8/16/2019 Compact Hex Exchanger
1/2
σa 0.697:= βa 751.3m
2
m3
:=
Fin length : Lf 0.436 0.120−( )
2in:= Lf 4.013 10
3−× m=
ma 20kg
s:= Th1 380K:= Th2 320K:= mw 50
kg
s:= Tc1 17 273+( )K:=
a)Rea 10
4:= Ga
Rea µa⋅
Dha
:= Ga 69.403kg
m2s
=
From fig 9.7 for Re=10^4 : j 0.0039:= f a 0.018:=
ha jG
acp
a⋅
Pra
2
3
⋅:= ha 343.803W
m2K
=
ME421 HEAT EXCHANGER AND STEAM GENERATOR DESIGN
MT 2 QUESTION 7 SOLUTION
SPRING 2006
Question 7 (35 pts): An air-to-water compact heat exchanger
is to be designed to serve as an intercooler in a
gas turbine plant. The geometrical details of the proposed
surface, 9.68-0.870-R for the air side is given in Figu
9.8 and in Table 9.1. Hot air with a flow rate 20 kg/s at 2 atm
and 380 K with Re=104 enters the matrix and ex
at 320 K. Water enters at 17 ºC and with a flow rate of 50 kg/s
flows inside the flat tubes. Water side geometricdetails are
Dh=0.373 cm, ów=0.129, βw=138 m2/m3.a) Obtain air side heat
transfer coefficient (ha)
b) Calculate fin efficiency (çf ) and overall surface
efficiency (ço) for the air side with the fins mof Aluminum
(kAl=250 W/mK). Here you may assume fins as flat rectangular
cross-section fins.
c) Find the free flow area (Aff,a) and frontal area (Afr,a) for
the air sided) By taking the frontal area of water side same as the
air side frontal area (Afr,w=Afr,a), calculate wa
heat transfer coefficient (hw) using Gnielinski correlation.
e) By neglecting wall resistance and fouling, calculate the
overall heat transfer coefficient based on air
surface (Ua)f) Using å-NTU method, obtain air side heat transfer
area (Aa) and the heat exchanger length.g) BONUS: Calculate the air
side pressure drop (∆pa).
Properties of air at 2 atm and Tf=(320+380)/2=350 K:
cp,a=1009 J/kg.K Pra=0.708 µa=20.8x10-6 N.s/m2
ρa=2.0 kg/m3Properties of water at 17 ºC:
cp,w=4184 J/kg.K Prw=7.65 µw=10.84x10-4 N.s/m2 kw=0.594
W/m.K ρw=999kg/m3
Solution :
cpa 1009 Jkg K⋅
⋅:= Pra 0.708:= µa 20.8 10 6−⋅ N s⋅
m2
⋅:= ρa 2 kg
m3
⋅:=
cpw 4184J
kg K⋅⋅:= Prw 7.65:= µw 10.84 10
4−⋅
N s⋅
m2
⋅:= k w 0.594W
m K⋅⋅:= ρw 999
kg
m3
⋅:=
Water side geometry: Dhw 0.373cm:= σw 0.129:= βw 138m
2
m3
:=
From the core geometry (Fig 9.8 and Table 9.1 for surface
9.68-0.870-R:
t 0.0102cm:= Af_to_At_ratio 0.795:= Dha 0.2997cm:=
8/16/2019 Compact Hex Exchanger
2/2
Ua1
ηo ha⋅
1
Aw_to_Aa_ratio hw⋅
+
1−
:=U
a218.721
W
m2K
=
f )Q ma cpa⋅ Th1 Th2−( )⋅:= Q 1.211 10
6× W=
Tc2 Tc1Q
mw cpw⋅+:= Tc2 295.788 K=
Cw mw cpw⋅:= Cw 2.092 105
×kg m
2
s3K
= Ca ma cpa⋅:= Ca 2.018 104
×kg m
2
s3K
=
Cmax Cw:= Cmin Ca:=
Cr
Cmin
Cmax
:= Cr 0.096= εQ
Cmin Th1 Tc1−( )⋅:= ε 0.667=
unmixed - unmixed cross flow heat exchanger: from Fig 2.15 (c) :
NTU 1.25:=
Aa
NTU Cmin⋅
Ua
:= Aa 115.33 m2
=
Volume : VAa
βa
:= V 0.154m3
= LV
Afra
:= L 0.371 m=
g ) assuming incompressible: ∆pa f a
Ga2
2 ρ a⋅⋅
4L
Dha
⋅:= ∆pa 1.074 104
× Pa=
b )
k Al 250W
m K⋅:= mf
2 ha⋅
k Al t⋅:= mf 164.21
1
m= ηf
tanh mf Lf ⋅( )mf Lf ⋅
:= ηf 0.877=
ηo 1 Af_to_At_ratio 1 ηf −( )⋅−:= ηo 0.902=
c ) Aff a
ma
Ga
:= Aff a 0.288m2
= Afra
Aff a
σa
:= Afra 0.413m2
=
d ) Afrw Afra:= Gw
mw
σw Afrw⋅:= Gw 937.475
kg
m2s
= Rew
Gw Dhw⋅
µw
:= Rew 3.226 103
×=
Using Gnielinski correlation:
f w 1.58 ln Rew( )⋅ 3.28−( )2−
:= f w 0.011=
Nuw
f w
2
Rew 1000−( )⋅ Prw⋅
1 12.7f w
2⋅ Prw
2
31−
⋅+
:= Nuw 25.378= hw
k w Nuw⋅
Dhw:= hw 4.041 10
3
×
W
m2K
=
e )
Aw_to_Aa_ratioβw
βa
:= Aw_to_Aa_ratio 0.184=
