Page 89 - Design and Operation of Heat Exchangers and their Networks

Steady-state characteristics of heat exchangers 77

By substitution of Eqs. (3.30)–(3.33) into Eq. (3.19), the temperature
distributions in the counterflow heat exchanger can be presented as
_ _
ð
ð
kA= _ C h + kA= _ C cÞ 1 zÞ
t h t 0 e C h =C c
c
¼ (3.34)
0
t t 0 ð kA= _ C h kA= _ C cÞ _ _
h c e C h =C c
h i
_ _ ð kA= _ C h kA= _ C cÞ 1 zÞ
ð
t c t 0 C h =C c 1 e
c
¼ (3.35)
t t 0 _ _ ð kA= _ C h kA= _ C cÞ
0
h c C h =C c e
The outlet temperatures of hot and cold fluids and the heat load of the
exchanger are
_ _ kA= _ C h kA= _ C cÞ
ð
00
t t 0 1 C h =C c e
h c
¼ (3.36)
0
t t 0 _ _ kA= _ C h kA= _ C cÞ
ð
h c 1 C h =C c e
h i
_ _ kA= _ C h kA= _ C cÞ
ð
C h =C c 1 e
00
t t 0
c c
¼ (3.37)
0
t t 0 _ _ kA= _ C h kA= _ C cÞ
ð
h c 1 C h =C c e
h i
_
ð
C h 1 e kA= _ C h kA= _ C cÞ t t c 0
0
h

_

_

Q ¼ C h t t 00 ¼ C c t t ¼
00
0
0
h h c c
_
_
ð
1 C h =C c e kA= _ C h kA= _ C cÞ
(3.38)
From Eqs. (3.36), (3.37), the ratio of the temperature difference at the
two ends of the heat exchanger can be written as
t t 00 c ð kA= _ C h kA= _ C cÞ
0
h
t t 0 c ¼ e (3.39)
00
h

_
_
0
00
Because kA=C h kA=C c ¼ kA t t 00 t t 0 =Q, Eq. (3.39) can
h h c c
be presented by the logarithmic mean temperature difference:

00
0
t t c 00 t t 0 c
h
h
Q ¼ kA (3.40)
00
0
ln t t = t t 0
00
h c h c
It is important to emphasize that the analytical solution, Eq. (3.11),is
valid only if all the eigenvalues are distinct. Obviously, it is not the case when
_
_
C h ¼ C c , in this case the eigenvalues have the double root, r 1 ¼r 2 ¼0. In
(m)
general, if the coefficient matrix A has a m-multiple eigenvalue r , the
m-independent solutions corresponding to r (m) can be expressed as
m
e
t i xðÞ ¼ d i x i 1 r ðÞ x ð i ¼ 1, 2, …, mÞ (3.41)