View
216
Download
3
Category
Preview:
Citation preview
Objectives of conduction analysis
To determine the temperature field, T(x,y,z,t), in a body(i.e. how temperature varies with position within the body) T(x,y,z,t) depends on:
- boundary conditions - initial condition
- material properties (k, cp, …) - geometry of the body (shape, size)
Why we need T(x,y,z,t) ? - to compute heat flux at any location (using Fourier’s eqn.) - compute thermal stresses, expansion, deflection due to temp. etc. - design insulation thickness - chip temperature calculation - heat treatment of metals
T(x,y,z)
0Area =
A xx x+x
= Internal heat generation per unit vol. (W/m3)
qx qx+xA
Solid bar, insulated on all long sides (1D heat conduction)
Unidirectional heat conduction (1D)
Unidirectional heat conduction (1D)
First Law (energy balance)
t
EqxAqq xxx
)(
stgenoutin EEEE )(
t
TxcA
t
uxA
t
E
uxAE
)(
xx
qqq
x
TkAq
xxxx
x
If k is a constantt
T
t
T
k
c
k
q
x
T
1
2
2
Unidirectional heat conduction (1D)(contd…)
t
T cq
x
Tk
x
t
TxAcqxAx
x
Tk
xA
x
TkA
x
TkA
Longitudinal conduction
Thermal inertiaInternal heat generation
Unidirectional heat conduction (1D)(contd…)
For T to rise, LHS must be positive (heat input is positive)
For a fixed heat input, T rises faster for higher In this special case, heat flow is 1D. If sides were not
insulated, heat flow could be 2D, 3D.
Boundary and Initial conditions:
The objective of deriving the heat diffusion equation is to determine the temperature distribution within the conducting body.
We have set up a differential equation, with T as the dependent variable. The solution will give us T(x,y,z). Solution depends on boundary conditions (BC) and initial conditions (IC).
Boundary and Initial conditions (contd…)
How many BC’s and IC’s ?
- Heat equation is second order in spatial coordinate. Hence, 2 BC’s needed for each coordinate.
* 1D problem: 2 BC in x-direction * 2D problem: 2 BC in x-direction, 2 in y-direction
* 3D problem: 2 in x-dir., 2 in y-dir., and 2 in z-dir.
- Heat equation is first order in time. Hence one IC needed
1- Dimensional Heat Conduction
The Plane Wall :
.. .. . . … . . . . . . .. . . . . . . . . . .
. . . . . .
.. . . . . . . . . . .. . . . . . . . . . . . . .
…. . . . . . .. .. .. . . . . .. . .. . . . . .. . . . . . .
. .
k
T∞,2
Ts,1Ts,2
x=0 x=LHot fluid
Cold fluid
0
dx
dTk
dx
d
kAL
TTTT
L
kA
dx
dTkAq
ssssx
/
2,1,2,1,
Const. K; solution is:
Thermal resistance(electrical analogy)
OHM’s LAW :Flow of Electricity
V=IR elect
Voltage Drop = Current flow×Resistance
Thermal Analogy to Ohm’s Law :
thermqRT
Temp Drop=Heat Flow×Resistance
1 D Heat Conduction through a Plane Wall
qx
T∞,1 Ts,1 Ts,2 T∞,2
Ak
LAh2
1
Ah1
1
.. .. . . … . . . . . . .. . . . . . . . . . .
. . . . . .
.. . . . . . . . . . .. . . . . . . . . . . . . .
…. . . . . . .. .. .. . . . . .. . .. . . . . .. . . . . . .
. .
k
T∞,2
Ts,1Ts,2
x=0 x=LHot fluid
Cold fluid
T∞,1
AhkA
L
AhRt
21
11(Thermal Resistance )
Resistance expressions
THERMAL RESISTANCES
Conduction Rcond = x/kA
Convection Rconv = (hA)-1
Fins Rfin = (h
Radiation(aprox) Rrad = [4AF(T1T2)
1.5]-1
Composite Walls :
A B CT∞,1
T∞,2
h1
h2K A K B K C
L A L B L C
q x
T∞,1 T∞,2
Ah1
1Ah2
1Ak
L
A
A
Ak
L
B
B
Ak
L
C
C
TUA
Ahk
L
kL
kL
Ah
TT
R
TTq
C
C
B
B
A
Atx
21
2,1,2,1,
11
AR
Uwheretot
1, Overall heat transfer coefficient
Overall Heat transfer Coefficient
21
total
h1
kL
h1
1
AR
1U
Contact Resistance :
A B
TA
TB T
xc,t
q
TR
21
111
hk
L
k
L
k
L
h
U
C
C
B
B
A
A
Series-Parallel :
AB
D
CT1 T2
K c
K DK A
K B
AB+AC=AA=AD
LB=LC
Series-Parallel (contd…)
Assumptions :
(1) Face between B and C is insulated.
(2) Uniform temperature at any face normal to X.
T1T2
Ak
L
A
A
Ak
L
D
DAk
L
B
B
Ak
L
C
C
Consider a composite plane wall as shown:
Develop an approximate solution for the rate of heat transfer through the wall.
Example:
kI = 20 W/mk
AI = 1 m2, L = 1m
kII = 10 W/mk
AII = 1 m2, L = 1m
T1 = 0°C Tf = 100°C
h = 1000 W/ m2 k
qx
1 D Conduction(Radial conduction in a composite
cylinder)h1
T∞,1
k1
r1
r2
k2T∞,2 r3
h2
T∞,1 T∞,2
)2)((
1
11 Lrh )2)((
1
22 Lrh
12
ln2
1
Lkrr
22
ln3
2
Lkrr
t
r R
TTq 1,2,
Critical Insulation Thickness :
r0ri
T∞
Tih
Insulation Thickness : r o-r i
hLrkLR ir
r
tot )2(
1
2
)ln(
0
0
Objective : decrease q , increases
Vary r0 ; as r0 increases ,first term increases, second term decreases.
totR
Critical Insulation Thickness (contd…)
Set 00
dr
dRtot
02
1
2
1
02
0
hLrLkr
h
kr 0
Max or Min. ? Take : 00
2
2
dr
Rd tot ath
kr 0
h
kr
tot
hLrLkrdr
Rd
0
02
02
02
2 1
2
1
02 3
2
Lk
h
Maximum – Minimum problem
Minimum q at r0 =(k/h)=r c r (critical radius)
good for steam pipes etc.good for electrical
cablesR c r=k/h
r0
R t o t
Critical Insulation Thickness (contd…)
1D Conduction in Sphere1D Conduction in Sphere
r1
r2
k
Ts,1
Ts,2
T∞,1
T∞,2
k
rrR
rr
TTk
dr
dTkAq
TTTrT
dr
dTkr
dr
d
r
condt
ssr
rr
rrsss
4
/1/1
/1/1
4
)(
01
21,
21
2,1,
/1
/12,1,1,
2
2
21
1
Inside Solid:
Applications: * current carrying conductors
* chemically reacting systems
* nuclear reactors
= Energy generation per unit volumeV
Eq
Conduction with Thermal Energy Generation
Conduction with Thermal Energy Generation
The Plane Wall :
k
Ts,1
Ts,2
x=0 x=+L
Hot fluid
Cold fluid
T∞,2T∞,1
x= -L
q Assumptions:
1D, steady state, constant k,
uniform q
Conduction With Thermal Energy Generation (contd…)
CxCxk
qTSolution
TTLx
TTLxcondBoundary
k
q
dx
Td
s
s
2:
,
,:.
0
21
2
2,
1,
2
2
Not linear any more
Derive the expression and show that it is not independent of x any more
Hence thermal resistance concept is not correct to use when there is internal heat generation
dxdT
kqfluxHeat
TT
LxTT
Lx
kLq
TsolutionFinal
CandCfindtoconditionsboundaryUse
x
ssss
:
221
2: 1,2,1,2,
2
22
21
Conduction with Thermal Energy Generation (cont..)
Cylinder with heat source
Assumptions:
1D, steady state, constant k, uniform
Start with 1D heat equation in cylindrical co-ordinates:
kq
drdT
rdrd
r
01
ro
r
Ts
T∞ h
q
q
rk
Trcond
04
,:.
s
s
Tr
rq
rTSolution
dr
dTr
TrBoundary
2
22
0
0
1)(:
0,0
Ts may not be known. Instead, T and h may be specified.
Exercise: Eliminate Ts, using T and h.
Cylinder With Heat Source
Cylinder with heat source(contd…)
Example:
A current of 100A is passed through a stainless steel wire having a thermal conductivity K=25W/mK, diameter 3mm, and electrical resistivity R = 2.0 . The length of the wire is 1m. The wire is submerged in a liquid at 100°C, and the heat transfer coefficient is 10W/m2K. Calculate the centre temperature of the wire at steady state condition.
Recommended