!"#$%&'( '**+,-./-0 $( 12340+2-* %5
Consider a house that has a 10-m x 20-m base and a 4-m-high wall. All four walls of the house have an R-value of 2.31 m2 ·°C/W. The two 10-m x 4-m walls have no windows. The third wall has five windows made of 0.5-cm-thick glass ( k = 0.78 W/mK), 1.2 m x 1.8 m in size. The fourth wall has the same size and number of windows, but they are double- paned with a 1.5-cm-thick stagnant air space ( k = 0.026 W/m · K) enclosed between two 0.5-cm-thick glass layers. The thermostat in the house is set at 22°C and the average temperature outside at that location is 8°C during the winter. Disregarding any direct radiation gain or loss through the windows and taking the heat transfer coefficients at the inner and outer surfaces of the house to be 7 and 15 W/m2 · °C, respectively, determine the average rate of heat transfer through each wall. If the cost of electric heating is Rs.7/kWh, estimate the annual heating cost for a 4 month long winter. 34*2#4".#0 The thermal conductivities are given to be k = = 0.026 W/m°C for air, and 0.78 W/m°C for glass. /+$5&0.0 The rate of heat transfer through each wall can be determined by applying thermal resistance network. The convection resistances at the inner and outer surfaces are common in all cases. Walls without windows :
Ri
R o
L wall
(7 W/m 2 .C)(10 4 m 2 )
R value
kA
ho A
(10 4 m 2 )
1 (15 W/m 2 .C)(10 4 m 2 )
R i R wall R o
T 1 T 2 Rtotal
6
0.05775 C/W
Q
0.001667 C/W
0.003571 0.05775 0.001667 0.062988 C/W
(22 8)C
Wall
0.003571 C/W
2.31 m 2 C/W
A
1
Q
Then
1
hi A
R wall
R total
1
0.062988C/W
222.3 W
Ri
Rwall
Ro
Wall with single pane windows:
Ri
R wall
Rglass
1
R total
Then
&5
1
2
(7 W/m .C)(20 4 m 2 )
hi A L wall
R value
kA
A
Lglass
kA
Reqv Ro
1
1 R wall
1 ho A
0.001786 C/W
2.31 m 2 C/W (20 4) 5(1.2 1.8) m 2
0.005 m (0.78 W/m 2 . o C)(1.2 1.8)m 2
5
1
1
Rglass
0.033382
5
0.002968
(15 W/m .C)(20 4 m 2 )
Ri R eqv R o
Q
0.033382 C/W
0.002968 C/W
1
1 2
R eqv
0.00058 o C/W Ri
0.000833 C/W
0.001786 0.000583 0.000833 0.003202 C/W
T 1 T 2 R total
(22 8)C 0.003202C/W
4372 W
R
lass
R
all
Ro
4th wall with double pane windows: R
Ri
R wall
R glass
1 Reqv R total
Then
Q
R value
kA
Lair
A
Lglass
kA
lass
Rwall
kA
Rair R window
L wall
Rair R
lass
Ro
2.31 m 2 C/W (20 4) 5(1.2 1.8)m 2
0.005 m (0.78 W/m 2 .C)(1.2 1.8) m 2
0.033382 C/W
0.002968 C/W
0.267094 C/W
0.015 m (0.026 W/m 2 . o C)(1.2 1.8)m 2
2 Rglass Rair 2 0.002968 0.267094 0.27303 C/W 1 R wall
5
1 R window
Ri R eqv Ro
T 1 T 2 R total
1
0.033382
5
1 0.27303
Reqv
0.020717 C/W
0.001786 0.020717 0.000833 0.023336 C/W
( 22 8)C 0.023336C/W
600 W
Total Heat loss = 2*222.3+4372+600=5416.6W=5.417KW TOTAL ANNUAL COST: 5.417 kW * 4 * 30 *24hr* 7Rs/kWh = Rs. 1,09,198.66
&5 o 2 Air flows at 120 C in a thin-wall stainless-steel (k =18W/m.K) tube with h=65W/m K. The inside diameter of the tube is 2.5 cm and the wall thickness is 0.4mm. The tube is 2 o exposed to an environment with h=6.5 W/m K and the ambient temperature is 15 C. Calculate the overall heat-transfer coefficient and the heat loss per meter length. What thickness of soft rubber (k =0.13W/m.K) as insulating material should be added to reduce the heat loss by 50%? What would be your result if rigid foam (k=0.026W/m.K) is used in place of rubber? Compare and comment on your results.
3. A rod of length L has one end maintained at temperature T 0 and is exposed to an environment of temperature T ∞ . An electrical heating element is placed in the rod so that heat is generated uniformly along the length at a rate q˙ . Derive an expression (a) for the temperature distribution in the rod and ( b) for the total heat transferred to the environment. Obtain an expression for the value of q˙ that will make the heat transfer zero at the end that is maintained at T 0 . '-*5
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One end of a copper rod 30 cm long is firmly connected to a wall that is maintained at 200◦ C. The other end is firmly connected to a wall that is maintained at 93 ◦ C. Air is blown across the rod so that a heat-transfer coefficient of 17 W /m2 · ◦ C is maintained. The diameter of the rod is 12.5 mm. The temperature of the air is 38 ◦ C. What is the net heat lost to the air in watts?
F5
A 5-mm-diameter spherical ball at 50°C is covered by a 1-mm-thick plastic insulation (k = 0.13 W/m · °C). The ball is exposed to a medium at 15°C, with a combined convection and radiation heat transfer coefficient of 20 W/m 2 · °C. Determine if the plastic insulation on the ball will help or hurt heat transfer from the ball. What should be the minimum thickness of the insulation if the heat transfer rate has to be reduced to 50% of the original heat transfer rate? Ans. Derive the expression for critical radius of insulation for sphere: it turns out to be rc=2k /h. For given values, critical radius is 13mm. Hence adding plastic insulation will increase heat transfer as long as the radius of insulation is less than 13mm. Further
G5
Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 180°C. Circular aluminum alloy fins ( k = 186 W/m · °C) of outer diameter 6 cm and constant thickness 1 mm are attached to the tube. The space between the fins is 3 mm, and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding air at 25°C, with a heat transfer coefficient of 40 W/m 2 · °C. Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.
/00,12".*+0 1 Steady operating conditions exist. 2 The heat transfer coefficient is constant and uniform over the entire fin surfaces. 3 Thermal conductivity is constant. 4 Heat transfer by radiation is negligible. 34*2#4".#0 The thermal conductivity of the fins is given to be k = 186 W/m°C. /+$5&0.0 In case of no fins, heat transfer from the tube per meter of its length is
Ano fin
. m D1 L (0.05 m)(1 m) 01571
Q no fin
. m 2 )(180 25) C 974 W hAno fin (Tb T ) (40 W / m2 . C)( 01571
180C
2
The efficiency of these circular fins is, from the efficiency curve,
L ( D2 r 2
D1 ) / 2 (0.06 0.05) / 2 0.005 m (t / 2) 0.03 (0.001 / 2) 1.22 r 1
t L 2
0.025
0.001 0.005 kt 2 h
40 W/m
2o
C
o
(186 W/m C)(0.001 m)
fin 0.97 0.08
25C
Heat transfer from a single fin is
Afin Q fin
2 (r 2 2 r 12 ) 2 r 2 t 2 (0.032 0.025 2 ) 2 (0.03)(0.001) 0.001916 m 2 fin Q fin,max fin hAfin (T b T ) 0.97(40 W/m 2 .C)(0.001916 m 2 )(180 25)C 11.53 W
Heat transfer from a single unfinned portion of the tube is
Aunfin Q unfin
D1 s (0.05 m)(0.003 m) 0.0004712 m 2 hAunfin (T b T ) (40 W/m 2 .C)(0.0004712 m 2 )(180 25)C 2.92 W
There are 250 fins and thus 250 interfin spacings per meter length of the tube. The total heat transfer from the finned tube is then determined from
Q total,fin
n(Q fin Q unfin ) 250(11.53 2.92) 3613 W
Therefore the increase in heat transfer from the tube per meter of its length as a result of the addition of the fins is
Q increase
Q total,fin Q no fin 3613 974 2639 W
7. A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating 0.04 W. The board is impregnated with copper fillings and has an effective thermal conductivity of 20 W/m·°C. All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at 40°C, with a heat transfer coefficient of 50 W/m2·°C. (a) Determine the temperatures on the two sides of the circuit board. ( b) Now a 0.2-cm-thick, 12-cm-high, and 18-cm-long aluminum plate (k = 237 W/m·°C) with 864 2-cm-long aluminum pin fins of diameter 0.25 cm is attached to the back side of the circuit board with a 0.02-cm-thick epoxy adhesive (k = 1.8 W/m·°C). Determine the new temperatures on the two sides of the circuit board.
/00,12".*+0 1 Stea y operat ng con t ons ex st. 2 T e temperature n t e oar an a ong t e ns var es in one direction only (normal to the board). 3 All the heat generated in the chips is conducted across the circuit board, and is dissipated from the back side of the board. 4 Heat transfer from the fin tips is negligible. 5 The heat transfer coefficient is constant and uniform over the entire fin surface. 6 The thermal properties of the fins are constant. 7 The heat transfer coefficient accounts for the effect of radiation from the fins. 34*2#4".#0 The thermal conductivities are given to be k = 20 W/m°C for the circuit board, k = 237 W/m°C for the aluminum plate and fins, and k = 1.8 W/m°C for the epoxy adhesive. /+$5&0.0 (a) The total rate of heat transfer dissipated by the chips is Q 80 ( 0.04 W)
2 cm
3.2 W
The individual resistances are R
Re
oard
RAluminum
ox
Rconv
T 1
T 2 T 2
A ( 0.12 m)( 0.18 m)
R board
Rconv
Rtotal
L
(20 W / m. C)( 0.0216 m2 )
kA
Rboard
1
(50 W / m 2 . C)( 0.0216 m2 )
hA
0.0216 m 2
0003 . m
1
Rconv
0.00694 0.9259
0.00694 C / W
0.9259 C / W
0.93284 C / W
The temperatures on the two sides of the circuit board are Q Q
T1 T 2
T1 T2 R board
T 2 QR total
T1
Rtotal
T2 T1 QR board
40 C ( 3.2 W)( 0.93284 C / W) 43.0 C
43.0 C ( 3.2 W)( 0.00694 C / W) 40.5 0. 02 43.0 C
Therefore, the board is nearly isothermal. (b) Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be a
hp
h D
2
kAc
fin
4h
kD
k D / 4
tanh aL
aL
4(50 W / m2 . C)
( 237 W / m. C)( 0.0025 m)
tanh(18.37 m -1 0.02 m) 18.37 m -1 0.02 m
1837 . m
-1
0957 .
The fins can be assumed to be at base temperature provided that the fin area is modified by multiplying it by 0.957. Then the various thermal resistances are Repoxy RAl
L
kA L
kA
Afinned
00002 . m (18 . W / m. C)( 0.0216 m2 ) 0002 . m
(237 W / m. C)( 0.0216 m2 )
fin nDL
Aunfinned
Atotal,with fins
1
0.0051 C / W
000039 . C / W
0.957 864 ( 0.0025 m)( 0. 02 m) 0130 . m2
0.0216 864 Afinned
D
2
4
Aunfinned
0.0216 864
(0.0025)
2
4
1
Rtotal
Rboard
0.00694 0.0051 0. 00039 0. 1361 0. 1484 C / W
hAtotal,with fins
Repoxy
2
(50 W / m . C) (0.147 m 2 )
Raluminum
00174 . m2
0. 017 0147 0130 . . m2
Rconv
01361 . C / W
Rconv
Then the temperatures on the two sides of the circuit board becomes Q
T1 T 2
Q
T1 T 2
Rtotal R board
T1
T 2
QR total
T2 T1 QR board
C / W) 40.5 C 40 C ( 3.2 W)( 01484 .
40.5 C ( 3.2 W)( 0.00694 C / W) 40.5 0. 02 40.5 C
8. Consider a house with a flat roof whose outer dimensions are 12 m x12 m. The outer
walls of the house are 6 m high. The walls and the roof of the house are made of 20cm thick concrete ( k = 0.75 W/m · °C). The temperatures of the inner and outer surfaces of the house are 15°C and 3°C, respectively. Accounting for the effects of the edges of adjoining surfaces, determine the rate of heat loss from the house through its walls and the roof. What is the error involved in ignoring the effects of the edges and corners and treating the roof as a 12 m x 12 m surface and the walls as 6 m x 12 m surfaces for simplicity? /00,12".*+0 1 Steady operating conditions exist. 2 Heat transfer at the edges and corners is two-or threedimensional. 3 Thermal conductivity of the concrete is constant. 4 The edge effects of adjoining surfaces on heat transfer are to be considered. 34*2#4".#0 The thermal conductivity of the concrete is given to be k = 0.75 W/m°C. /+$5&0.0 The rate of heat transfer excluding the edges and corners is first determined to be
Atotal (12 0.4)(12 0.4) 4(12 0.4)(6 0.2) 403.7 m2 Q
kAtotal
( T1 T 2 )
3C
L
2
(0.75 W / m. C)( 403.7 m )
(15 3) C 18,167 W L 0.2 m The heat transfer rate through the edges can be determined using the shape factor relations in Table 3-5, S corners+ edges 4 corners 4 edges 4 0.15 L 4 0.54 w
15C L
4 0.15(0.2 m) + 4 0.54(12 m) 26.04 m Q corners + edges S corners + edges k (T 1 T 2 ) ( 26.04 m )(0.75 W/m. C )(15 3)C 234 W
and
Q total 18,167 234 1.840 10 4 W 18.4 kW
Ignoring the edge effects of adjoining surfaces, the rate of heat transfer is determined from Atotal (12)(12) 4(12)(6) 432 m2 Q
kAtotal L
(T 1 T 2 )
(0.75 W/m.C)(432 m 2 ) 0.2 m
(15 3)C 1.94 10 4 19.4 kW
The percentage error involved in ignoring the effects of the edges then becomes %error
19.4 18.4 18.4
100 5.6%
9. Hot water at an average temperature of 80°C and an average velocity of 1.5 m/s is flowing through a 25-m section of a pipe that has an outer diameter of 5 cm. The pipe extends 2 m in the ambient air above the ground, dips into the ground ( k = 1.5 W/m · °C) vertically for 3 m, and continues horizontally at this depth for 20m more before it enters the next building. The first section of the pipe is exposed to the ambient air at 8°C, with a heat transfer coefficient of 22 W/m 2 · °C. If the surface of the ground is covered with snow at 0°C, determine ( a) the total rate of heat loss from the hot water and (b) the temperature drop of the hot water as it flows through this 25-m-long section of the pipe.
34*2#4".#0 The thermal conductivity of the ground is given to be k = 1.5 W/m°C. /+$5&0.0 (a) We assume that the surface temperature of the tube is equal to the temperature of the water. Then the heat loss from the part of the tube that is on the ground is
A s
DL (0.05 m)(2 m ) 0.3142 m 2
Q hA s (T s
8C
T )
0C
(22 W/m 2 .C)(0.3142 m 2 )(80 8)C 498 W Considering the shape factor, the heat loss for vertical part of the tube can be determined from S
2 L
3m
2 (3 m )
3.44 m 4(3 m) ln (0.05 m) . W / m. C)(80 0) C 413 W Q Sk (T1 T2 ) ( 3.44 m)(15 4 L ln D
20 m
80C
The shape factor, and the rate of heat loss on the horizontal part that is in the ground are S
2 L
2 (20 m )
22.9 m 4 z 4(3 m) ln ln D (0.05 m) . W / m. C)(80 0) C 2748 W Q Sk (T1 T2 ) (22.9 m)(15 and the total rate of heat loss from the hot water becomes Q total
498 413 2748 3659 W
(b) Using the water properties at the room temperature, the temperature drop of the hot water as it flows through this 25-m section of the wall becomes
C p T Q m
T
Q
C p m
Q
(V )C p
Q
(VAc )C p
3659 J/s
(0.05 m) 2 (1000 kg/m 3 )(1.5 m/s) (4180 J/kg.C) 4
0.30C
