hydraulic machinery 1
TRANSCRIPT
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Monroe L. Weber-ShirkSchool of Civil and
Environmental Engineering
Hydraulic Machinery
Pumps, Turbines...
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Hydraulic Machinery Overview
Types of Pumps
Dimensionless Parameters for Turbomachines
Power requirements Head-discharge curves
Pump Issues
Cavitation
NPSH
Priming
Pump selection
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Types of Pumps
Positive displacement
piston pump
peristaltic pump
gear pump two-lobe rotary pump
screw pump
Jet pumps
Turbomachines axial-flow (propeller pump)
radial-flow (centrifugal pump)
mixed-flow (both axial and radial flow)
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Positive Displacement Pumps
Piston pump
Diaphragm pump
Peristaltic pump
Rotary pumps
gear pump
two-lobe rotary pump
screw pump
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Rotary Pumps
Gear Pump
fluid is trapped between gear teeth and the
housingTwo-lobe Rotary Pump
(gear pump with two teeth on each gear
same principle as gear pump fewer chambers - more extreme pulsation
trapped fluid
http://www.koboldusa.com/onlinecatalog/flow/zdm/zdm1.html -
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Rotary Pumps
Disadvantages
precise machining
abrasives wear surfaces rapidlypulsating output
Uses
vacuum pumps
air compressors
hydraulic fluid pumps
food handling
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Screw Pump
Can handle debris
Used to raise thelevel of wastewater
Abrasive materialwill damage theseal between screwand the housing
Grain augers usethe same principle
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Positive Displacement Pumps
What happens if you close a valve on the
effluent side of a positive displacement pump?
What does flow rate vs. time look like for apiston pump?
0
0.2
0.4
0.6
0.8
1
1.2
0 2 4 6 8 10
rads
totalflow
1st piston
2nd piston
3rd piston
3 pistons
Thirsty Refugees
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(inefficient process)
Turbomachines
impeller
casing housing
( ) ( )2 12 1z t tT Q r V rV r = -
Demours centrifugal pump - 1730
Theory conservation of angular momentum
conversion of kinetic energy to potential energy in flow
expansion ___________ ________
Pump components rotating element - ___________
encloses the rotating element and seals the pressurized
liquid inside - ________ or _________
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Radial Pumps
ImpellerVanes
Casing
Suction Eye Impeller
Discharge
centrifugal
Flow Expansion
diameter rotational speed
also called _________ pumps
broad range of applicable flows and heads
higher heads can be achieved by increasing the_______ or the ________ ______ of the impeller
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Dimensionless Parameters for
Turbomachines
We would like to be able to compare pumps withsimilar geometry. Dimensional analysis to therescue...
To use the laws of similitude to compareperformance of two pumps we need
exact geometric similitude
all linear dimensions must be scaled identically
roughness must scale homologous - streamlines are similar
constant ratio of dynamic pressures at correspondingpoints
also known as kinematic similitude
3D
Q
same
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Kinematic Similitude:
Constant Force Ratio
VD
gl
V2
gl
V
lV2
V
c
viscous
gravity
surface-tension
elastic
Reynolds
ratio of inertial to _______ forces
Froude ratio of inertial to ________ force
Weber
ratio of inertial to _______ ______ forces
Mach
ratio of inertial to _______ forces
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Turbomachinery Parameters
2
2C
V
pp
2
C
V
HgH
2
flowD
Q
A
QV
2
4
C
Q
HgDflowH
4
2 3, , ,
flow flow
H
impeller flow flow
HgD D QC f Re
Q D D D
e
w
D= =
impeller (Impeller is better defined)
3, , , , , ,
flow
p
impeller flow flow
D QC f Re F W M
D D D
e
w
=
Where is the fluid?
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Shape Factor
Related to the ratio of flow passage
diameter to impeller diameter
Defined for the point of best efficiency
What determines the ideal shape for a
pump?
)(fS ,,, pQ
Exercise
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Impeller Geometry:
Shape Factor
0.18
0.37
1.25
2.33
3.67axial: high _______, low _______
mixed
radial
mixed
S Q
gHp 3 4
pressure flow
pressureflow
S
500
1000
3400
6400
10000
N
43psp
H
QNN
*N in rpm, Q in gpm, H in ft
*
Impeller
diameter
Nsp= 2732S
Radial: high _______, low ____
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Use of Shape Factor:
Specific Speed
The maximum efficiencies for all pumps occurswhen the Shape Factor is close to 1!
Flow passage dimension is close to impeller diameter!
Low expansion losses!
There must be an optimal shape factor given adischarge and a head.
Double suction (like two pumps in parallel) Multistage (pumps in series)
Use Q and H for each stage
S Q
gHp 3 4
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Additional Dimensionless
Parameters
CH Hg
2D2
53
D
PCP
CQ Q
D3
S
CQ1 2
CH3 4
impeller
power
Alternate equivalent way
to calculate S.
D is the _______ diameter
P is the _____
(defined at max efficiency)
HQPw
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Head-Discharge Curve
Theoreticalhead-dischargecurve
Actual head-
discharge curve
Q
H circulatory flow -inability of finite
number of blades to
guide flow
friction - ____
shock - incorrect angle
of blade inlet ___
other losses
bearing friction
packing friction
disk friction
internal leakage
V2
V2
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Pump Power Requirements
HQPw
s
wP
P
Pe
m
sm
P
Pe
mP
m
ee
HQP
water
pump
shaftmotor
Water power
Subscripts
w = _______
p = _______
s = _______m = _______
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Impeller Shape vs. Power Curves
S
1 - O.33
2 - 0.81
3 - 1.5
4 - 2.1
5 - 3.4
Discharge (% of design)Powe
r(%ofde
sign)
axial
radial
Implicationshttp://www.mcnallyinstitute.com/
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Affinity Laws
With speed, , held constant:
3
2
1
2
1
P
P2
2
1
2
1
H
H
2
1
2
1
Q
Q
With diameter, D, held constant:
5
1 1
2 2
P D
P D
=
2
2
1
2
1
D
D
H
H3
1 1
2 2
Q D
Q D
=
CH Hg
2
D2
CQQ
D3
HQP
53
D
PCP
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Pump Example
Given a pump with shape factor of 4.57, a
diameter of 366 mm, a 2-m head, a speed of
600 rpm, and dimensionless performancecurves (previous slide).
What will the discharge be?
How large a motor will be needed if motorefficiency is 95%?
Exercise
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0 0.02 0.04 0.06 0.08 0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D=0.366 m
CH
Hg
2D
2
3D
QCQ
Efficiency
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0 0.02 0.04 0.06 0.08 0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D=0.366 m
CH
Hg
2D
2
3D
QCQ
Efficiency
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Pumps in Parallel or in Series
Parallel
Flow ________
Head ________
Series
Flow ________
Head ________
Multistage
adds
same
same
adds
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Cavitation in Water Pumps
water vapor bubbles
form when the pressure
is less than the vaporpressure of water
very high pressures
(800 MPa or 115,000
psi) develop when thevapor bubbles collapse 0
1000
2000
3000
4000
5000
6000
7000
8000
0 10 20 30 40
Temperature (C)
Vaporpressure(
Pa)
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Net Positive Suction Head
1
2
2
2s s v
R
p V pNPSH
gg g= + -
s = suction
Total head -pv!
NPSHRincreases with Q2!
Elevation datum
Absolute pressure
2
2
eye eyevR
p VpNPSH
gg g= - + At cavitation!
z
How much total head in excess of vapor pressure is available?
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NPSHR
2
2atm v s s v
L
p p p V pz h
gg g g g - D - - = + -
2
2atm s s
reservoir L
p p Vz hgg g+ = + +
2
2atm s s
L
p p Vz h
gg g- D - = +
atm v
L A
p pz h NPSH
g g- D - - =
Subtract vapor pressure
2 2
1 1 2 21 2
2 2 L
p V p Vz z h
g gg g+ + = + + +
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NPSH problem
Determine the minimum
reservoir level relative to the
pump centerline that will beacceptable. The NPSHrfor
the pump is 2.5 m. Assume
you have applied the energyequation and found a head
loss of 0.5 m.
18C
?
Exercise
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Pumps in Pipe Systems
0
20
40
60
80
100
120
0 0.2 0.4 0.6 0.8Discharge (m3/s)
Head(m)
system operating point
Static head
Head vs. dischargecurve for ________pump
What happens as the static head changes (a tank fills)?
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Priming
CH
p2D2
p CH2D2
CH
Hg
2D2
density
1.225 kg/m3
The pressure increase created is
proportional to the _______ of the fluid
being pumped. A pump designed for water will be
unable to produce much pressure
increase when pumping air
Density of air at sea level is __________
Change in pressure produced by pump is
about 0.1% of design when pumping air
rather than water!
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Priming Solutions
Applications with water at less than atmospheric
pressure on the suction side of the pump require a
method to remove the air from the pump and the
inlet piping
Solutions
foot valve
priming tank vacuum source
self priming
foot valve
to vacuum pumppriming tank
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Self-Priming Centrifugal Pumps
Require a small volume of liquid in the
pump
Recirculate this liquid and entrain airfromthe suction side of the pump
The entrained air is separated from the
liquid and discharged in the pressure side ofthe pump
http://www.gouldspumps.com/download_files/3796/3796_priming.stm -
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Variable Flows?
How can you obtain a wide range of flows?
__________________________
__________________________
__________________________
__________________________
__________________________
Why is the flow from two identical pumps
usually less than the 2x the flow from one
pump?
Valve
Multiple pumps (same size)Multiple pumps (different sizes)
Variable speed motor
Storage tank
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RPM for Pumps
60 cycle
variable speed
belt drive
number of
poles sync full load rad/sec
2 3600 3500 367
4 1800 1750 183
6 1200 1167 1228 900 875 92
10 720 700 73
12 600 583 61
14 514 500 52
16 450 438 46
18 400 389 4120 360 350 37
22 327 318 33
24 300 292 31
26 277 269 28
28 257 250 26
30 240 233 24
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Estimate of Pump rpm
The best efficiency is obtained when S=1
Given a desired flow and head the
approximate pump rpm can be estimated!
S Q
gHp 3 4
( )3 4
pgH
Qw
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Pump Selection
Material Compatibility
Solids
Flow Head
NPSHa
Pump Selection software A finite number of pumps will come close to
meeting the specifications!
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Pump Selection Chart
Model X
Model M
http://www.pricepump.com/
http://www.pricepump.com/http://www.pricepump.com/http://www.pricepump.com/http://www.pricepump.com/ -
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End of Curve Operation
Right of the BEP
is sufficient NPSH available for the pump to operateproperly?
fluid velocities through the suction and dischargenozzles of the pump could be extremely high, resultingin increased pump and system noise
Left of BEP operation
high thrust loads on the pump bearings and mechanicalface seals result in premature failure.
The pump is oversized, resulting in lower efficiencyand higher operating and capital costs.
Q
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Goulds Pump Curves
BEP = 1836 L/s
S Q
gHp 3 4
Splitcase double suction890 rpm = 93.2 rad/s
S=0.787
Check the Power!
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Pump Installation Design
Why not use one big pump?
Can the system handle a power failure?
Can the pump be shut down for
maintenance?
How is the pump primed?
Are there enough valves so the pump can be
removed for service without disabling the
system?
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Pump Summary
Positive displacement vs. turbomachines
Dimensional analysis
Useful for scaling
Useful for characterizing full range of pumpperformance from relatively few data points
Turbomachines convert shaft work into increasedpressure (or vice versa for turbines)
The operating point is determined by where thepump and system curves intersect
NPSH
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Water problem?
Early in my college days I took a break and spent 17 months in Salvadoranrefugee camps in Honduras. The refugee camps were located high in themountains and for several of the camps the only sources of water large enoughto sustain the population of 6-10,000 were located at much lower elevations. So
it was necessary to lift water to the camps using pumps.
When I arrived at the camps the pumps were failing frequently and the pipeswere bursting frequently. Piston pumps were used. The refugees werecomplaining because they needed water. The Honduran army battalion wasnervous because they didnt want any refugees leaving the camp. There was
only one set of spare parts (valve springs and valves) for the pump and the lastset of parts only lasted a few days. The pump repair crew didnt want to startusing the pump until the real cause of the problem was fixed because spareparts have to be flown in from Miami.
Water problem:
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Water problem:
proposed solutions?
piston pump (80 L/min)
2 km pipeline (2galvanized and then3 PVC) with rise of100 m
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Pump Curve Solution
22D
HgCH
CQ
Q
D3
srevs
rev/8.62
2
60
min1
min
600
037.0366.0/8.62
/8.9222
2 ms
smmCH
068.0QC
3DCQ Q smmsQ /21.0366.0/8.62068.033
( )( )( )
( )( )
3 39800 / 0.21 / 25.55
0.78 0.95
N m m s mP kW= =
mP
m
ee
HQP
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Pump Curve Solution
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0 0.02 0.04 0.06 0.08 0.10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D=0.366 m
CH Hg
2
D2
CQ
Q
D3
Efficiency
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NPSH solution
18C
?A RNPSH NPSH=
atm vl R
p pz h NPSH
g
-D = - - Papv 2000
3
/9789 mN
101300atmp Pa=
mmmN
PaPaz 5.25.0/9789
20001013003
mz 14.7
atm vA L
p pNPSH z h
g
-= - D -
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Find Q
( ) ( )2 1
2 1z t tT Q r V rV r = -
2 2z tT QV r r=
1
V1
2
2gz1 hp
p2
V2
2
2gz2 hl
2 2z tT Q V r w r w=
2 2tz
p
V rTh
Q g
ww
g= =
2
22 2
22
tV r V
zg g
w
= +
work
Dimensional analysis
Neglect head loss
Datum is reservoir level
Let A = 10 cm2
2 2tz
V rT
VA g
ww
g=
H ld lift t
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How could we lift water more
efficiently?
2
22 2
2
tV r Vz
g g
w= + D
2 2 22 2tQ A V r g z AV w= - D =
Solve for Q=AV
Decrease V without decreasing Q! (
vt
r
cs
p zh T
z Q z
w
g=
D D
2 22 2t
p z
A z V r g zz
h T
g w
w
D - DD=
2 2z tT Q V r w r w=
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Lost energy
2
22 2
2
tV r Vz
g g
w= + D
2
22 2
2
tV r Vz
g g
w= + D
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0.1
1
10
100
1000
0.0001 0.001 0.01 0.1 1 10
Flow (m3/s)
Pumpinghead(m)
Axial
Mixed
Radial
Positive
displacement
20
4060
10
200
400600
100
2000
40006000
1000
1 2 4 6
Selection of Pump Type
P
ower(kW)