H.T through the Tubes with parameters.ppt

Fluid Flow and Heat
Transfer in Tubes

The pipe, duct, tube, and conduit are usually used interchangeably
for flow sections. Most liquids are transported in circular pipes.
This is because pipes with a circular cross section can withstand
large pressure differences between the inside and the outside
without undergoing any distortion.
For a fixed surface area, the circular tube gives the most heat
transfer for the least pressure drop,
 Therefore, circular tubes have devastating popularity in heat
transfer equipment.
Flow in a tube can be laminar or turbulent, depending on the flow
conditions.
Fluid Flow and Heat Transfer in Tubes

Velocity Boundary layer in Tube
Hydrodynamic Entry Length
In case of pipe/tube flow , the development of boundary layer in
the same fashion as flat plate boundary layer.
Fluid velocity in a tube changes from zero at surface to maximum
at the centre line.
Formed parabolic region known as hydro dynamically developed
region
The region from tube inlet to the centre at which the boundary
layers merged is called Hydrodynamic entry region/length.

 For calculation we use average/mean velocity (Uave).
In actual practice of heating and cooling, the mean velocity may
change due to change in density because of temperature change.
The thickness of boundary layer is depend on the radius of the
 tube.

Boundary layers from the tube walls meet at the centre of the
tube and entire flow acquires the characteristics of the B.L.
Once boundary layer thickness become equal to the radius of
the tube, there will not be any further change in the velocity
distribution
This is known as fully developed velocity profile flow
conditions

Thermal Boundary layer in Tube
 Thermal Entry Length
The region of flow in which thermal boundary layer develops and
merged in the tube centre is called thermal entry region
Its length is known as thermal entry length.
The temperature profile in thermally developed region may vary
with x-direction of the flow.
Therefore, temperature profile can be different at x-direction of
the tube in the developed region.
The dimensional less parameter Prandtl Number (Pr) shows effect
of both the boundary layers.

Prandtl number and boundary layers
(Thermal and Velocity)
For laminar flow in a tube, the magnitude of the Prandtl number
is measure of the relative growth of the velocity and thermal B.L
 If Pr=1
Both boundary layers are coincide to each other.
If Pr>1
The velocity boundary layer out grows the thermal boundary layer
If Pr<1
The thermal boundary layer is more dominant than the velocity
boundary layer.

Reynolds Number Re and flow in tubes

Reynolds Number Re and flow in tubes
It is certainly desirable to have precise values of Reynolds
numbers for laminar, transitional, and turbulent flows.
The transition from laminar to turbulent flow also depends on the
degree of disturbance of the flow such as,
surface roughness,
 pipe vibrations,
fluctuations in the flow.
Under most practical conditions,
 If Re<2300
Flow is laminar
If 2300  Re 10000
Flow is transitional
If Re>10000
Flow is turbulent

Hydrodynamic and Thermal
entry length of Tubes
Calculation formulae

Hydrodynamic and thermal entry length
For laminar flow:
For Turbulent flow:
Hydraulic diameter (Dh) of the tube
Friction Factor (f)
D
L
D
L
th
h
.
Pr
.
Re
.
05
.
0
.
Re
.
05
.
0


D
L
L th
h 10


P
A
D c
h /
4

2
.
0
Re
/
184
.
0
Re
/
64



f
f Laminar Flow
Turbulent Flow

The conservation of energy equation for steady flow of fluid in tube:
For average Temperature:
Conservation of heat flux at any location on the tube
For constant surface heat flux
For constant surface temperature
)
( i
e
p T
T
C
m
Q 
 

c
mean A
U
m 

 
m
pT
C
m
Q 
 
)
( m
s
s T
T
h
q 


)
( i
e
p
s T
T
C
m
A
q
Q 

 


)
( m
s
ave T
T
hA
T
hA
Q 




2
T
T
T e
i
m



2
T
T
T
T
2
T
T
T
T
T
2
T
T
T
T
e
i
s
am
e
s
i
s
am
e
i
ave
am
)
(
)
(
)
(
















2
T
T
T e
i
b



b
s
am T
T
T 



Logarithmic Mean Temperature Difference (LMTD)
For heating of fluid:
Ts>Ti and Te
For cooling of fluid:
Ts<Ti and Te

















i
s
e
s
i
s
e
s
m
T
T
T
T
T
T
T
T
T
T
ln
)
(
)
(
ln















i
e
i
e
m
T
T
T
T
T
ln

Problem
Water flows through a tube of ¾-inch internal diameter at the rate
of 0.4 gallon per minute. The viscosity of water may be taken as
2.36 lb /hr ft. Determine either the flow is laminar or Turbulent.
If it is laminar determine the entrance length.
Data
D=3/4=0.0625ft
V=0.4 gallon/min
=62.4 lb/hr ft.

 D
U
R ave
e 
D
Lh Re
05
.
0

 
2
2
2
ave
ft
003067
0
A
0625
0
4
14
3
D
4
Area
A
time
Volume
U
.
.
.
/






sec
/
.
/
)
.
(
.
/
3
3
ft
0008913
0
time
Volume
60
ft
1337
0
4
0
time
Volume

  
3
ft
1337
0
gal
1 .


sec
/
.
.
.
/
ft
29
0
U
003067
0
0008913
0
A
time
Volume
U
ave
ave



2300
1725
R
3600
36
2
0625
0
29
0
4
62
D
U
R
e
ave
e






/
.
.
.
.


This value of Reynolds number is lower than the 2300
Therefore, the flow is Laminar .
0625
.
0
1725
05
.
0
Re
05
.
0 


 D
Lh
ft
Lh 39
.
5


Problem: Heating of Water by Resistance Heaters in a Tube
Water is to be heated from 15°C to 65°C as it flows through a 3-cm
internal diameter 5-m-long tube as shown in Fig. The tube is
equipped with an electric resistance heater that provides uniform
heating throughout the surface of the tube. The outer surface of the
heater is well insulated, so that in steady operation all the heat
generated in the heater is transferred to the water in the tube. If the
system is to provide hot water at a rate of 10 L/min.
Determine:
i.The power rating of the resistance heater.
ii.The inner surface temperature of the pipe at the exit.

Solution
Properties: The properties of water at the bulk mean temp:
@ Tb = 40°C (From table A-9).
C
T
C
T
T
T
T
o
b
o
b
e
i
b
40
40
2
65
15
2






s
m
C
m
W
k
C
kg
J
C
m
kg
o
o
p
/
10
658
.
0
/
32
.
4
Pr
/
631
.
0
/
4179
/
1
.
992
2
6
3













For Cross-sectional area of the tube
For total surface area of the tube
Mass flow rate
 
2
4
c
2
2
c
m
10
069
7
A
03
0
4
14
3
D
4
A






.
.
.

2
s
s
m
471
0
A
5
03
0
14
3
DL
PL
A
.
.
.





 
sec
/
.
.
.
min
/
.
min
/
kg
1654
0
m
01
0
1
992
m
m
01
0
V
L
10
V
V
m
3












 

Heat supplied to water to increase the temperature from 15 to 65o
C
Power required for electric resistance heater is 34.6 Kw, because
Heater is totally insulated.
Surface temperature of the tube
Surface Heat Flux
 
 
Kw
6
34
Q
KJ
6
34
Q
15
65
179
4
1654
0
Q
T
T
C
m
Q i
e
p
.
sec
/
.
.
.












2
s
s
s
m
s
m
s
s
m
Kw
46
73
q
471
0
6
34
A
Q
q
h
q
T
T
T
T
h
q
/
.
.
.
)
(













For heat transfer coefficient (h)
Where Nusselt number (Nu)
For Reynolds number
Nu
D
k
h
k
hD
Nu


4
0
8
0
023
0
Nu .
.
Pr
.
Re
.

sec
/
.
.
.
Re
m
236
0
U
10
069
7
01
0
A
V
U
D
U
mean
4
c
mean
mean









10760
10
658
0
03
0
2360
0
6



 
Re
.
.
.
Re
This Reynolds number is greater than the 10000, therefore,
The flow is turbulent in this case and the entry lengths for
hydrodynamic and thermal are,
Now, substitute values in equation
m
3
0
L
L
03
0
10
D
10
L
L
th
h
th
h
.
.






4
0
8
0
023
0
Nu .
.
Pr
.
Re
.

C
m
W
1462
h
5
69
03
0
631
0
h
o
2
/
.
.
.



Nu
D
k
h 

For surface temperature
Discussion:
Note that the inner surface temperature of the pipe will be 50°C
higher than the mean water temperature at the pipe exit.
This temperature difference of 50°C between the water and the
surface will remain constant throughout the fully developed flow
region.
C
115
T
1462
460
73
65
T
h
q
T
T
o
S
S
s
m
S





.

Problem
Water enters a 2.5-cm-internal-diameter thin copper tube of a
heat exchanger at 15°C at a rate of 0.3 kg/s, and is heated by
steam condensing outside at 120°C. If the average heat transfer
coefficient is 800 W/m2o
C, determine the length of the tube
required in order to heat the water to 115°C.

SOLUTION:
Properties: The specific heat of water at the bulk mean temperature
C
T
C
T
T
T
T
o
b
o
b
e
i
b
65
65
2
115
15
2






C
kg
J
C o
p /
4187

The enthalpy of steam at 120°C is 2203 kJ/kg (Table A-9).
Heat transfer rate determine by
kW
Q
Q
T
T
C
m
Q i
e
p
6
.
125
)
15
115
(
187
.
4
3
.
0
)
(











The logarithmic mean temperature difference is
C
T
T
T
T
T
C
T
T
T
T
T
o
i
i
i
s
i
o
e
e
e
s
e
105
15
120
)
(
5
115
120
)
(































i
e
i
e
T
T
T
T
T
ln
ln
C
T o
85
.
32
105
5
ln
105
5
ln 










The heat transfer surface area is
Length of the pipe for heating
2
ln
ln
78
.
4
85
.
32
8
.
0
6
.
125
m
A
T
h
Q
A
T
hA
Q
S
S
S









m
L
D
A
L
DL
A
S
S
61
025
.
0
14
.
3
78
.
4








H.T through the Tubes with parameters.ppt

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