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EVALUACIÓN BIOLOGIA
3. ONE-D ONE-DIME IMENSIO NSIONA NAL L STE STEA A DY STA STA TE CONDUCTION Conduction in a Single Layer Plane Wall • Assume:
L k λ
(1) Steady state (2) One-dimensional [W/m3] =0 (3) Q& dr
0
& qQ x
x
• Find: (1) Temperature distribution (2) Heat transfer rate
Fig. 3.1 1
The Heat Conduction Equation Starting point: The heat conduction equation for 3-D ∂
∂T ∂ ∂T ∂ ∂T ∂T ( λ ) + ( λ ) + ( λ ) + Q& = ρc zdr ∂ x ∂ x ∂ y ∂ y ∂ z ∂ z ∂t
(3.1) becomes for 1D d
dT ( λ ) = 0 dx dx
(3.1)
(3.2)
• Assume: Constant λ 2
d T d
2
0
(3.3) 2
(3.3) is valid for all problems described by rectangular coordinates, subject to the four above assumptions.
General Solution Integrate (3.3)
dT dx
= C
1
Integrate again
T = C x + C 1
2
(3.4)
• C 1 and C 2 are constants of integration determined from B.C.
• Temperature distribution is linear 3
Application to Special Cases Apply solution (3.4) to special cases (different B.C.)
• Objective: (1) Determine the temperature distribution T ( x ) (2) Determine the heat transfer rate Q& x
(3) Construct the thermal circuit
4
• Case (i): Specified temperatures at both surfaces L k
Boundary conditions:
T (0)
T s1
(3.5)
T ( L)
T s 2
(3.6)
T s1
0
(1) Determine C 1, C 2 and T ( x ):
T = C x + C 1
2
T ( x ) T s 2
x R Rcd = cd
T s1
Solution is given by (3.4)
λ
L L Ak S λ
q x
Q&
T s 2
x
(3.4)
Fig. 3.2
5
Applying B.C., general solution becomes: Linear profile
T ( x )
T s1
x
(T s 2
T s1 ) L
(3.7)
(2) Determine q x : Apply Fourier's law (1.5)
q& = x
Q&
x
S
− λ
∂T
(1.5)
∂ x
6
∂T & Q = − λ S x
(3.8)
∂ x
Differentiate (3.7) and substitute into (3.8) Q& =
λ S (T - T ) s1
x
s2
(3.8a)
L
L k
(3) Thermal circuit. Rewrite (3.8a): Q& x =
(T - T s1
s2
T s1
λ
T ( x )
(3.8b)
L
0
S λ Define: Thermal resistance due to conduction, Rcd
R Rcd = cd
T s1
T s 2
x L L Ak S λ
T s 2
q x
Q&
x
Fig. 3.2
7
R
cd
=
L
(3.9)
L k
S λ
(3.8b) becomes (T - T Q& x = s1 s2 R
T s1
(3.10)
0
cd
λ
T ( x )
R Rcd = cd
Analogy with Ohm's law for electric circuits:
Q&
x
Rcd
L L Ak S λ
q x
Q&
T s 2
x
Fig. 3.2
current
(T s1 T s 2 )
T s1
T s 2
x
voltage drop
electric resistance
8
Conduction in a Multi-layer Plane Wall The Heat Equations and Boundary Conditions
9
Heat must go through all layers with no change (unless heat is generated – e.g. 1000W must get through all layers): T s2 − T s1 T s3 − T s2 T s4 − T s3 & Q x = − λ1 S = − λ2 S = − λ3 S L1
L2
L3
Or using conduction resistance: T s2 − T s1 T s3 − T s2 T s4 − T s3T 1 & Q x = − =− =− L1 L2 L3 λ1 S
λ2 S
L1 T s1
λ3 S
T 1 T T − s s 1 4 Q& x = = L2 L3 R1 + R2 + R3 L1 + + λ 1 S λ 2 S λ 3 S
T s1 − T s 4
k 1 λ
1
λk 2
2
T s 2
L3
k 3 λ
3 T s 4
T s 3 0
And summing up the resistances and exchanging temp. differences
L2
11
x
L1 L
Ah S α1
1
S λ1 Ak
T s1
q & x Q
1
T
L L2
L L
11
3
Ah λ 3 S S λ α4 Ak 2 S Ak 3
2
x T s 2 Fig. 3.5
T s 3
2
T s 4 10
ΔT & Q = x
(3.11)
∑ R
T
L1
1
T s 1
T = overall temperature difference across all resistances
k 1 λ
1
λk 2
2
T s 2
L3
k 3 λ
3 T s 4
T s 3 0
11
R = sum of all resistances
L2
T
L1 L
Ah S α1 1
x
1
T s 1
λ 1 S Ak 1
T
L L2
L L
11
3
Ah λ 3 S α4 S λ Ak 2 S Ak 3
2
q & x
Q
x T s 2
T s 3
2
T s 4
Fig. 3.5
Determining temperature at any point, for example at the point 2, apply equation for heat transfer rate for appropriate layer T s1 − T s 2 & Q x = L1 λ1 S
4
11
T
4
Radial Conduction in a Single Layer Cylindrical Wall The Heat Conduction Equation Assume: (1) Constant λ (2) Steady state: (3) 1-D:
0
T t 0
r 2 r r 1
0 Fig. 3 .6
z (4) No energy generation: Q& zdr = 0
12
Simplified Heat equation in cylindrical coordinates:
d
dT ( r ) dr dr
(3.12)
0
General solution T (r ) = C 1 ln r + C 2
(3.13)
(1) Determine temperature distribution - profile Specified temperatures at both surfaces r 1
B.C. T (r 1) = T s1 T (r 2) = T s 2
0
r r 2
T s1 T s 2 Fig.13 3 .7
T ( r ) =
T s1
T s 2
ln ( r 1/r 2 )
ln ( r /r 2 ) + T s 2
(3.14)
Logarithmic profile (2) Determine the radial heat transfer rate Q& r : Apply Fourier's law
Q& = − λ .S(r) r
dT
(3.15)
dr
For a cylinder of length L the area S (r ) is
S(r) = 2 π rL
(3.16)
Differentiate (3.14)
dT
T s1
T s 2 1
dr
ln( r 1 / r 2 ) r
(3.17) 14
Q& r =
T s1 − T s2
(3.18)
(1/2π λ L)ln(r 2 /r 1 )
(3) Thermal circuit: Define the thermal resistance for radial conduction, Rcd R
cd
=
ln ( r r ) 2
2 πλ L
1
r 1
(3.19)
0
T s1 T s 2
(3.19) into (3.18) T s1
Q& = r
T − T s1
s2
R
cd
r r 2
Rcd q&r r Q
T s 2
Fig. 3.7
(3.20) 15
Heat is transferred from inside to outside the tube Which profile is correct? 1 or 2? Q& r
Superheated steam
16
Radial Conduction in a Multi-layer Cylindrical Wall r 3
Assume:
r 2 k 2 r 1 k 1 λ1
(1) One-dimensional (2) Steady state
T 1 h1
(3) Constant conductivity
r 4 k 3 λ2
λ3
T
4
h4 T s1 T s2
T s3 T s4
(4) No heat generation (5) Perfect interface contact
&
qQ r r
T 1 Rcv1
T
Rcd 1 Rcd 2 Rcd 3 Rcv 4
Fig . 3.10
Three conduction resistances: 17
4
R
cd1
=
ln(r /r ) 2
1
2π λ L 1
R
cd2
=
ln(r /r ) 3
2
2π λ L 2
R
cd3
=
ln(r /r 4
3
2π λ L 3
Heat transfer rate: Ohm analogy Q& r =
ln(r 2 /r 1 ) 2π λ1 L
+
T s1 − T s4 ln(r 3 /r 2 ) 2π λ2 L
+
ln(r 4 /r 3 ) 2π λ3 L (3.21)
18
Contact Resistance • Perfect interface contact vs. actual contact (see Figure)
• Gaps act as a resistance to heat flow • The temperature drop depends on
T T ct
the contact resistance Rct
x
• Rct is determined experimentally Operational temperature
