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7/28/2019 03. Steady 1D Heat Conduction
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3. ONE-DIMENSIONAL STEADY STATE
CONDUCTION
Conduction in a Single Layer Plane
Wall
Find:(1) Temperature distribution
(2) Heat transfer rate
Assume:
(1) Steady state(2) One-dimensional
(3) 0Qdr=& [W/m3]
k
0
L
xq
x
3.1Fig.
Q&
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The Heat Conduction Equation
(3.1) becomes for 1D0)
dx
dT(
dx
d= (3.2)
Assume: Constant
02
2
=d
Td(3.3)
Starting point: The heat conduction equation for 3-D
t
T
cQ)z
T
(z)y
T
(y)x
T
(x zdr
=+
+
+
&
(3.1)
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(3.3) is valid for allproblems described by rectangular
coordinates, subject to the four above assumptions.
General Solution
Integrate (3.3)
1C
dx
dT=
Integrate again
(3.4)21
CxCT +=
C1 and C2 are constants of integration determinedfrom B.C.
Temperature distribution is linear
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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
xQ&
(3) Construct the thermal circuit
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Case (i): Specified temperatures at both
surfaces
Boundary conditions:
)0( 1sTT = (3.5))( 2sTLT = (3.6)
(1) Determine C1, C2 and T(x):
Solution is given by (3.4)
(3.4)21CxCT +=
k
0
L
x
)(xT
1sT
2sT
Ak
LRcd =
1sT 2sT
3.2Fig.
xq
SL=R
cd
xQ&
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(2) Determine xq : Apply Fourier's law (1.5)
(1.5)
x
T
S
Q
q
x
x
=
&
&
Linear profile
(3.7)LxTTTxT sss )()( 121
Applying B.C., general solution becomes:
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Differentiate (3.7) and substitute into (3.8)
L
)T-(TSQ s2s1x=& (3.8a)
(3) Thermal circuit. Rewrite (3.8a):
S
L
T-(T=Q s2s1
x& (3.8b)
Define: Thermal resistance due to
cdRconduction,
(3.8)x
T
SQx
=&
k
0
L
x
)(xT
1sT
2sT
Ak
LRcd =
1sT 2sT
3.2Fig.
xq
S
L=R
cd
xQ&
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S
L
=Rcd(3.9)
(3.8b) becomes
cd
s2s1x R
T-(T
=Q&
(3.10)
Analogy with Ohm's law for
electric circuits:
xQ& current
)(21 ss
TT voltage dropcdR electric resistance
k
0
L
x
)(xT
1sT
2sT
Ak
LRcd =
1sT 2sT
3.2Fig.
x
q
S
L=R
cd
xQ&
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Conduction in a Multi-layer Plane Wall
The Heat Equations and Boundary Conditions
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3
s3s43
2
s2s32
1
s1s21x
L
TTS
L
TTS
L
TTSQ
=
=
=&
Heat must go through all layers with no change (unless heat
is generated e.g. 1000W must get through all layers):
Or using conduction resistance:
S
L
TT
S
L
TT
S
L
TTQ
3
3
s3s4
2
2
s2s3
1
1
s1s2x
=
=
=&
And summing up the resistances and
exchanging temp. differences
S
L
S
L
S
L
TT
RRR
TTQ ssssx
3
3
2
2
1
1
41
321
41
++
=
++
=&
2sTxq
1T 33
Ak
L
22
Ak
L
11
Ak
L
4
1
Ah1
1
Ah
4sT3sT1sT
1T
T
1sT 2sT
3sT4sT
x0
3k2k1k
3L2L1L
3.5Fig.
1S
1
1S
L
2S
L
3S
L
2S
1
1
2
3
xQ&
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T= overall temperature differenceacross all resistances
R = sum of all resistances
(3.11)
=R
T
Qx&
Determining temperature at any point, for example
at the point 2, apply equation for heat transfer ratefor appropriate layer
S
L
TTQ
1
1
ssx
21 =&
2sTxq
4T1T3
3
Ak
L
2
2
Ak
L
1
1
Ak
L
4
1
Ah1
1
Ah
4sT3sT1sT
1T
4T
1sT 2sT
3sT4sT
x0
3k2k1k
3L2L1L
3.5Fig.
1S
1
1S
L
2S
L
3S
L
2S
1
1
2
3
xQ&
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Radial Conduction in a Single Layer
Cylindrical Wall
The Heat Conduction
Equation
Assume:
(1) Constant
(2) Steady state: 0
tT
(3) 1-D: 0=
z
(4) No energy generation: 0Qzdr=&
6.3Fig.
1r
2r
r
0
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(3.12)0)( =dr
dTr
dr
d
Simplified Heat equation in cylindrical coordinates:
General solution
(3.13)T(r) = C1
ln r + C2
B.C.
1sT
T(r2) =
T(r1) =
2sT
Specified temperatures at both surfaces
(1) Determine temperature distribution - profile
7.3Fig.
0
1r 2rr
1sT
2sT
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(2) Determine the radial heat transfer rater
Q& : Apply
Fourier's law
dr
dT.S(r)Q
r=& (3.15)
For a cylinder of lengthL the areaS(r) is
rL2S(r) = (3.16)
Differentiate (3.14)
rrr
TT
dr
dT ss 1
)/ln( 21
21 = (3.17)
(3.14)22
2
21 +)/(ln
)/(ln
=)( sss Trr
TTrT
rr
1
Logarithmic profile
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(3.18))/rL)ln(r(1/2
TTQ12
s2s1r
=&
(3) Thermal circuit: Define the thermal resistance for
cdRradial conduction,
L2
)rr(ln
R 12cd = (3.19)
(3.19) into (3.18)
(3.20)R
TT=Q
cd
s2s1
r
&
7.3Fig.
0
1r 2rr
1sT
2sT
rq
1sT 2sTcdR
rQ&
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Heat is transferred from inside to outside the tube
Which profile is correct? 1 or 2?
Superheated
steam
rQ&
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Radial Conduction in a Multi-layer
Cylindrical Wall
Assume:
(1) One-dimensional
(2) Steady state
(3) Constant conductivity
(4) No heat generation
(5) Perfect interface contact
Three conduction resistances:
1k
2k3k
1r
2r
3r4r
1h4h1T
4T
10.3.Fig
1cvR 4cvR3cdR2cdR1cdR
1T 4Trq
Ts1 Ts2 Ts3 Ts4
rQ&
1 2 3
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L2
)/rln(r
R1
12
cd1 =
L2
)/rln(rR
2
23
cd2
=
L2
/rln(rR
3
34
cd3=
Heat transfer rate: Ohm analogy
(3.21)
L2
)/rln(r+
L2
)/rln(r+
L2
)/rln(r TT=Q
3
34
2
23
1
124s1sr
&
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Contact Resistance Perfect interface contact vs. actual
contact (see Figure)
Gaps act as a resistance to heat flow
is determined experimentallyctR
The temperature drop depends on
ct
Rthe contact resistance
Operationaltemperature
Surfacetemperature
Fouriers law:
21 RRR
TQ
ctx
++=&
ctT
T
x3.11Fig.