bahan uliah perpindahan panas

1. Transient
Conduction

2. 2D Conduction
ME 309 2
Transient Conduction
• All heat transfer problems are time-dependent
– Slow ones may be considered steady-state
• Changing environment, operating conditions, and feedback control lead
time-dependent transient problems
• Real problems may include finite and semi-infinite solids, or complex
geometries, as well as two and three dimensional conduction
• Solution techniques involve the lumped capacitance method, exact and
approximate solutions, and finite difference methods.
 Lumped Capacitance Method can be used for solids within which temperature
gradients are negligible (Sections 5.1-5.2)

3. 2D Conduction
ME 309 3
i
T
x
T 
)
0
,
(
x
Lumped Capacitance Method
• A hot metal is initially at a uniform temperature, Ti , and at t=0 is
quenched by immersion in a cool liquid, of lower temperature
 The temperature of the solid decreases in time, t>0, due to
convection heat transfer at the solid-liquid interface, until it reaches

T
T

T

T

4. 2D Conduction
ME 309 4
Lumped Capacitance Method
• If the thermal conductivity of the solid is very high, resistance to
conduction within the solid will be small compared to resistance to
heat transfer between solid and surroundings.
• Temperature gradients within the solid will be negligible, i.e.. the
temperature of the solid is spatially uniform at any instant.
i
T
x
T 
)
0
,
(
x
T

5. 2D Conduction
ME 309 5





























































1
1
1
0
ln
ln
,
;
)
(
1
1
th
i
th
th
t
s
s
i
i
s
t
t
d
dt
d
hA
Vc
dt
dt
d
hA
Vc
T
T
T
T
dt
dT
Vc
T
T
hA
i

6. 2D Conduction
ME 309 6
Lumped Capacitance Method
Overall energy balance on the cooling solid:
dt
dT
Vc
T
T
hAs 


 )
(
The temperature of the solid at a specified time t is:
























t
Vc
hA
T
T
T
T s
i
i
exp
The total energy transfer, Q, occurring up to some time t is:
st
out E
E 
 

where 


 T
T
The time required for the solid to reach a temperature T is:



 i
s
hA
Vc
t ln



 T
Ti
i

 

t
t
s dt
hA
dt
q
Q
0
0















i
t
i
Vc
Q 

 exp
1
)
(

7. 2D Conduction
ME 309 7
Transient Temperature Response
Based on eq. (5.2), the temperature difference between solid and fluid
decays exponentially.
 Thermal time constant
t
t
s
t C
R
Vc
hA











 )
(
1
Rt is the resistance to
convection heat transfer,
Ct is the lumped thermal
capacitance of the solid
 Increase in Rt or Ct causes
solid to respond more slowly
and more time will be required
to reach thermal equilibrium.

8. 2D Conduction
ME 309 8
Validity of Lumped Capacitance Method
 A suitable criterion to determine validity of method:
relative magnitudes of temperature drop in the solid to the
temperature difference between surface and fluid.
Bi
k
hL
R
R
hA
kA
L
T
T
conv
cond
liquid
solid
solid






)
/
1
(
)
/
(
)
convection
to
due
(
/
)
conduction
to
due
(
? What should be the relative magnitude of T solid versus T
solid/liquid for the lumped capacitance method to be valid?

9. 2D Conduction
ME 309 9
Biot and Fourier Numbers
The lumped capacitance method is valid when
1
.
0


k
hL
Bi c where the characteristic length:
Lc=V/As=Volume of solid/surface area
We can also define a “dimensionless time”, the Fourier number:
c
k
L
t
Fo
c 



 ;
2
Nondimensional form of Eq. (5.2) valid for all bulk metal cooling
cases
 
Fo
Bi
T
T
T
T
i
i










exp (5.4)

10. 2D Conduction
ME 309 10
Charging a thermal energy storage system consisting of a packed bed of aluminum
spheres.
How long does it take to accumulate 90% of the maximum possible thermal energy?
Aluminum sphere
D = 75 mm, T = 25 C
i
o
Gas
T C
g,i
o
= 300
h = 75 W/m -K
2
= 2700 kg/m3
k = 240 W/m-K
c = 950 J/kg-K
Problem 5.11
Assumptions:
• Radiation between spheres?
• Conduction between spheres?
• Properties?
• Conduction within the sphere?
• Lumped capacitance?

11. 2D Conduction
ME 309 11
1.Validity of lumped capacitance
• Calculate the Bi number, hL/k = 75*(0.075)/240 =
0.02
• Time to accumulate 90% of the maximum
possible thermal energy
 
st
t
i
E
0.90 1 exp t /
cV

 

    
3
t s 2
2700 kg / m 0.075m 950 J / kg K
Vc / hA Dc / 6h 427s.
6 75 W / m K
  
  
   
 
 
t
t ln 0.1 427s 2.30 984s

    

12. 2D Conduction
ME 309 12
Example (Problem 5.6 Textbook)
The heat transfer coefficient for air flowing over a sphere is
to be determined by observing the temperature-time history
of a sphere fabricated from pure copper. The sphere, which is
12.7 mm in diameter, is at 66°C before it is inserted into an
air stream having a temperature of 27°C. A thermocouple on
the outer surface of the sphere indicates 55°C, 69 s after the
sphere is inserted in the air stream.
 Calculate the heat transfer coefficient, assuming that the
sphere behaves as a spacewise isothermal object. Is your
assumption reasonable?

13. 2D Conduction
ME 309 13

14. 2D Conduction
ME 309 14

15. 2D Conduction
ME 309 15
Heating of coated furnace wall
 
1
1
1 2 2 2
tot f 2
1 1
U R R 10 m K/W 20 W/m K.
h 25 W/m K


 
 
 
 
       
 
 
  
 
2
UL 20 W/m K 0.01 m
Bi 0.0033 1
k 60 W/m K
 
   

Heat transfer to the wall is determined by the total resistance to heat transfer
from the gas to the metal surface

16. 2D Conduction
ME 309 16
Other transient problems
• When the lumped capacitance analysis is not valid, we
must solve the partial differential equations analytically or
numerically
• Exact and approximate solutions may be used
• Tabulated values of coefficients used in the solutions of
these equations are available
• Transient temperature distributions for commonly
encountered problems involving semi-infinite solids can be
found in the literature

17. 2D Conduction
ME 309 17
General lumped capacitance analysis
gen
out
in
st
E
E
E
dt
dT
mc
dt
dE 

 




18. 2D Conduction
ME 309 18
Effect of the Bi number
T∞ ,h
T∞ ,h
Bi<<1 Bi ~ 1 Bi>>1
T

19. 2D Conduction
ME 309 19
• Analytical solution for convection + radiation at
the same time does not exist
• Numerical solution
– Discretize the time-derivative of temperature as
• Radiation + convection
t
t
T
t
t
T
dt
dT





)
(
)
(
   
   
 
4
4
4
4
4
4
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
walls
s
s
walls
s
walls
s
T
t
T
A
T
t
T
hA
t
t
T
t
t
T
T
t
T
T
t
T
hA
t
t
T
t
t
T
T
T
T
T
hA
dt
dT
cV































20. 2D Conduction
ME 309 20
Convection + Radiation from a metal sphere
convection
radiation
k,R
T∞,h
Tw

21. 2D Conduction
ME 309 21
MATLAB code
h = 10; % convection coeff
R = 1e-2; % 1 cm radius
V = 4/3*pi*R^3;
A = 4*pi*R^2;
Cp = 300; % specific heat
rho = 3000; % density
k = 200; % conductivity
Tinf = 300;
Twall = 20;
T0 = 20;
sig = 5.7e-8;
eps = 0.0;
%
tau_c = h*A/rho/Cp/V;% time constant
const_r = sig*eps*A/rho/Cp/V; % ??
t0 = 0; % initial time
tF = 3000; % final time
dt = 1; % delta t
% initialization
tt = zeros(ceil((tF-t0)/dt+1),1);
Tc = zeros(size(tt));
Tr = zeros(size(tt));
Tcr = zeros(size(tt));
Ttc = T0; % initial temperatures
Ttr = T0; % ‘’
Ttrc = T0; % ‘’
i = 0; % initialization of index i
% start numerical integration
% from t0 to tF
for t = t0:dt:tF
i = i+1;
tt(i) = t;
% convection term
conv = tau_c*(Ttc-Tinf);
% radiation term
rad = const_r*(Ttr^4-Twall^4);
convrad = tau_c*(Ttrc - Tinf)...
+const_r*(Ttrc^4-Twall^4);
% update temperatures
Tc(i) = Ttc - dt*conv;
Tr(i) = Ttr - dt*rad;
Tcr(i) = Ttrc - dt*convrad;
% store for the next step
Ttc = Tc(i);
Ttr = Tr(i);
Ttrc = Tcr(i);
end
plot(tt,Tc,tt,Tr,tt,Tcr)

22. 2D Conduction
ME 309 22
Lumped capacity solution w/ variable ambient temperature
(advanced exercise) see Lienhard
• A cold object is suddenly immersed into a bath whose temperature increases
linearly in time
T = Ti object’s initial temperature (constant)
T∞= Ti + bt bath temperature (b > 0)
Energy balance: (Hint: show this as an exercise)
Solution:
A fine exercise in differential algebra…
see the next page
th
th
th
i
th
bt
dt
d
T
bt
T
dt
dT













23. 2D Conduction
ME 309 23
Analytical solution to 1st order system’s response to a ramp input
  th
th
th
th
th
t
th
t
t
th
t
th
t
e
bt
dt
e
d
e
bt
e
e
dt
d














/
/
/
/
/
*
*
1
*
th
th
th
i
th
bt
dt
d
T
bt
T
dt
dT












Multiply both sides with et/τ
Regroup the left-hand-side

24. 2D Conduction
ME 309 24
( )
( )
{ { {
( )
/
/ / /
/ / / /
/ / /
/
1
* ( )
* *
th
th th th
th th th th
th th th
th
t
t t t
th th
t t t t
v
u uv
th du
dv
t t t
th
t
th
d e bt bt
e d e e dt
dt
e bt d e bt e e d bt
e bt e b e C
b t Ce
t
t t t
t t t t
t t t
t
q
q
t t
q
t
q t
q t -
= Þ =
æ ö
÷
ç ÷
Þ = = -
ç ÷
ç ÷
ç
è ø
= - +
Þ = - +
ò ò
ò ò
14442 4443
14444
42 4444
4
3
Analytical solution to 1st order system’s response to a ramp input
Use integration by parts for
the right-hand-side
Divide by et/T

25. 2D Conduction
ME 309 25
Temperature measurement of a chamber with linearly varying
temperature
• Consider a thermometer with a thermal time
constant, th, that measures the temperature of
a chamber. If the chamber temperature
changes linearly in time. How will the
thermometer report the temperature
variation?
T
time
thermometer’s response
T-Ti = b(t-th(1 - e-t/τ))
temperature lag, bth
T = Ti
T = Ti - bth

26. 2D Conduction
ME 309 26
Nondimensionalization
• For similar geometries objects behave the
same way
• Nondimensional numbers such as Bi and Fo
can be used to identify the conditions
– Bi: helps us to validate the lumped capacitance
assumption
– Fo: can be used to set the final time of a
simulation.
• Typically Fo = 10 is more than enough to reach a
steady-state.
 
  object
heating
object
cooling
Fo
Bi
T
T
T
T
Fo
Bi
T
T
T
T
i
i
i
i





















exp
1
exp
Fo
θ/θ
i

27. 2D Conduction
ME 309 27
Consecutive heating and cooling by convection
Ambient temperature
T
Temperature response, T
,
h T¥ , ,
m c T
T¥

28. 2D Conduction
ME 309 28
Nondimensionalization of unsteady problems
Governing equation for one-dimensional unsteady problem
Initial condition: T(x,0) = Ti
Convective boundary conditions:
Temperature distribution function:
T = T(x,t,Ti,T∞,L,k,α,h)
t
T
x
T





 1
2
2
 
 








 T
t
L
T
h
x
T
k
L
x
,
T∞,h
L
-L
x
T(x,0) = Ti

29. 2D Conduction
ME 309 29
• Buckingham-pi theorem:
Number of dimensionless groups =
number of parameters
- number of fundamental units used in the parameters
T = T(x,t,Ti,T∞,L,k,α,h)
Number of parameters = 8
Number of fundamental units used in the parameters = 4 (m,s,kg,K)
There are total of 4 dimensionless groups:
?
?
?
?

30. 2D Conduction
ME 309 30
Nondimensionless groups of unsteady convection
k
hL
Bi
L
t
Fo
L
x
x
T
T
T
T
i
i
/
/
/
2
*









 
Fo
Bi
T
T
T
T
i
i










exp
Lumped capacitance uses only 3 of them. Why?

31. 2D Conduction
ME 309 31
Spatial effects (role of temperature distribution in the solid)
t
T
x
T





 1
2
2
T∞,h
L
-L
x
T(x,0) = Ti
Nondimensional form of the governing equation
 
 








 T
t
L
T
h
x
T
k
L
x
,
Governing equation
Boundary condition
Fo
x 





 *
2
*
*
2
Nondimensional form of the boundary condition ???

32. 2D Conduction
ME 309 32
Spatial effects (role of temperature distribution in the solid)
t
T
x
T





 1
2
2
T∞,h
L
-L
x
T(x,0) = Ti
Nondimensional form of the governing equation
 
 








 T
t
L
T
h
x
T
k
L
x
,
Governing equation
Boundary condition
Fo
x 





 *
2
*
*
2
Nondimensional form of the boundary condition ???
*
*
*
? 






x

33. 2D Conduction
ME 309 33
Spatial effects (role of temperature distribution in the solid)
t
T
x
T





 1
2
2
T∞,h
L
-L
x
T(x,0) = Ti
Nondimensional form of the governing equation
 
 








 T
t
L
T
h
x
T
k
L
x
,
Governing equation
Boundary condition
Fo
x 





 *
2
*
*
2
Nondimensional form of the boundary condition
*
*
*
*
*
*















Bi
x
k
hL
x

34. 2D Conduction
ME 309 34
Convective cooling of slabs for large Bi
from A.F. Mills, Basic Heat & Mass Transfer
Governing equation
Initial condition: θ*(x*,0) = (T-Ts) /(Ti-Ts) = 1
Boundary conditions: θ*(1,t*) = θ*(-1,t*) = 0
Solution by separation of variables:
“drop the stars from notation for convenience”
θ = Z(t)*X(x)
Ts
Ts
L
-L
x
T(x,0) = Ti
Fo
t
t
x







 *
*
*
2
*
*
2
;

35. 2D Conduction
ME 309 35
   
    t
t
t
e
x
C
C
e
x
C
C
x
C
x
C
X
e
C
Z
dt
dZ
Z
dx
X
d
X
2
2
2
sin
cos
sin
cos
1
1
3
1
2
1
3
2
1
2
2
2























 Plug in θ = XZ into the governing
equation
Solve each homogeneous part
Put each solution into
θ to apply BC/IC

36. 2D Conduction
ME 309 36
θ = 0
L
-L
x
θ(x,0) = 1
Because of symmetry:
1. θ(x,t) = θ(-x,t)
2. dθ/dx = 0 at x = 0
    t
t
e
x
C
C
e
x
C
C
2
2
sin
cos 3
1
2
1













0
How many unknowns?
How many conditions left?
θ=0

37. 2D Conduction
ME 309 37
θ=0
L
-L
x
θ(x,0)=1
θ=0
 
?
0
cos
)
,
1
(
2
2
1








 
 t
e
C
C
t

38. 2D Conduction
ME 309 38
θ=0
L
-L
x
θ(x,0)=1
θ=0
 
?
;
)
,
(
)
,
(
2
1
cos
)
,
(
,...
2
,
1
,
0
;
2
1
0
cos
)
,
1
(
1
2
1
2
1
2
2
2
























































n
n
n
t
n
n
n
t
C
t
x
t
x
e
x
n
C
t
x
n
n
e
C
C
t

39. 2D Conduction
ME 309 39







































































)
2
/
1
(
)
1
(
2
2
1
cos
2
1
2
1
cos
)
0
,
(
2
1
cos
)
,
(
1
0
2
0
2
1 2
2
n
C
dx
x
n
C
x
n
C
x
e
x
n
C
t
x
n
n
n
n
n
t
n
n
n
Famous orthogonality condition: Multiply both sides by cos()
and integrate over x.

40. 2D Conduction
ME 309 40
θ(x,t)
x
increasing time
Temperature profile
0
1
Fo = 1
Fo = .01

41. 2D Conduction
ME 309 41
Semi-infinite solid
• A solid that is initially of
uniform temperature Ti and is
assumed to extend to infinity
from a surface at which
thermal conditions are altered.
• Examples:
1. Large objects with low k
2. Immediate vicinity of any object
in short term

42. 2D Conduction
ME 309 42
Similarity solution
;
2
2
t
x 






Governing equation:
Boundary condition: θ(0,t) = 1
Initial condition: θ(x,0) = 0
nondimensional form ‘*’s
are dropped for convenience
Step temperature input
x  ∞
t
x
4


Introduce a new variable:
Governing equation becomes?

43. 2D Conduction
ME 309 43
Similarity solution


















































d
d
d
d
t
x
d
d
t
x
t
d
d
t
x
d
d
t
x
d
d
x
t
x
t
x
2
4
4
1
2
1
2
;
2
2
2
2
2
/
3
2
2
2
2
2
2
Rewrite the governing equation
in terms of η
Final equation is
an ODE
t = Fo

44. 2D Conduction
ME 309 44
Similarity solution
   
0, ,0
s i
T t T T x T
  
 
, x
erf
2 t
s
i s
T x t T
T T 
  
  
  
 
s i
s
k T T
q
t


 
See Incropera for flux and convection Boundary Conditions at the
surface
Heat flux at the
surface