Chapt. 14 (Solved Problems) - philadelphia.edu.jo
Transcript of Chapt. 14 (Solved Problems) - philadelphia.edu.jo
Dr. Munzer Ebaid Dr. Munzer Ebaid ١١
CHAPTER (14)
TURBOMACHINARY
SOLVED PROBLEMS
DR. MUNZER EBAID
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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=
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Problem (14.15)
)(max,,: ηinDrpmnGiven 201100 ==
?=Q&
?=H ?)( =PPower
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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: =∆=
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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)
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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