MHMT10
Momentum Heat Mass Transfer
D
     source
Dt
Heat transferconduction
Multidimensional heat conduction problems. Fins and
heat conduction with internal sources or sinks. Unsteady
heat conduction in solids. Penetration theory.
Rudolf Žitný, Ústav procesní a
zpracovatelské techniky ČVUT FS 2010
MHMT10
Heat transfer - conduction
Thermal resistance of fluid (thermal boundary layer) can be expressed in terms of
heat transfer coefficients  added to the thermal resistances of solid layers.
For example resulting RT of serial resistances of fluid and
two concentric pipes can be expressed as (see previous
lecture)
1
2
Tf1
1
D1
Dm
Tf2
2
Tf 1  Tf 2
RT 


Q

D
1
D
1
1
ln m 
ln 2 

21 L D1 22 L Dm D11 L D2 2 L

 
 

 


 
 
1
inner tube
outer tube(insulation)
D2
that only the thickness of insulation (outer diameter D2) can be changed. Then
the thermal resistance (and effectiveness of insulation) depends only upon D 2,
see graph calculated for D1=0.02 m, Dm=0.021, 1=40 W/(m.K) (steel), 2=0.1
(insulation), 1=1000, 2=5 (natural convection) . Up to a critical D2 the thermal
resistance DECREASES with the increasing thickness of insulation!
heat transfer to
ambient fluid
3
2,9
RT [K/W]
Example (critical thickness of insulation): Let us assume
heat transfer
in pipe
2,8
2,7
2,6
0,02
0,03
0,04
0,05
D2 [m ]
0,06
0,07
MHMT10
Heat transfer - conduction
Thermal resistance of a composite tube with circular or spiral fins attached to outer
tube (fins can be also an integral part of the outer tube).
2
Resulting thermal resistance is calculated according to
almost the same expression as in the previous case
1
Tf1
1
D1
Dm
D2
Tf2
2
RT 
Tf 1  Tf 2
Q

Dm
D2
1
1
1

ln

ln


21 L D1
22 L Dm  D11 L S 2 2 L
1
inner tube
outer tube (without fins)
heat transfer
in pipe
heat transfer
from fin to
ambient fluid
with the only but very important difference: Instead of the outer surface of plain tube
D2L is used the overall outer surface of fins S2L. Such a modification assumes that
the 2 on the surface of fins is the same as on the surface of tube and first of all that
the thermal resistance of fin itself is negligible (simple speaking it is assumed that
the fin is made of material having infinite value of thermal conductivity ).
MHMT10
Heat transfer - conduction
The assumption of perfectly conductive fin is unacceptable and in reality the thermal
resistance of fin must be respected by multiplying the surface S2 by fin’s efficiency
fin which depends upon thermal conductivity of fin, its geometry (thickness and
height) and also upon the heat transfer coefficient 2.
, Tf
Tw
b
T(x)
Efficiency of a thin rectangular fin can be derived easily by solution of temperature
profile along the height of fin
d 2T 2
0 2 
(T f  T )
dx
b
This is FK equation with a source term, representing heat transfer
from both sides of surface (2dx) to the control volume (bdx)
with boundary conditions at the heel of fin (T(x=0)=Tw) and at top of fin dT/dx=0
(there is no heat flux at x=H)
dx
x
H
2
2
)
exp( x
)
b

b

T ( x)  T f  (Tw  T f )(

)
2
2
1  exp( H
) 1  exp(  H
)
b
b
exp( x
 fin 
Q
Q 
dT
|x  0
2b
 H 2 tanh Bi
dx


tanh

2
2 H (Tw  T f )
H
2b
Bi
b
efficiency
Efficiency of fin is calculated from temperature gradient at the heel of fin (the
gradient determines heat flux at the heel)
1
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
0
5
10
15
Bi 20
MHMT10
Conduction - nonstationary
MHMT10
Conduction - nonstationary
Temperature distribution in unsteady case generally depends upon time t and
coordinates x,y,z. Sometimes, when the temperature distribution is almost
homogeneous inside the whole body, the partial differential Fourier Kirchhoff
equation reduces to an ordinary differential equation. This simplification is correct
if the thermal resistance of solid is much less than the thermal resistance of fluid,
more specifically if Biot number is small enough
D
Bi 
here D is a characteristic diameter of a solid
object and s is thermal conductivity of solid
s
Fourier Kirchhoff equation can be integrated over the whole volume of solid
 cp
V
Ts
dv    qdv    n qds    (T f  Ts )ds
t
V
S
S
which reduces to ordinary dif. equation as soon as Ts depends only on time
 c pV
dTs
  S (T f  Ts )
dt
Ts  T f  (T0  T f ) exp(
S
t)
 c pV
 0 .1
MHMT10
Conduction - nonstationary
As soon as the Biot number is large (Bi>0.1, therefore if the solid body is too big, for
example semi-infinite space) it is necessary to solve the parabolic partial differential
Fourier Kirchhoff equation. For the case that the solid body is homogeneous (constant
thermal conductivity, density and specific heat capacity) and without internal heat
sources the FK equation reduces to
T
 2T  2T  2T
 a( 2  2  2 )
t
x
y
z
The coefficient of temperature diffusivity a=/cp is the ratio of
temperature conductivity and thermal inertia
with the boundary conditions of the same kind as in the steady state case and with
initial conditions (temperature distribution at time t=0).
This solution T(t,x,y,z) can be expressed for simple geometries in an analytical form
(heating brick, plate, cylinder, sphere) or numerically in case of more complicated
geometries.
MHMT10
Conduction - nonstationary
Start up flow of viscous liquid in halfspace (solved in lecture 4) was described by
equation which is identical with the Fourier Kirchhoff equation for one dimensional
temperature distribution in halfspace and with the step change of surface temperature
as a boundary condition:
T
 2T
T(t=0,x)=T0 T(t,x=0)=Tw
a 2
t
x
Exactly the same solution as for the start up flow
(complementary error function erfc) holds for
dimensionless temperature 
Tw
  2
 2,

x
t
T0
δ
x

T  T0
,
Tw  T0
  at
d 2
d
 2
0
2
d
d
x
2 
  1

2


2
exp(


)d  erfc(

0
x
)
2 at
MHMT10
Conduction - nonstationary
Erfc function describes temperature response to a unit step at surface (jump from
zero to a constant value 1). The case with prescribed time course of temperature at
surface Tw(t) can be solved by using the superposition principle and the response can
be expressed as a convolution integral.
t
T (t , x)   E (t   , x)
Tw()
Tw ( )d
0 impulse response short pulse (wall
temperature at time  )


d
x
Time course Tw(t) can be substituted by short pulses
t
Temperature at a distance x is the sum of
responses to short pulses Tw()d
The function E(t,,x)=E(t-,x) is the impulse function (response at a distance x to a
temperature pulse of infinitely short duration but unit area – Dirac delta function).
The impulse response can be derived from derivative of the erfc function
x
x2
E (t , x) 
exp( 
)
3/ 2
4at
2 at
MHMT10
Penetration theory
Still too complicated? Your pocket calculator is not equipped with the erffunction? Use the acceptable approximation by linear temperature profile,
(exactly the same procedure as with the start up flow in a half-space)

T
T
dx


a
|x  0
0 t
x
Tw

t
T0
Tw

Tdx

a
t 0

t+t
x
T
 Tw
(
)a w
t 2

Integrate Fourier equations
(up to this step it is
accurate)
Approximate
temperature
profile by line
δ
+Δ
Result is ODE for thickness  as a function of time
  4at
Using the exact temperature
profile predicted by erffunction, the penetration
depth slightly differs =(at)
MHMT10
Penetration theory
=at penetration depth. Extremely simple and important result, it
gives us prediction of how far the temperature change penetrates at the time t.
This estimate enables prediction of thermal and momentum boundary layers
thickness etc. The same formula can be used for calculation of penetration depth
in diffusion, replacing temperature diffusivity a by the diffusion coefficient DA .
Wire Cu
=0.11 m
=398 W/m/K
=8930 kg/m3
Cp=386 J/kg/K

MHMT10
Penetration theory and 
The penetration theory can be applied also in the case that the semi-infinite
space is in contact with fluid and surface temperature depends upon
temperature of fluid and the heat transfer coefficient

Tf
Tw  T0

  (T f  Tw )
Tw
Derive the result as a homework
t
T0
δ
+Δ
t+t
x

 2

  2  2( ) ln(1   )  4at



2
MHMT10
PLATE - finite depth
In case of a finite thickness plate the penetration theory can be used only for
short times (small Fourier number < 0.1)
Let us define Fourier number and Biot number
in terms of half thickness of plate H/2
Bi<0.1
Bi1
Bi
Tf
Tf
Tw=Tf
Fo>1
Fo>1
Tw
Fo<1
Tw
Fo<1
T0
T0
T0
H/2
x
T  Tw
H / 2 x


T0  Tw
4at
Penetration theory
4 t
Fo 
cp H 2
H
Bi 
2
H/2
H/2
Long times (large
Fourier) and finite Biot..
The most complicated
case/see next slide
Fourier method

T  Tf
T0  T f
 exp(
2 t
)
cp H
= exp( Bi  Fo)
Integral method
MHMT10
PLATE - finite depth Fourier method
Using dimensionless temperature , distance , time  (Fourier number) and
dimensionless heat transfer coefficient Bi (Biot number)

T  Tf
T0  T f

2x
H

4t
cp H 2
Bi 
H
2
the Fourier Kirchhoff equation, boundary and initial conditions
T
T
T
 2T
|x  0  0

|


(
T

T
)
T  T0
cp
 2
x  H /2
f
x
x
t
x
are transformed to


  2
| 0  0   1
| 1   Bi
 2




 
Bi1
Fourier method is based upon superposition of solutions
Tf
satisfying differential equation
2
2

F

Gi
1
1

F

G
2
i
i
i

(

,

)

F
(

)
G
(

)




G

F
i
i
i
i
2
Tw
i
i
Fi  Gi 

 2
t
T0
H/2
x
and boundary conditions
Gi
| 1   BiGi

depends
on  only
depends
on  only
Gi
| 0  0

MHMT10
PLATE - finite depth
Spatial component Gi() follows from
 2Gi
 i2Gi  0  Gi ( )  cos( i )
2

The function cos() automatically satisfies the boundary condition at =0 for
arbitrary . The boundary condition at wall is satisfied only for yet undetermined
values , roots of transcendental equation
i sin( i )  Bi cos( i )
and this equation must be solved numerically, giving infinite series of roots 1, 2,…
These eigenvalues i determine also the temporary components Fi
Fi
  i2 Fi  0  Fi ( )  exp(  i2 )

Final temperature distribution is the infinite series of these elementary solutions

 ( ,  )   ci cos( i ) exp( i2 )
i 1
each term satisfies FK equation
and both boundary conditions
MHMT10
PLATE - finite depth

The coefficients ci are determined by the initial condition 1   ci cos(  i )
i 1
To solve the coefficients ci from this identity (which should be satisfied for
arbitrary ) it is convenient to utilise orthogonality of functions Gi() and Gj() that
follows from original ordinary differential equations and boundary conditions
1
1
1
d 2G j
d 2Gi
2
2
0 cos( i ) cos(  j )d  0
0 (Gi d 2  G j d 2 )d   ( i   j ) 0 GiG j d
this is
for ij
nonzero
this is zero (integrate per parts and
use boundary conditions)
therefore
ci 
therefore this
must be zero
sin  i
1
 cos( i )d
0
1
 cos (   )d
2
i
0

i
1 sin 2 i

2
4 i

4sin i
2 i  sin 2 i
giving final temperature profile

 ( ,  )  
i 1
4sin i
cos( i ) exp( i2 )
2i  sin 2 i
MHMT10
PLATE/CYLINDER/SPHERE
Unsteady temperature profiles inside a sphere and infinitely long cylinders
can be obtained in almost the same way, giving temperature profiles in form

of infinite series
 ( ,  )   ci G (  i ) exp(  i2 )
i 1
where G (  i )  cos(  i )
 J 0 (  i )
plate
cylinder
 sin(  i ) /  sphere
and only eigenvalues I have to be calculated numerically. This form of
analytical solution is especially suitable for description of temperature field
at longer times, because exponential terms quickly decay and only few
terms in the series are necessary.
For shorter times the penetration theory can be applied effectively

T  T0
x
x  2
x

 erfc(
)  exp(
 ( ) at )erfc(

at )
T f  T0


2 at
2 at 
this is the previously
derived solution for
Bi 
correction term corresponding to the thermal resistance
of fluid (finite heat transfer coefficient  )
This analytical solution is presented in the book Carslaw H.S., Yeager J.C.: Conduction of Heat in Solids. Oxford Sci.Publ. 2 nd Edition, 2004
MHMT10
3D – brick,finite cylinder…
It is fantastic that an unsteady temperature field in the finite 2D or 3D bodies
can be obtained in the form of PRODUCT of 1D solutions expressed in terms
of dimensionless temperatures x y z !!!
y
For example the temperature distribution
x
in an infinitely long rod with rectangular
cross section hx x hy is calculated as
T (t , x, y )  T f
at x
at y
 (t , x, y ) 
  x ( 2 , ) y ( 2 , )
T0  T f
hx hx
hy hy
hy
hx
 y a
 (t , x, y )  x a
Proof:

 y  x
t
 x hx2
 y hy2
 2 y 1
 2  2  2 x 1


y 
x
x 2 y 2  x 2 hx2
 y 2 hy2
 y a
 2 y 1
 x a
 2 x 1
  x
 a ( 2 2 y 
 )
2 y
2
2
2 x
 x hx
 y hy
 x hx
 y hy

t
 2  2

x 2 y 2
…and you see that the FK
equation is satisfied if x y
are solutions of 1D problem.
MHMT10
3D – excercise
Calculate temperature in the center of a cube
Calculate temperature in the corner using erfc solution
y
x
MHMT10
EXAM
Fins and
Unsteady heat conduction
MHMT10
What is important (at least for exam)
Thermal resistance of finned tube
RT 
Tf 1  Tf 2
Q

Dm
D2
1
1
1

ln

ln


21 L D1
22 L Dm  D11 L S 2 2 L
1
inner tube
outer tube (without fins)
heat transfer
in pipe
heat transfer
from fin to
ambient fluid
Efficiency of a planar fin
 fin
Q
2b
 H 2 tanh Bi


tanh

2
Q 
H
2b
Bi
MHMT10
What is important (at least for exam)
Unsteady heat conduction
T
 2T  2T  2T
 a( 2  2  2 )
t
x
y
z
Transient heating of a semi-infinite space
T  T0
x
 erfc(
)
Tw  T0
2 at
Simplified solution by penetration depth
  4at
Temperature response to variable surface temperature
t
T (t , x)   Tw ( ) E (t   , x)d
0
MHMT10
What is important (at least for exam)
 2Gi
 i2Gi  0
Find solution of equation
2

for Dirichlet boundary condition at =1 and Neumann boundary
condition at =0
Why is it necessary to use dimensionless temperatures in the
“product” solution of 2D and 3D problems?