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3.31
25
°
C, 30
10
3
6
.
×=
Pa
MPa
125 kg s
(a) mass balance (steady-state)
0
125
12
=+
⇒=
−=
±±
MM
kg s
Energy balance (neglecting PE terms)
0
22
11
1
2
2
2
F
H
G
I
K
J
++
F
H
G
I
K
J
±
²
±
²
MH
v
v
;
±
Mv
Am
n
v
A
mass density,
n
=
molar density,
==
ρ
=
v
=
velocity,
A
=
pipe area,
m
=
molecular weight.
±
.
.
.
.
..
M
m
P
RT
vA
v
v
mv
=
×
×× ⋅
××
×
=
×
×⋅
=×
=
125
16
30
009
2283
2
16
21
4170
10
417
6
2
2
3
kg s
kg kmol
Pa
298.15 K
8.314
10 Pa m
ms
m
kg kmol
m s
kg m Ns
J kmol
kJ kmol
33
2
2
af
a
f
π
Back to energy balance, now on a molar basis
HH
mv
mv
CT T
p
2
2
1
2
−=−= −
As a first guess, neglect kinetic energy terms . . .
T T
p
C
1 2
02
−=⇒==
°
5
Now check this assumption
v
nv
n
Pv
P
v
2
2
2
6
1
6
20
34 24
===
×
×
=
.
.
.
m
s
Recalculate including the kinetic energy terms
m
vv
p
Jkmo
l
1
2
2
2
2
16
2
34 24
5209
−
=
ch
c
h
TT
T
1
5209
368
1000
014
=−
×
=− °
l
Jmo
l
mo
lkmo
l
C
.
.
Thus the kinetic energy term makes such a small contribution, we can safely ignore it.
(b)
Mass balance on compressor (steady-state)
0
NN
3.0
×
10
6
Pa
T
2
= ?
compressor

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