This preview shows pages 1–3. Sign up to view the full content.
1
ME 416
Computer Assisted Design of
Thermal Systems
Solutions to Practice Problem Set #2
Turbomachinery
1. We begin by determining the actual power.
The actual power will be
determined from the generic centrifugal compressor graph. The inlet volume flow
rate must be calculated from
m
v
=
V
1
1
where
v =
RT
P
=
(0.287)(295)
98
= 0.864 m / kg
1
1
1
3
Calculating
V = v m = (0.864)(16.4) = 14.2 m / s
= (14.2 m / s)(60 s / min)(3.281 ft / m) = 30.1 x 10 cfm
1
1
3
3
3
3
Reading from the graph for a pressure ratio of 7, we find
Basic BHP = 7,000 hp
Correcting for the inlet pressure, since it’s not 14.5 psia gives
[ ]
/ .
)
.
W =
(P
Basic BHP =
(98 / 6.895)
14.5
(7000) = 6862 hp
= (6862 hp)(0.7455 kW / hp) = 5,115 kW
act
1
6895
145
The isentropic efficiency is given by
m
/
W
w
=
w
w
=
act
ideal
act
ideal
s
η
This requires the ideal work.
Assuming a constant specific heat or 1.005 J/kg, we
can write
w
ideal
= c
P
(T
2i
 T
1
)
where T
2i
comes from our isentropic ratio
K
514
)
7
)(
295
(
P
P
T
T
4
.
1
/
)
1
4
.
1
(
/
)
1
(
1
2
1
i
2
=
=
=

γ

γ
so that
w
ideal
= (1.005)(514295) = 220 kJ/kg
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentME 416 CAD of Thermal Systems
2
η
s
ideal
act
ideal
act
=
w
w
=
w
W
m
=
220
5115/ 16.4
= 0.705
/
2. We begin with the single stage compression system.
With a compression ratio
of 12:1, we have
P
2
= 12P
1
= 1200 kPa
The ideal exit temperature is calculated from
K
610
)
12
)(
300
(
P
P
T
T
4
.
1
/
)
1
4
.
1
(
/
)
1
(
1
2
1
i
2
=
=
=

γ

γ
The ideal work is calculated from
w
ideal
= c
P
(T
2i
 T
1
) = (1.005)(610300) = 312 kJ/kg
Then using the adiabatic efficiency, we calculate the actual work
w
act
= w
ideal
/
η
s
= 312/0.85 = 367 kJ/kg
For the two stage compression with intercooling system, we have for stage #1
P
2
= 6.5P
1
= 650 kPa
The ideal exit temperature is calculated from
K
512
)
5
.
6
)(
300
(
P
P
T
T
4
.
1
/
)
1
4
.
1
(
/
)
1
(
1
2
1
i
2
=
=
=

γ

γ
The ideal work is calculated from
w
ideal
= c
P
(T
2i
 T
1
) = (1.005)(512300) = 213 kJ/kg
Then using the adiabatic efficiency, we calculate the actual work
w
act,1
= w
ideal
/
η
s
= 213/0.85 = 251 kJ/kg
For the intercooler analysis we will need to calculate the actual exit temperature
for the first stage, which comes from
T
2a
= T
1
+ w
actl
/c
P
= 300 + (251)/1.005 = 550 K
The intercooler is assumed to be a constant pressure device, so that
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '07
 SOMERTON

Click to edit the document details