3.
ONE-DIMENSIONAL STEADY STATE CONDUCTION Conduction in a Single Layer Plane Wall Assume:
(1) Steady state (2) One-dimensional & =0 [W/m3] (3) Q
zdr
L k
0
& Q q xx
x
Fig. 3.1
1
Find:
(1) Temperature distribution (2) Heat transfer rate
The Heat Conduction Equation
Starting point: The heat conduction equation for 3-D
T T T T & ) + ( ) + Q = c ( ) + ( zdr x x y y z z t
(3.1) becomes for 1D d dT ( )=0 dx dx
(3.1)
(3.2)
Assume: Constant
d 2T dx
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 =C 1 dx
Integrate again
T = C1 x + C 2
(3.4)
C1 and C2 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
Case (i): Specified temperatures at both
surfaces
Boundary conditions:
L k
T ( x)
Ts 2
L L Ak S
T (0) = Ts1 T ( L) = Ts 2
(3.5) (3.6)
Ts1 0
cd
x
RRcd= =
(1) Determine C1, C2 and T(x):
Ts1
Solution is given by (3.4)
Q
Fig. 3.2
q& x
Ts 2
T = C1 x + C 2
(3.4)
Applying B.C., general solution becomes: Linear profile
x T ( x ) = Ts1 + (Ts 2 Ts1 ) L
(3.7)
(2) Determine q x : Apply Fourier's law (1.5)
T q &x = S x
x
& Q
(1.5)
T & Q = S x x
Differentiate (3.7) and substitute into (3.8)
(3.8)
& = Q
x
S (T - T )
s1 s2
L
Ts1 0
cd
(3.8a)
L k
T ( x)
Ts 2
L L Ak S
(3) Thermal circuit. Rewrite (3.8a):
(Ts1 - Ts2 ) & = Q x L S
(3.8b)
x
RRcd= =
Define: Thermal resistance due to conduction, Rcd
Ts1
Q
Fig. 3.2
q& x
Ts 2
x
7
L R = cd S
(3.8b) becomes (Ts1 - Ts2 ) & = Q x R
cd
(3.9)
Ts1
L k
T ( x)
Ts 2
L L Ak S
(3.10)
0
cd
x
RRcd= =
Analogy with Ohm's law for electric circuits: & current Q
x
Ts1
Q
Fig. 3.2
q& x
Ts 2
(Ts1 Ts 2 ) voltage drop
Rcd electric resistance
8
Conduction in a Multi-layer Plane Wall
The Heat Equations and Boundary Conditions
Heat must go through all layers with no change (unless heat is generated e.g. 1000W must get through all layers):
Ts2 Ts1 Ts3 Ts2 Ts4 Ts3 & Qx = 1 S = 2 S = 3 S L3 L1 L2
Or using conduction resistance: Ts2 Ts1 Ts3 Ts2 Ts4 Ts3T1 & Qx = = = Ts1 L1 L2 L3 1 S 2 S 3 S And summing up the resistances and exchanging temp. differences
0
L1
k 1
L2 k2
L3
Ts 2
k3
Ts 3
Ts4
&x = Q
Ts1 Ts 4 Ts1 Ts 4 = R1 + R2 + R3 L1 + L2 + L3 1S 2 S 3 S
T1
1 1 Ah S1
L L 1 S 1 Ak
1
Ts1 Qx x Ts 2
Fig. 3.5
q &
L 1 LL2 1 L3 Ah 33 S S 4 Ak2 S Ak 2 2 T
Ts 3
Ts 4
10
T & Q = x R
(3.11)
T 1
L1
T = overall temperature difference
across all resistances
T s 1
k 1
L2 k2 2
Ts 3
L3
Ts 2
k3
Ts 4
T 4
R = sum of all resistances
T 1
1 1 Ah S 1
L L 1 S 1 Ak
1
Ts1
q &x Q
L LL2 1 1 L3 Ah 3 S S 2 Ak 2 S Ak 4 3 2 T 4
Ts 2
Fig. 3.5
Ts 3
Ts 4
Determining temperature at any point, for example at the point 2, apply equation for heat transfer rate for appropriate layer Ts1 Ts 2 & Qx = L1 1 S
11
Radial Conduction in a Single Layer Cylindrical Wall
The Heat Conduction Equation
Assume: (1) Constant T (2) Steady state: =0 t (3) 1-D: = =0 z & =0 (4) No energy generation: Q zdr
0
r2 r r1
Fig. 3 .6
12
Simplified Heat equation in cylindrical coordinates:
d dT (r )=0 dr dr
General solution
(3.12)
T(r) = C1 ln r + C2
(3.13)
(1) Determine temperature distribution - profile Specified temperatures at both surfaces B.C.
r1
r
r2
T
T(r1) = Ts1 T(r2) = Ts 2
Ts1
s2
Fig.13 3 .7
Ts1 Ts 2 T (r ) = ln ( r/r2 ) + Ts 2 (3.14) ln ( r1/r2 ) Logarithmic profile
& : Apply (2) Determine the radial heat transfer rate Q r Fourier's law
dT & Q = .S(r) r dr
For a cylinder of length L the area S(r) is (3.15)
S(r) = 2 rL
Differentiate (3.14) dT Ts1 Ts 2 1 = dr ln( r1 / r2 ) r
(3.16)
(3.17)
14
&r = Q
Ts1 Ts2 (1/2 L)ln(r2 /r1 )
(3.18)
(3) Thermal circuit: Define the thermal resistance for radial conduction, Rcd Rcd = ln ( r r )
2 1
2 L
(3.19)
r1
0
r2
T
Ts1
s2
(3.19) into (3.18)
Ts1
Rcd
&rr q Q
Fig. 3.7
Ts 2
& = Q r
Ts1 Ts2 Rcd
(3.20)
15
Heat is transferred from inside to outside the tube Which profile is correct? 1 or 2?
&r Q
Superheated steam
16
Radial Conduction in a Multi-layer Cylindrical Wall
r3
r4 k3
3
Assume: (1) One-dimensional (2) Steady state (3) Constant conductivity (4) No heat generation (5) Perfect interface contact
r2 k2 r1 k11 2 T1 h1
T 4 h 4
Ts1 Ts2 Ts3 Ts4
T1
& Q q r r
Rcv1
Rcd 1 Rcd 2 Rcd 3 Rcv 4
T 4
Fig . 3.10
Three conduction resistances:
17
Rcd1 = Rcd2 =
Rcd3 =
ln(r /r )
2 1 1
2 L ln(r /r )
3 2 2
2 L
ln(r4 /r3 ) 2 3 L
Heat transfer rate: Ohm analogy
& r= Q
T s1 T s4 ln(r2 /r1 ) ln(r3 /r2 ) ln(r4 /r3 ) + + 2 1 L 2 2 L 2 3 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
the contact resistance Rct
T
Tct
Rct is determined experimentally
Fouriers law:
&x = Q T R1 + Rct + R2
Operational temperature
x
Fig. 3.11
Surface temperature
19
