Chapter 4
Heat Exchangers
Effectiveness & NTU
Definition & Problems
1
HEAT EXCHANGER
Effectiveness of a heat exchanger
Ratio of actual heat transfer to the maximum p
possible heat transfer
Effectiveness, ε =
Actual heat transfer
Q
=
Max possible heat transfer Qmax
Where
Q = mcCpc((t2 – t1) = mhCph((T1 – T2)
Qmax = Cmin (T1 – t1)
Cmin = minimum heat capacity
Heat capacity, C = m x Cp
mh – Mass flow rate of hot fluid
mc – Mass
M
fl
flow rate
t off cold
ld fluid
fl id
Cph – Specific heat of hot fluid
Cpc – Specific heat of cold fluid
T1 – Entry temperature of hot fluid
T2 – Exit
E it temperature
t
t
off hot
h t fluid
fl id
t1 – Entry temperature of cold fluid
t2 – Exit temperature of cold fluid
2
HEAT EXCHANGER
Number of Transfer units (NTU)
NTU =
UA
C min
Where
U = Overall
O erall heat transfer coefficient
Cmin = minimum heat capacity
A = Heat exchanger area
Heat capacity, C = m x Cp
Application of Effectiveness – NTU method
¾
Usually, a heat exchanger can be designed by the LMTD method when the inlet
and outlet temperatures of both fluids are known.
¾
However, when the outlet temperatures of the fluids are not known and if the
problem is to determine the exit temperature of heat exchanger, effectiveness
– NTU method is used.
¾
The plot between effectiveness and NTU for various values of (Cmin/Cmax) is
available in the HMT databook Pg No: 163 to 168 for different heat exchangers.
3
NTU Problems
1.
A parallel flow heat exchanger is used to cool 4.2 kg/min of hot liquid of
specific heat 3.5 kJ/kg K at 130 oC. A cooling water of specific heat 4.18
kJ/kg K is used for cooling purpose at a temperature of 15 oC. The mass
flow rate of cooling water is 17 kg/min.
kg/min calculate the following.
following
a. Outlet temperature of hot liquid
b. Outlet temperature of water
c. Effectiveness of heat exchanger.
Take Overall
O
heat transfer
f co-efficient
ff
as 1100 W/m
/ 2K. Heat exchanger
area is 0.30 m2
Given:
Mass flow rate of hot fluid, mh= 4.2 kg/min = 0.07 kg/s
Specific heat of hot fluid, Cph = 3.5 kJ/kg K = 3.5 x 103 J/kg K
Inlet temperature of hot fluid, T1 = 130o C
Mass flow rate of cooling water, mc= 17 kg/min = 0.28 kg/s
Specific heat of water, Cpc = 4.18 kJ/kg K = 4.18 x 103 J/kg K
Inlet temperature of cooling water, t1 = 15o C
Overall heat transfer co-efficient, U = 1100 W/m2K
Area, A = 0.30 m2
4
NTU Problems
To find:
a. Outlet temperature of liquid,(T2) b. Outlet temperature of water,(t2)
c. Effectiveness of heat exchanger,(ε).
Solution:
Capacity rate of hot liquid, Ch = mh x Cph = 245 W/K
Capacity rate of water,
Cc = mc x Cpc = 1170.4 W/K
Æ (1)
Æ (2)
From (1) and (2),
Cmin = 245 W/K
Cmax = 1170.4
1170 4 W/K
=>
Cmin / Cmax = 0.209
Number of transfer units, NTU = UA / Cmin
=>
NTU = 1.34
Æ (3)
[HMT data book, Pg no: 152]
Æ (4)
T find
To
fi d effectiveness
ff ti
ε, refer
f HMT data
d t book
b k page no 163
[parallel flow heat exchanger]
5
NTU Problems
From graph,
Xaxis Æ NTU =1.34
Curve Æ Cmin / Cmax = 0.209
Maximum possible heat transfer
Qmax = Cmin (T1 – t1) = 28175 W
Corresponding Yaxis value is 64%
i.e., Effectiveness, ε = 0.64
Actual heat transfer rate
Q = ε x Qmax = 18032 W
We know that,
Heat transfer, Q = mc Cpc (t2 – t1)
=>
18032 = 1170.4 t2 - 17556
=>
t2 = 30.40o C
Outlet temperature of cold water, t2 = 30.40 oC
We know that,
Heat transfer, Q = mh Cph (T1 – T2)
=>
18032 = 31850 – 245 T2
=>
T2 = 56.4o C
Outlet temperature of hot liquid, T2 = 56.4 oC
6
NTU Problems
2.
In a counter flow heat exchanger, water at 20 oC is flowing at the rate of
1200 kg/h. It is heated by oil of specific heat 2100 J/kg K flowing at the rate
of 520 kg/h at inlet temperature of 95 oC. Determine the following.
a Total heat transfer
a.
b Outlet temperature of water
b.
c. Outlet temperature of oil.
Take Overall heat transfer co-efficient is 1000 W/m2K.
Heat exchanger area is 1 m2
Given:
Hot fluid – oil
(T1 , T2)
Cold fluid – water
(t1 , t2)
Mass flow rate of water, mc= 1200 kg/h = 0.33 kg/s
Inlet temperature of water, t1 = 20o C
Mass flow rate of oil, mh= 520 kg/h = 0.144 kg/s
Specific heat of oil, Cph = 2100 J/kg K
Inlet temperature of oil, T1 = 95o C
Overall heat transfer co-efficient, U = 1000 W/m2K
Heat exchanger area, A = 1 m2
7
NTU Problems
To find:
a. Total heat transfer, (Q)
c. Outlet temperature of oil, (T2)
b. Outlet temperature of water, (t2)
Solution:
Capacity rate of hot oil, C = mh x Cph = 302.4 W/K
Æ (1)
Capacity rate of water, C = mc x Cpc = 1381.3 W/K
Æ (2)
[Specific heat of water Cpc = 4186 J/kg K]
From (1) and (2),
Cmin = 302.4 W/K
Cmax = 1381.3
1381 3 W/K
=>
Cmin / Cmax = 0.218
Number of transfer units, NTU = UA / Cmin
=>
NTU = 3.3
Æ (3)
[HMT data book, Pg no: 152]
Æ (4)
T find
To
fi d effectiveness
ff ti
ε, refer
f HMT data
d t book
b k page no 164
[Counter flow heat exchanger]
8
NTU Problems
From graph,
Xaxis Æ NTU = 3.3
Curve Æ Cmin / Cmax = 0.218
Corresponding Yaxis value is 0.95
i.e., Effectiveness,ε = 0.95
Maximum possible heat transfer
Qmax = Cmin (T1 – t1) = 22680 W
Actual heat transfer rate
Q = ε x Qmax = 21546 W
We know that,
Heat transfer, Q = mc Cpc (t2 – t1)
=>
21546 = 1381
1381.38
38 t2 - 27627.6
27627 6
=>
t2 = 35.5o C
Outlet temperature of water, t2 = 35.5 oC
We know that,
Heat transfer, Q = mh Cph (T1 – T2)
=>
21546 = 28728 – 302.4 T2
=>
T2 = 23
23.75
75o C
Outlet temperature of oil, T2 = 23.75 oC
9
NTU Problems
3.
In a counter flow heat exchanger, water is heated from 20 oC to 80 oC by
an oil with a specific heat of 2.5 kJ/kg K and mass flow rate of 0.5 kg/s.
The oil is cooled from 110 oC to 40 oC. If the overall heat transfer
co-efficient is 1400 W/m2K,
K find the following by using NTU method.
method
a. Mass flow rate of water
b. Effectiveness of heat exchanger
c. Surface area.
Given:
Gi
Hot fluid – oil
(T1 , T2)
Cold fluid – water
(t1 , t2)
Inlet temperature of water, t1 = 20o C
Outlet temperature of water, t2 = 80o C
Mass flow rate of oil
oil, mh= 0
0.5
5 kg/s
Specific heat of oil, Cph = 2.5 kJ/kg-K = 2.5 x 103 J/kg-K
Inlet temperature of oil, T1 = 110o C
p
of oil, T2 = 40o C
Outlet temperature
Overall heat transfer co-efficient, U = 1400 W/m2K
10
NTU Problems
T find:
To
fi d
a. Mass flow rate of water, mc
b. Effectiveness of heat exchanger, ε
c. Surface area, A
Solution:
We know that,
Heat lost by oil = Heat gained by the water
=>
Qh = Qc
=>
mh Cph (T1 – T2) = mc Cpc (t2 – t1)
[Specific heat of water Cpc = 4186 J/kg K]
=>
M
Mass
fl
flow rate
t off water,
t mc = 0.348
0 348 k
kg/s
/
Capacity rate of oil,
C = mh x Cph = 1250 W/K
Capacity rate of water, C = mc x Cpc = 1456.73 W/K
Æ (1)
Æ (2)
From (1) and (2),
Cmin = 1250 W/K
Cmax = 1456.73 W/K
=>
Cmin / Cmax = 0.858
Æ (3)
11
NTU Problems
We know that,
Effectiveness, ε = (T1 – T2) / (T1 – t1)
[since mhCph = Cmin]
[HMT data book, Pg no: 152]
Effectiveness ε = 0.77
Effectiveness,
0 77
To find NTU, refer HMT data book page no 164 [Counter flow]
From graph,
Yaxis Æ ε = 0.77
Curve Æ Cmin / Cmax = 0.858
We know that,
NTU = UA / Cmin
Substituting for NTU, U and Cmin
=>
Corresponding Xaxis value is 2.9
i.e.,
NTU = 2.9
[[HMT data book,, Pg
g no: 152]]
Surface area, A = 2.59 m2
12
HEAT EXCHANGER
Fouling of a heat exchanger
Surface of the heat exchangers do not remain clean after it has been in use for some
time
Surface becomes fouled with scales or deposits which in turns affects the value of
heat transfer coefficient (U)
This fouling effect is taken care by introducing an additional thermal resistance
called the fouling resistance or fouling factor (Rf)
Uouter =
1
⎛r ⎞ 1
1
r ⎛r ⎞ ⎛r ⎞
+ Rfo + o ln⎜⎜ o ⎟⎟ + ⎜⎜ o ⎟⎟Rfi + ⎜⎜ o ⎟⎟
ho
k ⎝ ri ⎠ ⎝ ri ⎠
⎝ ri ⎠ hi
U inner =
1
⎛r ⎞ 1
r ⎛r ⎞ ⎛ r ⎞
1
+ R fi + i ln⎜⎜ o ⎟⎟ + ⎜⎜ i ⎟⎟ R fo + ⎜⎜ i ⎟⎟
k ⎝ ri ⎠ ⎝ ro ⎠
hi
⎝ ro ⎠ ho
Causes for fouling
1.
2.
3.
4.
5.
Presence of inherent salt & scales in the fluids used in the heat exchangers
Condensation of inorganic vapor components
Usage
g of flue g
gas which contains ash & effluents with it
Chemical reaction resulting from the fluids & the heat exchanger components
Usage of low melting matter
13