Thermal Engineering Page 3
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1
2
3
44-99 CO
gas flows through a pipe. The volume flow rate and the density at the inlet and
2
the volume flow rate at the exit of the pipe are to be determined.
3 MPa
500 K
CO
450 K
2
2 kg/s
Properties The gas constant, the critical pressure, and the critical temperature of CO
are
2
(Table A-1)
3
R = 0.1889 kPa·m
/kg·K,
T
= 304.2 K,
P
= 7.39 MPa
cr
cr
Analysis (a) From the ideal gas equation of state,
⋅
⋅
3
m &
RT
(2
kg/s)(0.18
89
kPa
m
/kg
K)(500
K)
&
=
=
=
V
3
1
0.06297
m
/kg
(2.1%
error)
1
P
(3000
kPa)
1
P
(3000
kPa)
ρ
=
=
=
3
1
31.76
kg/
m
(2.1%
error)
1
⋅
⋅
3
RT
(0.1889
kPa
m
/kg
K)(500
K)
1
⋅
⋅
3
m &
RT
(2
kg/s)(0.18
89
kPa
m
/kg
K)(450
K)
&
=
=
=
V
2
3
0.05667
m
/kg
(3.6%
error)
2
P
(3000
kPa)
2
(b) From the compressibility chart (EES function for compressibility factor is used)
⎫
P
3
MPa
=
=
=
1
P
0.407
⎪
R
⎪
P
7.39
MPa
=
cr
⎬
Z
. 0
9791
1
T
500
K
⎪
=
=
=
1
T
1.64
⎪
R
1 ,
⎭
T
304.2
K
cr
⎫
P
3
MPa
=
=
=
2
P
0.407
⎪
R
⎪
P
7.39
MPa
=
cr
⎬
Z
. 0
9656
2
T
450
K
⎪
=
=
=
2
T
1.48
⎪
R
2 ,
⎭
T
304.2
K
cr
⋅
⋅
Z &
3
m
RT
(0.9791)(2
kg/s)(0.18
89
kPa
m
/kg
K)(500
K)
&
=
=
=
V
1
1
3
Thus,
0.06165
m
/kg
1
P
(3000
kPa)
1
P
(3000
kPa)
ρ
=
=
=
3
1
32.44
kg/
m
1
⋅
⋅
3
Z
RT
(0.9791)(0
.1889
kPa
m
/kg
K)(500
K)
1
1
⋅
⋅
3
Z &
m
RT
(0.9656)(2
kg/s)(0.18
89
kPa
m
/kg
K)(450
K)
&
=
=
=
V
3
2
2
0.05472
m
/kg
2
P
(3000
kPa)
2
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