Page 99 - Design and Operation of Heat Exchangers and their Networks

Steady-state characteristics of heat exchangers 87

values of the outlet fluid temperatures. Also the heat transfer coefficients in
the hot and cold fluid sides may be evaluated from suitable heat transfer cor-
relations and fluid properties, together with the thermal resistance of the wall
separating the two fluids and fouling resistances, giving the overall heat
transfer coefficient k by Eq. (2.56) as
1 1 R f ,h δ w R f,c 1
¼ + + + + (3.94)
kA α h A h A h λ w A m A c α c A c

in which A m is the mean wall area perpendicular to conductive heat flux
through the wall. For a given heat exchanger with specified flow rates
and inlet fluid temperatures, the values of NTU and R can be evaluated.
Then, we can use the ε-NTU relation to calculate the effectiveness to obtain
the outlet fluid temperatures.

Example 3.1 Rating a counterflow heat exchanger
Consider a problem of rating a counterflow heat exchanger with surface area
2
of A¼38m , and the overall heat transfer coefficient has been estimated to
2
be k¼180W/m K. The hot fluid has a mass flow rate of 1.6kg/s, specific
thermal capacity of 1.2kJ/kgK, and the inlet temperature of 230°C. The
cold fluid has a mass flow rate of 1.0kg/s, specific thermal capacity of
4.2kJ/kgK, and the inlet temperature of 50°C. Determine the heat load
of the exchanger and the outlet temperatures of both fluids.

Solution
From Eq. (3.64), we have

_
NTU h ¼ kA=C h ¼ 180 38= 1:6 1:2 10 3 ¼ 3:56
3 3
R h ¼ _ C h = _ C c ¼ 1:6 1:2 10 = 1:0 4:2 10 ¼ 0:457
The temperature coefficient of the counterflow heat exchanger is
calculated with Eq. (3.70):
ð
1 e NTU h 1 R h Þ 1 e 3:56 1 0:457Þ
ð
ε h ¼ ¼ ¼ 0:916
1 R h e NTU h 1 R h Þ 1 0:457e 3:56 1 0:457Þ
ð
ð
Then, the outlet fluid temperatures are ready to be solved with
Eq. (3.60):

00 0 0 0
t ¼ t ε h t t ¼ 230 0:916 230 50ð Þ ¼ 65:1°C
h h h c
_ 0 0
Q ¼ C h ε h t t ¼ 1:6 1:2 0:916 230 50ð Þ ¼ 317 kW
h c
00 0
t ¼ t + Q= _ C c ¼ 50 + 317= 1:0 4:2ð Þ ¼ 125:4°C
c c