Transcript Document

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:
T(x,y,z)
- 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
Unidirectional heat
conduction (1D)
Area =
A
0
Solid bar, insulated on all
long sides (1D heat
conduction)
qx
x
x+x
x
A
q = Internal heat generation per unit vol. (W/m3)
qx+x
Unidirectional heat
conduction (1D)
First Law (energy balance)
( E in - E out ) + E gen = E st
q x - q x + x
E = (  Ax)u
E
u
T
=  Ax = Axc
t
t
t
q
q
x
E
+ A (  x ) q =
t
T
x
= - kA
x +  x
= q
x
q x
+
 x
x
Unidirectional heat conduction
(1D)(contd…)
T
T
  T
- kA
+ kA
+ A k
x
x
x  x
  T 
T
k
 + q =  c
x  x 
t
Longitudinal
conduction
Internal heat
generation
If k is a constant
T

 x + Axq =  Acx
t

Thermal inertia
 2T q  c
+ =
2
x
k
k
T 1 T
=
t a t
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 a
 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 :
Ts,1
Hot
fluid
….
. . ... ...
.. .. .. .. . …
k
..
. .............. ......... ..
.. .. ....... ...... .. .. .. ..
. . . ..... . .
x=0
Ts,2
Cold
fluid
T∞,2
x=L
d 
dT 
k
=0
dx 
dx 
Const. K; solution is:
dT
kA
Ts ,1 - Ts , 2
Ts ,1 - Ts , 2  =
qx = - kA
=
dx
L
L / kA
Thermal resistance
(electrical analogy)
OHM’s LAW :Flow of Electricity
V=IR elect
Voltage Drop = Current flow×Resistance
Thermal Analogy to Ohm’s
Law :
T = qRtherm
Temp Drop=Heat Flow×Resistance
1 D Heat Conduction through a
Plane Wall
T∞,1
Ts,1
Hot
fluid
….
. . ... ...
.. .. .. ... …
k
..
. .............. ......... ..
.. .. ....... ...... .. .. .. ..
. . . ..... . .
x=0
T∞,1
qx
Ts,1
1
h1 A
Ts,2
T∞,2
x=L
Ts,2
L
k A
Cold
fluid
T∞,2
1
h2 A
1
L
1
 Rt = h A + kA + h A
1
2
(Thermal Resistance )
Resistance expressions
THERMAL RESISTANCES
 Conduction
Rcond = x/kA
 Convection
Rconv = (hA)
 Fins
-1
Rfin = (h-
 Radiation(aprox)
1.5 -1
Rrad = [4AF(T1T2) ]
Composite Walls :
T∞,1
h1
A
B
KA
KB
C
KC
h2
T∞,2
T∞,1
qx
qx =
1
h1 A
T ,1 - T , 2
R
=
t
where,U =
1
Rtot A
LA
LB
LA
kA A
LB
kB A
T∞,2
LC
LC
kC A
1
h2 A
T ,1 - T , 2
= UAT
L
1
L
L
1
+ A + B + C +
h1 A
kA
kB
kC
h2 A
= Overall heat transfer coefficient
Overall Heat transfer Coefficient
1
1
U=
=
1
L
1
RtotalA
+
+
h1
k
h2
Contact Resistance :
TA
TB
A
B
Rt , c
T
=
qx
T
U =
LA
1
+
h1
kA
1
LC
LB
1
+
+
+
kB
kC
h2
Series-Parallel :
A
T1
B
KB
KA
C
Kc
AB+AC=AA=AD
D
KD
T2
LB=LC
Series-Parallel
(contd…)
T1
LA
kA A
LB
kB A
LD
kD A
LC
kC A
T2
Assumptions :
(1) Face between B and C is insulated.
(2) Uniform temperature at any face normal to X.
Example:
Consider a composite plane wall as shown:
kI = 20 W/mk
AI = 1 m2, L = 1m
qx
T1 =
0°C
Tf = 100°C
kII = 10 W/mk
h = 1000 W/ m2 k
AII = 1 m2, L = 1m
Develop an approximate solution for the rate of heat
transfer through the wall.
1 D Conduction(Radial
conduction in a composite
cylinder)
h1
r1
T∞,1
r2
h2
T∞,2 r k
3
2
k1
qr =
T∞,1
T∞,2
1
(h1 )(2r1 L)
ln
1
(h2 )(2r2 L)
r1
r2
2Lk1
ln
r2
r3
2Lk 2
T , 2 - T ,1
R
t
Critical Insulation
Thickness :
T∞
h
Insulation Thickness : r o-r i
ri
Ti
Objective :
r0
Rtot =
ln( rr0i )
2kL
+
1
( 2r0 L) h
decrease q , increases Rtot
Vary r0 ; as r0 increases ,first term
increases, second term decreases.
Critical Insulation
Thickness (contd…)
Maximum – Minimum problem
Set
dRtot
=0
dr0
1
1
=0
2
2kr0 L 2hLr 0
r0 =
k
h
Max or Min. ?
Take :
d 2 Rtot
=0
2
dr 0
at
r0 =
k
h
d 2 Rtot
-1
1
=
+
dr 2 0
2kr 2 0 L r 2 0 hL r0 = k
h
h2
=
0
2Lk 3
Critical Insulation
Thickness (contd…)
Minimum q at r0 =(k/h)=r c r (critical radius)
R tot
good for
electrical
cables
good for steam pipes etc.
R c r=k/h
r0
1D Conduction in Sphere
r2
r1
T∞,2
k
Inside Solid:
Ts,2
Ts,1
T∞,1
1 d 
 kr
2
r dr 
2
dT 
=0
dr 
 T ( r ) = Ts ,1 - Ts ,1 - T
 qr = - kA
 Rt ,cond

 1-  r / r  
1

s,2 
1

r
/
r



1
2 

dT
4 k Ts ,1 - Ts , 2 
=
1 / r1 - 1 / r2 
dr
1 / r1 - 1 / r2
=
4 k
Conduction with Thermal
Energy Generation
E
q =
= Energy generation per unit volume
V
Applications: * current carrying conductors
* chemically reacting systems
* nuclear reactors
Conduction with Thermal
Energy Generation
The Plane Wall :
k
Ts,1
q
T∞,1
Ts,2
Assumptions:
T∞,2
Hot
fluid
Cold
fluid
x= -L
x=0
x=+L
1D, steady state,
constant k,
uniform q
Conduction With Thermal
Energy Generation (contd…)
2
d T
dx
2
+
q
=0
k
Boundary
x = -L,
cond . :
x = +L,
Solution :
T = -
q
2k
x
2
T = Ts , 1
T = Ts , 2
+C x +C
1
2
Conduction with Thermal
Energy Generation (cont..)
Use boundary conditions to find C1 and C2
2
 x2  Ts , 2 -Ts ,1 x Ts , 2 +Ts ,1
q
L

1 - 2  +
+
Final solution : T =
2k  L 
2
L
2
Not linear any more
Heat flux :
dT


=
qx
k
dx
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
Cylinder with heat source
T∞ h
Assumptions:
1D, steady state, constant
k, uniform q
ro
r
Start with 1D heat equation in cylindrical
co-ordinates:
Ts
q
1 d  dT
r
r dr  dr
 q
 + =0
 k
Cylinder With Heat Source
Boundary cond. : r = r0 ,
T = Ts
dT
=0
r = 0,
dr
2 

q 2  r 
=
Solution : T (r )
r0 1 - 2  +Ts
4k  r0 
Ts may not be known. Instead, T and h may be specified.
Exercise: Eliminate Ts, using T and h.
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.