Dr. Munzer Ebaid Dr. Munzer Ebaid ١١

CHAPTER (14)

TURBOMACHINARY

SOLVED PROBLEMS

DR. MUNZER EBAID

Dr. Munzer Ebaid Dr. Munzer Ebaid ٢٢

Thrust Coefficient of a propeller:

Power Coefficient of a propeller:

Efficiency of a propeller:

Head Coefficient of an axial Pump:

gnDHCC TH 22

4 ∆==

π

Discharge Coefficient of an axial Pump:

Power Coefficient of an axial Pump :

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)(),(),( feetNPSHgpmQrpmNWhere

Critical value for Cavitation to occur )8500( ≤SSN

Specific speed:

Theoretical Adiabatic Power

with no cooling

Efficiency of a compressor with no water cooling=SHAFTactual

theoComp P

P)(

=η

Specific speed for Turbines

Wind turbine max. theoretical power produced by a

( ) 43

21

NPSHnQNSS =

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Theoretical Isothermal Power

with cooling

Specific speed for Turbines

Then: Max. Power of the Turbine2

2JVQP ρ=

Power for wind Turbines

Power for Reaction Turbines

Torque for Reaction Turbines

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PROBLEMS ON

PROPELLERS

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Fig. (14.2)

Coefficient of ThrustCoefficient of Power

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Problem (14.1)

305114003 mkgrpmnmD .,, === ρ

00 =V 00 =nDV

From rest

From rest

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Problem (14.8)

max? ηω At=

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Problem (14.8)

hkmVmD

402

0 ==

max? ηω At=

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PROBLEMS ON

AXIAL PUMPS

Dr. Munzer Ebaid Dr. Munzer Ebaid ١١١١Dimensionless performance curves for a typical axial-flow pump

Fig. (14.6)

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Performance curves for a typical axial-flow pump

Fig. (14.7)

cmD 6.35=

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Problem (14.11)

?=Q&

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Problem (14.11)

mHcmDrpmnGiven 3401000 =∆== ,,:

?=Q&

gnDHCC TH 22

4 ∆==

π

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Problem (14.15)

?=Q&

?=H?)( =PPower

cmD 6.35=

Dr. Munzer Ebaid Dr. Munzer Ebaid ١٦١٦

Problem (14.15)

)(max,,: ηinDrpmnGiven 201100 ==

?=Q&

?=H ?)( =PPower

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Dr. Munzer Ebaid Dr. Munzer Ebaid ١٨١٨

Problem (14.20)

?)( =PPower

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Problem (14.20)

rpmn 1800=

?)( =PPower

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PROBLEMS ON

CENTRIFUGAL PUMPS

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Performance curves for a typical centrifugal pump; D = 37.1 cm

Fig. (14.9)

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Dimensionless performance curves for a typical centrifugal pump from data given in Fig. 14.9

Fig. (14.10)

Dr. Munzer Ebaid Dr. Munzer Ebaid ٢٣٢٣Fig. (14.10) Fig. (14.9)

Problem (14.22)

?=Q&fthrpmnGiven 150,1600: =∆=

Dr. Munzer Ebaid Dr. Munzer Ebaid ٢٤٢٤

Problem (14.22)

?=Q&

fthrpmnGiven 150,1600: =∆=

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Problem (14.24)

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Problem (14.24)

rpsnGiven 30: = Pump Characteristics (Fig.14.9)

104,6.35)( == HeadoffshuttherpsnFor

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PROBLEMS ON

SPECIFIC SPEED

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Used in USA

Fig. (14.14)

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Problem (14.30)

)( 43HgQn

nSpeedSpecific S ∆×==

&

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Problem (14.34)

Find type of pump?

smQrpmNGiven 301600 .,: ==

Loss

partMech

T

partMech

P hg

Vgz

gp

hg

Vz

gp

h +

+++=

+++

..22

22

222

21

111 α

ρα

ρ

Apply Energy Equation between (1) & (2)

1

2

)( 43HgQn

nSpeedSpecific S ∆×==

&

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smQrpmNGiven 301600 .,: ==

42DQV

π

&=

01.0=f

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PROBLEMS ON

NPSH

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Problem (14.33)

?max =N

tempVapourpptempliquidgivenpressurevapourpressureSuctionNPSH

@2

.@−=

−=

( ) 4343

21

NPSHgNQNSS =

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T=60 F

5910462

1442560 ..

. =×=pressurevapourh

Axial Flow Pump

gpmQ 5000=&

Problem (14.33)

?max =N

tempVapourpptempliquidgivenpressurevapourpressureSuctionNPSH

@2

.@−=

−=

atmossuction Pftp += 5

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Problem (14.38)

SHAFTactual

theoComp P

P)(

=η

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Problem (14.38)

SHAFTactual

theoComp P

P)(

=η

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Problem (14.38)

Dr. Munzer Ebaid Dr. Munzer Ebaid ٣٨٣٨

Problem (14.38)

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PROBLEMS ON

WIND TURBINES

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Problem (14.48)

U: Wind speed

A: Area captured by the wind turbine

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Problem (14.48)

?=CaptureA

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END OF SOLVED PROBLEMS