Page 453 - Design and Operation of Heat Exchangers and their Networks

436 Design and operation of heat exchangers and their networks

The outlet temperatures of the jth exchanger ( j=E1, E2, E3, HU, and
CU) are obtained from Eq. (6.2):

t 00 ¼ 1 ε j t 0 þε j t 0 (9.13)
h, j h, j c, j

t 00 c, j ¼ R j ε j t 0 h, j þ 1 R j ε j t 0 c, j (9.14)
where
_
NTU j ¼ k j A E, j =C h,E, j (9.15)
_ _
R j ¼ C h,E, j =C c,E, j (9.16)
ð
1 e NTU j 1 R jÞ
ε j ¼ (9.17)
ð
1 R j e NTU j 1 R jÞ
Attention should be payed to the following special cases:
For NTU j →∞, R j 1 : ε j ¼ 1 (9.18)
For NTU j →∞, R j > 1 : ε j ¼ 1=R j (9.19)
NTU j
For R j ¼ 1 : ε j ¼ (9.20)
1þNTU j
For R j ¼ 0 : ε j ¼ 1 e NTU j (9.21)
For R j →∞ : ε j ¼ 0, R j ε j ¼ 1 e NTU c, j (9.22)
where
_
NTU c, j ¼ k j A E, j =C c,E, j (9.23)
The optimal control vector c can be found by minimizing the utility
cost C(d, c, u):

4 _
min 171:428 10 C C2 t c,out,HU t c,in,HU Þ
ð
c2R c
4 _
þ60:576 10 C H2 t h,in,CU t h,out,CU Þg (9.24)
ð
s:t: t 00 t h,out,E1 ¼ 0
H1
t 00 t h,in,CU 0
H2
t 00 t h,out,CU ¼ 0
H2
_ _
C c,E1 t c,out,E1 þc 5 t 0 =C C1 t 00 ¼ 0
C1 C1
t c,in,HU t 00 0
C2
00
t c,out,HU t C2 ¼ 0
c 2 þc 3 _ C C2 0

R c ¼ 0 c 1 _ C H2 ,0 c 2 _ C C2 ,0 c 3 _ C C2 ,0 c 4 _ C H1 ,0 c 5 _ C C1
By solving Eq. (9.24) at the nominal operation point, the minimum total
utility cost can be obtained as C min (d,u N )=0.8117 $/h. The corresponding