pumps turbines

45

Transcript of pumps turbines

Page 1: pumps turbines
Page 2: pumps turbines

Vapor-cycle Vapor-cycle Power PlantsPower Plants

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Ideal Regenerative Rankine Cycle Ideal Regenerative Rankine Cycle with Open Feedwater Heaterwith Open Feedwater Heater

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Open Systems - Control VolumeOpen Systems - Control VolumeOpen Systems - Control VolumeOpen Systems - Control Volume

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Axial-flow Axial-flow Hydraulic Hydraulic TurbineTurbine

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Compressors, pumps, and fansCompressors, pumps, and fans

CompressorSide view End view of pump of pump

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Inflow and Outflow Velocity TrianglesInflow and Outflow Velocity Triangles

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Radial Flow Pump (Fan)Radial Flow Pump (Fan)

WV

U = r

Relative Velocity – tangent to blade

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Radial Flow TurbineRadial Flow Turbine

WV

U = r

Absolute velocity – tangent to stator blade

Relative velocity – tangent to rotor blade

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Axial Flow Compressor (Pump)Axial Flow Compressor (Pump)

U = r

Relative Velocity – tangent to rotor blade

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Axial Flow TurbineAxial Flow Turbine

U = r

Relative Velocity – tangent to rotor blade

Absolute velocity – tangent to

stator blade

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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.

Page 24: pumps turbines

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

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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

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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

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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

Page 28: pumps turbines

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

Page 29: pumps turbines

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

Page 30: pumps turbines

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.

Page 31: pumps turbines

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

,

,

Page 32: pumps turbines

Velocity TrianglesVelocity Triangles

U1 = r1

U2 = r2

U2 < U1

Relative velocity – tangent to turbine blade

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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

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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,

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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=

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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

Page 37: pumps turbines

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)

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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

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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

Page 40: pumps turbines

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)

Page 41: pumps turbines

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)

Page 42: pumps turbines

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

Page 43: pumps turbines

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.

Page 44: pumps turbines

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

Page 45: pumps turbines

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