Effectiveness and number of transfer units for Parallel flow, Heat Transfer (2151909)

1. Gandhinagar institute
of technology(012)
ALA
Subject:- HT(2151909)
Topic name:- Effectiveness and number of transfer
units for “Parallel flow”
Prepared by:- jani Parth u. (150120119051)
guided by :-Prof. Nikita gupta
1

2. content
1.Introduction
2.NTU (Number of transfer unit) method
3.effectiveness for parallel method
4.Example
2

3. 1. introduction
 Heat exchanger may be defined as an equipment which transfer the
energy from a hot fluid to cold fluid, with maximum rate and minimum
investment
 Type of heat exchanger :- classification according nature of heat
exchanger, relative direction of fluid motion ,design and construction
features.
1. Nature of heat exchange process
i. Direct contact heat exchanger
ii. Indirect contact heat exchanger
(A) Regenerators (B) Recuperators
3

4. ...
2. Relative direction of fluid motion
i. Parallel flow
ii. Counter flow
iii. Cross flow
3. Design and construction features
i. Concentric tubes
ii. Shell and tubes
4. Physical state of fluid
(A) Condensers (B) Evaporators
4

5. 2. NTU method (Number of transfer unit)
 The LMTD Is used in equation if the inlet and output
temperatures of the two fluids are known
 When the problem is determine the inlet or exit temperature for
particular heat exchanger are not known, the analysis is performed
more easily, by using a method based on effectiveness of the heat
exchanger and number of transfer unit.
 Heat exchanger effectiveness is used which does not involve any of the
outlet temperatures.
UdAQ 
5

6. 3. Heat exchanger effectiveness
The heat transfer effectiveness is defined as the ratio of actual heat
transfer to the maximum possible heat transfer.
ε =
𝑎𝑐𝑡𝑢𝑎𝑙 ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 ℎ𝑒𝑎𝑡 𝑟𝑎𝑛𝑠𝑓𝑒𝑟
=
𝑄
𝑄 𝑀𝐴𝑋
)(
)(
11min
21
ch
hhh
ttc
ttC



)(
)(
11min
12
ch
ccc
ttC
ttC



dtccmdQ
dthcmdQ
UdAdQ
cc
hh


 
6

7. …
)(
)(
12min
12
ch
ccc
ttC
ttC



1
11min
2
)(
c
C
c t
C
tcthc
t 



h
ch
hh
C
ttC
tt
)( 11min
12



)(
)(
11min
21
ch
hhh
ttc
ttC



7

8. ...
 By compiling non- dimensional grouping, ε can be expressed as function
of three non dimensional parameter ,this is know as NTU method.







ch
ch
cc
dQdtdt
11







ch
chch
cc
ttUdAttd
11
)()(



















c
h
hch
ch
chch
ch
C
C
C
UA
tt
tt
cc
UdA
tt
ttd
1ln
11)(
11
22
8

9. …















c
h
hch
ch
C
C
C
UA
tt
tt
1exp
11
22
  
























 c
h
hch
chch
ch C
C
C
UA
cc
ttCtt
tt
1exp
11
)(
1
11min11
11

9































ch
c
h
CC
C
C
C
C
UA
11
1exp1
min
min


10. Example 1
Steam condenses at atmospheric pressure on the external surface of the tubes of the
steam condenser. The tubes are 12 in number and each is 30mm in diameter and 10m
long .the inlet and outlet temperatures of cooling water flowing inside the tubes are
25 c and 60 c respectively.if the flow rate is 1.1 kg/s, calculate following:
1. The rate of condensation of steam
2. The Mean overall heat transfer coefficient based on the inner surface area
3. The NTU
4. The effectiveness of he condenser
SOLUTION :- given : N = 12; 𝑑𝑖= 30 mm = 0.03 m ; L=10 m ; 𝑡 𝑐1=25 c, 𝑡 𝑐2=60
c,𝑡ℎ1=𝑡ℎ2= 100 c , 𝑚 𝑤=𝑚 𝑐=1.1 kg/s
i. The rate of condensation of steam, 𝑚 𝑠 (= 𝑚ℎ)
10

11. …
heat lost by steam = heat gained by water
𝑚 𝑠 × ℎ 𝑓𝑔= 𝑚 𝑐 × 𝑐 𝑝𝑐 (𝑡 𝑐2 - 𝑡 𝑐1)
𝑚 𝑠 × 2257= 1.1× 4.187 (60-25)
𝑚 𝑠 = 0.0714 kg/s
ii. The mean overall heat transfer coefficient
total heat transfer rate is given by
Q = 𝑚 𝑐× 𝑐 𝑝𝑐× (𝑡 𝑐2- 𝑡 𝑐1)
=1.1× 4187× (60-25)
=161199.5 j/s
ϴ 𝑚=
ϴ1 − ϴ2
ln
ϴ1
ϴ2
= 55.68 c
11

12. ...
Substituting the value in the above equation , we get
191199.5= U× 11.31 ×55.68
U= 255.9 w/𝑚2
c
iii. The number of transfer units
in condenser 𝑐 𝑚𝑎𝑥 refer to the hot fluids which remains temperatures.
Therefore,𝑐 𝑚𝑖𝑛 refers to water,
𝑐 𝑚𝑖𝑛refers to water;
𝑐 𝑚𝑖𝑛= m×𝑐 𝑝𝑐 = 1.1 × 4187 = 4605.7 𝑤/𝑐
NTU =
𝑈𝐴
𝐶 𝑚𝑖𝑛
=
255.9×11.31
4605.7
=0.628
12

13. …
iv. the effectiveness of the condenser
ε = 1 - exp (NTU)
ε = 1 - exp (-0.628)
= 0.47
13

14. 14