forced convection heat transfer chapter 3.ppt

3.1 INTRODUCTION
Convection is one of the three basic mechanisms of heat transfer. It
is a heat transfer that takes place between moving fluid and solid
surface. Like conduction heat transfer, convection heat transfer
requires material medium. Heat transfer by convection combines
heat transfer by conduction and bulk flow of fluid.
Convection heat transfer depends on several parameters like:
 Fluid properties
 Dynamic viscosity, μ.
 Thermal conductivity, k.
 Density, ρ.
 Specific heat capacity, cp.
 Fluid velocity, v.
 Geometry of solid surface.
 Roughness of solid surface.
 Type of fluid flow (turbulent or laminar).

3.1 INTRODUCTION…
 The dependence of convection heat transfer on so many
variables makes it very complex and it is usually
determined experimentally.
 The convection heat transfer rate is given by Newton’s law
of cooling,
(3.1)
Where
h=convection heat transfer coefficient,
A=heat transfer surface area,
Ts=surface temperature,
T∞=fluid temperature far away from surface.
)
(
.
.


 T
T
hA
Q s
conv

3.1 INTRODUCTION…
Non dimensional convection coefficient
 The convection heat transfer coefficient, h, is
nondimensionalized to obtain the Nusselt number,
Nu.
(3.2)
Where k = thermal conductivity,
 = characteristic length.
k
h
Nu



3.1 INTRODUCTION…
 The Nusselt number can also be given as the ratio of
convection heat transfer rate to conduction heat
transfer rate.
(3.3)
rate
transfer
heat
Conduction
rate
transfer
heat
Convection
Nu 
.
.
.
.
cond
conv
Q
Q
Nu 

3.2 VELOCITY BOUNDARY LAYER
Fig. 3.1 Velocity boundary layer development over flat plate

 u
u 99
.
0
 is the value of y for
which

3.2 VELOCITY BOUNDARY LAYER…
 For external flows the velocity boundary layer concept provides
the basis for determining the local friction coefficient.
(3.4)
 Where the surface shear stress for Newtonian fluid is obtained
from the relation,
(3.5)
 The drag or friction force over the surface is determined by
(3.6)
2
2
1


u
C s
f


0




y
s
y
u


2
2


u
A
C
F f
D


3.3 THERMAL BOUNDARY LAYER
Fig. 3.2 Thermal boundary layer development over isothermal flat plate
t is the value of y for which
the ratio 99
.
0
)
/(
)
( 

 
T
T
T
T s
s

3.3 THERMAL BOUNDARY LAYER…
 At any distance x from the leading edge, the local heat flux may be obtained by
applying Fourier’s law to the fluid at y= 0. That is,
(3.7)
 By combining equation (3.7) with Newton’s law of cooling we then obtain
(3.8)
 The relative thickness of the velocity and thermal boundary layers is
described by Prandtl number,
 The hydrodynamic and thermal boundary layers are related in terms of
Prandtl number as
(3.9)
0





y
f
s
y
T
k
q







T
T
y
T
k
h
s
y
f
0
k
C
y
diffusivit
Heat
y
diffusivit
Momentum P






Pr
1
Pr
026
.
1
1 


t

3.3 THERMAL BOUNDARY LAYER…
Fig. 3.3 Relative thickness of the velocity and thermal boundary layers

3.4 LAMINAR AND TURBULENT FLOW
Fig.3.4 Laminar, transition and turbulent flow regions

3.4 LAMINAR AND TURBULENT FLOW…
The location where the flow turns to turbulent is
determined by a dimensionless grouping of variables
called the Reynolds number.
(3.10)
Where u∞=free stream velocity
=characteristic length of the geometry
/=kinematic velocity of the fluid
The critical Reynolds number at which flow turns to
turbulent is about 5x105.





u
forces
Vicous
forces
Inertia
Re

3.5 FLOW OVER FLAT PLATES
 The friction and heat transfer coefficients for a flat plate can be
determined by solving the conservation of mass, momentum
and energy equations. The average Nusselt number can be
expressed as
(3.11)
Where C, m and n are constants and L is length of the plate.
 Generally, properties of the fluid vary with temperature. To
make the heat transfer analysis simple, properties are evaluated
at film temperature given by
(3.12)
 In heat transfer analysis, we are usually interested in the heat
transfer and drag force on the entire surface of the plate which
are determined using the average heat transfer and friction
coefficients.
n
m
L
C
k
hL
Nu Pr
Re


2



T
T
T s
f

3.5 FLOW OVER FLAT PLATES…
 The local and average Nusselt number and friction
coefficient are determined for laminar and turbulent
flows separately,
Laminar flow
 The boundary layer thickness
(3.13)
 Local friction coefficient
(3.14)
2
1
Re
5
x
x


2
1
2
,
Re
664
.
0
2
/
1
x
x
x
f
u
C 





 Combining equations (3.9) and (3.13), the thermal
boundary layer thickness is given by
(3.15)
 Local Nusselt number
(3.16)
 Average friction coefficient
(3.17)
)(Pr)
(Re
026
.
1
5
Pr
026
.
1
1
2
/
1
1
x
t
x

 


)
6
.
0
(
Pr
Re
332
.
0 3
1
2
1


 pr
k
x
h
Nu x
x
x
2
1
Re
328
.
1
L
f
C 

 Average Nusselt number
(3.18)
 The critical Reynolds number
)
6
.
0
(
Pr
Re
664
.
0 3
1
2
1


 pr
k
hL
Nu L
5
10
5
Re 

 

Cr
Cr
x
u

Turbulent flow
 Local friction coefficient
(3.19)
 Local Nusselt number
(3.20)
 Average friction coefficient
(3.21)
 Average Nusselt number
(3.22)
 
7
5
5
1
, 10
Re
10
5
Re
0592
.
0



 x
x
x
f
C














 7
5
3
1
5
4
10
Re
10
5
60
6
.
0
Pr
Re
0296
.
0
x
x
x
x
pr
k
x
h
Nu
 
7
5
5
1
10
Re
10
5
Re
074
.
0



 L
L
f
C














 7
5
3
1
5
4
10
Re
10
5
60
6
.
0
Pr
Re
037
.
0
L
L
pr
k
hL
Nu

For combined laminar and turbulent flow
(3.23)
(3.24)
 The heat transfer rate is obtained from
(3.25)
Where
 The drag force is calculated as
(3.26)
 
7
5
5
1
10
Re
10
5
Re
1742
Re
074
.
0




 L
L
L
f
C
  














 7
5
3
1
5
4
10
Re
10
5
60
6
.
0
Pr
871
Re
037
.
0
L
L
pr
k
hL
Nu
)
(
.
.


 T
T
hA
Q s
conv
 
L
k
Nu
h 
2
2


u
A
C
F f
D


Example 3.1
An electric air heater consists of a horizontal array of thin metal
strips that are each 10mm long in the direction of an airstream
that is in parallel flows over the top of the strips. Each strip is
0.2m wide, and 25 strips are arranged side by side, forming a
continuous and smooth surface over which the air flows at
2m/s. During operation each strip is maintained at 5000C and
the air is at 250C.
a. What is the rate of convection heat transfer from the first
strip? The fifth strip? The tenth strip? All the strips?
b. For air velocities at 3, 5, and 10 m/s, determine the
convection heat rates for all the locations of part (a).
Represent your results in tabular or bar graph form.

Solution

Properties of air (Tf=535K, 1atm): =43.54x10-6m2/s,
k=0.0429W/m.K, Pr=0.683.
a)The location of transition is determined from
Since xc >>L=0.25m, the air flow is laminar over the entire
heater. For the first strip,
where h1 is obtained from
Then
m
u
xc 9
.
10
2
10
54
.
43
10
5
10
5
6
5
5









  




 T
T
w
L
h
Q s
1
1
.
K
m
w
L
k
h x
2
3
1
2
1
1 /
8
.
53
Pr
Re
664
.
0 


w
Q 1
.
51
1
.


for the fifth strip,
In similar manner for the tenth strip and entire 25 strips,
4
0
.
5
0
5
.

 
 Q
Q
Q
     



 






 T
T
w
L
h
T
T
w
L
h
Q s
s 4
5 4
0
5
0
5
.
K
m
w
L
k
h
and
K
m
w
L
k
h
x
x
2
3
1
2
1
04
.
0
4
0
2
3
1
2
1
05
.
0
5
0
/
9
.
26
Pr
Re
664
.
0
4
/
1
.
24
Pr
Re
664
.
0
5










w
Q 2
.
12
5
.


      w
T
T
w
L
h
T
T
w
L
h
Q s
s 3
.
8
9
10 9
0
10
0
10
.








 



   w
T
T
w
L
h
Q s 3
.
255
25
0
25
0
.



 



b) velocity,
m/s
heat transfer rate, w
first strip fifth strip tenth strip entire strip
2 51.08 12.06 8.29 255.39
5 80.76 19.06 13.11 403.80
10 114.20 26.96 18.53 571.06

3.6 FLOW ACROSS CYLINDERS AND SPHERES
 The average Nusselt number for flow over a
cylinder is given empirically as proposed by
Churchill and Bernstein:
(3.27)
 Equation (3.26) is applicable for conditions where
Re.Pr > 0.2. The fluid properties are evaluated at
film temperature .
 
5
4
8
5
4
1
3
2
3
1
2
1
28200
Re
1
Pr
4
.
0
1
Pr
Re
62
.
0
3
.
0

























k
hD
Nu
2



T
T
T s
f

3.6 FLOW ACROSS CYLINDERS AND SPHERES…
 The average Nusselt number for flow over a cylinder can
be expressed in compact form as
(3.28)
Where the constants C and m are obtained from table 3.1.
 Heat transfer to or from a bank (or bundle) of tubes in
cross flow is relevant to numerous industrial applications,
such as steam generation in a boiler or air cooling in the
coil of an air conditioner.
3
1
Pr
Rem
C
k
hD
Nu 


Table 3.1 Constants used in equation (3.28)

The tube rows of a bank are either staggered or aligned in
the direction of the fluid velocity V (Fig. 3.6)
Fig. 3.5 Tube bank in cross flow

Fig. 3.6 Aligned and staggered tube arrangement

 For airflow across tube bundles composed of 10 or
more rows (NL10), the average Nusselt number can
be obtained by the Grimison correlation,
(3.29)
 The constants C1 and m are obtained from table 3.2.
 The maximum Reynolds number is given by
(3.30)
















7
.
0
Pr
000
,
40
Re
2000
10
Re max
,
max
,
1 D
L
m
D
N
C
k
hD
Nu

 D
V
D
max
max
,
Re 

 For the aligned arrangement the maximum velocity, Vmax
occurs at the transverse plane A1 of Fig. 3.6a and is given for
an incompressible fluid as
(3.31)
 For the staggered configuration, the maximum velocity
may occur at either the transverse plane A1 or the diagonal
plane A2 . It will occur at A2 if the rows are spaced such that
Where SD is given by
V
D
S
S
V
T
T


max
2
)
(
)
(
2
D
S
S
or
D
S
D
S T
D
T
D





2
1
2
2
2 














 T
L
D
S
S
S

 And
(3.32)
 If SD>(ST+D)/2, the maximum velocity occurs at A1 and is
given by equation (3.31).
 For flow of fluids other than air equation (3.28) is modified
by inserting 1.13Pr1/3.
(3.33)
 All the properties in these equations are evaluated at film
temperature. If NL<10 a correction factor given in table 3.3 is
used as
(3.34)
 
V
D
S
S
V
D
T


2
max
















7
.
0
Pr
000
,
40
Re
2000
10
Pr
Re
13
.
1 max
,
3
1
max
,
1 D
L
m
D
N
C
k
hD
Nu
10
2
10 
 
L
L N
N Nu
C
Nu

Table 3.2 Constants used in equations (3.29) and (3.33)

Table 3.3 Constant C2 used in equation (3.34)

 The heat transfer rate could be more predicted by using the log
mean temperature difference instead of using T=Ts-T∞.
(3.35)
 Where Ti and To are temperatures of the fluid as it enters and leaves
the bank, respectively.
 The outlet temperature, which is needed to determine Tlm may
be estimated from
(3.36)
 Where N is the total number of tubes in the bank , NT is the number
of tubes in the transverse plane and V is the speed at inlet.
   
   
 
o
s
i
s
o
s
i
s
lm
T
T
T
T
T
T
T
T
T







/
ln












P
T
T
i
s
o
s
c
S
VN
DNh
T
T
T
T


exp

 Once Tlm is known, the heat transfer rate may be
computed from
(3.37)
 For flows over a sphere, the average Nusselt number
can be obtained by the Whitaker correlation,
(3.38)
 Equation (3.37) is valid for 3.5≤Re≤ 80,000 and
0.7≤Pr≤380. The fluid properties are evaluated at film
temperature except s which is evaluated at the surface
temperature, Ts.
 
lm
T
DL
h
N
Q 
 
.
4
1
4
.
0
3
2
2
1
Pr
Re
06
.
0
Re
4
.
0
2 












 


 
s
k
hD
Nu



Example 3.2
 A preheater involves the use of condensing steam at
1000C on the inside of bank of tubes to heat air that
enters at 1 atm and 250C. The air moves at 5m/s in
cross flow over the tubes. Each tube is 1 m long and
has an outside diameter of 10mm. The bank consists of
196 tube in a square, aligned array for which
ST=SL=15mm. What is the total rate of heat transfer to
the air?

Solution

 Atmospheric air (T∞=298K):=15.8x10-6m2/s,
k=0.0263W/mK, Pr=0.707, cp=1007J/kgK, =1.17kg/m3;
(Ts=373K): Pr=0.695
 The total heat transfer rate
 From tables 3.2 and 3.3, C = 0.27, m = 0.63 and
C2=0.99.
   
   
  lm
o
s
i
s
o
s
i
s
T
DL
N
h
T
T
T
T
T
T
T
T
DL
N
h
Q 






 

/
ln
.
9494
10
8
.
15
01
.
0
15
Re
,
/
15
5
5
15
6
max
,
max 







 
D
T
T
s
m
V
D
S
S
V

 From the Zhukauskas correlation
Hence,
     
 
C
c
S
VN
h
DN
T
T
T
T
K
m
W
D
k
Nu
h
Nu
p
T
T
i
s
o
s
D
D
0
2
4
/
1
36
.
0
63
.
0
7
.
27
1007
015
.
0
14
5
17
.
1
200
196
01
.
0
exp
75
exp
.
/
200
01
.
0
/
0263
.
0
9
.
75
/
9
.
75
695
.
0
/
707
.
0
707
.
0
9494
27
.
0
99
.
0






































 
kW
T
DL
N
h
Q lm 5
.
58
7
.
27
/
75
ln
7
.
27
75
1
01
.
0
196
200
.







 


3.7 FLOW IN TUBES
Fig. 3.7 Hydrodynamic boundary layer development in tube

3.7 FLOW IN TUBES…
Fig. 3.8 Thermal boundary layer development in

 The Reynolds number for flow in a circular tube is
defined as
(3.39)
Where ρ is the fluid density
D is tube diameter
 is fluid dynamic viscosity
um is the average fluid velocity given by
(3.40)
Where is mass flow rate of fluid.

 D
um
D 
Re
A
m
um

.

.
m

 In a fully developed flow the critical Reynolds
number is 2300.
 The pressure drop during the flow is given by
(3.41)
Where f is the friction factor and L is tube length
 For laminar and turbulent flows the hydrodynamic
entry lengths may be obtained from
(3.42)
(3.43)
2
2
m
u
D
L
f
P



D
x D
lam
h
fd Re
05
.
0
, 
D
x
turb
h
fd 10
, 

 For laminar and turbulent flows the thermal entry
lengths may be obtained from
(3.44)
(3.45)
Laminar flow
 For hydrodynamically developed laminar flow in a
tube, the velocity profile is profile is parabolic and
given by
(3.46)
D
x D
lam
t
fd Pr
Re
05
.
0
, 
D
x
turb
t
fd 10
, 
)
1
(
2
)
( 2
2
o
m
r
r
u
r
u 


 And the surface shear stress is
(3.47)
 The surface shear stress can also be written as
(3.48)
 It follows from equations (3.46) and (3.47) that the
friction coefficient Cf can be given as
(3.49)
D
u
dr
du m
r
r
s
o


8
2 



2
2
m
f
s
u
C

 
D
f
C
Re
16


 The friction factor f used in the pressure drop
calculation in laminar flow is given by
(3.50)
 The average Nusselt number for the hydrodynamically
or thermally developed laminar flow is given by Sider
and Tate
(3.51)
 All the properties are evaluated at bulk mean fluid
temperature, except for s, which is evaluated at the
surface temperature.
D
f
Re
64

14
.
0
3
1
Pr
Re
86
.
1 















s
b
L
D
k
hD
Nu



Turbulent flow
 The friction factor for fully developed turbulent flow in a smooth
tube is given by
(3.52)
 For flows in smooth or rough tubes the friction factor is obtained
from Moody diagram of Fig. 3.9.
 The friction factor can be obtained from the Colebrook equation
for flows in smooth or rough tubes in transition and turbulent
flows.
(3.53)
 The average Nusselt number for turbulent flow in a tube is given
from the Chilton-Colburn correlation as
(3.54)
2
.
0
Re
184
.
0 
 D
f











f
D
f Re
51
.
2
7
.
3
.
log
0
.
2
1 
3
1
Pr
Re
125
.
0 f
NuD 

Fig. 3.9 The Moody Diagram

Material Condition ε (mm)
Steel Sheet metal, new
Stainless, new
Commercial, new
Riveted
Rusted
0.05
0.002
0.046
3.0
2.0
Iron Cast, new
Wrought, new
Galvanized, new
Asphalted cast
0.26
0.046
0.15
0.12
Brass Drawn, new 0.002
Plastic Drawn tubing 0.0015
Glass - 0 (smooth)
Concrete Smoothed
Rough
0.04
2.0
Rubber Smoothed 0.01
Wood Stave 0.5
Table 3.4 Roughness Values for Commercial Ducts

Example 3.3
An engine oil cooler consists of a bundle of 25 smooth
tubes, each of length L=2.5m and diameter of
D=110mm.
a) If oil at total flow rate of 24kg/s is in fully developed
flow through the tubes, what are the power drop and
the pump power requirements?
b) Compute and plot the pressure drop and pump power
requirement as a function of flow rate for 10-30kg/s.

Solution
Properties of Engine oil (300 K): =884kg/m3,
μ=0.486kg/s.m.

a) Considering flow through a single tube
Hence, the flow is laminar and
5
.
251
486
.
0
01
.
0
25
24
4
4
Re
.








D
m
D
2545
.
0
5
.
251
64
Re
64



D
f
   
  s
m
D
m
um /
8
.
13
4
/
01
.
0
884
25
/
24
4
/ 2
2
.
1







The pressure drop is,
The pump power required is
Mpa
m
N
L
D
u
f
p m
38
.
5
/
10
38
.
5
5
.
2
01
.
0
2
8
.
13
884
2545
.
0
2
2
6
2
2











kw
m
p
V
p
P 146
884
24
10
38
.
5 6
.
.










b) Plot of pressure drop and pump power

Example 3.4
Water at 150C (=999.1kg/m3 and =1.138x10-3 kg/ms) is
flowing in a 4cm diameter and 30m long horizontal
pipe made of new stainless steel steadily at a rate of
5L/s. Determine (a) the pressure drop and (b) the
pumping power requirement to overcome this pressure
drop.
L = 30 m
D = 4 cm
Water
5 L/s

Solution
The density and dynamic viscosity of water are given to be
(=999.1kg/m3 and =1.138x10-3 kg/ms, respectively. The
roughness of new stainless steel is 0.002 mm (Table 3.4).
First we calculate the mean velocity and the Reynolds
number to determine the flow regime:
which is greater than 10,000. Therefore, the flow is turbulent.
5
3
3
2
3
2
10
40
.
1
s
kg/m
10
138
.
1
m)
m/s)(0.04
98
.
3
)(
kg/m
1
.
999
(
Re
/
m
98
.
3
4
/
m)
(0.04
/
m
0.005
4
/















D
s
s
D
V
A
V
m
c
m
V
V



The relative roughness of the pipe is
The friction factor can be determined from the Moody
chart, but to avoid the reading error, we determine it
from the Colebrook equation using an equation solver
(or an iterative scheme),
10
5
m
04
.
0
m
10
2
/ 5
6






D



























f
f
f
D
f 5
5
10
40
.
1
51
.
2
7
.
3
10
5
log
0
.
2
1
Re
51
.
2
7
.
3
/
log
0
.
2
1 

It gives f = 0.0171. Then the pressure drop and the required
power input become
(a)
(b)
Therefore, useful power input in the amount of 0.508 kW is
needed to overcome the frictional losses in the pipe.
101.5kPa
kN/m
1
kPa
1
ms
kg
1000
kN
1
2
m/s
)(3.98
kg/m
(999.1
04
.
0
m
30
0.0171
2
ρV
D
L
f
ΔP
2
2
3
2
m

























m
kW
0.508











/s
m
kPa
1
kW
1
)
kPa
5
.
101
)(
/
m
005
.
0
( 3
3
u
pump, s
P
V
W 


forced convection heat transfer chapter 3.ppt

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to forced convection heat transfer chapter 3.ppt

Similar to forced convection heat transfer chapter 3.ppt (20)

Recently uploaded

Recently uploaded (20)

forced convection heat transfer chapter 3.ppt