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Transcript of pumps turbines
Vapor-cycle Vapor-cycle Power PlantsPower Plants
Ideal Regenerative Rankine Cycle Ideal Regenerative Rankine Cycle with Open Feedwater Heaterwith Open Feedwater Heater
Open Systems - Control VolumeOpen Systems - Control VolumeOpen Systems - Control VolumeOpen Systems - Control Volume
Axial-flow Axial-flow Hydraulic Hydraulic TurbineTurbine
Compressors, pumps, and fansCompressors, pumps, and fans
CompressorSide view End view of pump of pump
Inflow and Outflow Velocity TrianglesInflow and Outflow Velocity Triangles
Radial Flow Pump (Fan)Radial Flow Pump (Fan)
WV
U = r
Relative Velocity – tangent to blade
Radial Flow TurbineRadial Flow Turbine
WV
U = r
Absolute velocity – tangent to stator blade
Relative velocity – tangent to rotor blade
Axial Flow Compressor (Pump)Axial Flow Compressor (Pump)
U = r
Relative Velocity – tangent to rotor blade
Axial Flow TurbineAxial Flow Turbine
U = r
Relative Velocity – tangent to rotor blade
Absolute velocity – tangent to
stator blade
Lawn SprinklerLawn SprinklerWater enters a rotating lawn sprinkler through its base at the steady rate of 16 gal/min as shown in the figure. The exit cross-sectional area of each of the two nozzles is 0.04 in.2, and the flow leaving each nozzle is tangential. The radius from the axis of rotation to the centerline of each nozzle is 8 in. (a) Determine the resisting torque required to hold the sprinkler head stationary. (b) Determine the resisting torque associated with the sprinkler rotating with a constant speed of 500 rpm. (c) Determine the angular velocity of the sprinkler if no resisting torque is applied.
Lawn SprinklerLawn Sprinkler
e r VUVeWW :velocity relative
er U :veloecity nozzle
eVV :velocity absolute
222
2
2
Continuity equation
s/ft17.64
ft
in144
gal48.7
ft
s60
min
in04.02
min/gal16
A2
QW
s
slugs0692.0
gal48.7
ft
s60
min
min
gal16
ft
slugs94.1
AW2AVQm
0eAeW2eAeVdAnW
2
23
22
2
3
3
2211
22z1CS z1
Relative velocity W2
Lawn SprinklerLawn SprinklerMoment-of-momentum equation
z22z222
m
222r2
m
z1z1z1r1
CSmd
zshaftshaft
eVrmeVrm0
eAeW2 eVereAeV eVer
dAnWVre TT
ft-lb96.2s
ftslugs96.2
s
ft17.64
12
ft8
s
slugs0692.0 VrmT
2
2
22shaft
(a) U2 = 0 ,
V2 = W2
Lawn SprinklerLawn Sprinkler
(b) = 500 rpm
tf-lb 35.1s
ft26.29
12
ft8
s
slugs0692.0 VrmT
s
ft26.29
s
ft91.3417.64UWV
s
ft91.34ft
12
8
s
rad2500r U
22shaft
222
2
(c) Tshaft = 0
min
rev920
rad
rev
2
1
min
s60
ft12/8
s/ft17.64
r
Ws
ft17.64rUW
0V0T
2
2
22
2shaft
Lawn SprinklerLawn Sprinkler
The two-nozzle law sprinkler discharge water at a rate of 1 ft3/min and rotates at 1 rev/s. The length from pivot to nozzle exit is 12 in., and the nozzles produce jets that are ¼ in. in diameter. Determine the resisting torque due to friction in the bearing.
W = Vrel
Lawn SprinklerLawn Sprinkler
e sinWe cosWW :velocity relative
er U :velocity nozzle
eVeVV :velocity absolute
2r2
2
rr
Continuity equation
s
ft45.24
s
min
60
1
ft48
1
min
ft12
D
Q2
/4πD2
Q
A2
QW
AW2Qm
0nAnW2QdAnW
2
3
222
2
22
22CS
Lawn SprinklerLawn Sprinkler
Moment-of-Momentum equation
ft-lb 131.0s
ftslugs131.0
ft12
12
s
rad225sin
s
ft45.24ft
12
12
s
ft
60
1
ft
slugs94.1
r sinWQrr sinWrmT
e r sinWrmee r sinWrm
eAeW2 e r sinWe cosWer
dAnWVre TT
e r sinWe cosWWUV
2
2
3
3
222222shaft
z222r222
m
2222r2r2
CSmd
zshaftshaft
22r2222
Hydraulic TurbineHydraulic TurbineA simplified sketch of a hydraulic turbine runner is shown in the figure. Relative to the rotating runner, water enters at section (1) (cylindrical cross section area A1 at r1 = 1.5m) at an angle of 100o
and leaves at an angle of 50o from the tangential direction. The blade height at sections (1) and (2) is 0.45 m and the volume flow rate through the turbine is 30 m3/s. The runner speed is 130 rpm in the direction shown. Determine the shaft power developed.
Hydraulic TurbineHydraulic Turbine
Rotating speed
s
m6.11m85.0
s
rad2
s
min
60
1
min
rev130r U
s
m4.20m5.1
s
rad2
s
min
60
1
min
rev130r U
22
11
Continuity equation
s
m512
m450m8502
sm30
hr2
Q
A
QV
s
m077
m450m512
sm30
hr2
Q
A
QV
s
kg00030
s
m30
m
kg1000Qmm
AVAeVnAVm
3
2222R
3
1111R
3
321
Rr
...
/
...
/
,
ˆˆ
,
,
Velocity TrianglesVelocity Triangles
U1 = r1
U2 = r2
U2 < U1
Relative velocity – tangent to turbine blade
Velocity TriangleVelocity TriangleInflow 100o
50o
V2V1
U1 = r1
U2 = r2
VR,2VR,1
V,1
V,2
W1
W2
Outflow
40o10o
o
1111,R
o11111,
10cosWsinVV
10sinWUcosVV
o
2221,R
o22222,
40cosWsinVV
40sinWUcosVV
Turbine blade tip Turbine blade root
Relative velocity – tangent to turbine blade
Shaft PowerShaft PowerInflow 100o
50o
V2V1
U1 = r1
U2 = r2
VR,2VR,1
V,1
V,2
W1
W2
Outflow
40o10o
MW83.12W10283.1
s
m6.21
s
m4.20
s
m11.1
s
m6.11
s
kg000,30VUVUmW
s
m11.1
s
m40tan5.126.1140tanVr 40sinWUV
s
m6.21
s
m10tan07.74.2010tanVr 10sinWUV
7
1,12,2shaft
oo2,R2
o222,
oo1,R1
o111,
Prob. 5.75: Axial Flow PumpProb. 5.75: Axial Flow Pump
An axial flow gasoline pump consists of a rotating row of blades (rotor) followed downstream by a stationary row of blades (stator). The gasoline enters the rotor axially (without any angular momentum) with an absolute velocity of 3 m/s. The rotor blade inlet and exit angles are 60o and 45o from the axial direction. .The pump annulus passage cross-sectional area is constant Consider the flow as being tangent to the blades involved. Sketch velocity triangles for flow just upstream and downstream of the rotor. How much energy is added to each kilogram of gasoline?
V1=
Velocity TrianglesVelocity Triangles
V1
W1 U1=rm
W2
V2
0V
60cosWVV
1,
o111,x
222,
o2222,x
sinVV
45cosWcosVV
2
45o
V1
U2=rm
60o
U1 = U2
V1
kg
mN44.11
s/mkg
N1
s
m20.2
s
m20.5VU
m
Ww
s
m72.3
25.36cos
s/m3
cos
VV
25.36s/m3
s/m20.2tan
V
Vtan
s
m20.245sin
s
m24.4
s
m20.545sinWUV
s
m20.5UU
s
m24.4
45cos
s/m3
45cos
V
45cos
VW
s
m20.560tan
s
m360tanV60sinWU
s
m6
60cos
s/m3
60cos
VW
22,2shaft
shaft
o2
2,x2
o1
2,x
2,12
oo222,
12
oo1
o
2,x2
oo1
o11
oo1
1
m/s3VVcosVcosVnV 2,x1,x2211
Continuity equation
Shaft Power
Velocity Triangle
s(same arithmetic mean radius)
(outflow direction)
Turbine Power OutputTurbine Power OutputThe hydraulic turbine has an efficiency of 90 percent. The 10oC water flow rate is 10,000 m3/min. Determine the turbine output power in watts for frictionless pipe flow.
(1)
(2)
Control Volume Analysis
Turbine Power OutputTurbine Power OutputAssume steady state and constant water density = 1000 kg/m3 at 10oC.
Energy equation
in netshaft
in
2
in
out
2
out
innetinout
innetCS in
net
Wgz2
Vpmgz
2
Vpm loss No
0quuloss WQdAnVe
Continuity equation Qmmm 21
Turbine Power OutputTurbine Power Output
Mw2.196mN10962.1s/mkg
N
s
mkg10962.1
m120m0s
m81.9
s60
m000,10
m
kg1000
zzQggzgzmW
0VV
0pp
8
22
28
2
3
3
1212innet
21
21
Turbine output power
Mw6.176Mw2.1969.0WWinnett
(Head water and tail water)
Head LossHead LossProblem 5.20R: A hydroelectric power plant operates under the condition illustrated in the figure. The head loss associated with flow from the water level upstream of the dam, section (1), to the turbine discharge at atmospheric pressure, section (2), is 20 m. How much power is transferred from the water to the turbine blade?
(1)
(2)
Mw5.23s/mN105.23
2
s/m2m20m100
s
m81.9
s
m30
m
kg999
2
VhzzgQwmW
ghloss loss2
Vzzgww
s/m2V ,0V
ppp
losswzzg2
VVpp
WQzzg2
VVppuu m
6
2
2
3
3
22
L21out net
shaftout net
shaft
L21
22
21in net
shaftout net
shaft
21
atm21
21in net
shaft12
21
2212
in netshaft
innet12
21
2212
12
Head loss hL = 20 m, reduces the available elevation head from 100 m to 80 m.
Energy equation
Shaft power
Energy EquationEnergy EquationProblem 5.120: A liquid enters a fluid machine at section (1) and leaves at sections (2) and (3) as shown in the figure. The density of the fluid is constant at 2 slugs/ft3. All of the flow occurs in a horizontal plane and is frictionless and adiabatic. For the above-mentioned and additional conditions indicated in the figure, determine the amount of shaft power involved.
Solution: Continuity EquationSolution: Continuity Equation
s
slugs125.3
s
slugs125.325.6mmm
s
slugs125.3ft
144
5
s
ft45
ft
slugs2AVm
s
slugs25.6ft
144
30
s
ft15
ft
slugs2AVm
0mmmAVAVAVdAnV
312
2
3333
2
3111
3213322CS 11
Note: in general 32 mm
Solution: Energy EquationSolution: Energy Equation
hp03.31s/ftlb
hp
550
1
s
ftlb5.17067
s/ftslugs
lb1
s
ft45
2
1
ft
in12
ft/slugs2
in/lb7.14
s
slugs125.3
s/ftslugs
lb1
s
ft35
2
1
ft
in12
ft/slugs2
in/lb50
s
slugs125.3
s/ftslugs
lb1
s
ft15
2
1
ft
in12
ft/slugs2
in/lb80
s
slugs25.6
2
Vpm
2
Vpm
2
VpmW
0umumumloss & 0Q
0emememeAVeAVeAVdAnVe
2
22
3
2
2
22
3
2
2
22
3
2
233
3
222
2
211
1in net
shaft
332211innet
332211333222CS 111
(adabatic & frictionless)
Net shaft power is out of the CV