1) The document discusses heat transfer by conduction through plane walls and cylindrical pipes. It presents the general heat conduction equations and derives the equations for the temperature distribution in plane walls and cylindrical pipes. 2) It also discusses heat transfer by convection and defines the overall heat transfer coefficient (U) for a plane wall subjected to convection on both sides. Expressions are developed for the surface temperatures and overall heat transfer coefficient. 3) An example problem is presented to calculate the surface temperatures and overall heat transfer coefficient for a plane wall subjected to a uniform heat flux on one side and convection on the other.

1. Chapter 2: Overall Heat Transfer
Coefficient

2. Heat transfer by conduction:
Fourier’s law: “The heat flux is directly proportional
to the temperature gradient”.
k = thermal conductivity of material [W/m.K]
dx
dT
q 
dx
dT
kq 
A
Q
q
.

ordinates)-co(cartesian
.
dx
dT
kAQ 
ordinates)-coal(cylindric
.
dr
dT
kAQ 

3. General heat conduction equation: z
a) In Cartesian co-ordinates (x,y,z):
x
y
b) In cylindrical polar co-ordinates (r,,z):

 









 T
k
q
z
T
y
T
x
T v 1
.
2
2
2
2
2
2
 












 T
k
q
z
TT
rr
T
rr
T v 111
.
2
2
2
2
22
2
][W/mvolumeunitpergenerationheat 3
.
vq
]/[m2
sydiffusivitthermaloftcoefficien
[s]time

4. Plane wall:
For one dimensional steady-state heat transfer by conduction
in a plane wall without heat generation, the general heat
conduction is reduced to
By integration
By another integration
The boundary conditions are the known temperatures. That is
and,
02
2

dx
Td
1C
dx
dT

21 CxCT 
.
Q
0xat1  TT
Lxat2  TT

5. When these boundary conditions are applied to the equation
for temperature distribution, we obtain
Accordingly, the expression for the temperature distribution
(temperature profile) becomes:
The temperature distribution is thus linear (straight line) across
the plane wall.
12 TC 
xCTT 11 
LCTT 112 
L
TT
C 21
1


x
L
TT
TT 




 
 21
1

6. Rate of heat Transfer:
Fourier’s law
Evidently for heat conduction in a plane wall, the thermal
resistance takes the form
dx
dT
kAQ 
.
L
TT
C
dx
dT 21
1














 

kA
L
TT
L
TT
kAQ 2121
.
kA
L
R wallplane 
.
Q 1T 2T
kALR wallplane 

7. Cylindrical pipes:
For one dimensional steady-state heat transfer by conduction
in a cylindrical pipe without heat generation, the general heat
conduction is reduced to
By integration
By another integration
0
1
2
2

dr
dT
rdr
Td
02
2

dr
dT
dr
Td
r
0






dr
dT
r
dr
d
r
C
dr
dT
C
dr
dT
r 1
1 or 

8. Incorporating the relevant boundary conditions that
1)
2)
The constants are determined as follows
By subtraction
11 at rrTT 
22 at rrTT 
21 ln CrCT 
2111 ln CrCT 
2212 ln CrCT 







2
1
121 ln
r
r
CTT
















1
2
21
2
1
21
1
lnln
r
r
TT
r
r
TT
C

9. The substitution in the equation of gives
When the values of the constants are substituted into the
equation of the temperature distribution, one obtains the
following expression for temperature distribution in the pipe
wall
1
1
2
21
12 ln
ln
r
r
r
TT
TC








2C
1
1
2
21
1
1
2
21
ln
ln
ln
ln
r
r
r
TT
Tr
r
r
TT
T






























1
1
2
21
1 ln
ln
r
r
r
r
TT
TT

10. Rate of heat flow:
Fourier’s law
evaluated at r1or r2
Then, the thermal resistance for heat conduction in a
cylindrical pipe takes the form
21
.
rrrr dr
dT
Ak
dr
dT
AkQ














r
r
r
TT
r
C
dr
dT 1
ln
1
2
211








 
      kLrr
TT
rr
TT
kl
rrr
TT
lrkQ


2lnln
2
1
ln
2
12
21
12
21
112
21
1
.








 

 
kl
rr
Rpipe
2
ln 12


11. Equivalent thermal circuit for heat flow by conduction through
the walls of a cylindrical pipe is shown in the following figure:
Heat transfer by convection:
Newton-Rikhman’s law of convection:
“The heat flux is directly proportional to the temperature
difference between the wall and the fluid”.
.
Q 1T 2T
 
kl
rr
Rpipe
2
ln 12

Tq 
 TTT w

12. In case of heat convection from/to a cylindrical pipe, the
thermal resistance takes the form
The equivalent thermal resistance circuit for heat transfer by
convection is shown in the following figure:
  TThq w
 
hA
TT
TThAQ w
w
1
.




.
Q
1T 2T
hARconvection 1
hA
Rconvection
1


13. Overall heat transfer coefficient through a plane
wall (U):
Consider the plane wall, shown in the figure, exposed to a hot
fluid A on one side and a cooler fluid B on the other side. The
rate of heat transfer is expressed by
these three expressions
1) Heat transfer by convection from
fluid A to wall surface (1):
2) Heat transfer by conduction
the plane wall:
 11
.
TTAhQ A 
kAL
TT
Q 21
.



14. 3) Heat transfer by convection from wall surface (2) to fluid B:
The three previous equations can be rewritten as follows:
By addition
 BTTAhQ  22
.
Ah
QTTA
1
.
1
1

kA
L
QTT 
.
21
Ah
QTT B
2
.
2
1








AhkA
L
Ah
QTT BA
21
.
11

15. The equivalent thermal resistance circuit for heat transfer
through the plane wall with convective boundaries is shown in
the following figure:
Let
U = overall heat transfer coefficient [ W/m²K]
Comparing these two equations, one obtains
AhkA
L
Ah
TT
Q BA
21
.
11



.
Q
Ah1
1
kA
L
Ah2
1 BA TTUAQ 
.

16. h1 = heat transfer coefficient of surface (1) [W/m2.K]
= heat transfer film coefficient of wall surface (1)
= individual heat transfer coefficient of wall surface (1)
h2 = heat transfer coefficient of surface (2) [W/m2.K]
= heat transfer film coefficient of wall surface (2)
= individual heat transfer coefficient of wall surface (2)
k = thermal conductivity of the wall material [W/m.K]
L = wall thickness [m]
21
11
1
hk
L
h
U



17. The wall conduction term may often can be neglected, since a
thin wall of large thermal conductivity is generally used in heat
exchangers. Also, one of the convection coefficients is often much
smaller than the other and hence dominates determination of the
overall heat transfer coefficient. For example, if one of the fluids is a
gas and the other is a liquid or a liquid-vapor mixture such as boiling
or condensation, the gas-side convection coefficient is much smaller.
h is small, in case of gases (low viscosity and low specific heat) and in
case of laminar flow (low velocity).
h is big, in case of liquids (high viscosity and high specific heat) and
in case of turbulent flow (high velocity).
 kL
small
bigsmall
h
hk
L
h
U 


11
1

18. Example 1:
An iron plate of thickness L with thermal conductivity k is subjected
to a constant, uniform heat flux qo (W/m²) at the boundary surface
at x = 0. From the other boundary surface at x = L, heat is
dissipated by convection into a fluid at temperature T∞ with a heat
transfer coefficient h. The figure shows the geometry and the
nomenclature.
Develop an expression for the determination of the surface
temperatures T1 and T2 at the surfaces x = 0 and x = L. Also, develop
an expression for the overall heat transfer coefficient U.
Calculate the surface temperatures T1 and T2 and the overall heat
transfer coefficient U for L = 2 cm, k = 20 W/m.K, qo = 105 W/m2 , T∞
= 50oC, and h = 500 W/m2.K.

19. Data: L = 2 cm, k = 20 W/m2 °C, qo = 105 W/m2 , T∞ = 50oC, and
h = 500 W/m2 °C.
Find: T1, T2, U
Solution: Applying the thermal
resistance concept:
.
Q
kA
L
hA
1















hAkA
L
TT
hA
TT
kA
L
TT
Q
1
1
1
221
.
11
1221












 
hk
L
TT
h
TT
k
L
TT
qo

20. By equating the first and the last expression, T1 is found
and by equating the first and the third expressions, T2 is found:
and since there is no convective heat transfer on surface (1),
 T
h
q
T o
2
21
11
1
hk
L
h
U


h0
1
2
1
 hand
h








hk
L
U
1
1
 





 T
U
q
Tq
hk
L
T o
o
1
1

21. Introducing The numerical values of various quantities in the
above results, we obtains
Note if the wall thickness is 2 mm, then T1 = 260 °C, T2 = 250°C
and U = 476.2 W/m2 °C.
CTq
hk
L
T o
o 3505010
500
1
20
02.01 5
1 











 
CT
h
q
T oo
25050
500
105
2  
CmW
hk
L
U o
./33.333
500
1
20
02.0
1
1
1 2


















22. Overall heat transfer coefficient in pipes:
Consider a pipe exposed to a hot fluid on the inner side and a
cooler fluid on the outer side, as shown in the figure. The area of
convection is not the same for both fluids in this case, these areas
depend on the inside pipe diameter and wall thickness.
The heat transfer is expressed by
the following relations:
1) Heat transfer by convection from
the hot fluid on the inner side to
the inner wall surface of the pipe:
ii
i
Ah
TT
Q
1
1
.



23. 2) Heat transfer by conduction through the pipe wall itself:
3) Heat transfer by convection from the outer wall surface of the
pipe to the cold fluid on the outer side:
The three previous equations can be rewritten as follows:
 
kL
rr
TT
Q
io
2
ln
21
.


oo
o
Ah
TT
Q
1
2
.


ii
i
Ah
QTT
1.
1 
 
kL
rr
QTT io
2
ln.
21 

24. By addition
The equivalent thermal resistance circuit for heat transfer
through the pipe wall with convective boundaries is shown in the
following figure:
oo
o
Ah
QTT
1.
2 
 







oo
io
ii
oi
AhkL
rr
Ah
QTT
1
2
ln1.

  








oo
io
ii
oi
AhkL
rr
Ah
TT
Q
1
2
ln1
.

.
Q
ii Ah1   kLrr io 2ln oo Ah1

25. Let
Uo = overall heat transfer coefficient based on the
outer area of pipe.
Ui = overall heat transfer coefficient based on the
inner area of pipe.
Ao = 2ro L, is the outer surface area of the pipe.
Ai = 2ri L, is the inner surface area of the pipe.
Upon comparing the two equations of , one obtains
   oiiioioo TTAUTTAUQ 
.
.
Q
 
oo
io
ii
oo
AhkL
rr
Ah
AU
1
2
ln1
1



iioo AUAU 

26. and
 
o
ioo
ii
o
o
hk
rrr
rh
r
U
1ln
1


 
oo
io
ii
ii
AhkL
rr
Ah
AU
1
2
ln1
1



 
oo
iioi
i
i
rh
r
k
rrr
h
U


ln1
1

27. Example 2:
Steam at 120oC flows in an insulated pipe. The pipe is made of
mild steel (kp =45 W/m.K) and has an inside radius of 5 cm and
an outside radius of 5.6 cm. The pipe is covered with 2.5 cm layer
of magnesia insulation (kin = 0.071 W/m.K). The inside heat
transfer coefficient is 85 W/m2.K and the outside heat transfer
coefficient is 12.5 W/m2.K. Determine the overall heat transfer
coefficients Uo and Ui and the heat transfer rate from the steam
per meter of pipe length, if the surrounding air temperature is
20oC.
Data: Ti = 120oC, kp = 45 W/m.K, r1 = 0.05 m, r2 = 0.056 m,
r3 = 0.081 m, kin = 0.071 W/m.K, hi = 85 W/m2.K,
ho = 12.5 W/m2.K, To = 20oC.
Find: Uo , Ui , LQ
.

28. Solution:
   
oin
233
p
123
1i
3
o
h
1
k
rrlnr
k
rrlnr
rh
r
1
U


   
5.12
1
071.0
056.0081.0ln081.0
45
05.0056.0ln081.0
05.085
081.0
1


oU
08.04211.000020.001906.0
1

oU
KmWUo ./944.1
5144.0
1 2

iioo AUAU 
KmWU
r
r
U
A
A
U oo
i
o
i ./149.3944.1
05.0
081.0 2
1
3


29.  oio TTrU
L
Q
 3
.
2
   mW
L
Q
/94.9820120081.02944.1
.
 
 oioo TTAUQ 
.

 oio TTLrUQ  3
.
2

30. Values of the overall heat transfer coefficient
U (W/m².K)Fluid combination
850 – 1700Water to water heat exchanger
110 – 350Water to oil heat exchanger
1000 – 6000Steam condensers (water in tubes)
800 – 1400Ammonia condenser (water in tubes)
250 – 700Alcohol condenser (water in tubes)
25 - 50Finned-tube heat exchanger (water in
tubes, air in cross flow)
10 – 40Gas to gas heat exchanger