14-Practice Problems.pdf
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Chapter 14
Turbomachinery
Problem 14.1
A propeller is to be selected for a light airplane with a mass of 1500 kg which
will cruise at 100 m/s at an altitude where the density is 1 kg/m
3
The lift-to-drag
ratio at cruise conditions is 30:1, and the engine rpm is 3500. The thrust coe
!
-
cient at maximum e
!
ciency is 0.03, and the maximum e
!
ciency is 60%. Find the
diameter of the propeller, the advance ratio at maximum e
!
ciency, and the power
output required by the engine.
Solution
At cruise conditions, the drag is equal to the thrust.
! = " = #$(#$") = 1500×9%81$30 = 490%5
N
The thrust is given by
! = &
!
'(
2
"
4
Μ
¶
3500
60
2
490%5 = 0%03×1×
"
4
= 102%1"
4
Solving for diameter
"
4
= 4%805
" =
1%48
m
The advance ratio is
) =
*
"
100
58%3×1%48
=
1%16
125
("
=
126
CHAPTER 14. TURBOMACHINERY
The power output is
+ =
!*
"
,
=
490%5×100
0%6
= 81%75
kW
=
110
hp
Problem 14.2
A pump delivers 0.25 m
3
/s of water against a head of 250 m at a rotational speed of
2000 rpm. Find the speci
Þ
c speed, and recommend the appropriate type of pump.
Solution
The speci
Þ
c speed of the pump is
(
#
=
(-
1$2
(.!/)
3$4
=
2000
60
×0%25
1$2
(9%81×250)
3$4
= 0%048
From Fig. 14.14, a mixed
ß
ow pump is recommended.
Problem 14.3
A Francis turbine is being designed for a hydroelectric power system. The
ß
ow
rate of water into the turbine is 5 m
3
$
s. The outer radius of the blade is 0.8 m, and
the inner radius is 0.5 m. The width of the blade is 15 cm. The inlet vane angle
is 80
o
%
The turbine rotates at 10 rps. Find the inlet angle of the
ß
ow with respect
to the turbine to ensure nonseparating
ß
ow, the outlet vane angle to maximize the
power, and the power delivered by the turbine.
Solution
The radial component of velocity into the turbine is
*
%&1
=
-
0
1
=
-
212
1
3
=
5
21 ×0%8×0%15
= 6%63
m/s
127
The rotational speed is
10×21 = 62%8
rad/s. The angle for nonseparating
ß
ow is
4
1
= ARCCOT(
2
1
5
*
%&1
+COT6
1
)
= ARCCOT(
0%8×62%8
6%63
+COT80
o
)
=
7%35
o
The power produced by the turbine is
+ = '-5(2
1
*
1
COS4
1
!2
2
*
2
COS4
2
)
At maximum power the outlet angle,
4
2
, should be
1$2
. In other words, the
ß
ow
would exit radially inward. The outlet angle for the exit is
4
2
= ARCCOT(
2
2
5
*
%&2
+COT6
2
)
so
COT4
2
= COT
1
2
= 0 =
2
2
5
*
%&2
+COT6
2
The radial velocity for the inner radius is
*
%&2
=
-
0
2
=
-
212
2
3
5
21 ×0%5×0%15
= 10%6
m/s
=
Thus
COT6
2
= !
0%5×62%8
10%6
= !2%96
6
2
=
161
o
The inlet velocity is
*
1
=
*
%&1
SIN4
1
=
6%63
SIN7%35
o
= 51%8
m/s
The power output is
+ = '-52
1
*
1
COS4
1
= 1000×5×62%8×0%8×51%8×COS7%35
o
=
1%29
MW
128
CHAPTER 14. TURBOMACHINERY
Problem 14.4
A pump is being used to pump water at 80
o
F (
7
'
= 0%506
psia) from a supply
reservoir at 20 psia. The inlet to the pump is a 3-inch pipe. The NPSH for the
pump is 10 ft. Find the maximum
ß
ow rate in gpm to avoid cavitation. Neglect
head losses associated with the inlet and supply pipe.
Solution
The net positive suction head (NPSH) is de
Þ
ned as the di
"
erence between the local
head at the entrance to the pump and the vapor pressure.
NPSH
=
7
!
7
'
8
The energy equation between the supply reservoir (1) and the entrance to the pump
(2) is
8
+
*
2.
+9
1
=
7
2
8
+
*
2.
+9
2
+:
(
Simplifying
8
=
7
1
!
*
2.
8
The head at the pump entrance must be
7
2
8
=
NPSH
+
7
'
8
= 10+
0%506×144
62%4
= 11%17
ft
The velocity head must be
*
2.
=
20×144
62%4
!11%17
= 35%0
ft
So the velocity is
"
*
2
=
2×32%2×35 = 47%5
ft/s
The corresponding
ß
ow rate is
1
4
×
Μ
3
12
¶
2
- = *0 = 47%5×
= 2%33
cfs
= 140
cfm
=
1047
gpm
7
1
7
2
129
Problem 14.5
A wind tunnel is being designed as shown. The air is drawn in through a se-
ries of screens and
ß
ow straighteners at a diameter of 1.5 m. The test section of
the tunnel is 1 m. A fan is mounted downstream of the test section. The head loss
coe
!
cient for the screens and straighteners is 0.2 and the head loss coe
!
cient for
the rest of the tunnel is 0.05 based on the velocity in the test section. The axial
fan has a pressure-
ß
ow rate curve represented by
"
Μ
-
100
¶
2
#
!7 = 1000
1!
Pa
where
-
is in m
3
/s. Find the velocity in the test section. Take
' = 1%2
kg/m
3
%
Solution
A system curve has to be generated and combined with the pressure-discharge
characteristics of the fan. Writing the energy equation from the intake of the wind
tunnel to the exit
8
+
*
2.
+9
1
+:
)
=
7
2
8
+
*
2.
+9
2
+:
*
+:
(
which simpli
Þ
es to
:
)
=
*
2.
+:
(
!7
)
= '
*
2
+!7
(
The velocity at the exit can be expressed in terms of discharge as
*
2
=
-
0
=
4
×1
2
= 1%27-
-
+
The pressure loss is given by
!7
(
= 0%2'
*
,
2
+0%05'
*
2
7
1
or in terms of pressure
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