Page 101 - Design and Operation of Heat Exchangers and their Networks

Steady-state characteristics of heat exchangers 89

all of those solutions were mathematically equivalent and all of the expres-
sions obtained required a numerical evaluation.
The energy equations for the crossflow with both fluids unmixed can be
expressed as follows:

_ ∂t h kA
C h ¼ ð t c t h Þ (3.97)
∂x L x
kA
_ ∂t c
C c ¼ ð t h t c Þ (3.98)
∂y L y
Normalizing the temperature of each fluid and the length of the
exchanger in its flow direction,

_
x ¼ kA=C h x=L x ¼ NTU h x=L x (3.99)

_
y ¼ kA=C c y=L y ¼ NTU c y=L y ¼ R h NTU h y=L y (3.100)
t t 0
t ¼ c (3.101)
t t 0
0
h c
we can present the energy equations in the dimensionless form as
∂t h
¼ t c t h (3.102)
∂x
∂t c
¼ t h t c (3.103)
∂y
with the boundary condition
x ¼ 0 : t h ¼ 1; y ¼ 0 : t c ¼ 0 (3.104)
We will solve this equation system by means of the Laplace transform.
Applying the Laplace transform respecting to y, we obtain the ordinary dif-
ferential equation:

de t h s
¼ e t h (3.105)
dx s +1
with the boundary condition

x ¼ 0 : e t h ¼ 1=s (3.106)
The solution in the Laplace domain is derived as

1 sx

e t h ¼ e s +1 (3.107)
s