Page 64 - Design and Operation of Heat Exchangers and their Networks

Basic thermal design theory for heat exchangers 51

We use Eq. (2.108) to size the counterflow heat exchanger:
Q 350 10 3
ð kAÞ ¼ ¼ ¼ 8948 W=K
i Δt m,d 39:11
To get the heat exchanger area of tube inside A i , we shall evaluate the
overall heat transfer coefficient based on the area of tube inside. Using the
same calculation procedure for Example 2.4, we have

t h,m ¼ 100 + 80ð Þ=2 ¼ 90°C
c p,h ¼ 4:206 kJ=kgK,λ h ¼ 0:6728 W=mK,μ ¼ 3:142 10 4 sPa,
h
Pr h ¼ c p,h μ =λ h ¼ 1:964:
h

Q= c p,h t t 00
0
ð
½
_ m h h h 350= 4:206 100 80ފ
G h ¼ ¼ 2 ¼ 2
A c,h N tube πd =4 53 π 0:016 =4
i
2
¼ 390:5kg=m s
Re h ¼ G h d i =μ ¼ 390:5 0:016=3:142 10 4 ¼ 19;886
h
2 2
½
f =8 ¼ 1:82lg Re h Þ 1:64½ ð Š =8 ¼ 1:82lg 19886Þ 1:64Š =8
ð
¼ 0:003269
" #
2=3
ð
ð f =8Þ Re h 1000ÞPr h d i 0:11
Nu h ¼ 1+ ð Pr=Pr w Þ
p ffiffiffiffiffiffiffi 2=3 L
1+12:7 f =8 Pr 1
h
" #
2=3
0:003269 19, 886 1000Þ 1:964 0:016 0:11
ð
ð
¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1+ 1:964=Pr w Þ
1+12:7 0:003269 1:964 2=3 1 2:701 |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}
¼1
¼ 88:65
2
α h ¼ Nu h λ h =d i ¼ 88:65 0:6728=0:016 ¼ 3728 W=m K
d o ¼ d i +2δ w ¼ 0:016 + 2 0:001 ¼ 0:018 m:
ð
ð
d i ln d o =d i Þ 0:016 ln 0:018=0:016Þ 5 2
R w,i ¼ ¼ ¼ 2:356 10 m K=W
2λ w 2 40

ð kAÞ i 1 d i
L ¼ + R w +
N tube πd i α h α c d o

8948 1 5 0:016
¼ +2:356 10 + ¼ 2:971m
53 π 0:016 3728 1500 0:018
and
ð kAÞ i 8948 2
k i ¼ ¼ ¼ 1131 W=m K
N tube πd i L 53 π 0:016 2:971
According to the energy equation q¼α h (t h t h,w )¼k i Δt d , we have