Lecture Week 2 2
description
Transcript of Lecture Week 2 2
-
Week_2_2
Objectives:
1. 1-D Heat Conduction in a Plane Wall
2. Electrical Resistance Analogy
3. Interface Resistance
4. 1-D Heat Flow in Cylindrical/ Spherical Coordinates
5. Electrical Resistance Analogy in Cylindrical Coords.
ME 3345 Heat Transfer
Reading Assignment: 3.1 3.4
-
1.With Uniform Volumetric Energy Generation; Assumptions ?
Solution:
2. What if there is no heat generation?
Steady State Temperature Distribution in a Plane Wall System
2 2 2
2 2 2
1T T T q T
k tx y z
21 2( )
2
qT x x C x C
k
1 2( )T x C x C
02
2
k
q
x
T
x
qq
-
Note: The form of the equation giving the steady state temperature
distribution is independent of boundary conditions. However, the
Temperature magnitude does depend on B.C.s.
1. Prescribed Temperature at Boundary: (e.g. T = To)
2. Uniform Heat Flux at boundary: (q = qo)
3. Adiabatic at Boundary: dT/dx = 0
4. Convection and Radiation at Boundary:
44
surTTTThdx
dTk
-
Example: What is the 1-D temperature distribution in a Plane Wall with
no energy generation subjected convection on both surfaces?
Assume that the thermal conductivity, k, is constant, no heat generation
uniform, fluid temperature and convective heat transfer coefficient ,
T and h, respectively.
K
x
0q
0 L
h, T h, T
T1 T2
-
R
I
V
I = V/R
We may also use an electrical resistance analogy to
help solve heat conduction problems.
Electrical Analog of Heat Conduction
T1 T2
Thermal resistance:
th
LR
kA
LR
A
Electrical resistance:
xth
kA Tq T
L R
xq
V= T
I = qx
-
In order to develop the electrical resistance for each
mode of heat transfer, we must look at the linear
heat transport coefficients:
1. Conduction:
2. Convection:
3. Radiation:
th
kA Lq T R
L kA
hARThAq
th
1
AhRTAhq
r
thr
1
(no energy generation!!!)
-
Heat Transfer Through a Plane Wall
L
A
,1sT
,2 2,T h
,2sT
,1 1,T h
Hot
Fluid
Cold
Fluid
,1 ,2
1 2 3
( )
( )x
T Tq
R R R
AhR
2
1
3AhR
1
1
1 kA
LR
2
qx
T,1 T,2 Ts,1 Ts,2
-
Thermal Circuits
1 2
1 1, where is the .x tot
tot
T Lq R total thermal resist
R h A kA h A
1, where ( ) is the .x totq UA T U R A overall heat transfer coeff
-
Thermal Contact Resistance
(Very Small)
TA TB
Unit Area
q"x
2, [m K/W]
A Bt c
x
T TR
q
Contact resistance values range from
10 6 10 3 m2K/W, depending on
(1) materials, (2) surface roughness,
and (3) contact pressure.
Effective thermal conductivity: ,/c t ck R
, , /t c t cR R A
-
A B
A B
A B
L L
k k
1T
2T
Example: A composite wall with contact resistance .
What is the heat flux if the temperatures are known. cR
1 2 and T T
1 2
1 2
/ /
xtot
A A c B B
T Tq
R
T T
L k R L k
Answer:
xq1T AT BT 2
T
/A AL k /B BL kcR
-
How to improve Contact Resistance ??
-
Apply Pressure
How to improve Contact Resistance ??
Use filler material
Soft metal (indium, lead, tin, silver)
Thermal grease
Epoxy materials
Soldering
Very active research area
-
Tables 3-1 and 3-2 give values for some solid-solid contact
resistance.
Types of contact: metal-metal, metal-insulators, etc.
Interfacial materials: vacuum, air, soft metal, solder, grease,
plastic, etc.
Contact resistance is often important in practice but only
empirical theories exist.
Some Comments Regarding Contact Resistance
-
Thermal Contact Resistance
(Very Small)
TA TB
Unit Area
q"x
Thermal contact resistance gives rise
to a temperature discontinuity at the
interface. A surface energy balance
would be:
qx is continuous =
dx
dTk
dx
dTk ba
Temperature is discontinuous =
BActx TTRq"
,
"
-
Radial Heat Transfer
For 1-D, steady state
constant k k(T)
w/o heat generation, we have in the cylindrical coordinates,
0d dT
rdr dr
1
dTr C
dr
1 2( ) lnT r C r C
(2 )r cdT dT
q kA k rLdr dr r r+dr
r
qr
qr+dr
/r r cdT
q q A kdr
-
For a cylindrical shell with known surface temperatures
T2T1
r2
r1
1 2( ) lnT r C r C
1 1 1 2ln( )T C r C
2 1 2 2ln( )T C r C
(1)
(2)
Solving for C1 and C2 yields the following temperature distribution:
2
2
2
1
21 ln
ln
)( Tr
r
rr
TTrT
-
T2T1
r2
r1
T
r
T1
T2
r1 r2
T1 > T2 T
r
T1
T2
r1 r2
T1 < T2
The temperature distributions
What about temperature gradient?
q= k = constant
A and dT/dr not constant
Note that / is getting smaller as the radius is getting larger.dT dr
How to get resistance ??
-
T2T1
r2
r1
Heat transfer rate and thermal resistance
12
21
/ln22
rr
TTLk
dr
dTrLk
dr
dTkAqr
Rate of heat transfer:
Heat flux:
12
21
/ln rr
TT
r
k
dr
dTk
A
qr
2
2
2
1
21 ln
ln
)( Tr
r
rr
TTrT
Constant
-
T2T1
r2
r1
Thermal Resistance:
12
21
/ln2
rr
TTLkqr
Lk
rrT
Lk
rr
TTqr
2
/ln
1
2
/ln 1212
12
Lk
rrR condt
2
/ln 12,
-
The cross-sectional areas are different (inner radius vs. outer radius)
-
Composite Cylindrical Walls
1 1 2 2(2 ) (2 ) ...
rtot
Tq UA
R
UA U r L U r L
Different overall heat
transfer coefficients.
-
Composite Cylindrical Walls
-
Spherical shell
T2T1
r2
r1
24rdT dT
q kA krdr dr
2
4
rq dr dTk r
2 2
1 1
1 22
1 2
4 ( )
4 1/ 1/
r Trrr T
t
q k T Tdr TdT q
k r r Rr
1 1
12
1
4 ( )
4 1/ 1/
r Trrr T
q k T TdrdT q
k r rr
11 1 2
1 2
1/ 1/( ) ( )
1/ 1/
r rT r T T T
r r