th "area of a lamination" R_ad_on=(2/A_lam)*(1/(2*L/k_p+R``_c)+1/(2*L/k_Cu+R``_c))^(-1) "ad. limit in the on position"
which leads to Rad,on = 119.1 K/W. The resistance associated with the adiabatic approximation in the “off” state, Rad,off from Figure 3(b) is: −1
Rad ,off
L L ⎡ ⎤ + + Rc′′ ⎢ ⎥ k p kCu 2 ⎢ 1 1 ⎥ = = + Alam ⎢ L + L + R′′ L + L + R′′ ⎥ Alam c c ⎥ ⎢k kCu k p ⎣ p kCu ⎦
R_ad_off=(L/k_p+L/k_Cu+R``_c)/A_lam
"ad. limit in the off position"
The resistance ratio for the adiabatic limit is: RRad =
Rad ,off Rad ,on
RR_ad=R_ad_off/R_ad_on
"resistance ratio estimated using adiabatic limit"
which leads to RRad = 34.0. The “isothermal” approximation provides no resistance to heat transfer in the y-direction (i.e., perpendicular to the laminations). Therefore, the heat can spread without penalty at the interface and the resistance network for this approximation is shown in Figure 4(a) for the “on” position and Figure 4(b) “off” position.
(a) (b) Figure 4: Resistance network for the “isothermal” approximation in the (a) “on” and (b) “off” positions.
By inspection, the two resistance networks shown in Figures 4(a) and 4(b) will yield the same resistance and therefore the “isothermal” assumption will predict a resistance ratio, RRiso, of 1.0. Clearly the company’s claim of a resistance ratio of 100 is not possible as it does not lie between the two bounding quantities associated with the isothermal and adiabatic approximations. b) Provide one or more suggestions for design changes that would improve the performance of the switch (i.e., increase the resistance ratio). Justify your suggestions. The resistance ratio would be increased by any change that causes the two resistance networks in Figure 3 to be more different. Possible improvements include using materials with a larger ratio of conductivities (i.e., lower conductivity plastic and, to a lesser degree, higher conductivity copper) or eliminating the contact resistance. From a more practical standpoint, any design change that causes the actual the device to behave more like the adiabatic approximation in Figure 3 than the isothermal approximation in Figure 4 will improve the performance; for example, increasing the contact resistance between adjacent laminations would be important. c.) Sketch the temperature distribution through the two parallel paths associated with the adiabatic limit of the switch’s operation in the “off” position. Do not worry about the quantitative details of the sketch, just make sure that the qualitative features are correct.
Figure 5: Qualitative temperature distribution through paths 1 and 2 consistent with the adiabatic approximation when the switch is in the “off” position.
d.) Sketch the temperature distribution through the two parallel paths associated with the adiabatic limit in the “on” position. Again, do not worry about the quantitative details of your sketch, just make sure that the qualitative features are correct.
Figure 6: Qualitative temperature distribution through paths 1 and 2 consistent with the adiabatic approximation when the switch is in the “on” position.
Problem 2.8-2 (2-13 in text): Resistance of a Bus Bar Figure P2.8-2 illustrates a thermal bus bar that has width W = 2 cm (into the page). H1 = 5 cm
L2 = 7 cm 2 h = 10 W/m -K T∞ = 20°C
TH = 80°C
L1 = 3 cm
H2 = 1 cm
k = 1 W/m-K
Figure P2.8-2: Thermal bus bar.
The bus bar is made of a material with conductivity k = 1 W/m-K. The middle section is L2 = 7 cm long with thickness H2 = 1 cm. The two ends are each L1 = 3 cm long with thickness H1 = 3 cm. One end of the bar is held at TH = 80ºC and the other is exposed to air at T∞ = 20ºC with h = 10 W/m2-K. a.) Use FEHT to predict the rate of heat transfer through the bus bar. The geometry is entered in FEHT, a mesh is generated, and the boundary conditions and material properties are specified (Figure 2).
Figure 2: FEHT model.
The mesh is refined several times and then the heat transfer at the convective boundary is obtained; this leads to q ′ = 5.40 W/m or q = 0.108 W. b.) Obtain upper and lower bounds for the rate of heat transfer through the bus bar using appropriately defined resistance approximations. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" L_1=3 [cm]*convert(cm,m)
"length of edge pieces"
H_1=5 [cm]*convert(cm,m) L_2=7 [cm]*convert(cm,m) H_2=1 [cm]*convert(cm,m) W=2 [cm]*convert(cm,m) k=1 [W/m-K] h_bar=10 [W/m^2-K] T_infinity=20 [C] T_H=80 [C]
"height of edge pieces" "length of center piece" "height of center piece" "width" "conductivity" "heat transfer coefficient at right side" "ambient temperature at right side" "left side temperature"
An upper bound on the heat transfer rate is obtained using the isothermal limit. The isothermal limit is calculated according to:
qiso =
(TH − T∞ ) L1 L2 1 +2 + h H1 W k H1 W k H 2 W
Rconv ,iso
Rcond 1,iso
Rcond 2,iso
"Isothermal limit" R_conv_iso=1/(h_bar*W*H_1) R_cond_1_iso=L_1/(k*W*H_1) R_cond_2_iso=L_2/(k*W*H_2) R_total_iso=(R_conv_iso+2*R_cond_1_iso+R_cond_2_iso) q_dot_iso=(T_H-T_infinity)/R_total_iso
which leads to qiso = 0.118 W. A lower bound on the heat transfer rate is obtained using the adiabatic limit. The adiabatic limit is calculated according to:
qad =
(TH − T∞ ) L1 L2 1 +2 + h H2 W k H2 W k H2 W
Rconv ,ad
Rcond 1,ad
"Adiabatic limit" R_conv_ad=1/(h_bar*W*H_2) R_cond_1_ad=L_1/(k*W*H_2) R_cond_2_ad=L_2/(k*W*H_2) R_total_ad=R_conv_ad+2*R_cond_1_ad+R_cond_2_ad q_dot_ad=(T_H-T_infinity)/R_total_ad
which leads to qad = 0.052 W.
Rcond 2,ad
Problem 2.8-3: Heat Switch Figure P2.8-3 illustrates a design for a superconducting heat switch. superconducting strands, km,normal = 50 W/m-K km,sc = 0.2 W/m-K
magnet a = 5 mm
TC
b = 1 mm
x
TH b = 1 mm
polymer matrix kp = 2.5 W/m-K
Figure P2.8-3: Superconducting heat switch.
The heat switch is made by embedding eight square superconducting strands in a polymer matrix. The width of switch is W = 10 mm (into the page). The size of the strands are a = 5 mm and the width of polymer that surrounds each strand is b = 1 mm. The conductivity of the polymer is kp = 2.5 W/m-K. The heat switch is surrounded by a magnet. When the heat switch is on (i.e., the thermal resistance through the switch in the x-direction is low, allowing heat flow from TH to TC), the magnet is on. Therefore, the magnetic field tends to drive the superconductors to their normal state where they have a high thermal conductivity, km,normal = 50 W/m-K. To turn the heat switch off (i.e., to make the thermal resistance through the switch high, preventing heat transfer from TH to TC), the magnet is deactivated. The superconductors return to their superconducting state, where they have a low thermal conductivity, km,sc = 0.2 W/m-K. The edges of the switch are insulated. a.) Develop a model using a 1-D resistance network that provides a lower bound on the resistance of the switch when it is in its off state (i.e., km = km,sc). The inputs are entered in EES: $UnitSystem SI MASS RAD PA C J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" W=10 [mm]*convert(mm,m) a=5 [mm]*convert(mm,m) b=1 [mm]*convert(mm,m) k_p=2.5 [W/m-K] k_m_normal=50 [W/m-K] k_m_sc=0.2 [W/m-K]
"width of switch" "size of strands" "width of polymer layer" "polymer conductivity" "metal conductivity in its normal state" "metal conductivity in its superconducting state"
The conductivity of the material is set to its superconducting value:
k_m=k_m_sc
"metal conductivity"
The isothermal limit provides a lower bound on the resistance. The isothermal limit is shown in Figure P2.8-3-2. Riso,2 Riso,1 TH
TC
Riso,3
Figure P2.8-3-2: Isothermal limit resistance network.
The resistance Riso,1 represents the conduction through the polymer layers that stretch across the width of the switch: Riso ,1 =
5b k p W ( 2 a + 3b)
(1)
The resistance Riso,2 represents the conduction through the polymer layers that stretch between adjacent superconducting fibers: Riso ,2 =
4a k p W 3b
(2)
The resistance Riso,3 represents the conduction through the superconducting fibers:
Riso ,3 = "isothermal model" R_iso_1=5*b/(k_p*W*(2*a+3*b)) R_iso_2=4*a/(k_p*W*3*b) R_iso_3=4*a/(k_m*W*2*a)
4a km W 2 a
(3)
"conduction through polymer layers" "conduction through polymer between superconducting strands" "conduction through superconducting strands"
The resistance predicted in the isothermal limit is: ⎛ 1 1 ⎞ Riso = Riso ,1 + ⎜ + ⎟⎟ ⎜R R ,2 ,3 iso iso ⎝ ⎠ R_iso=R_iso_1+(1/R_iso_2+1/R_iso_3)^(-1)
which leads to Riso = 225.9 K/W.
"isothermal limit"
−1
(4)
b.) Develop a model using a 1-D resistance network that provides an upper bound on the resistance of the switch when it is in its off state (i.e., km = km,sc). The adiabatic limit provides an upper bound on the resistance. The adiabatic limit is shown in Figure P2.8-3-3. Rad,3
Rad,2
TH
TC
Rad,1
Figure P2.8-3-3: Adiabatic limit resistance network.
The resistance Rad,1 represents the conduction through the polymer layers that stretch across the entire length of the switch in the x-direction: Rad ,1 =
(4 a + 5b)
(5)
k p W 3b
The resistance Rad,2 represents the conduction through the polymer layers that lie between adjacent superconducting fibers in the x-direction: Rad ,2 =
5b kp W 2 a
(6)
The resistance Rad,3 represents the conduction through the superconducting fibers: Rad ,3 = "adiabatic model" R_ad_1=(4*a+5*b)/(k_p*W*3*b) R_ad_2=5*b/(k_p*W*2*a) R_ad_3=4*a/(k_m*W*2*a)
4a km W 2 a
(7)
"conduction through polymer layers in x-direction" "conduction through polymer between strands" "conduction through superconducting strands"
The resistance predicted in the adiabatic limit is: ⎡ 1 ⎤ 1 + Rad = ⎢ ⎥ ⎢⎣ Rad ,1 ( Rad ,2 + Rad ,3 ) ⎥⎦ R_ad=(1/R_ad_1+1/(R_ad_2+R_ad_3))^(-1)
which leads to Rad = 251.2 K/W.
"adiabatic limit"
−1
(8)
c.) Plot your answers from parts (a) and (b) as a function of km for km,sc < km < km,normal. Figure P2.8-3-4 illustrates Riso and Rad as a function of km. Notice that Rad is always larger than Riso and that regardless of which model you use, the resistance of the switch decreases substantially when the superconducting strands go from their superconducting to their normal state.
Thermal resistance (K/W)
500 adiabatic limit
100
isothermal limit
10 0.2
1
10
50
Conductivity of strands (W/m-K)
Figure P2.8-3-4: Thermal resistance as a function of km.
d.) The performance of a heat switch is provided by the resistance ratio; the ratio of the resistance of the switch in its off state to its resistance in the on state. Use your model to provide an upper and lower bound on the resistance ratio of the switch. Setting km = km,sc leads to Riso = 225.9 K/W and Rad = 251.2 K/W. Setting km = km,normal leads Riso = 19.33 K/W and Rad = 22.39 K/W. Therefore, the maximum possible value of the resistance ratio is 251.2/19.33 = 12.995 and the minimum possible value of the resistance ratio is 225.9/22.39 = 10.089. e.) Plot the ratio of your answer from part (b) to your answer from part (a) as a function of km for km,sc < km < km,normal. Explain the shape of your plot. Figure P2.8-3-5 illustrates Rad/Riso as a function of km.
Adiabatic to isothermal resistance
1.18 1.16 1.14 1.12 1.1 1.08 1.06 1.04 1.02 1 0.2
1
10
70
Strand conductivity (W/m-K)
Figure P2.8-3-5: Thermal resistance as a function of km.
Notice that the adiabatic and isothermal models predict the same value when km = kp (i.e., Rad/Riso = 1 at km = 2.5 W/m-K in Figure P2.8-3-5). This is because the problem is 1-D in this limit and therefore both resistance network representations are exactly correct.
Problem 2.9-1: Composite Equivalent Conductivity A composite material is formed from laminations of high conductivity material (khigh = 100 W/m-K) and low conductivity material (klow = 1 W/m-K) as shown in Figure P2.9-1. Both laminations have the same thickness, th. low conductivity laminations klow = 1 W/m-K
th th y x
high conductivity laminations khigh = 100 W/m-K
Figure P2.9-1: Composite material formed from high and low conductivity laminations.
a.) Do you expect the equivalent conductivity of the composite to be higher in the x or y directions? Note from Figure P2.9-1 that the x-direction is parallel to the laminations while the y direction is perpendicular to the laminations. The heat can flow entirely through high conductivity laminations in the x direction whereas it must pass through the low conductivity laminations in order to travel in the y direction. Therefore, the equivalent conductivity of the composite will be much higher in the x direction. b.) Estimate the equivalent conductivity of the composite in the x direction. You should not need to any calculations to come up with a good estimate for this quantity. You are essentially forcing the heat to flow through a block of high conductivity material with half the area of the composite in order to move in the x direction. Therefore, the equivalent conductivity of the composite in the x direction will be approximately half the conductivity of the high conductivity laminations, keff,x ~ 50 W/m-K.
Problem 2.9-2 (2-14 in text) A laminated stator is shown in Figure P2.9-2. The stator is composed of laminations with conductivity klam = 10 W/m-K that are coated with a very thin layer of epoxy with conductivity kepoxy = 2.0 W/m-K in order to prevent eddy current losses. The laminations are thlam = 0.5 mm thick and the epoxy coating is 0.1 mm thick (the total amount of epoxy separating each lamination is thepoxy = 0.2 mm). The inner radius of the laminations is rin= 8.0 mm and the outer radius of the laminations is ro,lam = 20 mm. The laminations are surrounded by a cylinder of plastic with conductivity kp = 1.5 W/m-K that has an outer radius of ro,p = 25 mm. The motor casing surrounds the plastic. The motor casing has an outer radius of ro,c = 35 mm and is composed of aluminum with conductivity kc = 200 W/m-K. laminations, thlam = 0.5 mm, klam = 10 W/m-K epoxy coating, thepoxy = 0.2 mm, kepoxy = 2.0 W/m-K kp = 1.5 W/m-K kc = 200 W/m-K T∞ = 20°C 2 h = 40 W/m -K
4 2 q ′′ = 5x10 W/m
Rc′′ = 1x10 K-m /W -4
rin = 8 mm ro,lam = 20 mm
2
ro,p = 25 mm ro,c = 35 mm Figure P2.9-2: Laminated stator.
A heat flux associated with the windage loss associated with the drag on the shaft is q ′′ = 5x104 W/m2 is imposed on the internal surface of the laminations. The outer surface of the motor is exposed to air at T∞ = 20°C with a heat transfer coefficient h = 40 W/m2-K. There is a contact resistance Rc′′ = 1x10-4 K-m2/W between the outer surface of the laminations and the inner surface of the plastic and the outer surface of the plastic and the inner surface of the motor housing. a.) Determine an upper and lower bound for the temperature at the inner surface of the laminations (Tin). The inputs are entered in EES: $UnitSystem SI MASS RAD PA C J $Tabstops 0.2 0.4 0.6 0.8 3.5 klam=10 [W/m-K] kepoxy=2 [W/m-K] kp=1.5 [W/m-K] kc=200 [W/m-K] tlam=0.5 [mm]*convert(mm,m)
"conductivity of laminations" "conductivity of epoxy" "conductiity of plastic" "conductivity of casing" "thickness of lamination"
tepoxy=0.2 [mm]*convert(mm,m) rin=8 [mm]*convert(mm,m) rolam=20 [mm]*convert(mm,m) rop=25 [mm]*convert(mm,m) roc=35 [mm]*convert(mm,m) qflux=5e4 [W/m^2] Rc=1e-4 [K-m^2/W] Tair=converttemp(C,K,20[C]) h=40 [W/m^2-K]
"thickness of epoxy" "inner radius of laminations" "outer radius of laminations" "outer radius of plastic" "outer radius of casing" "heat flux" "contact resistance" "air temperature" "heat transfer coefficient"
An upper bound on the temperature allows no heat spreading; therefore, the heat must take two parallel paths that pass through the iron and the epoxy. The total resistance associated with one lamination/epoxy pair is:
1 Rtotal ,ad
+
=
1 ⎛r ⎞ ⎛ r ⎞ ⎛r ⎞ ln ⎜ o ,c ⎟ ln ⎜⎜ o , p ⎟⎟ ln ⎜ o ,lam ⎟ ⎜ ⎟ ro ,lam ⎠ Rc′′ Rc′′ 1 ⎝ rin ⎠ + ⎝ ro , p ⎠ + + ⎝ + + 2 π klam thlam 2 π ro ,lam thlam 2 π k p thlam 2 π ro , p thlam 2 π kc thlam 2 π ro, p thlam h 1
⎛r ⎞ ⎛ r ⎞ ⎛r ⎞ ln ⎜ o ,c ⎟ ln ⎜⎜ o , p ⎟⎟ ln ⎜ o ,lam ⎟ ⎜r ⎟ ro ,lam ⎠ Rc′′ Rc′′ 1 ⎝ ⎝ o, p ⎠ + ⎝ rin ⎠ + + + + 2 π kepoxy thepoxy 2 π ro ,lam thepoxy 2 π k p thepoxy 2 π ro , p thepoxy 2 π kc thepoxy 2 π ro, p thepoxy h (1)
and the upper bound on the air temperature is:
Tupper = T∞ + Rtotal ,ad q ′′ 2 π rin ( thlam + thepoxy ) "Upper bound on temperature" Rlam=ln(rolam/rin)/(2*pi*tlam*klam) "resistance of laminations" Repoxy=ln(rolam/rin)/(2*pi*tepoxy*kepoxy) "resistance of epoxy" Rci1=Rc/(2*pi*rolam*tlam) "contact resistance" Rci2=Rc/(2*pi*rolam*tepoxy) Rp1=ln(rop/rolam)/(2*pi*tlam*kp) "resistance of plastic" Rp2=ln(rop/rolam)/(2*pi*tepoxy*kp) Rco1=Rc/(2*pi*rop*tlam) "contact resistance" Rco2=Rc/(2*pi*rop*tepoxy) Rcs1=ln(roc/rop)/(2*pi*tlam*kc) "resistance of casing" Rcs2=ln(roc/rop)/(2*pi*tepoxy*kc) Rcv1=1/(h*2*pi*roc*tlam) "convection resistance" Rcv2=1/(h*2*pi*roc*tepoxy) 1/Rtotal1=1/(Rlam+Rci1+Rp1+Rco1+Rcs1+Rcv1)+1/(Repoxy+Rci2+Rp2+Rco2+Rcs2+Rcv2) "total resistance" qdot=qflux*pi*rin*(tlam+tepoxy) "heat transfer" Tin1=Tair+qdot*Rtotal1 "upper bound on temperature"
which leads to Tupper = 502.7 K.
(2)
A lower bound on the temperature allows heat spreading; therefore, the heat must take two parallel paths that pass through the iron and the epoxy but then a single path to the air. The total resistance associated with one lamination/epoxy pair is:
Rtotal ,iso
⎡ ⎢ ⎢ ⎢ 1 =⎢ ⎛r ⎞ ⎢ ln ⎛ ro ,lam ⎞ ln ⎜ o ,lam ⎟ ⎜ ⎟ ⎢ r ⎝ rin ⎠ ⎢ ⎝ in ⎠ + ⎣⎢ 2 π klam thlam 2 π kepoxy thepoxy
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦⎥
−1
⎛ r ⎞ ln ⎜⎜ o , p ⎟⎟ Rc′′ ⎝ ro ,lam ⎠ + + + 2 π ro ,lam ( thlam + thepoxy ) 2 π k p ( thlam + thepoxy )
(3)
⎛r ⎞ ln ⎜ o ,c ⎟ ⎜r ⎟ Rc′′ 1 ⎝ o, p ⎠ + + 2 π ro , p ( thlam + thepoxy ) 2 π kc ( thlam + thepoxy ) 2 π ro , p ( thlam + thepoxy ) h
and the lower bound on the air temperature is: Tlower = T∞ + Rtotal ,iso q ′′ 2 π rin ( thlam + thepoxy ) "Lower bound on temperature" Rci=Rc/(2*pi*rolam*(tlam+tepoxy)) Rp=ln(rop/rolam)/(2*pi*(tlam+tepoxy)*kp) Rco=Rc/(2*pi*rop*(tlam+tepoxy)) Rcs=ln(roc/rop)/(2*pi*(tlam+tepoxy)*kc) Rcv=1/(h*2*pi*roc*(tlam+tepoxy)) Rtotal2=(1/Rlam+1/Repoxy)^(-1)+Rci+Rp+Rco+Rcs+Rcv Tin2=Tair+qdot*Rtotal2
(4)
"contact resistance" "plastic resistance" "contact resistance" "casing resistance" "convection resistance" "total resistance" "lower bound on temperature"
which leads to Tlower = 491.7 K. b.) You need to reduce the internal surface temperature of the laminations and there are a few design options available, including: (1) increase the lamination thickness (up to 0.7 mm), (2) reduce the epoxy thickness (down to 0.05 mm), (3) increase the epoxy conductivity (up to 2.5 W/m-K), or (4) increase the heat transfer coefficient (up to 100 W/m-K). Which of these options do you suggest and why? Examination of the Solution Window (Figure 2) shows that the resistance of the lamination is much less than the resistance of the epoxy; therefore, the resistance of the lamination dominates the resistance of the epoxy. The resistance due to convection is much larger than the resistance
of the lamination, contact resistance, or conduction through the plastic and casing. Therefore, the resistance to convection dominates the problem and the most effective mechanism for reducing the temperature is to increase the heat transfer coefficient.
Figure 2: Solution Window
Problem 2.9-3 Your company manufactures a product that consists of many small metal bars that run through a polymer matrix, as shown in Figure P2.9-3. The material can be used as a thermal path, allowing heat to transfer efficiently in the z-direction (the direction that the metal bars run) because the heat can travel without interruption through the metal bars. However, the material blocks heat flow in the x- and y-directions because the energy must be conducted through the low conductivity polymer. Because the scale of the metal bars is small relative to the size of the composite structure, it is appropriate to model the material as a composite with an effective conductivity that depends on direction. w = 0.2 mm w = 0.2 mm s = 1.0 mm L = 10 cm
s = 1.0 mm km = 100 W/m-K kp = 2 W/m-K z y
x
b = 2 cm
Figure P2.9-3: Composite material.
The metal bars are square with edge width w = 0.2 mm and are aligned with the z-direction. The bars are arrayed in a regularly spaced matrix with a center-to-center distance of s = 1.0 mm. The conductivity of the metal is km = 100 W/m-K. The length of the material in the z direction is L = 10 cm. The polymer fills the space between the bars and has a thermal conductivity kp = 2.0 W/m-K. The cross-section of the material in the x-y plane is square with edge width b = 2.0 cm. a.) Determine the effective conductivity in the x- , y- , and z-directions. The known information is entered in EES: $UnitSystem SI MASS RAD PA C J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" w = 0.2 [mm]*convert(mm,m) s = 1.0 [mm]*convert(mm,m) k_m=100 [W/m-K] L = 10 [cm]*convert(cm,m) k_p=2.0 [W/m-K] b=2.0 [cm]*convert(cm,m)
"size of metal bars" "center-to-center distance between bars" "metal conductivity" "length of composite" "polymer conductivity" "size of composite"
The effective conductivity in the x and y directions will be the same as the energy must be transferred through the same structure. The effective conductivity is calculated by considering a unit cell of the structure, shown in Figure 2.
Figure 2: Unit cell of the structure in the x or y direction
The conduction through the unit cell can be approximated using three resistances, shown in Figure 2. These resistances are:
Rm, x =
R p ,1, x =
R p ,2, x =
w km L w
(1)
( s − w)
(2)
kp L w
s k p L ( s − w)
(3)
The total resistance of the composite is:
Reff , x
⎡ 1 1 ⎤ =⎢ + ⎥ ⎣⎢ Rm, x + R p ,1, x R p ,2, x ⎦⎥
−1
(4)
The resistance must be equal to the resistance of a homogeneous material with the same size that has the effective thermal conductivity, keff,x:
s keff , x L s
= Reff , x
(5)
These calculations are entered in EES: "Effective conductivity in the x or y directions" R_m_x=w/(k_m*L*w) R_p_1_x=(s-w)/(k_p*L*w) R_p_2_x=s/(k_p*L*(s-w)) R_eff_x=(1/(R_m_x+R_p_1_x)+1/R_p_2_x)^(-1) R_eff_x=s/(s*L*k_eff_x)
and indicate that keff,x= keff,y = 2.1 W/m-K (the matrix behaves essentially like a polymer with respect to conduction in the x or y directions because the energy must transfer, primarily, through the polymer). The same type of analysis is accomplished for conduction in the z direction using the same unit cell shown in Figure 2. Conduction in the z direction occurs through the metal bar and surrounding polymer in parallel. The resistances of these two paths are:
L k m w2
(6)
L k p ( s − w2 )
(7)
Rm , z =
Rp, z =
2
The total resistance of the composite in the z direction is:
Reff , z
⎡ 1 1 ⎤ =⎢ + ⎥ ⎣⎢ Rm, z R p , z ⎦⎥
−1
(8)
The resistance must be equal to the resistance of a homogeneous material with the same size that has the effective thermal conductivity, keff,z: L keff , z s 2
These calculations are entered in EES: "Effective conductivity in the z direction" R_m_z=L/(k_m*w^2) R_p_z=L/(k_p*(s^2-w^2)) R_eff_z=(1/R_m_z+1/R_p_z)^(-1) R_eff_z=L/(k_eff_z*s^2)
= Reff , z
(9)
and indicate that keff,z = 5.9 W/m-K (which is approximately equal to the conductivity of the metal multiplied by the ratio of the metal area to the total area). b.) The outer edges of the material are insulated and the faces of the material at z = 0 and z = L are exposed to a convective boundary condition with h = 10 W/m2-K. Is it appropriate to treat this problem as a lumped capacitance problem? The Biot number is the ratio of the conductive resistance associated with getting energy from within the structure to the surface (Rcond) to the convective resistance associated with transferring the energy from the surface to the surrounding fluid (Rconv). The conduction resistance is associated with transferring energy from z = L/2 to z = L: Rcond , z =
L/2 keff , z b 2
(10)
1 h b2
(11)
The convection resistance is:
Rconv , z = The Biot number is therefore:
Bi = "Biot number in z-direction" h=10 [W/m^2-K] R_cond_z=(L/2)/(k_eff_z*b^2) R_conv_z=1/(h*b^2) Bi=R_cond_z/R_conv_z
Rcond , z Rconv , z
(12)
"heat transfer coefficient"
which leads to a Biot number of 0.08; the lumped capacitance assumption is justified for this problem.
Problem 2.1-1 (2-1 in text): Buried Tubes Figure P2.1-1 illustrates two tubes that are buried in the ground behind your house that transfer water to and from a wood burner. The left hand tube carries hot water from the burner back to your house at Tw,h = 135°F while the right hand tube carries cold water from your house back to the burner at Tw,c = 70°F. Both tubes have outer diameter Do = 0.75 inch and thickness th = 0.065 inch. The conductivity of the tubing material is kt = 0.22 W/m-K. The heat transfer coefficient between the water and the tube internal surface (in both tubes) is hw = 250 W/m2-K. The center to center distance between the tubes is w = 1.25 inch and the length of the tubes is L = 20 ft (into the page). The tubes are buried in soil that has conductivity ks = 0.30 W/m-K. ks = 0.30 W/m-K th = 0.065 inch
kt = 0.22 W/m-K
Tw,c = 70°F 2 hw = 250 W/m -K
Tw,h = 135°F 2 hw = 250 W/m -K w = 1.25 inch Do = 0.75 inch
Figure P2.1-1: Tubes buried in soil.
a.) Estimate the heat transfer from the hot water to the cold water due to their proximity to one another. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in D_o=0.75 [inch]*convert(inch,m) "outer diameter of tube" th=0.065 [inch]*convert(inch,m) "thickness of tube" T_hw=converttemp(F,K,135) "hot water temperature" T_cw=converttemp(F,K,70) "cold water temperature" L=20 [ft]*convert(ft,m) "length of tubes" w=1.25 [inch]*convert(inch,m) "center to center distance" k_t=0.22 [W/m-K] "conductivity of teflon" k_s=0.30 [W/m-K] "conductivity of sand" h_w=250 [W/m^2-K] "heat transfer coefficient between the water and tube inner surfaces"
The heat transfer is resisted by convection within the tube, conduction through the tube, and conduction in the soil. The convection resistance is calculated according to: Rconv =
1 hw L π ( Do − 2 th )
The conduction resistance associated with the tube is calculated according to:
(1)
Rcond
⎛ Do ⎞ ln ⎜ ⎟ Do − 2 th ⎠ ⎝ = 2 kt L π
R_conv=1/(h_w*L*pi*(D_o-2*th)) R_cond=ln(D_o/(D_o-2*th))/(2*pi*k_t*L)
(2) "convection resistance" "conduction resistance"
The shape factor for parallel, buried tubes (SF) is obtained using EES’ internal library and used to compute the resistance due to conduction through the soil:
Rtubetotube =
1 SF ks
SF=SF_4(D_o,D_o,w,L) R_tubetotube=1/(SF*k_s)
(3) "shape factor" "tube to tube resistance"
The heat transfer rate is therefore: q =
Tw,h − Tw,c 2 Rconv + 2 Rcond + Rtubetotube
q_dot=(T_hw-T_cw)/(2*R_conv+2*R_cond+R_tubetotube)
(4)
"heat transfer rate"
which leads to q = 137.3 W. b.) To do part (a) you should have needed to determine a shape factor; calculate an approximate value of the shape factor and compare it to the accepted value. The shape factor can be thought of as the ratio of the effective area for conduction, Aeff, to the effective length required for conduction, Leff. The effective area for conduction can be estimated according to:
Aeff ≈ w L
(5)
while the length for conduction is approximately:
Leff ≈ w − Do
(6)
and the approximate value of the shape factor should be:
SFapp ≈ A_eff=L*w
Aeff
(7)
Leff "approximate area"
L_eff=(w-D_o) SF_app=A_eff/L_eff
"approximate length" "approximate shape factor"
The exact value of the shape factor returned by the function is SF = 17.4 m while the approximate value is SFapp = 15.2 m; these are sufficiently close for a sanity check. c.) Plot the rate of heat transfer from the hot water to the cold water as a function of the center to center distance between the tubes. The heat transfer rate is shown in Figure 2 as a function of the center to center distance between the tubes.
Figure 2: Heat transfer rate as a function of the center to center distance between the tubes
Problem 2.1-3 (2-2 in text) A solar electric generation system (SEGS) employs molten salt as both the energy transport and storage fluid. The molten salt is heated to Tsalt = 500°C and stored in a buried semi-spherical tank. The top (flat) surface of the tank is at ground level. The diameter of the tank before insulation is applied Dt = 14 m. The outside surfaces of the tank are insulated with tins = 0.30 m thick fiberglass having a thermal conductivity of kins = 0.035 W/m-K. Sand having a thermal conductivity of ksand = 0.27 W/m-K surrounds the tank, except on its top surface. Estimate the rate of heat loss from this storage unit to the Tair = 25°C surroundings. You may neglect the resistance due to convection. The inputs are entered in EES: $UnitSytem SI K J Pa $TabStops 0.2 0.4 3.5 in "known information" t_ins=0.3 [m] D_t=14 [m] k_sand=0.27 [W/m-K] k_ins=0.035 [W/m-K] T_air=convertTemp(C,K,25 [C]) T_salt=converttemp(C,K,500 [C])
"insulation thickness" "tank diameter" "sand thermal conductivity" "insulation thermal conductivity" "air temperature in K" "salt temperature in K"
The resistance between the bottom (buried, hemispherical) surface of the tank and the air is due to the conduction through the spherical shell of insulation
Rins ,bottom
⎛ 1 ⎞ 1 + ⎜ ⎟ Dt / 2 ( Dt / 2 + tins ) ⎠ ⎝ = 2 π kins
(1)
and the sand:
Rsand =
1 k sand S
(2)
where S is the shape factor, obtained using the EES function SF_22. The resistance from the top, unburied surface of the tank is only due to conduction through the insulation: Rins ,top =
tins D2 kins π t 4
The heat transfer from the bottom and top surfaces are calculated according to:
(3)
qbottom =
(Tsalt − Tair ) Rsoil + Rins ,bottom
(4)
and
qtop =
(Tsalt − Tair ) Rins ,top
(5)
The total heat loss is:
qtotal = qtop + qbottom R_ins_bottom=(1/(D_t/2)-1/(D_t/2+t_ins))/(2*pi*k_ins) "resistance to conduction through insulation on bottom" S=SF_22(D_t+2*t_ins) "shape factor" R_sand=1/(k_sand*S) "resistance of sand" R_ins_top=t_ins/(k_ins*pi*D_t^2/4) "resistance to conduction through insulation on top" q_dot_bottom=(T_salt-T_air)/(R_sand+R_ins_bottom) "heat loss from bottom surface of tank" q_dot_top=(T_salt-T_air)/(R_ins_top) "heat loss from top surface of tank" q_dot_total=q_dot_bottom+q_dot_top "total heat loss"
which leads to qtotal = 12.95 kW.
(6)
Problem 2.2-3 (2-3 in text): Heat Transfer Coefficient Measurement Device You are the engineer responsible for a simple device that is used to measure heat transfer coefficient as a function of position within a tank of liquid (Figure P2.2-3). The heat transfer coefficient can be correlated against vapor quality, fluid composition, and other useful quantities. The measurement device is composed of many thin plates of low conductivity material that are interspersed with large, copper interconnects. Heater bars run along both edges of the thin plates. The heater bars are insulated and can only transfer energy to the plate; the heater bars are conductive and can therefore be assumed to come to a uniform temperature as a current is applied. This uniform temperature is assumed to be applied to the top and bottom edges of the plates. The copper interconnects are thermally well-connected to the fluid; therefore, the temperature of the left and right edges of each plate are equal to the fluid temperature. This is convenient because it isolates the effect of adjacent plates from one another which allows each plate to measure the local heat transfer coefficient. Both surfaces of the plate are exposed to the fluid temperature via a heat transfer coefficient. It is possible to infer the heat transfer coefficient by measuring heat transfer required to elevate the heater bar temperature a specified temperature above the fluid temperature. top and bottom surfaces exposed to fluid 2 T∞ = 20°C, h = 50 W/m -K copper interconnet, T∞ = 20°C
a = 20 mm b = 15 mm
plate: k = 20 W/m-K th = 0.5 mm
heater bar, Th = 40°C
Figure P2.2-3: Device to measure heat transfer coefficient as a function of position.
The nominal design of an individual heater plate utilizes metal with k = 20 W/m-K, th = 0.5 mm, a = 20 mm, and b = 15 mm (note that a and b are defined as the half-width and half-height of the heater plate, respectively, and th is the thickness as shown in Figure P2-3). The heater bar temperature is maintained at Th = 40ºC and the fluid temperature is T∞ = 20ºC. The nominal value of the average heat transfer coefficient is h = 50 W/m2-K. a.) Develop an analytical model that can predict the temperature distribution in the plate under these nominal conditions. The problem can be simplified and tackled using the quarter-symmetry model of a single plate, shown in Figure P2.2-3, is specified mathematically in Figure 2(a).
(a)
(b) Figure 2: Quarter symmetry model of a plate.
The lines of symmetry (x =0 and y = 0) are adiabatic, the top edge (y = b) is held at the heater temperature (Th) and the right edge (x = a) is held at the fluid temperature (T∞). These boundary conditions are expressed below: ∂T ∂x
=0 x =0
Tx =a = T∞
∂T ∂y
=0 y =0
Ty =b = Th The differential control volume and associated first law balance are shown in Figure 2(a):
q x + q y = q x + dx + q y + dy + qconv or, after expanding the x + dx and y + dy terms: 0=
∂q y ∂q x dx + dy + qconv ∂x ∂y
Substituting the rate equations into the energy balance leads to: 0=
∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ −k th dy dx + ⎢ − k th dx dy + 2 dx dy h ( T − T∞ ) ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂y ⎣ ∂y ⎥⎦
or
∂ 2T ∂ 2T 2 h + − T − T∞ = 0 ∂x 2 ∂y 2 k th
(
)
(1)
Equation (1) is not homogeneous; however, a transformation variable can be identified by inspection:
θ = T − T∞
(2)
Substituting Eq. (2) into Eq. (1) transforms the governing partial differential equation into a homogeneous equation; i.e., any multiple of θ will satisfy transformed partial differential equation: ∂ 2θ ∂ 2θ + 2 − m2 θ = 0 2 ∂x ∂y
where
m=
2h k th
(3)
Substituting Eq. (2) into the boundary conditions leads to: ∂θ ∂x
=0
θ x =a = 0 ∂θ ∂y
(4)
x =0
=0
(5) (6)
y =0
θ y =b = θ h
(7)
where
θ h = Th − T∞ The transformed problem is shown in Figure 2(b); note that the transformed problem is linear and consists of a homogeneous partial differential equation and homogeneous boundary conditions in the x-direction. Therefore, we can apply a separation of variables solution to the problem and x is our homogeneous direction. The solution is assumed to be the product of θX which is a function of x and θY which is a function of y:
θ ( x, y ) = θ X ( x ) θ Y ( y ) Substituting the product solution into the differential equation leads to:
θY
d 2θ X d 2θ Y θ + X − m2 θ X θY = 0 2 2 dx dy
Dividing by the product θX θY: 2 d 2θ X d θ Y 2 dx 2 + dy − m 2 = 0 θ Y
X θ
−λ 2
λ2
Recall that x is our homogeneous direction; therefore, we need the θX group in the equation to equal a negative constant (-λ2); the remaining part of the equation must equal the positive value
of the same constant (λ2). Therefore, the two ordinary differential equations that result from the separation process are:
d 2θ X + λ2 θ X = 0 2 dx
(8)
d 2θ Y − ( λ 2 + m2 )θY = 0 dy 2
(9)
and
The next step is to solve the eigenproblem; the solution to the ordinary differential equation for θX is:
θ X = C1 cos ( λ x ) + C2 sin ( λ x )
(10)
The 1st boundary conditions in the homogeneous direction, Eq. (4), leads to: dθ X dx
=0
(11)
x =0
Substituting Eq. (10) into Eq. (11) leads to:
dθ X dx
x =0
= −C1 λ sin ( λ 0 ) + C2 λ cos ( λ 0 ) = 0
=0
=1
which can only be true if C2 = 0. the 2nd boundary condition in the x-direction, Eq. (5), leads to:
θ x =a = C1 cos ( λ a ) = 0 which can only be true if the argument of the cosine is 0. Therefore, the eigenfunctions and eigenvalues for the problem are:
θ X i = C1,i cos ( λi x ) where λi =
( 2 i − 1) π 2
a
The solution to the ordinary differential equation in the non-homogeneous direction, Eq. (9) can be obtained using Maple: > restart; > assume((lambda^2+m^2),positive); > ODEy:=diff(diff(thetaY(y),y),y)-(lambda^2+m^2)*thetaY(y)=0;
2 ⎞ ⎛d ODEy := ⎜ 2 thetaY( y ) ⎟⎟ − ( λ∼ 2 + m~ 2 ) thetaY( y ) = 0 ⎜ dy ⎝ ⎠
> qYs:=dsolve(ODEy);
qYs := thetaY( y ) = _C1 e
λ∼ 2 + m~2 y )
(
+ _C2 e
( − λ∼2 + m~2 y )
Note that Maple can convert this exponential form to an equivalent form involving hyperbolic sines and cosines with the convert command. > convert(qYs,'trigh');
thetaY( y ) = ( _C1 + _C2 ) cosh( λ∼ 2 + m~ 2 y ) + ( _C1 − _C2 ) sinh( λ∼ 2 + m~ 2 y )
θ Yi = C3,i cosh
(
)
λi2 + m 2 y + C4,i sinh
(
λi2 + m 2 y
)
The solution associated with the ith eigenvalue is therefore:
θi = θ X i θ Yi = cos ( λi x ) ⎡⎢C3,i cosh ⎣
(
)
λi2 + m 2 y + C4,i sinh
(
)
λi2 + m 2 y ⎤⎥ ⎦
(12)
where the constant C1,i is absorbed into the constants C3,i and C4,i. It is good practice to verify that Eq. (12) satisfies both boundary conditions in the x-direction and the partial differential equation for any value of i: > restart; > assume(i,integer); > lambda:=(2*i-1)*Pi/(2*a);
λ :=
( 2 i~ − 1 ) π 2a
> theta:=(x,y)->cos(lambda*x)*(C_3*cosh(sqrt(lambda^2+m^2)*y)+C_4*sinh(sqrt(lambda^2+m^2)*y));
θ := ( x, y ) → cos( λ x ) ( C_3 cosh( λ 2 + m 2 y ) + C_4 sinh( λ 2 + m 2 y ) ) > eval(diff(theta(x,y),x),x=0);
0 > theta(a,y);
0 > diff(diff(theta(x,y),x),x)+diff(diff(theta(x,y),y),y)-m^2*theta(x,y);
1 ( 2 i~ − 1 ) π x ⎞ 2 2 − cos⎛⎜⎜ ⎟⎟ ( 2 i~ − 1 ) π 2a 4 ⎝ ⎠ ⎛ ⎛ ⎛ ⎞ ( 2 i~ − 1 ) 2 π 2 ⎜ ⎜ ⎜ + 4 m 2 y ⎟⎟ ⎜ ⎜ ⎜ 2 a ⎜ ⎜ ⎟ ⎜ ⎜⎜ C_3 cosh⎜⎜ ⎟⎟ + C_4 sinh⎜⎜ 2 ⎝ ⎝ ⎠ ⎝
⎞⎞ ( 2 i~ − 1 ) 2 π 2 + 4 m 2 y ⎟⎟ ⎟⎟ 2 a ⎟⎟ ⎟⎟ ⎟⎟ 2 ⎠⎠
⎛ ⎜ ⎜ ( 2 i~ − 1 ) π x ⎞ ⎜⎜ a 2 + cos⎛⎜⎜ ⎟⎟ ⎜ 2a ⎝ ⎠⎝ 2 2 ⎛ ⎞ ( 2 i~ − 1 ) π ⎜ + 4 m 2 y ⎟⎟ ⎜ 2 1 a ⎜ ⎟ ⎛ ( 2 i~ − 1 ) 2 π 2 ⎞ ⎜ ⎟⎟ ⎜ + 4 m 2 ⎟⎟ C_3 cosh⎜ 2 ⎜ 2 4 a ⎝ ⎠⎝ ⎠ 2 2 ⎞ ⎛ ⎞ ( 2 i~ − 1 ) π 2 ⎟ ⎜ ⎟ + 4 m y ⎟ ⎜ ⎟ 1 a2 ⎟ ⎜ ⎟ ⎛ ( 2 i~ − 1 ) 2 π 2 ⎞ 2 ⎟⎟ ⎜ + C_4 sinh⎜⎜ + 4 m ⎟⎟ ⎟⎟ − m 2 2 ⎜ 2 4 a ⎝ ⎠⎝ ⎠⎠ ( 2 i~ − 1 ) π x ⎞ cos⎛⎜⎜ ⎟⎟ 2 a ⎝ ⎠ ⎛ ⎞ ⎞⎞ ⎛ ⎛ ( 2 i~ − 1 )2 π2 ( 2 i~ − 1 ) 2 π 2 2 ⎜ ⎜ ⎟ ⎜ + 4 m y + 4 m 2 y ⎟⎟ ⎟⎟ ⎜ ⎜ ⎟ ⎜ 2 2 a a ⎜ ⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ C_3 cosh⎜⎜ ⎟⎟ + C_4 sinh⎜⎜ ⎟⎟ ⎟⎟ 2 2 ⎝ ⎝ ⎠ ⎝ ⎠⎠ > simplify(%);
0
The general solution is the sum of the solution for each eigenvalue: ∞
∞
θ = ∑θi = ∑ cos ( λi x ) ⎡⎢C3,i cosh i =1
⎣
i =1
(
)
λi2 + m 2 y + C4,i sinh
(
)
λi2 + m 2 y ⎤⎥ ⎦
(13)
The general solution must satisfy the boundary conditions in the non-homogeneous direction. Substituting Eq. (13) into Eq. (6) leads to: ∂θ ∂y
⎡ ⎤ 2 2 2 2 2 2 2 2 ⎢ = ∑ cos ( λi x ) C3,i λi + m sinh λi + m 0 + C4,i λi + m cosh λi + m 0 ⎥ = 0 ⎢ i =1
⎥ =0 =1 ⎣⎢ ⎦⎥
y =0
)
(
∞
(
)
or ∞
∑C i =1
4,i
λi2 + m 2 cos ( λi x ) = 0
which can only be true if C4,i = 0: ∞
θ = ∑ Ci cosh i =1
(
)
λi2 + m 2 y cos ( λi x )
(14)
where the subscript 3 has been removed from C3,i because it is the only remaining undetermined constant. Substituting Eq. (14) into Eq. (7) leads to: ∞
θ y =b = ∑ Ci cosh i =1
(
)
λi2 + m 2 b cos ( λi x ) = θ h
This equation is multiplied by cos(λj x) and integrated from 0 to a: ∞
∑ Ci cosh i =1
(
)
a
a
0
0
λi2 + m 2 b ∫ cos ( λi x ) cos ( λ j x ) dx = θ h ∫ cos ( λ j x ) dx
Orthogonality causes all of the terms in the summation to integrate to zero except for the one in which i = j: Ci cosh
(
λ +m b 2 i
2
) ∫ cos (λ x ) dx = θ ∫ cos (λ x ) dx a
a
2
i
h
0
i
0
Maple is used to carry out the integrations: > restart; > assume(i,integer); > lambda:=(2*i-1)*Pi/(2*a);
λ := > int((cos(lambda*x))^2,x=0..a);
( 2 i~ − 1 ) π 2a a 2
> int(cos(lambda*x),x=0..a);
( 1 + i~ )
2 ( -1 ) a ( 2 i~ − 1 ) π
which leads to an equation for each of the constants.
θ h 4 ( −1)
1+ i
Ci =
The inputs are entered in EES: $UnitSystem SI MASS RAD PA C J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs"
cosh
(
)
λi2 + m 2 b ( 2 i − 1) π
th_mm=0.5 [mm] th=th_mm * convert(mm,m) k=20 [W/m-K] a_mm=20 [mm] a= a_mm * convert(mm,m) b= 15 [mm] *convert(mm,m) T_h = converttemp(C,K,40 [C]) T_infinity=converttemp(C,K,20 [C]) h_bar=50 [W/m^2-K]
"thickness of plate in mm" "thickness of plate" "conductivity of plate" "half-width of plate in mm" "half-width of plate" "half-height of plate" "heater temperature" "fluid temperature" "heat transfer coefficient"
The position is specified using dimensionless variables x_bar and y_bar: "position" x_bar=0.5 y_bar=0.5 x=x_bar*a y=y_bar*b
The solution is evaluated: theta_h=T_h-T_infinity "heater temperature difference" m=sqrt(2*h_bar/(k*th)) "solution constant" N=100 "number of terms" duplicate i=1,N lambda[i]=(2*i-1)*pi/(2*a) "eigenvalues" C[i]=theta_h*(4*(-1)^(1+i)/(2*i-1)/Pi)/(cosh(sqrt(lambda[i]^2+m^2)*b)) theta[i]=C[i]*cos(lambda[i]*x)*cosh(sqrt(lambda[i]^2+m^2)*y) end T=sum(theta[1..N])+T_infinity T_C=converttemp(K,C,T)
A parametric table is created and used to generate the contour plot shown in Figure 3.
Figure 3: Contour plot of temperature on the plate.
b.) The measured quantity is the rate of heat transfer to the plate from the heater ( qh ) and therefore the relationship between qh and h (the quantity that is inferred from the heater power) determines how useful the instrument is. Determine the heater power. The heater power can be computed by integrating the conduction heat transfer along the top surface according to: a
qh = 4 ∫ k 0
∂θ ∂y
th dx
(15)
y =b
where the factor of 4 comes from the quarter symmetry of the model. substituted into Eq. (15): ∞
qh = 4 k th ∑ Ci λi2 + m 2 sinh i =1
(
λi2 + m 2 b
Equation (14) is
) ∫ cos (λ x ) dx a
i
0
which leads to: ∞
qh = 2 k th a ∑ Ci λi2 + m 2 sinh i =1
Equation (16) is evaluated in EES: "heater power"
(
λi2 + m 2 b
)
(16)
duplicate i=1,N q_dot_h[i]=2*k*th*a*C[i]*sqrt(lambda[i]^2+m^2)*sinh(sqrt(lambda[i]^2+m^2)*b) end q_dot_h=sum(q_dot_h[1..N])
c.) If the uncertainty in the measurement of the heater power is δ qh = 0.01 W, estimate the uncertainty in the measured heat transfer coefficient ( δ h ). A parametric table is used to explore the relationship between h and qhtr ; the two columns in the table are the variables h_bar and q_dot_h. Figure 4 illustrates the heat transfer coefficient as a function of the heater power, all else held constant.
Figure 4: Heat transfer coefficient as a function of the heater power.
Note that a good measuring device would show a strong relationship between the physical quantity being measured ( h ) and the measurement ( qh ); this would be indicated by a large partial derivative of heat transfer coefficient with respect to heater power. Given the uncertainty in the heater power measurement, δ qh (related to, for example, the resolution of the data acquisition system, noise, etc.), it is possible to estimate the uncertainty in the measurement of the heat transfer coefficient, δ h , according to:
δh =
∂h δ qh ∂qh
For example, if the uncertainty in the heater power is 0.01 W then Fig. 3 suggests that the partial derivative of heat transfer coefficient with respect to heater power is approximately 125 W/m2-
K-W and the uncertainty in the heat transfer coefficient measurement would be δ h = 1.8 W/m2K. As an engineer designing this measurement device, you would like to calculate not the heater power but rather the partial derivative in the heat transfer coefficient with respect to heater power in order to automate the process of computing δ h and therefore evaluate how design changes affect the instrument. This would be difficult to accomplish analytically; note that both m and the constants Ci in Eq. (16) are functions of h . It is more convenient to determine the partial derivative numerically. That is, set the heat transfer coefficient at its nominal value plus a small amount ( h + Δh ) and evaluate the heater power ( qh , h +Δh ); note that Δh should be small relative to the nominal value of the heat transfer coefficient. Then set the heat transfer coefficient at its nominal value less a small amount ( h − Δh ) and evaluate the heater power ( qh , h −Δh ). The partial derivative is approximately: ∂h 2 Δh ≈ ∂qh qh ,h +Δh − qh ,h −Δh
Since qh needs to be evaluated at several values of the heat transfer coefficient, it is convenient to have the computation of the heater power occur within a function that is called twice within the equation window. Because the solution is in the form of an EES code, this is an ideal problem to use a MODULE. A MODULE is a stand-alone EES program that can be called from the main EES equation window. The MODULE is provided with inputs and it calculates outputs. The protocol of a call to a MODULE involves the name of the MODULE followed by a series of inputs separated by a colon from a series of outputs. For example, we will create a MODULE Heaterpower that calculates the value of the variable q_dot_h. It is important to note that all of the variables from the main equation window that are required by the MODULE must be passed to the MODULE; the MODULE can only access variables that are passed to it as parameters or using the $COMMON directive. The MODULE Heaterpower must be placed at the top of the EES code and is a small selfcontained EES program; we create the MODULE by copying those lines of the main EES code that are required to calculate the variable q_dot_h. MODULE Heaterpower(th,k,a,b,T_h,T_infinity,h_bar:q_dot_h) theta_h=T_h-T_infinity "heater temperature difference" m=sqrt(2*h_bar/(k*th)) "solution constant" N=100 "number of terms" duplicate i=1,N lambda[i]=(2*i-1)*pi/(2*a) "eigenvalues" C[i]=theta_h*(4*(-1)^(1+i)/(2*i-1)/Pi)/(cosh(sqrt(lambda[i]^2+m^2)*b)) q_dot_h[i]=2*k*th*a*C[i]*sqrt(lambda[i]^2+m^2)*sinh(sqrt(lambda[i]^2+m^2)*b) end q_dot_h=sum(q_dot_h[1..N]) end
The calling protocol for the MODULE consists of a series of inputs (the variables d, k, a, b, T_htr, T_f, and h) that are separated by a series outputs (in this case only the variable q_dot_htr) by a colon. MODULES are most useful where a certain sequence of code must be executed multiple times; in this case, the MODULE Heaterpower enables the partial derivative to be easily computed. delta_q_dot_htr=0.01 [W] dh=1 [W/m^2-K]
"heater power resolution" "perturbation of heat transfer coefficient"
CALL Q_dot_heater(d,k,a,b,T_htr,T_f,h-dh:q_dot_htr_minus) CALL Q_dot_heater(d,k,a,b,T_htr,T_f,h+dh:q_dot_htr_plus) dhdqdot=2*dh/(q_dot_htr_plus-q_dot_htr_minus) delta_h=dhdqdot*delta_q_dot_htr
The modifications to the EES code verifies that the uncertainty in heat transfer coefficient, delta_h, is 1.26 W/m2-K and provides a convenient tool for assessing the impact of the various design parameters on the performance of the measurement system. For example, Figure 5 illustrates the uncertainty in the heat transfer coefficient as a function of the plate thickness (th) for various values of its half-width (a).
Figure 5: Uncertainty in the measured heat transfer coefficient as a function of the plate thickness for various values of the plate half-width.
Problem 2.2-6 (2-4 in text) A laminated composite structure is shown in Figure P2.2-6. H = 3 cm
2 q ′′ = 10000 W/m
Tset = 20°C W = 6 cm
Tset = 20°C
kx = 50 W/m-K ky = 4 W/m-K Figure P2.2-6: Composite structure exposed to a heat flux.
The structure is anisotropic. The effective conductivity of the composite in the x-direction is kx = 50 W/m-K and in the y-direction it is ky = 4 W/m-K. The top of the structure is exposed to a heat flux of q ′′ = 10,000 W/m2. The other edges are maintained at Tset = 20°C. The height of the structure is H = 3 cm and the half-width is W = 6 cm. a.) Develop a separation of variables solution for the 2-D steady-state temperature distribution in the composite. The known information is entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" W=6 [cm]*convert(cm,m) H=3 [cm]*convert(cm,m) q``=10000 [W/m^2] k_x=50 [W/m-K] k_y=4 [W/m-K] T_set=converttemp(C,K,20[C])
"width of laminate" "height of laminate" "heat flux" "conductivity in the x-direction" "conductivity in the y-direction" "specified edge temperatures"
A half-symmetry model of the problem shown in Figure P2.2-6 is governed by the transformed partial differential equation: kx
∂ 2θ ∂ 2θ + k =0 y ∂x 2 ∂y 2
(1)
with the boundary conditions: ∂θ ∂x
=0
(2)
x =0
θ x =W = 0
(3)
ky
∂θ ∂y
= q ′′
(4)
y=H
θ y =0 = 0
(5)
θ = T − Tset
(6)
θ = θ X θY
(7)
where
The separated solution is assumed:
and substituted into Eq. (1):
kx θY
d 2θ X d 2θ Y k θ X + y =0 dx 2 dx 2
(8)
Dividing through by θY θX leads to:
d 2θ X d 2θ Y dx 2 + k y dx 2 = 0 k θY θX
x
−λ2
(9)
λ2
The process leads to two ordinary differential equations:
d 2θ X + λ2 θ X = 0 2 dx
(10)
d 2θ Y − β 2 λ 2 θY = 0 dy 2
(11)
where
β2 =
kx ky
(12)
beta=sqrt(k_x/k_y)
Notice that the sign of the constant is selected so that the solution in the homogeneous direction (x) is sines and cosines. The solution to Eq. (10) is:
θ X = C1 sin ( λ x ) + C2 cos ( λ x )
(13)
The boundary condition at x = 0, Eq. (2) requires that C1 = 0 and therefore:
θ X = C2 cos ( λ x )
(14)
The boundary condition at x = L, Eq. (3), leads to the eigencondition for the problem: cos ( λi W ) = 0
(15)
which requires that:
λi =
(1 + 2 i ) π 2W
for i = 0,1,...∞
N_term=11 [-] duplicate i=0,N_term lambda[i]=(1+2*i)*pi/(2*W) end
(16)
"number of terms" "eigenvalues"
The eigenfunctions are:
θ X i = C2,i cos ( λi x )
(17)
The solution to Eq. (11) for each eigenvalue is:
θ Yi = C3,i sinh ( β λi y ) + C4,i cosh ( β λi y )
(18)
The general solution for for each eigenvalue is:
θi = θ X i θ Yi = C2,i cos ( λi x ) ⎡⎣C3,i sinh ( β λi y ) + C4,i cosh ( β λi y ) ⎤⎦
(19)
The sum of these solutions is, itself, a solution (note that the constant C2,i is absorbed into the other constants): ∞
∞
i =1
i =1
θ = ∑ θi = ∑ cos ( λi x ) ⎡⎣C3,i sinh ( β λi y ) + C4,i cosh ( β λi y ) ⎤⎦
(20)
The boundary condition at y = 0, Eq. (5), leads to C4,i = 0: ∞
∞
i =1
i =1
θ = ∑ θi = ∑ Ci cos ( λi x ) sinh ( β λi y )
(21)
Equation (21) is substituted into the boundary condition at y = H, Eq. (4): ∞
β λi k y ∑ Ci cos ( λi x ) cosh ( β λi H ) = q ′′
(22)
i =1
Equation (22) is multiplied by an eigenfunction and integrated from x = 0 to x = W: L
L
0
0
β λi k y cosh ( β λi H ) Ci ∫ cos 2 ( λi x ) dx = q ′′∫ cos ( λi x ) dx The integrals are carried out in Maple: > restart; > assume(i,integer); > lambda:=(1+2*i)*Pi/(2*W);
λ :=
( 1 + 2 i~ ) π 2W
> int((cos(lambda*x))^2,x=0..W);
W 2 > int(cos(lambda*x),x=0..W);
2 ( -1 )i~ W ( 1 + 2 i~ ) π
and used to complete the computation of the constants in EES: duplicate i=0,N_term k_y*C[i]*beta*lambda[i]*cosh(beta*lambda[i]*H)*W/2=q``*2*(-1)^i*W/((1+2*i)*Pi) end
The solution is evaluated at an arbitary position: x_bar=0.5 [-] y_bar=0.5 [-] x=x_bar*W y=y_bar*H duplicate i=0,N_term theta[i]=C[i]*cos(lambda[i]*x)*sinh(beta*lambda[i]*y) end theta=sum(theta[0..N_term]) T=theta+T_set T_C=converttemp(K,C,T)
b.) Prepare a contour plot of the temperature distribution.
"dimensionless x-position" "dimensionless y-position" "x-position" "y-position"
(23)
Figure 2 illlustrates a contour plot of the temperature distribution. 1
Dimensionless y position, y/H
20 23.12
0.8
26.24 29.36 32.48
0.6
35.6 38.72
0.4
41.84 44.96 48.08
0.2
51.2 0 0
0.2
0.4
0.6
0.8
Dimensionless x position, x/W Figure 2: Contour plot of the temperature distribution.
1
Problem 2.3-1 (2-5 in text): Cryogen Transfer Pipe Figure P2.3-1 illustrates a pipe that connects two tanks of liquid oxygen on a spacecraft. The pipe is subjected to a heat flux, q′′ = 8,000 W/m2, which can be assumed to be uniformly applied to the outer surface of the pipe and entirely absorbed. Neglect radiation from the surface of the pipe to space. The inner radius of the pipe is rin = 6 cm, the outer radius of the pipe is rout = 10 cm, and the half-length of the pipe is L = 10 cm. The ends of the pipe are attached to the liquid oxygen tanks and therefore are at a uniform temperature of TLOx = 125 K. The pipe is made of a material with a conductivity of k = 10 W/m-K. The pipe is empty and therefore the internal surface can be assumed to be adiabatic. rout = 10 cm rin = 6 cm
L = 10 cm TLOx = 125 K
2 k = 10 W/m-K q′′s = 8,000 W/m Figure P2.3-1: Cryogen transfer pipe connecting two liquid oxygen tanks.
a.) Develop an analytical model that can predict the temperature distribution within the pipe. Prepare a contour plot of the temperature distribution within the pipe. A differential control volume leads to the energy balance: qx + qr = qx + dx + qr + dr or
0=
∂qx ∂q dx + r dr ∂x ∂r
Substituting the rate equations: qx = − k 2 π r dr
∂T ∂x
qr = − k 2 π r dx
∂T ∂r
and
into the differential energy balance leads to:
0=
∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ dx + ⎢ −k 2 π r dx ⎥ dr − k 2 π r dr ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂r ⎣ ∂r ⎦
or r
∂ 2T ∂ ⎡ ∂T ⎤ + r =0 ∂x 2 ∂r ⎢⎣ ∂r ⎥⎦
A half-symmetry model of the pipe will be generated; the boundary conditions are therefore: ∂T ∂x
=0 x =0
Tx = L = TLOx
k
∂T ∂r
r = rin
∂T ∂r
r = rout
=0
= q′′
As stated, there are two non-homogeneous boundary conditions; however, the boundary condition at x = L can be made homogeneous by defining the temperature difference:
θ = T − TLOx The partial differential equation and boundary conditions are written in terms of θ: r
∂ 2θ ∂ ⎡ ∂θ ⎤ + r =0 ∂x 2 ∂r ⎢⎣ ∂r ⎥⎦ ∂θ ∂x
=0
(2)
x =0
θ x= L = 0 ∂θ ∂r
(1)
=0 r = rin
(3) (4)
k
∂θ ∂r
= q′′
(5)
r = rout
Note that the two homogeneous boundary conditions are in the x-direction and so the eigenfunctions of the problem will be in this direction. We assume that the solution is separable; that is, the solution is the product of a function only of x (θX) and r (θR):
θ ( x, y ) = θ X ( x ) θ R ( r ) Substituting the product solution into the governing partial differential equation, Eq. (1), leads to: rθ R
d 2θ X d ⎡ dθ R ⎤ +θ X r =0 2 dx dr ⎢⎣ dr ⎥⎦
Dividing by the product r θX θR leads to: d ⎡ dθ R ⎤ d 2θ X ⎢r ⎥ 2 dx + dr ⎣ dr ⎦ = 0 rθ R θX ±λ2
∓λ2
Note that the 1st term is a function only of x while the 2nd term is a function only of r; these two quantities must be equal and opposite constants (±λ2). The choice of the sign is again important; the eigenfunctions must be in x and therefore the two ordinary differential equations must be: d 2θ X + λ2 θ X = 0 dx 2
(6)
d ⎡ dθ R ⎤ r − λ2 rθ R = 0 ⎢ ⎥ dr ⎣ dr ⎦
(7)
The eigenproblem will be solved first; the solution to Eq. (6) is:
θ X = C1 cos ( λ x ) + C2 sin ( λ x ) The boundary condition at x = 0, Eq. (2), eliminates the sine term. The boundary condition at x = L, Eq. (3), leads to: C1 cos ( λ L ) = 0
which provides the eigenvalues and the eigenfunctions:
θ X i = C1,i cos ( λi x ) where λi =
( 2 i − 1) π 2L
for i = 1, 2,..∞
The ordinary differential equation for θR, Eq. (7), is solved by Bessel functions, as discussed in Section 1.8. Equation (7) is a form of Bessel's equation: d ⎛ p dθ ⎞ 2 s ⎜x ⎟±c x θ =0 dx ⎝ dx ⎠ where s = 1 and p = 1; the quantity s – p +2 is therefore not equal to zero and we are directed toward the left-side of the Bessel function flow chart presented in Section 1.8.4 where the parameters n = 0, a = 1, and n/a = 0 are computed. The last term is negative and therefore the general solution to Eq. (7) is:
θ Ri = C3,i BesselI ( 0,λi r ) + C4,i BesselK ( 0,λi r ) This general solution could have been obtained using Maple: > restart; > ODEr:=diff(r*diff(thetar(r),r),r)-lambda^2*r*thetar(r)=0; 2 d ⎛d ⎞ ⎛ ⎞ ⎜ ODEr := ⎜⎜ thetar( r ) ⎟⎟ + r ⎜ 2 thetar( r ) ⎟⎟ − λ 2 r thetar( r ) = 0 ⎝ dr ⎠ ⎝ dr ⎠
> thetars:=dsolve(ODEr);
thetars := thetar( r ) = _C1 BesselI( 0, λ r ) + _C2 BesselK( 0, λ r )
The general solution for each eigenvalue is therefore:
θi = θ X i θ Ri = cos ( λi x ) ⎡⎣C3,i BesselI ( 0,λi r ) + C4,i BesselK ( 0,λi r ) ⎤⎦ The general solution is entered as a function of x and y in Maple: > restart; > assume(i,integer); > lambda:=(2*i-1)*Pi/(2*L);
λ :=
( 2 i~ − 1 ) π 2L
> theta:=(x,r)->cos(lambda*x)*(C3*BesselI(0,lambda*r)+C4*BesselK(0,lambda*r));
θ := ( x, r ) → cos( λ x ) ( C3 BesselI( 0, λ r ) + C4 BesselK( 0, λ r ) )
Verify that it solves both boundary conditions in the x-directions, Eqs. (2) and (3): > eval(diff(theta(x,r),x),x=0);
0
> theta(L,r);
0
and the partial differential equation, Eq. (1): > r*diff(diff(theta(x,r),x),x)+diff(r*diff(theta(x,r),r),r);
1 ( 2 i~ − 1 ) π x ⎞ 2 2 − r cos⎛⎜⎜ ⎟⎟ ( 2 i~ − 1 ) π 2 L 4 ⎝ ⎠ ⎛ ( 2 i~ − 1 ) π r ⎞ + C4 BesselK⎛ 0, ( 2 i~ − 1 ) π r ⎞ ⎞ L 2 + ⎛ ⎜⎜ C3 BesselI⎜⎜ 0, ⎟⎟ ⎜⎜ ⎟⎟ ⎟⎟ 2L 2L ⎝ ⎝ ⎠ ⎝ ⎠⎠ ( 2 i~ − 1 ) π r ⎛ ⎛ ⎞ ( 2 i~ − 1 ) π ⎜ C3 BesselI⎜⎜ 1, ⎟⎟ ⎜ ( 2 i~ − 1 ) π x 1 2 L ⎝ ⎛ ⎞ ⎠ cos⎜⎜ ⎟⎟ ⎜⎜ 2 L 2 L ⎝ ⎠⎝
⎛ ⎜ ⎜ ( 2 i~ − 1 ) π r ⎞ ⎞ ⎛ ⎜ C4 BesselK⎜⎜ 1, ⎟⎟ ( 2 i~ − 1 ) π ⎟ ⎜ ⎟ 2L 1 ( 2 i~ − 1 ) π x ⎜ ⎛ ⎞ ⎝ ⎠ ⎟ + r cos⎜ − ⎟ ⎟ ⎜ ⎟ ⎜⎜ L 2 2 L ⎠ ⎝ ⎠⎝ ( 2 i~ − 1 ) π r ⎞ ⎞ ⎛ 2 L BesselI⎛⎜⎜ 1, ⎜ ⎟⎟ ⎟ ⎜ ( 2 i~ − 1 ) π r 2L ⎞ ⎛ ⎝ ⎠ ⎟⎟ ( 2 i~ − 1 ) 2 π 2 ⎜ C3 ⎜ BesselI⎜⎜ 0, ⎟⎟ − ⎟ 1 2L ( 2 i~ − 1 ) π r ⎠ ⎝ ⎠ ⎝ − 2 4 L ( 2 i~ − 1 ) π r ⎞ ⎞ ⎛ 2 L BesselK⎛⎜⎜ 1, ⎜ ⎟⎟ ⎟ ⎜ ( 2 i~ − 1 ) π r ⎞ 2L ⎝ ⎠ ⎟⎟ ( 2 i~ − 1 ) 2 π 2 − C4 ⎜⎜ −BesselK⎛⎜⎜ 0, ⎟⎟ ⎟ 1 2L ( 2 i~ − 1 ) π r ⎝ ⎠ ⎠ ⎝ 2 4 L
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎟ ⎠
> simplify(%);
0
The general solution for θ is the series: ∞
∞
i =1
i =1
θ = ∑ θi = ∑ cos ( λi x ) ⎡⎣C3,i BesselI ( 0,λi r ) + C4,i BesselK ( 0,λi r ) ⎤⎦
(8)
The solution must satisfy the boundary conditions in the non-homogeneous direction; the boundary condition at r = rin, Eq. (4), leads to:
∂θ ∂r
∞
r = rin
= ∑ cos ( λi x ) i =1
d ⎡C3,i BesselI ( 0,λi r ) + C4,i BesselK ( 0,λi r ) ⎤⎦ r = r = 0 in dr ⎣
(9)
The derivatives of the Bessel functions may either be evaluated using the equations provided in Section 1.8.4 or, more conveniently, using Maple:
> restart; > diff(BesselI(0,lambda*r),r);
BesselI( 1, λ r ) λ
> diff(BesselK(0,lambda*r),r);
−BesselK( 1, λ r ) λ
Therefore Eq. (9) can be written as: ∞
∑ cos ( λ x ) ⎡⎣C i
i =1
3,i
λi BesselI (1,λi rin ) − C4,i λi BesselK (1,λi rin ) ⎤⎦ = 0
which can only be true if: C3,i λi BesselI (1,λi rin ) − C4,i λi BesselK (1,λi rin ) = 0
Unlike most of the problems we have previously encountered, neither of the constants is eliminated; instead we see that there is a relationship between the two constants: C4,i = C3,i
BesselI (1,λi rin ) BesselK (1,λi rin )
(10)
Substituting Eq. (10) into Eq. (8) leads to: ∞
∞
⎡
i =1
i =1
⎣
θ = ∑ θi = ∑ Ci cos ( λi x ) ⎢ BesselI ( 0,λi r ) +
⎤ BesselI (1,λi rin ) BesselK ( 0,λi r ) ⎥ BesselK (1,λi rin ) ⎦
(11)
Equation (11) is substituted into the remaining non-homogeneous boundary condition at r = rout , Eq. (5), in order to obtain: k
∂θ ∂r
r = rout
∞ ⎡ ⎤ BesselI (1,λi rin ) BesselK (1,λi rout ) ⎥ = q′′ = k ∑ Ci λi cos ( λi x ) ⎢ BesselI (1,λi rout ) − BesselK (1,λi rin ) i =1 ⎣ ⎦
This equation is multiplied by an arbitrary eigenfunction, cos(λj x), and integrated between the homogeneous boundary conditions (from x = 0 to x= L); using the orthogonality property of the eigenfunctions we obtain: L ⎡ ⎤L BesselI (1,λi rin ) k Ci λi ⎢ BesselI (1,λi rout ) − BesselK (1,λi rout ) ⎥ ∫ cos 2 ( λi x ) dx = q′′∫ cos ( λi x ) dx BesselK (1,λi rin ) 0 ⎣ ⎦0
which leads to an expression for each constant: ⎤ BesselI (1,λi rin ) sin ( λi L ) k Ci λi L ⎡ BesselK (1,λi rout ) ⎥ = q′′ ⎢ BesselI (1,λi rout ) − 2 BesselK (1,λi rin ) λi ⎣ ⎦ Solving for Ci: Ci =
2 q′′ sin ( λi L ) ⎡
λi2 k L ⎢ BesselI (1,λi rout ) − ⎣
⎤ BesselI (1,λi rin ) BesselK (1,λi rout ) ⎥ BesselK (1,λi rin ) ⎦
The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" q``_dot=8000 [W/m^2] r_in=6.0 [cm]*convert(cm,m) r_out=10.0 [cm]*convert(cm,m) L = 10.0 [cm]*convert(cm,m) T_LOx=125 [K] k=10 [W/m-K]
"Heat flux on pipe surface" "Pipe inner radius" "Pipe outer radius" "Pipe half-length" "Pipe end temperature" "Pipe conductivity"
A position is specified in terms of dimensionless coordinates: "dimensionless position" r_bar=0.5 x_bar=0.5 r=r_in+(r_out-r_in)*r_bar x=x_bar*L
The solution is implemented for the 1st N terms of the series: N=100 "Number of terms" duplicate i=1,N lambda[i]=(2*i-1)*pi/(2*L) C[i]=2*q``_dot*sin(lambda[i]*L)/(lambda[i]^2*k*L*(BesselI(1,lambda[i]*r_out)-& BesselI(1,lambda[i]*r_in)*BesselK(1,lambda[i]*r_out)/BesselK(1,lambda[i]*r_in))) theta[i]=C[i]*cos(lambda[i]*x)*(BesselI(0,lambda[i]*r)+BesselI(1,lambda[i]*r_in)*& BesselK(0,lambda[i]*r)/BesselK(1,lambda[i]*r_in)) end T=sum(theta[1..N])+T_LOx
The temperature distribution is computed over a range of positions and the results are used to generate the contour plot shown in Figure 2.
Figure 2: Contour plot of temperature.
Problem 2.3-2 (2-6 in text) Figure P2.3-2 illustrates a cylinder that is exposed to a concentrated heat flux at one end. extends to infinity k = 168 W/m-K rout = 200 μm Ts = 20°C rexp = 21 μm q ′′ = 1500 W/cm
2
adiabatic
Figure P2.3-2: Cylinder exposed to a concentrated heat flux at one end.
The cylinder extends infinitely in the x-direction. The surface at x = 0 experiences a uniform heat flux of q′′ = 1500 W/cm2 for r < rexp = 21 μm and is adiabatic for rexp < r < rout where rout = 200 μm is the outer radius of the cylinder. The outer surface of the cylinder is maintained at a uniform temperature of Ts = 20ºC. The conductivity of the cylinder material is k = 168 W/m-K. a.) Develop a separation of variables solution for the temperature distribution within the cylinder. Plot the temperature as a function of radius for various values of x. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 0.8 3.5 "Inputs" r_out=200 [micron]*convert(micron,m) q``_dot=1500 [W/cm^2]*convert(W/cm^2,W/m^2) k=168 [W/m-K] r_exp=21 [micron]*convert(micron,m)
A differential control volume leads to the energy balance: qx + qr = qx + dx + qr + dr or
0= Substituting the rate equations:
∂qx ∂q dx + r dr ∂x ∂r
"outer radius of domain" "exposure flux" "conductivity of work piece" "radius of exposure zone"
qx = − k 2 π r dr
∂T ∂x
qr = − k 2 π r dx
∂T ∂r
and
into the differential energy balance leads to: 0=
∂ ⎡ ∂T ⎤ ∂ ⎡ ∂T ⎤ dx + ⎢ −k 2 π r dx ⎥ dr − k 2 π r dr ⎢ ⎥ ∂x ⎣ ∂x ⎦ ∂r ⎣ ∂r ⎦
or r
∂ 2T ∂ ⎡ ∂T ⎤ + r =0 ∂x 2 ∂r ⎢⎣ ∂r ⎥⎦
The boundary conditions are: −k
∂T ∂x
x =0
⎧⎪q′′ for r < rexp =⎨ ⎪⎩0 for r > rexp
Tx →∞ = Ts Tr =0 must be finite
Tr = rout = Ts As stated, there are two non-homogeneous boundary conditions; however, the boundary condition at r = rout can be made homogeneous by defining the temperature difference:
θ = T − Ts The partial differential equation and boundary conditions are written in terms of θ: r
−k
∂ 2θ ∂ ⎡ ∂θ ⎤ + r =0 ∂x 2 ∂r ⎢⎣ ∂r ⎥⎦
∂θ ∂x
x =0
⎧⎪q′′ for r < rexp =⎨ ⎪⎩0 for r > rexp
(1)
(2)
θ x →∞ = 0
(3)
θ r =0 must be finite
(4)
θ r =r = 0
(5)
out
Note that the two homogeneous boundary conditions are in the x-direction and so the eigenfunctions of the problem will be in this direction. We assume that the solution is separable; that is, the solution is the product of a function only of x (θX) and r (θR):
θ ( x, y ) = θ X ( x ) θ R ( r ) Substituting the product solution into the governing partial differential equation, Eq. (1), leads to: d 2θ X d ⎡ dθ R ⎤ rθ R +θ X r =0 2 dx dr ⎢⎣ dr ⎥⎦
Dividing by the product r θX θR leads to: d ⎡ dθ R ⎤ d 2θ X ⎢r ⎥ 2 dx + dr ⎣ dr ⎦ = 0 rθ R θX ±λ2
∓λ2
Note that the 1st term is a function only of x while the 2nd term is a function only of r; these two quantities must be equal and opposite constants (±λ2). The choice of the sign is again important; the eigenfunctions must be in r and therefore the two ordinary differential equations must be: d 2θ X − λ2 θ X = 0 dx 2
(6)
d ⎡ dθ R ⎤ r + λ2 rθ R = 0 ⎢ ⎥ dr ⎣ dr ⎦
(7)
The eigenproblem will be solved first; the solution to Eq. (7) is:
θ R = C1 BesselJ ( 0,λ r ) + C2 BesselY ( 0,λ r ) The boundary condition at r = 0, Eq. (4), requires that C2 = 0.
θ R = C1 BesselJ ( 0,λ r )
The boundary condition at r = rout, Eq. (5), leads to: C1 BesselJ ( 0,λ rout ) = 0
(8)
The 0th order Bessel function of the 1st kind (i.e., Bessel_J(0,x)) oscillates about zero every time the argument changes by 2π in the same way that sine and cosine do; therefore, there are an infinite number of eigenvalues λi that will satisfy Eq. (8) associated with an infinite number of eigenfunctions. The eigencondition for this problem cannot be used to explicitly solve for the eigenvalues; rather, an implicit equation for the eigenvalues results from Eq. (8): BesselJ ( 0,λi rout ) = 0 where i = 1, 2,...∞
(9)
The eigenfunctions for this problem are:
θ Ri = C1,i BesselJ ( 0,λi r ) where BesselJ ( 0,λi b ) = 0 for i = 1, 2,...∞ N=51 duplicate i=1,N lowerlimit[i]=(i-1)*pi/r_out upperlimit[i]=i*pi/r_out guess[i]=lowerlimit[i]+pi/(2*r_out) end
(11)
"number of terms in solution" "lower limit of eigenvalue" "upper limit of eigenvalue" "guess value for eigenvalue"
duplicate i=1,N BesselJ(0,lambda[i]*r_out)=0 end
"solve for eigenvalues"
The arrays lowerlimit[], upperlimit[], and guess[] are used to constrain the solution for the array lambda[] in the Variable Information window. The solution to the ordinary differential equation for θX, Eq. (6), is:
θ X i = C3,i exp ( λi x ) + C4,i exp ( −λi x ) The boundary condition at x → ∞, Eq. (3), requires that C3,i = 0:
θ X i = C4,i exp ( −λi x ) and so the general solution for each eigenvalue is:
θ i = θ Ri θ X i = Ci BesselJ ( 0,λi r ) exp ( −λi x ) The series solution for θ is:
∞
∞
i =1
i =1
θ = ∑ θi = ∑ Ci BesselJ ( 0,λi r ) exp ( −λi x )
(10)
Substituting Eq. (10) into the boundary condition at x = 0, Eq. (2), leads to: ∞ ⎧⎪q′′ for r < rexp k ∑ Ci λi BesselJ ( 0,λi r ) = ⎨ i =1 ⎪⎩0 for r > rexp
(11)
Equation (11) is multiplied by r BesselJ ( 0,λi r ) and integrated from r = 0 to r = rout: k Ci λi
rout
∫
r BesselJ ( 0,λi r ) dr = q′′ 2
0
rexp
∫ r BesselJ ( 0,λ r ) dr
(12)
i
0
Integral 1
Integral 2
The integrals in Eq. (12) are carried out in Maple: > restart; > Integral1:=int(r*(BesselJ(0,lambda[i]*r))^2,r=0..r_out);
Integral1 :=
2
2
1 r_out ( π λ i r_out BesselJ( 0, λ i r_out ) + π λ i r_out BesselJ( 1, λ i r_out ) ) 2 πλ i
> Integral2:=int(r*BesselJ(0,lambda[i]*r),r=0..r_exp);
Integral2 :=
r_exp BesselJ( 1, r_exp λ i ) λi
and used to complete the solution in EES: duplicate i=1,N Integral1[i]=1/2*r_out/Pi^(1/2)/lambda[i]*(Pi^(1/2)*lambda[i]*r_out*BesselJ(0,lambda[i]*r_out)^2+& Pi^(1/2)*lambda[i]*r_out*BesselJ(1,lambda[i]*r_out)^2) Integral2[i]= 1/lambda[i]*r_exp*BesselJ(1,r_exp*lambda[i]) k*lambda[i]*C[i]*Integral1[i]=q``_dot*Integral2[i] end
The solution is obtained at an arbitrary position according to: x=0 [micron]*convert(micron,m) r=0 [micron]*convert(micron,m) duplicate i=1,N theta[i]=C[i]*BesselJ(0,lambda[i]*r)*exp(-lambda[i]*x) end theta=sum(theta[1..N])
"x-position" "r-position" "i'th term"
Temperature difference relative to Ts (K)
Figure 2 shows the temperature elevation relative to Ts as a function of radius for various values of x. 1.2 x=0 x = 5 μm x = 10 μ m x = 25 μ m x = 50 μ m
1 0.8 0.6 0.4 0.2 0 0
20
40
60
80
100
120
140
160
180
200
Radius (μm) Figure 2: Temperature difference as a function of r for various values of x.
b.) Determine the average temperature of the cylinder at the surface exposed to the heat flux. The average temperature of the cylinder over the region x = 0 and 0 < r < rexp is: 1 T = 2 π rexp
rexp
∫ 2π r T
x =0
(13)
dr
0
or: rexp
1 T = Ts + 2 π rexp
∫ 2π r θ
x =0
dr
(14)
0
Substitutinge Eq. (10) into Eq. (14) leads to: T = Ts +
1 2 π rexp
rexp
∫
∞
2 π r ∑ Ci BesselJ ( 0,λi r ) dr
(15)
i =1
0
which can be rearranged: 2 T = Ts + 2 rexp
rexp
∞
∑ C ∫ r BesselJ ( 0,λ r ) dr i =1
i
i
0
Integral 2
(16)
"compute average temperature in flux region" duplicate i=1,N theta_bar[i]=C[i]*2*Integral2[i]/r_exp^2 end theta_bar=sum(theta_bar[1..N])
c.) Define a dimensionless thermal resistance between the surface exposed to the heat flux and Ts. Plot the dimensionless thermal resistance as a function of rout/rin. A dimensionless thermal resistance is defined by normalizing the actual resistance against the reistance to axial conduction through a cylinder that is rout in radius and rout long: 2 2 R π rout k (T − Ts ) π rout k (T − Ts ) ⎛ rout R= k⎜ = = 2 rout q′′ π rexp rout q′′ rout ⎜⎝ rexp
⎞ ⎟⎟ ⎠
2
(17)
actual resistance
R_bar=theta_bar*k/(q``_dot*r_exp)
"dimensionless thermal resistance"
Figure 3 illustrates the dimensionless thermal resistance as a function of rout/rexp. Dimensionless thermal resistance
80
10
1
0.1 1
10
100
Ratio of cylinder to exposure radii, rout/rexp Figure 3: Dimensionless thermal resistance as a function of rout/rexp.
d.) Show that your plot from (c) does not change if the problem parameters (e.g., Ts, k, etc.) are changed. The values of various parameters were changed and did not affect Figure 3 (the plots are overlaid onto Figure 3).
Problem 2.4-1 (2-7 in text) The plate shown in Figure P2.4-1 is exposed to a uniform heat flux q ′′ = 1x105 W/m2 along its top surface and is adiabatic at its bottom surface. The left side of the plate is kept at TL = 300 K and the right side is at TR = 500 K. The height and width of the plate are H = 1 cm and W = 5 cm, respectively. The conductivity of the plate is k = 10 W/m-K. 5 2 q ′′ = 1x10 W/m
W = 5 cm
TL = 300 K
H = 1 cm
y
TR = 500 K
x k = 10 W/m-K Figure P2.4-1: Plate.
a.) Derive an analytical solution for the temperature distribution in the plate. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" H=1 [cm]*convert(cm,m) W=5 [cm]*convert(cm,m) k=10 [W/m-K] q``=100000 [W/m^2] T_R=500 [K] T_L=300 [K]
"height of plate" "thickness of plate" "conductivity" "heat flux" "temperature of right hand surface" "temperature of left hand surface"
A mathematical statement of the transformed problem is: ∂ 2θ ∂ 2θ + =0 ∂x 2 ∂y 2
(1)
θ x =0 = 0
(2)
θ x =W = θ R
(3)
∂θ ∂y
(4)
with boundary conditions:
=0 y =0
k
∂θ ∂y
= q ′′
(5)
y=H
where
θ = T − TL
(6)
θ R = TR − TL
(7)
and
The problem has two, non-homogeneous boundary condition and therefore must be solved using superposition.
θ = θ A + θB
(8)
Problem θA retains the non-homogeneous boundary condition in the x-direction: ∂ 2θ A ∂ 2θ A + 2 =0 ∂x 2 ∂y
(9)
θ A, x = 0 = 0
(10)
θ A, x =W = θ R
(11)
∂θ A ∂y
=0
(12)
=0
(13)
with boundary conditions:
∂θ A ∂y
y =0
y=H
By inspection, the solution for θA is 1-D in x and given by:
θ A = θR x_bar=0.5 [-] x=x_bar*W y_bar=0.5 [-] y=y_bar*H
x L
(14) "dimensionless x location" "x location" "dimensionless y location" "y location"
"sub-problem A solution" theta_R=T_R-T_L surface" theta_A=theta_R*x/W
"temperature difference of right hand "solution for sub-problem A"
Problem θB retains the non-homogeneous boundary condition in the y-direction: ∂ 2θ B ∂ 2θ B + 2 =0 ∂x 2 ∂y
(15)
θ B , x =0 = 0
(16)
θ B , x =W = 0
(17)
∂θ B ∂y
=0
(18)
= q ′′
(19)
with boundary conditions:
k
∂θ B ∂y
y =0
y=H
The solution for θB is 2-D and can be obtained using separation of variables. The eigenfunctions are:
θ X B ,i = sin ( λB ,i x )
(20)
where the eigenvalues are:
λB , i =
iW
π
for i = 1, 2,..∞
"sub-problem B solution" N_term=11 [-] duplicate i=1,N_term lambda_B[i]=i*pi/W end
(21)
"number of terms" "i'th eigenvalue"
The solution in the non-homogeneous direction is:
θ YB ,i = C2,i cosh ( λB ,i y ) The series solution for θB is:
(22)
∞
θ B = ∑ Ci sin ( λB ,i x ) cosh ( λB ,i y )
(23)
i =1
Subsituting Eq. (23) into Eq. (19) leads to: ∞
k ∑ λB ,i Ci sin ( λB ,i x ) sinh ( λB ,i H ) = q ′′
(24)
i =1
Equation (24) is multiplied by an eigenfunction and integrated from x = 0 to x = W: W
W
k λB ,i Ci sinh ( λB ,i H ) ∫ sin ( λB ,i x ) dx = q ′′ ∫ sin ( λB ,i x ) dx 2
0
0
The integrals in Eq. (25) are evaluated in Maple: > restart; > assume(i,integer); > lambda_B:=i*Pi/W;
lambda_B :=
i~ π W
> int((sin(lambda_B*x))^2,x=0..W);
W 2 > int(sin(lambda_B*x),x=0..W);
W ( −1 + ( -1 )i~ ) − i~ π
and used to evaluate the constants: duplicate i=1,N_term C[i]*k*lambda_B[i]*sinh(lambda_B[i]*H)*W/2=q``*(-W*(-1+(-1)^i)/i/Pi) "i'th constant" end
The solution for θB is obtained: duplicate i=1,N_term theta_B[i]=C[i]*sin(lambda_B[i]*x)*cosh(lambda_B[i]*y) end theta_B=sum(theta_B[1..N_term])
"i'th term" "solution to sub-problem B"
and used to obtain the solution for T: theta=theta_A+theta_B T=theta+T_L
"superposition of solutions" "temperature"
(25)
b.) Prepare a contour plot of the temperature. Figure 2 illustrates a contour plot of the temperature distribution in the plate. 1
Dimensionless y-position, y/H
300 345.5
0.8
390.9 436.4 481.8
0.6
527.3 572.7
0.4
618.2 663.6 709.1
0.2
754.5 0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Dimensionless x-position, x/W Figure 2: Contour plot of the temperature distribution.
Problem 2.5-1 (2-8 in text): A Heating Element Figure P2.5-1 illustrates an electrical heating element that is affixed to the wall of a chemical reactor. The element is rectangular in cross-section and very long (into the page). The temperature distribution within the element is therefore two-dimensional, T(x, y). The width of the element is a = 5.0 cm and the height is b = 10.0 cm. The three edges of the element that are exposed to the chemical (at x = 0, y = 0, and x = a) are maintained at a temperature Tc = 200°C while the upper edge (at y = b) is affixed to the well-insulated wall of the reactor and can therefore be considered adiabatic. The element experiences a uniform volumetric rate of thermal energy generation, g ′′′ = 1x106 W/m3. The conductivity of the material is k = 0.8 W/m-K. reactor wall k = 0.8 W/m-K 6 3 g ′′′ = 1x10 W/m
Tc = 200°C a = 5 cm y x Tc = 200°C
Tc = 200°C b = 10 cm
Figure P2.5-1: Electrical heating element.
a.) Develop a 2-D numerical model of the element using EES. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" a=5.0 [cm]*convert(cm,m) b=10.0 [cm]*convert(cm,m) k=0.8 [W/m-K] T_c=converttemp(C,K,200 [C]) g```_dot=1e6 [W/m^3] L=1 [m]
"width of element" "height of element" "conductivity" "chemical temperature" "rate of volumetric generation" "unit length of element into the page"
The computational domain of the element with the regularly spaced grid of nodes is shown in Figure 2.
Figure 2: The regularly spaced grid used to obtain a numerical solution.
The first step in obtaining a numerical solution is to position the nodes throughout the computational domain. We will use grid with nodes placed on the edges and distributed uniformly throughout. The x and y distance between adjacent nodes (Δx and Δy) are: Δx =
L ( m − 1)
(1)
Δy =
b ( n − 1)
(2)
and the x and y positions of any node i,j are given by:
xi =
( i − 1) a ( m − 1)
(3)
yj =
( j − 1) b ( n − 1)
(4)
where m and n are the number of nodes used in the x and y directions. "Setup nodes" m=11 [-] "number of nodes in the x-direction"
n=11 [-] "number of nodes in the y-direction" Dx=a/(m-1) "distance between nodes in the x-direction" Dy=b/(n-1) "distance between nodes in the y-direction" duplicate i=1,m x[i]=(i-1)*Dx end duplicate j=1,n y[j]=(j-1)*Dy end
"x-position of each node"
"y-position of each node"
The next step in the solution is to write an energy balance for each node. Figure 3 illustrates a control volume and the associated energy transfers for an internal node (see Figure 2) which include conduction from each side ( q RHS and q LHS ), the top ( qtop ), and the bottom ( qbottom ). Note that the direction associated with these energy transfers is arbitrary (i.e., they could have been taken as positive if energy leaves the control volume), but it is important to write the equation in a manner consistent with the chosen directions.
Figure 3: Energy balance for an internal node
The energy balance suggested by Figure 3 is: q RHS [i, j ] + q LHS [i, j ] + qtop [i, j ] + qbottom [i, j ] + g [i, j ] = 0
(5)
The next step is to approximate each of the terms in the energy balance; the material separating the nodes is assumed to behave as a plane wall resistance and therefore: q RHS [i, j ] =
k Δy L (T [i + 1, j ] − T [i, j ]) Δx
(6)
where L is the length of the element (assumed to be 1 m in order to do the problem on a unit length basis); therefore, Δy L is the area for conduction and Δx is the distance over which the conduction heat transfer occurs. Note that the temperature difference is consistent with the direction of the arrow in Figure 3; if Ti+1,j is greater than Ti,j then energy is leaving the node and q RHS is positive. The other heat transfers are approximated using a similar model:
k Δy L (T [i − 1, j ] − T [i, j ]) Δx
(7)
k ΔxW (T [i, j + 1] − T [i, j ]) Δy
(8)
q LHS [i, j ] = qtop [i, j ] =
qbottom [i, j ] =
k ΔxW (T [i, j − 1] − T [i, j ]) Δy
(9)
The generation is the product of the volume of the control volume and the volumetric rate of generation: g [i, j ] = g ′′′ Δx Δy L
(10)
These equations are entered in EES using a nested duplicate statement: "Internal node control volumes" duplicate i=2,(m-1) duplicate j=2,(n-1) q_dot_LHS[i,j]+q_dot_RHS[i,j]+q_dot_top[i,j]+q_dot_bottom[i,j]+gen[i,j]=0 q_dot_LHS[i,j]=k*L*Dy*(T[i-1,j]-T[i,j])/Dx q_dot_RHS[i,j]=k*L*Dy*(T[i+1,j]-T[i,j])/Dx q_dot_top[i,j]=k*L*Dx*(T[i,j+1]-T[i,j])/Dy q_dot_bottom[i,j]=k*L*Dx*(T[i,j-1]-T[i,j])/Dy gen[i,j]=Dx*Dy*L*g```_dot end end
Note that each time the outer duplicate statement iterates once (i.e., i is increased by 1), the inner duplicate statement iterates (n-1) times (i.e., j runs from 2 to n-1). Therefore, all of the internal nodes are considered with these two nested duplicate loops. Also note that the unknowns are placed in an array rather than a vector. The entries in the array T is accessed using two indices contained in square brackets. The boundary nodes have to be treated separately. The left, right, and bottom boundaries are easy as the temperature is specified:
These equations are entered in EES:
T [1, j ] = Tc for j = 1...n
(11)
T [ m, j ] = Tc for j = 1...n
(12)
T [i,1] = Tc for i = 1...m
(13)
"Temperature along right edge" duplicate j=1,n T[1,j]=T_c end "Temperature along left edge" duplicate j=1,n T[m,j]=T_c end "Temperature along bottom edge" duplicate i=2,(m-1) T[i,1]=T_c end
The upper boundary nodes must be considered using energy balances. Figure 4 illustrates an energy balance associated with a node that is located on the top, insulated boundary (see Figure 2).
Figure 4: Energy balance for a node on the top boundary
The energy balance suggested by Figure 4 is: q RHS [i, n ] + q LHS [i, n ] + qbottom [i, n ] + g [i, n ] = 0
(14)
The conduction terms in the x direction must be approximated slightly differently:
q RHS [i, n ] =
k Δy L (T [i + 1, n] − T [i, n]) 2Δx
(15)
q LHS [i, n ] =
k Δy L (T [i − 1, n] − T [i, n]) 2 Δx
(16)
The factor of 2 in the denominator appears because there is half the available area for conduction through the sides of the control volume on the top boundary. The other conduction term is approximated as before: qbottom [i, n ] =
k ΔxW (T [i, n − 1] − T [i, n]) Δy
(17)
The generation term is: g [i, n ] =
Δx Δy L g ′′′ 2
(18)
These equations are entered in EES using a single duplicate statement: "Upper edge" duplicate i=2,(m-1) q_dot_LHS[i,n]+q_dot_RHS[i,n]+q_dot_bottom[i,n]+gen[i,n]=0 q_dot_LHS[i,n]=k*(Dy/2)*L*(T[i-1,n]-T[i,n])/Dx q_dot_RHS[i,n]=k*(Dy/2)*L*(T[i+1,n]-T[i,n])/Dx q_dot_bottom[i,n]=k*Dx*L*(T[i,n-1]-T[i,n])/Dy gen[i,n]=Dx*Dy*L*g```_dot/2 end
Notice that the control volumes at the top left corner (i.e., i =1, j =n) and the top right corner (i.e., i =m, j =n) have already been taken addressed by the equations for the left and right boundaries, Eqs. (11) and (12). Therefore, we have to make sure not to write additional equations related to this node or the problem will be over-specified and so the equations for the top boundary can only be written for i = 2...(m-1). We have derived a total of m x n equations in the m x n unknown temperatures; these equations completely specify the problem and they have now all been entered in EES. Therefore, a solution can be obtained by solving the EES code. The solution is contained in the Arrays Window; each column of the table corresponds to the temperatures associated with one value of i and all of the value of j (i.e., the temperatures in a column are at a constant value of y and varying values of x). b.) Plot the temperature as a function of x at various values of y. What is the maximum temperature within the element and where is it located? The solution is obtained for n=11 and the columns Ti,1 (corresponding to y =0) through Ti,11 (corresponding to y = 10 cm) are plotted in Figure 5 as a function of x.
Figure 5: Temperature as a function of x for various values of y.
The hottest spot in the element is at the adiabatic wall (y = 10 cm) and the center (x = 2.5 cm); the hottest temperature is about 860 K. c.) Prepare a reality check to show that your solution behaves according to your physical intuition. That is, change some aspect of your program and show that the results behave as you would expect (clearly describe the change that you made and show the result). There are several possible answers to this; I increased the conductivity by a factor of 10 and examined the temperature distribution. Figure 6 illustrates the temperature as a function of x for the original conductivity (k = 0.8 W/m-K) and the increased conductivity (k = 8.0 W/m-K); notice that the increased conductivity has had the expected effect of reducing the temperature rise.
Figure 6: Temperature as a function of x for y = 10 cm and two values of conductivity.
Problem 2.6-1 (2-9 in text): Model of Welding Process (revisited) Figure P2.6-1 illustrates a cut-away view of two plates that are being welded together. Both edges of the plate are clamped and effectively held at temperatures Ts = 25°C. The top of the plate is exposed to a heat flux that varies with position x, measured from joint, according to: qm′′ ( x ) = q ′′j exp ( − x / L j ) where q ′′j =1x106 W/m2 is the maximum heat flux (at the joint, x = 0)
and Lj = 2.0 cm is a measure of the extent of the heat flux. The back side of the plates are exposed to liquid cooling by a jet of fluid at Tf = -35°C with h = 5000 W/m2-K. A halfsymmetry model of the problem is shown in Figure P2.6-1. The thickness of the plate is b = 3.5 cm and the width of a single plate is W = 8.5 cm. You may assume that the welding process is steady-state and 2-D. You may neglect convection from the top of the plate. The conductivity of the plate material is k = 38 W/m-K. both edges are held at fixed temperature heat flux joint
qm′′
impingement cooling
k = 38 W/m-K W = 8.5 cm
Ts = 25°C
b = 3.5 cm y x 2 h = 5000 W/m -K, T f = −35°C Figure P2.6-1: Welding process and half-symmetry model of the welding process.
a.) Develop a separation of variables solution to the problem (note, this was done previously in Problem 2.2-1). Implement the solution in EES and prepare a plot of the temperature as a function of x at y = 0, 1.0, 2.0, 3.0, and 3.5 cm. b.) Prepare a contour plot of the temperature distribution. See the solution for Problem 2.2-1 for parts (a) and (b). c.) Develop a numerical model of the problem. Implement the solution in MATLAB and prepare a contour or surface plot of the temperature in the plate. The input parameters are entered in the MATLAB function P2p2_1; the input arguments are m and n, the number of nodes in the x and y coordinates while the output arguments are the x and y positions of each node and the predicted temperature at each node. function[xm,ym,T]=P2p6_1(m,n) W=0.085;
%width of plate (m)
b=0.035; k=38; T_s=298.1; T_f=238.2; h=5000; L=1;
%thickness of plate (m) %conductivity of plate material (W/m-K) %side temperature (K) %fluid temperature (K) %heat transfer coefficient (W/m^2-K) %per unit length (m)
end
A sub-function is defined to provide the heat flux on the upper surface: function[qflux]=qf(x) L_j=0.02; qf_j=1e6;
%length scale (m) %heat flux at center (W/m^2)
qflux=qf_j*exp(-x/L_j); end
A 2-D numerical model will be generated using a grid in which the x and y coordinates of each node are:
xi =
( i − 1)W ( m − 1)
for i = 1..m
(1)
yi =
( j − 1)W ( m − 1)
for j = 1..n
(2)
The distance between adjacent nodes is:
%Setup grid for i=1:m x(i,1)=(i-1)*W/(m-1); end Dx=W/(m-1); for j=1:n y(j,1)=(j-1)*W/(n-1); end Dy=b/(n-1);
Δx =
W ( m − 1)
(3)
Δy =
W ( n − 1)
(4)
The problem will be solved by setting up and inverting a matrix containing the algebraic equations that enforce the conservation of energy for each control volume. A control volume for an internal node includes conduction from the left and right sides ( q LHS and q RHS ) and top and bottom ( qtop and qbottom ). The energy balance is:
q RHS + q LHS + qtop + qbottom = 0 The conduction terms are approximated according to: q RHS =
k Δy L (Ti−1, j − Ti, j ) Δx
q LHS =
k Δy L (Ti+1, j − Ti, j ) Δx
qbottom =
k Δx L (Ti, j −1 − Ti, j ) Δy
qtop =
k Δx L (Ti, j +1 − Ti, j ) Δy
where L is the depth of the plate (into the page); L is set to 1.0 m which is consistent with doing the problem on a per unit length basis. Combining these equations leads to: k Δy L k Δy L k Δx L k Δx L Ti −1, j − Ti , j ) + Ti +1, j − Ti , j ) + Ti , j −1 − Ti , j ) + ( ( ( (Ti, j +1 − Ti, j ) = 0 Δx Δx Δy Δy
for i = 2.. ( m − 1) and j = 2.. ( n − 1)
(5)
The equation is rearranged to make it clear what the coefficient is for each unknown temperature: ⎡ k Δy L ⎡ k Δx L ⎤ ⎡ k Δx L ⎤ k Δx L ⎤ ⎡ k Δy L ⎤ ⎡ k Δy L ⎤ Ti , j ⎢ −2 −2 + Ti −1, j ⎢ + Ti +1, j ⎢ + Ti , j −1 ⎢ + Ti , j +1 ⎢ ⎥ ⎥ ⎥=0 ⎥ ⎥ Δx Δy ⎦ ⎣ Δx ⎦ ⎣ Δx ⎦ ⎣ ⎣ Δy ⎦ ⎣ Δy ⎦ for i = 2.. ( m − 1) and j = 2.. ( n − 1) (6) The control volume equations must be placed into the matrix equation: AX =b
where the equation for control volume i,j is placed into row m(j-1)+i of A and Ti,j corresponds to element Xm (j-1)+i in the vector X . Therefore, each coefficient in Eq. (6) (i.e., each term multiplying an unknown temperature on the left side of the equation) must be placed in the row of A corresponding to the control volume being examined and the column of A corresponding to the unknown in X . The matrix assignments consistent with Eq. (6) are: k Δy L k Δx L −2 for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δx Δy
(7)
Am( j −1)+i ,m( j −1)+i −1 =
k Δy L for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δx
(8)
Am( j −1)+i ,m( j −1)+i +1 =
k Δy L for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δx
(9)
Am( j −1)+i ,m( j −1−1)+i =
k Δx L for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δy
(10)
Am( j −1)+i ,m( j +1−1)+i =
k Δx L for i = 2.. ( m − 1) and j = 2.. ( n − 1) Δy
(11)
Am( j −1)+i ,m( j −1)+i = −2
A sparse matrix is allocated in MATLAB for A and the equations derived above are implemented using a nested for loop. The spalloc command requires three arguments, which are the number of rows and columns and the maximum number of elements in the matrix. Note that there are at most 5 non-zero entries in each row of A , corresponding to Eqs. (7) through (11); thus the last argument in the spalloc command which corresponds to the maximum number of non-zero entries in the sparse matrix. A=spalloc(m*n,m*n,5*m*n); %allocate a sparse matrix for A bm=zeros(m*n,1); %allocate a matrix for b %energy balances for internal nodes for i=2:(m-1) for j=2:(n-1) A(m*(j-1)+i,m*(j-1)+i)=-2*k*Dy*L/Dx-2*k*Dx*L/Dy; A(m*(j-1)+i,m*(j-1)+i-1)=k*Dy*L/Dx; A(m*(j-1)+i,m*(j-1)+i+1)=k*Dy*L/Dx; A(m*(j-1)+i,m*(j-1-1)+i)=k*Dx*L/Dy; A(m*(j-1)+i,m*(j+1-1)+i)=k*Dx*L/Dy; end end
The nodes on the right side have a specified temperature: Tm, j = Ts for j = 1..n
The matrix assignments suggested by these equations are: Am( j −1)+ m, m( j −1)+ m = 1 for j = 1..n bm( j −1)+ m = Tst for j = 1..n
These assignments are implemented in MATLAB: %right side temperature is specified for j=1:n A(m*(j-1)+m,m*(j-1)+m)=1; bm(m*(j-1)+m,1)=T_s; end
The nodes along the upper edge must be considered separately, leading to: ⎡ k Δy L ⎤ ⎡ k Δy L ⎤ ⎡ k Δy L k Δx L ⎤ ⎡ k Δx L ⎤ Ti ,n ⎢ − − + Ti −1, n ⎢ + Ti +1, n ⎢ + Ti ,n −1 ⎢ ⎥ ⎥ ⎥ ⎥ = − qm′′ Δx L Δx Δy ⎦ ⎣ ⎣ Δy ⎦ ⎣ 2 Δx ⎦ ⎣ 2 Δx ⎦ for i = 2.. ( m − 1)
(12)
which is expressed in matrix form as: k Δy L k Δx L − for i = 2.. ( m − 1) Δx Δy
(13)
Am( n −1)+i ,m( n −1)+i −1 =
k Δy L for i = 2.. ( m − 1) 2 Δx
(14)
Am( n −1)+i ,m( n −1)+i +1 =
k Δy L for i = 2.. ( m − 1) 2 Δx
(15)
Am( n −1)+i , m( n −1−1)+i =
k Δx L for i = 2.. ( m − 1) Δy
(16)
Am( n −1)+i ,m( n −1)+i = −
bm( n −1)+i ,1 = −qm′′ Δx L for i = 2.. ( m − 1) %upper edge for i=2:(m-1) A(m*(n-1)+i,m*(n-1)+i)=-k*Dy*L/Dx-k*Dx*L/Dy; A(m*(n-1)+i,m*(n-1)+i-1)=k*Dy*L/(2*Dx); A(m*(n-1)+i,m*(n-1)+i+1)=k*Dy*L/(2*Dx); A(m*(n-1)+i,m*(n-1-1)+i)=k*Dx*L/Dy; bm(m*(n-1)+i,1)=-qf(x(i))*Dx*L; end
(17)
The node at the upper left corner must be considered separately, leading to: ⎡ k Δy L k Δx L ⎤ ⎡ k Δy L ⎤ ⎡ k Δx L ⎤ Δx L − + T2,n ⎢ + T1,n −1 ⎢ = − qm′′ T1,n ⎢ − ⎥ ⎥ ⎥ 2 Δy ⎦ 2 ⎣ 2 Δx ⎣ 2 Δx ⎦ ⎣ 2 Δy ⎦
(18)
which is expressed in matrix form as: k Δy L k Δx L − 2 Δx 2 Δy
(19)
Am( n −1) +1,m( n −1)+1+1 =
k Δy L 2 Δx
(20)
Am( n −1)+1, m( n −1−1)+1 =
k Δx L 2 Δy
(21)
Δx L 2
(22)
Am( n −1)+1,m( n −1)+1 = −
bm( n −1)+1,1 = −qm′′
%upper left corner A(m*(n-1)+1,m*(n-1)+1)=-k*Dy*L/(2*Dx)-k*Dx*L/(2*Dy); A(m*(n-1)+1,m*(n-1)+1+1)=k*Dy*L/(2*Dx); A(m*(n-1)+1,m*(n-1-1)+1)=k*Dx*L/(2*Dy); bm(m*(n-1)+1,1)=-qf(x(1))*Dx*L/2;
The node along the left edge must be considered separately, leading to: ⎡ k Δy L ⎤ ⎡ k Δx L ⎤ ⎡ k Δx L ⎤ ⎡ k Δy L k Δx L ⎤ T1, j ⎢ − − + T2 +1, j ⎢ + T1, j −1 ⎢ + T1, j +1 ⎢ ⎥ ⎥ ⎥=0 ⎥ Δx Δy ⎦ ⎣ ⎣ 2 Δx ⎦ ⎣ 2 Δy ⎦ ⎣ 2 Δy ⎦ for j = 2.. ( n − 1)
(23)
which is expressed in matrix form as: k Δy L k Δx L − for j = 2.. ( n − 1) Δx Δy
(24)
Am( j −1)+1,m( j −1)+1+1 =
k Δy L for j = 2.. ( n − 1) Δx
(25)
Am( j −1) +1,m( j −1−1)+1 =
k Δx L for j = 2.. ( n − 1) 2 Δy
(26)
Am( j −1)+1,m( j −1)+1 = −
Am( j −1)+1,m( j +1−1)+1 =
k Δx L for j = 2.. ( n − 1) 2 Δy
(27)
%left edge for j=2:(n-1) A(m*(j-1)+1,m*(j-1)+1)=-k*Dy*L/Dx-k*Dx*L/Dy; A(m*(j-1)+1,m*(j-1)+1+1)=k*Dy*L/Dx; A(m*(j-1)+1,m*(j-1-1)+1)=k*Dx*L/(2*Dy); A(m*(j-1)+1,m*(j+1-1)+1)=k*Dx*L/(2*Dy); end
The node at the lower left corner must be considered separately, leading to: ⎡ k Δy L k Δ x L ⎡ k Δy L ⎤ ⎡ k Δx L ⎤ Δx L ⎤ Δx L T1,1 ⎢ − Tf − −h + T2,1 ⎢ + T1,2 ⎢ = −h ⎥ ⎥ ⎥ 2 Δy 2 ⎦ 2 ⎣ 2 Δx ⎣ 2 Δx ⎦ ⎣ 2 Δy ⎦
(28)
which is expressed in matrix form as: k Δy L k Δx L Δx L − −h 2 Δx 2 Δy 2
(29)
Am(1−1)+1, m(1−1)+1+1 =
k Δy L 2 Δx
(30)
Am(1−1)+1, m(1+1−1)+1 =
k Δx L 2 Δy
(31)
Am(1−1) +1,m(1−1)+1 = −
bm(1−1)+1,1 = −h
Δx L Tf 2
(32)
%lower left corner A(m*(1-1)+1,m*(1-1)+1)=-k*Dy*L/(2*Dx)-k*Dx*L/(2*Dy)-h*L*Dx/2; A(m*(1-1)+1,m*(1-1)+1+1)=k*Dy*L/(2*Dx); A(m*(1-1)+1,m*(1+1-1)+1)=k*Dx*L/(2*Dy); bm(m*(1-1)+1,1)=-h*L*Dx*T_f/2;
The nodes along the bottom edge must be considered separately, leading to: ⎡ k Δy L ⎤ ⎡ k Δy L ⎤ ⎡ k Δy L k Δx L ⎤ ⎡ k Δx L ⎤ Ti ,1 ⎢ − − − h Δx L ⎥ + Ti −1,1 ⎢ + Ti +1,1 ⎢ + Ti ,1+1 ⎢ ⎥ ⎥ ⎥ = − h Δx LT f Δx Δy (33) ⎣ ⎦ ⎣ Δy ⎦ ⎣ 2 Δx ⎦ ⎣ 2 Δx ⎦ for i = 2.. ( m − 1) which is expressed in matrix form as:
k Δy L k Δx L − - h Δx L for i = 2.. ( m − 1) Δx Δy
(34)
Am(1−1) +i , m(1−1)+i −1 =
k Δy L for i = 2.. ( m − 1) 2 Δx
(35)
Am(1−1) +i , m(1−1)+i +1 =
k Δy L for i = 2.. ( m − 1) 2 Δx
(36)
Am(1−1) +i , m(1+1−1)+i =
k Δx L for i = 2.. ( m − 1) Δy
(37)
Am(1−1)+i ,m(1−1)+i = −
bm(1−1)+i ,1 = − h Δx LT f for i = 2.. ( m − 1)
(38)
%lower edge for i=2:(m-1) A(m*(1-1)+i,m*(1-1)+i)=-k*Dy*L/Dx-k*Dx*L/Dy-h*Dx*L; A(m*(1-1)+i,m*(1-1)+i-1)=k*Dy*L/(2*Dx); A(m*(1-1)+i,m*(1-1)+i+1)=k*Dy*L/(2*Dx); A(m*(1-1)+i,m*(1+1-1)+i)=k*Dx*L/Dy; bm(m*(1-1)+i,1)=-h*Dx*L*T_f; end
The vector of unknowns X is obtained through matrix manipulation and then placed into matrix format. Matrices for the x and y positions of each node are also created. X=A\bm; for i=1:m for j=1:n xm(i,j)=x(i); ym(i,j)=y(j); T(i,j)=X(m*(j-1)+i); end end
A surface plot of the result is obtained by typing: >> [x,y,T]=P2p6_1(20,40); >> surf(x,y,T); The plot is shown in Figure 3.
Figure 3: Surface plot of temperature in the plate.
b.) Plot the temperature as a function of x at y = 0, b/2, and b and overlay on this plot the separation of variables solution obtained in part (a) evaluated at the same locations. The comparison is shown in Figure 4.
Figure 4: Temperature as a function of axial position for y = 0, b/2, and b predicted using the separation of variables solution and the numerical solution.
Problem 2.7-1 (2-10 in text): A Double Paned Window Figure P2.7-1(a) illustrates a double paned window. The window consists of two panes of glass each of which is tg = 0.95 cm thick and W = 4 ft wide by H = 5 ft high. The glass panes are separated by an air gap of g = 1.9 cm. You may assume that the air is stagnant with ka = 0.025 W/m-K. The glass has conductivity kg = 1.4 W/m-K. The heat transfer coefficient between the inner surface of the inner pane and the indoor air is hin = 10 W/m2-K and the heat transfer coefficient between the outer surface of the outer pane and the outdoor air is hout = 25 W/m2-K. You keep your house heated to Tin = 70°F. width of window, W = 4 ft tg = 0.95cm tg = 0.95 cm g = 1.9 cm
Tin = 70°F 2 hin = 10 W/m -K H = 5 ft
Tout = 23°F 2 hout = 25 W/m -K ka = 0.025 W/m-K kg = 1.4 W/m-K
casing shown in P2.10(b)
Figure P2.7-1(a): Double paned window.
The average heating season in Madison lasts about time = 130 days and the average outdoor temperature during this time is Tout = 23°F. You heat with natural gas and pay, on average, ec = 1.415 $/therm (a therm is an energy unit =1.055x108 J). a.) Calculate the average rate of heat transfer through the double paned window during the heating season. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" tg=0.375*convert(inch,m) g=0.75*convert(inch,m) k_g=1.4 [W/m-K] k_a=0.025 [W/m-K] H=5 [ft]*convert(ft,m) W=4[ft]*convert(ft,m) T_in=converttemp(F,K,70 [F]) h_in=10 [W/m^2-K] T_out=converttemp(F,K,23 [F]) h_out=25 [W/m^2-K] time=130 [day]*convert(day,s) ec=1.415 [$/therm]*convert($/therm,$/J)
"glass thickness" "air gap" "glass conductivity" "air conductivity" "height of window" "width of window" "indoor air temperature" "heat transfer coefficient on inside of window" "outdoor air temperature" "heat transfer coefficient on outside of window" "heating season duration" "cost of energy"
The heat transfer is resisted by convection on the inner and outer surfaces:
Rconv ,in =
1 hin W H
(1)
Rconv ,out =
1 hout W H
(2)
and conduction through the glass panes and the air:
R_conv_in=1/(h_in*W*H) R_cond_g=tg/(k_g*W*H) R_cond_a=g/(k_a*W*H) R_conv_out=1/(h_out*W*H)
Rcond , g =
tg kg W H
(3)
Rcond ,a =
g ka W H
(4)
"convection resistance on inside of window" "conduction resistance of glass pane" "conduction resistance of air gap" "convection resistance on outside of window"
The heat transfer rate through the window is: q =
Tin − Tout Rconv ,in + 2 Rcond , g + Rcond ,a + Rconv ,out
(5)
q_dot=(T_in-T_out)/(R_conv_in+2*R_cond_g+R_cond_a+R_conv_out) "rate of heat transfer through the window"
which leads to q = 53.0 W. b.) How much does the energy lost through the window cost during a single heating season? The total amount of energy lost over the course of a heating season is: Q = q time
(6)
cost = ec Q
(7)
and the associated cost is:
Q=q_dot*time cost=ec*Q
which leads to cost = $7.98/heating season.
"total energy loss" "cost to heat house per window"
There is a metal casing that holds the panes of glass and connects them to the surrounding wall, as shown in Figure P2.7-1(b). Because the metal casing is high conductivity, it seems likely that you could lose a substantial amount of heat by conduction through the casing (potentially negating the advantage of using a double paned window). The geometry of the casing is shown in Figure P2.7-1(b); note that the casing is symmetric about the center of the window. glass panes 1.9 cm Tin = 70°F 2 hin = 10 W/m -K
0.95 cm air
Tout = 23°F 2 hout = 25 W/m -K
2 cm 4 cm
metal casing km = 25 W/m-K
0.5 cm
3 cm 0.4 cm
wood
Figure P2-10(b) Metal casing.
All surfaces of the casing that are adjacent to glass, wood, or the air between the glass panes can be assumed to be adiabatic. The other surfaces are exposed to either the indoor or outdoor air. c.) Prepare a 2-D thermal analysis of the casing using FEHT. Turn in a print out of your geometry as well as a contour plot of the temperature distribution. What is the rate of energy lost via conduction through the casing per unit length (W/m)? The geometry from Figure 3 is entered approximately, as shown in Figure 4.
Figure 4: Approximate geometry.
Each of the points are selected individually and their exact coordinates are entered, which leads to the more precise geometry shown in Figure 4.
Figure 5: Geometry.
The material properties are specified by selecting the outline and selecting Material Properties from the Specify menu. The boundary conditions are specified by selecting each type of boundary and then selecting Boundary Conditions from the Specify menu. A crude mesh is generated, as shown in Figure 6.
Figure 6: Crude Mesh.
The temperature contours are shown in Figure 7.
Figure 7: Temperature Contours.
The heat transfer per unit length is obtained by selecting Heat Flow from the View menu and selecting all of the boundaries that are exposed to either the indoor or outdoor air. The heat flow is 10.0 W/m. d.) Show that your numerical model has converged by recording the rate of heat transfer per length for several values of the number of nodes. The mesh is refined several times and each time the heat transfer rate per unit length of casing is recorded; the results are shown in Figure 8.
Figure 8: Heat transfer through the casing per unit length as a function of the number of nodes.
e.) How much does the casing add to the cost of heating your house?
′ ) is entered in EES: The result from FEHT, the heat transfer per unit length ( qcasing q`_dot_casing=10.0 [W/m]
"casing heat transfer per unit meter"
The total heat transfer through the casing is: ′ qcasing = qcasing 2 (W + H )
(8)
The total heat lost through the casing is: Qcasing = qcasing time
and the associated cost is:
(9)
costcasing = ec Qcasing q_dot_casing=q`_dot_casing*2*(W+H) Q_casing=q_dot_casing*time cost_casing=ec*Q_casing "cost to heat house per window due to casing"
(10) "heat transfer rate through casing" "total energy flow through casing"
which leads to an additional cost of $8.27 per heating season.
P2.7-3 (2-11 in text): A Spacecraft Radiator A radiator panel extends from a spacecraft; both surfaces of the radiator are exposed to space (for the purposes of this problem it is acceptable to assume that space is at 0 K); the emittance of the surface is ε = 1.0. The plate is made of aluminum (k = 200 W/m-K and ρ = 2700 kg/m3) and has a fluid line attached to it, as shown in Figure 2.7-3(a). The half-width of the plate is a=0.5 m wide while the height of the plate is b=0.75m. The thickness of the plate is a design variable and will be varied in this analysis; begin by assuming that the thickness is th = 1.0 cm. The fluid lines carry coolant at Tc = 320 K. Assume that the fluid temperature is constant although the fluid temperature will actually decrease as it transfers heat to the radiator. The combination of convection and conduction through the panel-to-fluid line mounting leads to an effective heat transfer coefficient of h = 1,000 W/m2-K over the 3.0 cm strip occupied by the fluid line. k = 200 W/m-K ρ = 2700 kg/m3 ε = 1.0
space at 0 K
a = 0.5 m
3 cm th = 1 cm
b = 0.75 m
fluid at Tc = 320 K
half-symmetry model of panel, Figure P2-11(b)
Figure 2.7-3(a): Radiator panel
The radiator panel is symmetric about its half-width and the critical dimensions that are required to develop a half-symmetry model of the radiator are shown in Figure 2.7-3(b). There are three regions associated with the problem that must be defined separately so that the surface conditions can be set differently. Regions 1 and 3 are exposed to space on both sides while Region 2 is exposed to the coolant fluid one side and space on the other; for the purposes of this problem, the effect of radiation to space on the back side of Region 2 is neglected. Region 1 (both sides exposed to space) Region 2 (exposed to fluid - neglect radiation to space) Region 3 (both sides exposed to space) (0.50,0.75) (0.50,0.55) (0.50,0.52)
y x (0.50,0)
(0,0) (0.22,0)
(0.25,0)
line of symmetry
Figure 2.7-3(b): Half-symmetry model.
a.) Prepare a FEHT model that can predict the temperature distribution over the radiator panel.
The radiator panel can be modeled as 2-D problem because it is thin and has high conductivity. The Biot number compares the ratio of the internal conduction resistance to the external resistance; in this case due to radiation.
Rcond Rrad
(1)
th 2 k As
(2)
1 2 As σ ε s (T + Tsur )(Ts + Tsur )
(3)
Bi = where
Rcond = Using the concept of a radiation resistance: Rrad =
2 s
The Biot number is therefore: Bi =
2 th σ ε s (Ts2 + Tsur )(Ts + Tsur )
2k
(4)
The surrounding temperature in Eq. (4), Tsur corresponds to space which is essentially 0 K and therefore Eq. (4) reduces to: Bi =
th σ ε s Ts3 2k
(5)
We don’t know the value of the panel surface temperature, Ts in Eq. (5), but we can assume that it will be near the fluid temperature for any well-designed radiator. Using Ts = Tf in Eq. (5) results in a Biot number of 4.6x10-5 which is very small; clearly for any reasonable value of Ts the problem will be 2-D. FEHT can simulate 2-D problems of this type using the Extended Surface mode. Start FEHT and select Extended Surface from the Subject menu. Each of the three regions must be drawn separately using the Outline option from the Draw menu. It is easiest to set an appropriate grid using the Scale and Size selection from the Setup menu (Figure 2).
Figure 2: Scale and Size dialog window
Generate outlines for the 3 regions that are close to the correct scale and then double-click on each node and position it exactly (Figure 3).
Figure 3: Geometry definition.
The material properties for all three regions can be set by selecting each while holding down the shift key and then selecting Specify Properties from the Specify menu. Select Aluminum from the list of materials and then modify the conductivity to be 200 W/m-K and the thickness to be 0.01 m to match the problem statement (Figure 4).
Figure 4: Specify material properties.
The surface conditions for Regions 1 and 3 can be set by selecting these regions with the shift key held down and then selecting Surface Conditions from the Specify menu. These regions are radiating to space but there is no option listed in the Extended Surface Conditions for a radiative surface condition. The radiation heat flux is:
′′ = ε σ T 4 qrad
(6)
Equation (6) can be rewritten according to: ′′ = ε σ T 3 (T − 0 ) qrad
(7)
Equation (7) is similar to a convection equation:
′′ = hrad (T − T f ) qrad
(8)
where fluid temperature (Tf) is 0 K and the convection coefficient (hrad) is a function of temperature:
hrad = ε σ T 3
(9)
FEHT allows the specification of the surface conditions in terms of position (X and Y) as well as temperature (T); therefore, the radiative surface condition can be modeled as shown in Figure 5.
Figure 5: Extended Surface Conditions Dialog window for Regions 1 and 3.
Region 2 experiences a convection boundary condition with the coolant, but only from one side (as opposed to the double sided condition assumed by FEHT). The heat transfer from a differential element in Region 2 is given by: q = h dA (T − Tc )
(10)
whereas the convection surface condition in FEHT assumes convection from both sides of the plate and therefore: q = hF 2 dA (T − Tc )
(11)
where hF is the heat transfer coefficient that should be set in FEHT in order to simulate the single-sided convection coefficient represented by Eq. (10). Comparing Eqs. (10) and (11) leads to:
hF =
h 2
(12)
Click on Region 2 and select Surface Condition from the Specify menu; specify the surface conditions as shown in Figure 6.
Figure 6: Extended Surface dialog window for Region 2.
Finally, the boundary conditions along each edge of the computational domain must be specified; the line of symmetry is adiabatic and the remaining edges are also assumed to be adiabatic. Generate a reasonable but crude mesh, as shown in Figure 7(a) and then refine it. Note that the mesh within Region 2 will start relatively refined due to its small width and need not be refined as much as the mesh in Regions 1 and 3. However, it is possible to refine the mesh in a single
region by selecting that region and then selecting Reduce Mesh from the Draw menu. The result should be similar to Figure 7(b).
(a)
(b) Figure 7: (a) coarse and (b) refined mesh.
Solve the problem by selecting Calculate from the Run menu; the problem is non-linear due to the temperature dependent heat transfer coefficient and therefore the solution process will be iterative. Plot the temperature distribution by selecting Temperature Contours from the View menu (Figure 8).
Figure 8: Temperature distribution for a 1 cm thick plate.
b.) Export the solution to EES and calculate the total heat transferred from the radiator and the radiator efficiency (defined as the ratio of the radiator heat transfer to the heat transfer from the radiator if it were isothermal and at the coolant temperature). Select Tabular Output from the View menu and then Select All and Save As. Save the data as a file called ‘1 cm’. Open EES and select Open Lookup Table from the Tables menu; navigate the file ‘1 cm’ and open it; the solution (the temperature at each node together with the locations of the nodes) is contained in the lookup table. Using the technique discussed in EXAMPLE 16-1 it is possible to use the Interpolate2D function to obtain the temperature at an arbitrary x and y location on the radiator plate. $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "INPUTS" e=1.0 [-] a=0.5 [m] b=0.75 [m] T_f=320 [K] th=0.01 [m] rho=2700 [kg/m^3] k=200 [W/m-K] "location on plate" x=0.1 [m] y=0.5 [m] "2-D interpolation from table of nodal data"
"emittance of panel surface" "half-width of panel" "height of panel" "fluid temperature" "thickness" "density" "conductivity"
T=Interpolate2D('1 cm',X,Y,T,X=x,Y=y)
"temperature interpolated from data"
In order to obtain the total heat transferred from the panel it is necessary to integrate the heat flux over the entire area of the plate. This process can be accomplished manually by dividing the plate into many small integration areas, calculating the heat flux within each area, and summing the result. The plate is divided into Nx segments in the x direction, each with width: Δx =
a Nx
(13)
The x position of each segment is: xi = ( i − 0.5 ) Δx
(14)
The plate is divided into Ny segments in the y direction, each with height: Δy =
b Ny
(15)
The y position of each segment is: y j = ( j − 0.5 ) Δy
(16)
The corresponding EES code is: "Double integration manually" Nx=25 [-] "number of integration areas in x" Ny=10 [-] "number of integration areas in y" "size of an integration area" Dx=a/Nx Dy=b/Ny "setup position of x and y areas" duplicate i=1,Nx x[i]=(i-0.5)*Dx end duplicate j=1,Ny y[j]=(j-0.5)*Dy end
The total heat transferred from the plate ( q ) is calculated using the double integral: b a
q = ∫ ∫ q ′′x , y dx dy 0 0
where the heat flux is:
(17)
q ′′x , y = ε σ Tx4, y
(18)
The numerical summation that approximates the integration in Eq. (17) is: N y Nx
q = ∑∑ q ′′xi , y j Δx Δy
(19)
q ′′xi , y j = ε σ Tx4i , y j
(20)
j =1 i =1
where
The corresponding EES code is: duplicate i=1,Nx duplicate j=1,Ny "2-D interpolation from table of nodal data" T[i,j]=Interpolate2D('1 cm',X,Y,T,X=x[i],Y=y[j]) "temperature interpolated from data" q``_dot[i,j]=2*e*sigma#*T[i,j]^4 "heat flux" end end q_dot=sum(q``_dot[1..Nx,1..Ny])*Dx*Dy
Note that the same process can be accomplished using EES’ native Integral command. The Integral command carries out numerical integration using a more sophisticated algorithm than simply assuming a constant value over each step. EES’ Integral command requires 4 arguments and allows a 5th, optional argument; the protocol for calling the function is: F=Integral(Integrand,VarName,LowerLimit,UpperLimit,StepSize) where Integrand is the EES variable or expression that must be integrated, VarName is the integration variable, and LowerLimit and UpperLimit define the limits of integration. StepSize is optional and defines the numerical step that is used to accomplish the integration; a small value of StepSize will lead to more accurate results but take longer to calculate. The double integral in Eq. (17) can be accomplished by calling the Integral function twice; Eq. (17) is rewritten as: b
q = ∫ q′y dy 0
where
(21)
a
q ′y = ∫ q ′′x , y dx
(22)
0
Equation (22) is evaluated using the Integral command as shown in the EES code below: "Double integration using EES' Integral command" "2-D interpolation from table of nodal data" T=Interpolate2D('1 cm',X,Y,T,X=x,Y=y) q``_dot=2*e*sigma#*T^4 y=0 q`_dot=INTEGRAL(q``_dot,x,0,a,0.02)
"temperature interpolated from data" "heat flux" "heat transfer per unit length"
Note that the value of y was set as q ′y is a function of y but the value of x is not set as x is the integration variable. To evaluate Eq. (21), integrate q ′y from y=0 to y=b: "Double integration using EES' Integral command" "2-D interpolation from table of nodal data" T=Interpolate2D('1 cm',X,Y,T,X=x,Y=y) q``_dot=2*e*sigma#*T^4 q`_dot=INTEGRAL(q``_dot,x,0,a,0.02) q_dot=INTEGRAL(q`_dot,y,0,b,0.02)
"temperature interpolated from data" "heat flux" "heat transfer per unit length" "total heat transfer rate"
Note that the value of y is no longer set as y is the integration variable in the 2nd integral. The results of the manual integration should agree with the use of the Integral function; either should yield 339 W. The radiator efficiency (η) and mass (M) are computed according to:
η=
q 2 a b ε σ T f4
(23)
and
M = a b th ρ
(24)
c.) Explore the effect of thickness on the radiator efficiency and mass. The solution procedure described above is repeated for several values of the radiator thickness. Figure 10 shows the predicted temperature distribution for several values of the thickness. Note that the same contour levels were selected for each plot by selecting User in the Temperature Contour Information dialog window and specifying the range from 200 K to 320 K (Figure 9); this allows the different cases to be compared directly and shows clearly that the reduced thickness increases the resistance to conduction along the plate and therefore leads to progressively larger temperature gradients through the plate.
Figure 9: Temperature Contour Information dialog window.
Figure 10: Temperature distribution in the panel for thicknesses of (a) 1.0 cm, (b) 0.75 cm, (c) 0.5 cm, (d) 0.3 cm, (e) 0.2 cm, and (f) 0.1 cm.
Figure 11 illustrates the radiator efficiency and mass as a function of the thickness. As the thickness is reduced, the efficiency drops but so does the mass; clearly there must be a trade-off between these effects.
Figure 11: Efficiency and mass as a function of thickness.
Figure 12 shows the efficiency as a function of mass and makes the trade-off clearer; above a panel with a mass of nominally 2 kg there is a region of diminishing return where additional mass provides only a small gain in efficiency.
Figure 12: Efficiency as a function of mass.
A complete analysis would require more information then is given; specifically, how should the radiator performance be compared with its mass to determine a true optimal thickness. Barring additional constraints related to, for example, structural stability, Figure 12 suggests that any optimization process will result in a panel that is approximately 0.2 cm thick with a mass of 2 kg and an efficiency of 60%.
Problem 2.8-1 (2-12 in text): Cryogenic Thermal Switch There are several cryogenic systems that require a “thermal switch”, a device that can be used to control the thermal resistance between two objects. One class of thermal switch is activated mechanically and an attractive method of providing mechanical actuation at cryogenic temperatures is with a piezoelectric stack; unfortunately, the displacement provided by a piezoelectric stack is very small, typically on the order of 10 microns. A company has proposed an innovative design for a thermal switch, shown in Figure P2.81(a). Two blocks are composed of th = 10 μm laminations that are alternately copper (kCu = 400 W/m-K) and plastic (kp = 0.5 W/m-K). The thickness of each block is L = 2.0 cm in the direction of the heat flow. One edge of each block is carefully polished and these edges are pressed together; the contact resistance associated with this joint is Rc′′ = 5x10-4 K-m2/W. th = 10 μm plastic laminations kp = 0.5 W/m-K L = 2 cm L = 2 cm TH
direction of actuation
TC Figure P2.8-1(b) “on” position
“off” position -4 2 contact resistance, Rc′′ = 5x10 m -K/W
th = 10 μm copper laminations kCu = 400 W/m-K
Figure P2.8-1(a): Thermal switch in the “on” and “off” positions.
Figure P2.8-1(a) shows the orientation of the two blocks when the switch is in the “on” position; notice that the copper laminations are aligned with one another in this configuration which provides a continuous path for heat through high conductivity copper (with the exception of the contact resistance at the interface). The vertical location of the right-hand block is shifted by 10 μm to turn the switch "off". In the “off” position, the copper laminations are aligned with the plastic laminations; therefore, the heat transfer is inhibited by low conductivity plastic. Figure P2.8-1(b) illustrates a closer view of half (in the vertical direction) of two adjacent laminations in the “on” and “off” configurations. Note that the repeating nature of the geometry means that it is sufficient to analyze a single lamination set and assume that the upper and lower boundaries are adiabatic.
L = 2 cm
L = 2 cm TC
TH th/2 = 5 μm th/2 = 5 μm
kp = 0.5 W/m-K kCu = 400 W/m-K
Rc′′ = 5x10 m -K/W -4
2
“on” position TC
TH “off” position
Figure P2.8-1(b): A single set consisting of half of two adjacent laminations in the “on” and "off” positions.
The key parameter that characterizes a thermal switch is the resistance ratio (RR) which is defined as the ratio of the resistance of the switch in the “off” position to its resistance in the “on” position. The company claims that they can achieve a resistance ratio of more than 100 for this switch. a) Estimate upper and lower bounds for the resistance ratio for the proposed thermal switch using 1-D conduction network approximations. Be sure to draw and clearly label the resistance networks that are used to provide the estimates. Use your results to assess the company’s claim of a resistance ratio of 100. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $Tabstops 0.2 0.4 0.6 3.5 in "Inputs" th = 10 [micron]*convert(micron,m) k_Cu=400 [W/m-K] k_p=0.5 [W/m-K] L = 2.0 [cm]*convert(cm,m) R``_c=5e-4 [K-m^2/W] W=1 [m]
"thickness of laminations" "conductivity of copper laminations" "conductivity of plastic laminations" "lengt of laminations" "area specific contact resistance of interface" "unit depth into page"
The isothermal and adiabatic models provide two limiting cases. Both result in a 1-D problem that can be represented using a resistance network. The “adiabatic” approximation does not allow heat transfer in the y-direction (i.e., perpendicular to the laminations). The resistance network for the adiabatic approximation is shown in Figure 3(a) for the “on” position and in Figure 3(b) for the “off” position.
(a) (b) Figure 3: Resistance network for the “adiabatic” approximation in the (a) “on” and (b) “off” positions.
Note that the parameter Alam in Figure 3 is the cross-sectional area associated a single lamination: Alam = W th where W is the depth into the page (assumed to be 1 m). The resistance associated with the adiabatic approximation in the “on” state, Rad,on from Figure 3(a) is:
Rad ,on
⎡ ⎤ ⎢ ⎥ 2 ⎢ 1 1 ⎥ = + Alam ⎢ 2 L + R′′ 2 L + R′′ ⎥ c c ⎥ ⎢k kCu ⎣ p ⎦
−1
A_lam=W*th "area of a lamination" R_ad_on=(2/A_lam)*(1/(2*L/k_p+R``_c)+1/(2*L/k_Cu+R``_c))^(-1) "ad. limit in the on position"
which leads to Rad,on = 119.1 K/W. The resistance associated with the adiabatic approximation in the “off” state, Rad,off from Figure 3(b) is: −1
Rad ,off
L L ⎡ ⎤ + + Rc′′ ⎢ ⎥ k p kCu 2 ⎢ 1 1 ⎥ = = + Alam ⎢ L + L + R′′ L + L + R′′ ⎥ Alam c c ⎥ ⎢k kCu k p ⎣ p kCu ⎦
R_ad_off=(L/k_p+L/k_Cu+R``_c)/A_lam
"ad. limit in the off position"
The resistance ratio for the adiabatic limit is: RRad =
Rad ,off Rad ,on
RR_ad=R_ad_off/R_ad_on
"resistance ratio estimated using adiabatic limit"
which leads to RRad = 34.0. The “isothermal” approximation provides no resistance to heat transfer in the y-direction (i.e., perpendicular to the laminations). Therefore, the heat can spread without penalty at the interface and the resistance network for this approximation is shown in Figure 4(a) for the “on” position and Figure 4(b) “off” position.
(a) (b) Figure 4: Resistance network for the “isothermal” approximation in the (a) “on” and (b) “off” positions.
By inspection, the two resistance networks shown in Figures 4(a) and 4(b) will yield the same resistance and therefore the “isothermal” assumption will predict a resistance ratio, RRiso, of 1.0. Clearly the company’s claim of a resistance ratio of 100 is not possible as it does not lie between the two bounding quantities associated with the isothermal and adiabatic approximations. b) Provide one or more suggestions for design changes that would improve the performance of the switch (i.e., increase the resistance ratio). Justify your suggestions. The resistance ratio would be increased by any change that causes the two resistance networks in Figure 3 to be more different. Possible improvements include using materials with a larger ratio of conductivities (i.e., lower conductivity plastic and, to a lesser degree, higher conductivity copper) or eliminating the contact resistance. From a more practical standpoint, any design change that causes the actual the device to behave more like the adiabatic approximation in Figure 3 than the isothermal approximation in Figure 4 will improve the performance; for example, increasing the contact resistance between adjacent laminations would be important. c.) Sketch the temperature distribution through the two parallel paths associated with the adiabatic limit of the switch’s operation in the “off” position. Do not worry about the quantitative details of the sketch, just make sure that the qualitative features are correct.
Figure 5: Qualitative temperature distribution through paths 1 and 2 consistent with the adiabatic approximation when the switch is in the “off” position.
d.) Sketch the temperature distribution through the two parallel paths associated with the adiabatic limit in the “on” position. Again, do not worry about the quantitative details of your sketch, just make sure that the qualitative features are correct.
Figure 6: Qualitative temperature distribution through paths 1 and 2 consistent with the adiabatic approximation when the switch is in the “on” position.
Problem 2.8-2 (2-13 in text): Resistance of a Bus Bar Figure P2.8-2 illustrates a thermal bus bar that has width W = 2 cm (into the page). H1 = 5 cm
L2 = 7 cm 2 h = 10 W/m -K T∞ = 20°C
TH = 80°C
L1 = 3 cm
H2 = 1 cm
k = 1 W/m-K
Figure P2.8-2: Thermal bus bar.
The bus bar is made of a material with conductivity k = 1 W/m-K. The middle section is L2 = 7 cm long with thickness H2 = 1 cm. The two ends are each L1 = 3 cm long with thickness H1 = 3 cm. One end of the bar is held at TH = 80ºC and the other is exposed to air at T∞ = 20ºC with h = 10 W/m2-K. a.) Use FEHT to predict the rate of heat transfer through the bus bar. The geometry is entered in FEHT, a mesh is generated, and the boundary conditions and material properties are specified (Figure 2).
Figure 2: FEHT model.
The mesh is refined several times and then the heat transfer at the convective boundary is obtained; this leads to q ′ = 5.40 W/m or q = 0.108 W. b.) Obtain upper and lower bounds for the rate of heat transfer through the bus bar using appropriately defined resistance approximations. The inputs are entered in EES: $UnitSystem SI MASS RAD PA K J $TABSTOPS 0.2 0.4 0.6 0.8 3.5 in "Inputs" L_1=3 [cm]*convert(cm,m)
"length of edge pieces"
H_1=5 [cm]*convert(cm,m) L_2=7 [cm]*convert(cm,m) H_2=1 [cm]*convert(cm,m) W=2 [cm]*convert(cm,m) k=1 [W/m-K] h_bar=10 [W/m^2-K] T_infinity=20 [C] T_H=80 [C]
"height of edge pieces" "length of center piece" "height of center piece" "width" "conductivity" "heat transfer coefficient at right side" "ambient temperature at right side" "left side temperature"
An upper bound on the heat transfer rate is obtained using the isothermal limit. The isothermal limit is calculated according to:
qiso =
(TH − T∞ ) L1 L2 1 +2 + h H1 W k H1 W k H 2 W
Rconv ,iso
Rcond 1,iso
Rcond 2,iso
"Isothermal limit" R_conv_iso=1/(h_bar*W*H_1) R_cond_1_iso=L_1/(k*W*H_1) R_cond_2_iso=L_2/(k*W*H_2) R_total_iso=(R_conv_iso+2*R_cond_1_iso+R_cond_2_iso) q_dot_iso=(T_H-T_infinity)/R_total_iso
which leads to qiso = 0.118 W. A lower bound on the heat transfer rate is obtained using the adiabatic limit. The adiabatic limit is calculated according to:
qad =
(TH − T∞ ) L1 L2 1 +2 + h H2 W k H2 W k H2 W
Rconv ,ad
Rcond 1,ad
"Adiabatic limit" R_conv_ad=1/(h_bar*W*H_2) R_cond_1_ad=L_1/(k*W*H_2) R_cond_2_ad=L_2/(k*W*H_2) R_total_ad=R_conv_ad+2*R_cond_1_ad+R_cond_2_ad q_dot_ad=(T_H-T_infinity)/R_total_ad
which leads to qad = 0.052 W.
Rcond 2,ad
Problem 2.9-2 (2-14 in text) A laminated stator is shown in Figure P2.9-2. The stator is composed of laminations with conductivity klam = 10 W/m-K that are coated with a very thin layer of epoxy with conductivity kepoxy = 2.0 W/m-K in order to prevent eddy current losses. The laminations are thlam = 0.5 mm thick and the epoxy coating is 0.1 mm thick (the total amount of epoxy separating each lamination is thepoxy = 0.2 mm). The inner radius of the laminations is rin= 8.0 mm and the outer radius of the laminations is ro,lam = 20 mm. The laminations are surrounded by a cylinder of plastic with conductivity kp = 1.5 W/m-K that has an outer radius of ro,p = 25 mm. The motor casing surrounds the plastic. The motor casing has an outer radius of ro,c = 35 mm and is composed of aluminum with conductivity kc = 200 W/m-K. laminations, thlam = 0.5 mm, klam = 10 W/m-K epoxy coating, thepoxy = 0.2 mm, kepoxy = 2.0 W/m-K kp = 1.5 W/m-K kc = 200 W/m-K T∞ = 20°C 2 h = 40 W/m -K
4 2 q ′′ = 5x10 W/m
Rc′′ = 1x10 K-m /W -4
rin = 8 mm ro,lam = 20 mm
2
ro,p = 25 mm ro,c = 35 mm Figure P2.9-2: Laminated stator.
A heat flux associated with the windage loss associated with the drag on the shaft is q ′′ = 5x104 W/m2 is imposed on the internal surface of the laminations. The outer surface of the motor is exposed to air at T∞ = 20°C with a heat transfer coefficient h = 40 W/m2-K. There is a contact resistance Rc′′ = 1x10-4 K-m2/W between the outer surface of the laminations and the inner surface of the plastic and the outer surface of the plastic and the inner surface of the motor housing. a.) Determine an upper and lower bound for the temperature at the inner surface of the laminations (Tin). The inputs are entered in EES: $UnitSystem SI MASS RAD PA C J $Tabstops 0.2 0.4 0.6 0.8 3.5 klam=10 [W/m-K] kepoxy=2 [W/m-K] kp=1.5 [W/m-K] kc=200 [W/m-K] tlam=0.5 [mm]*convert(mm,m)
"conductivity of laminations" "conductivity of epoxy" "conductiity of plastic" "conductivity of casing" "thickness of lamination"
tepoxy=0.2 [mm]*convert(mm,m) rin=8 [mm]*convert(mm,m) rolam=20 [mm]*convert(mm,m) rop=25 [mm]*convert(mm,m) roc=35 [mm]*convert(mm,m) qflux=5e4 [W/m^2] Rc=1e-4 [K-m^2/W] Tair=converttemp(C,K,20[C]) h=40 [W/m^2-K]
"thickness of epoxy" "inner radius of laminations" "outer radius of laminations" "outer radius of plastic" "outer radius of casing" "heat flux" "contact resistance" "air temperature" "heat transfer coefficient"
An upper bound on the temperature allows no heat spreading; therefore, the heat must take two parallel paths that pass through the iron and the epoxy. The total resistance associated with one lamination/epoxy pair is:
1 Rtotal ,ad
+
=
1 ⎛r ⎞ ⎛ r ⎞ ⎛r ⎞ ln ⎜ o ,c ⎟ ln ⎜⎜ o , p ⎟⎟ ln ⎜ o ,lam ⎟ ⎜ ⎟ ro ,lam ⎠ Rc′′ Rc′′ 1 ⎝ rin ⎠ + ⎝ ro , p ⎠ + + ⎝ + + 2 π klam thlam 2 π ro ,lam thlam 2 π k p thlam 2 π ro , p thlam 2 π kc thlam 2 π ro, p thlam h 1
⎛r ⎞ ⎛ r ⎞ ⎛r ⎞ ln ⎜ o ,c ⎟ ln ⎜⎜ o , p ⎟⎟ ln ⎜ o ,lam ⎟ ⎜r ⎟ ro ,lam ⎠ Rc′′ Rc′′ 1 ⎝ ⎝ o, p ⎠ + ⎝ rin ⎠ + + + + 2 π kepoxy thepoxy 2 π ro ,lam thepoxy 2 π k p thepoxy 2 π ro , p thepoxy 2 π kc thepoxy 2 π ro, p thepoxy h (1)
and the upper bound on the air temperature is:
Tupper = T∞ + Rtotal ,ad q ′′ 2 π rin ( thlam + thepoxy ) "Upper bound on temperature" Rlam=ln(rolam/rin)/(2*pi*tlam*klam) "resistance of laminations" Repoxy=ln(rolam/rin)/(2*pi*tepoxy*kepoxy) "resistance of epoxy" Rci1=Rc/(2*pi*rolam*tlam) "contact resistance" Rci2=Rc/(2*pi*rolam*tepoxy) Rp1=ln(rop/rolam)/(2*pi*tlam*kp) "resistance of plastic" Rp2=ln(rop/rolam)/(2*pi*tepoxy*kp) Rco1=Rc/(2*pi*rop*tlam) "contact resistance" Rco2=Rc/(2*pi*rop*tepoxy) Rcs1=ln(roc/rop)/(2*pi*tlam*kc) "resistance of casing" Rcs2=ln(roc/rop)/(2*pi*tepoxy*kc) Rcv1=1/(h*2*pi*roc*tlam) "convection resistance" Rcv2=1/(h*2*pi*roc*tepoxy) 1/Rtotal1=1/(Rlam+Rci1+Rp1+Rco1+Rcs1+Rcv1)+1/(Repoxy+Rci2+Rp2+Rco2+Rcs2+Rcv2) "total resistance" qdot=qflux*pi*rin*(tlam+tepoxy) "heat transfer" Tin1=Tair+qdot*Rtotal1 "upper bound on temperature"
which leads to Tupper = 502.7 K.
(2)
A lower bound on the temperature allows heat spreading; therefore, the heat must take two parallel paths that pass through the iron and the epoxy but then a single path to the air. The total resistance associated with one lamination/epoxy pair is:
Rtotal ,iso
⎡ ⎢ ⎢ ⎢ 1 =⎢ ⎛r ⎞ ⎢ ln ⎛ ro ,lam ⎞ ln ⎜ o ,lam ⎟ ⎜ ⎟ ⎢ r ⎝ rin ⎠ ⎢ ⎝ in ⎠ + ⎣⎢ 2 π klam thlam 2 π kepoxy thepoxy
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦⎥
−1
⎛ r ⎞ ln ⎜⎜ o , p ⎟⎟ Rc′′ ⎝ ro ,lam ⎠ + + + 2 π ro ,lam ( thlam + thepoxy ) 2 π k p ( thlam + thepoxy )
(3)
⎛r ⎞ ln ⎜ o ,c ⎟ ⎜r ⎟ Rc′′ 1 ⎝ o, p ⎠ + + 2 π ro , p ( thlam + thepoxy ) 2 π kc ( thlam + thepoxy ) 2 π ro , p ( thlam + thepoxy ) h
and the lower bound on the air temperature is: Tlower = T∞ + Rtotal ,iso q ′′ 2 π rin ( thlam + thepoxy ) "Lower bound on temperature" Rci=Rc/(2*pi*rolam*(tlam+tepoxy)) Rp=ln(rop/rolam)/(2*pi*(tlam+tepoxy)*kp) Rco=Rc/(2*pi*rop*(tlam+tepoxy)) Rcs=ln(roc/rop)/(2*pi*(tlam+tepoxy)*kc) Rcv=1/(h*2*pi*roc*(tlam+tepoxy)) Rtotal2=(1/Rlam+1/Repoxy)^(-1)+Rci+Rp+Rco+Rcs+Rcv Tin2=Tair+qdot*Rtotal2
(4)
"contact resistance" "plastic resistance" "contact resistance" "casing resistance" "convection resistance" "total resistance" "lower bound on temperature"
which leads to Tlower = 491.7 K. b.) You need to reduce the internal surface temperature of the laminations and there are a few design options available, including: (1) increase the lamination thickness (up to 0.7 mm), (2) reduce the epoxy thickness (down to 0.05 mm), (3) increase the epoxy conductivity (up to 2.5 W/m-K), or (4) increase the heat transfer coefficient (up to 100 W/m-K). Which of these options do you suggest and why? Examination of the Solution Window (Figure 2) shows that the resistance of the lamination is much less than the resistance of the epoxy; therefore, the resistance of the lamination dominates the resistance of the epoxy. The resistance due to convection is much larger than the resistance
of the lamination, contact resistance, or conduction through the plastic and casing. Therefore, the resistance to convection dominates the problem and the most effective mechanism for reducing the temperature is to increase the heat transfer coefficient.
Figure 2: Solution Window
