Transcript Document 7322561

Hydraulic Machinery
Pumps, Turbines...
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Hydraulic Machinery Overview
 Types of Pumps
 Dimensionless Parameters for Turbomachines
 Power requirements
 Head-discharge curves
 Pump Issues
 Cavitation
 NPSH
 Priming
 Pump selection
Types of Pumps
Positive
displacement
piston pump
Diaphragm pump
peristaltic pump
Rotary pumps
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)
Reciprocating action pumps
 Piston pump
 can produce very high pressures
 hydraulic fluid pump
 high pressure water washers
diaphragm pump
Peristaltic Pump
 Fluid only contacts tubing
ID and roller
 Tubing ___
_______
velocity with respect to the
tubing determine flow rate
 Tubing eventually fails from
fatigue and abrasion
 Fluid may leak past roller at
high pressures
 Viscous fluids may be
pumped more slowly
Rotary Pumps
Gear Pump
fluid is trapped between gear teeth and the
housing
Two-lobe Rotary Pump
(gear pump with two “teeth” on each gear)
same principle as gear pump
fewer chambers - more extreme pulsation
trapped fluid
Rotary Pumps
 Disadvantages
 precise machining
 abrasives wear surfaces rapidly
 pulsating output
 Uses
 vacuum pumps
 air compressors
 hydraulic fluid pumps
 food handling
Screw Pump
 Can handle debris
 Used to raise the
level of wastewater
 Abrasive material
will damage the
seal between screw
and the housing
 Grain augers use
the same principle
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 a
piston pump?
2.5
total flow
2
1st piston
1.5
2nd piston
3rd piston
1
3 pistons
0.5
0
0
0.5
1
1.5
revolutions
Thirsty Refugees
Jet Pump
“eductor”
 A high pressure, high velocity jet discharge is
used to pump a larger volume of fluid.
 Advantages
 no moving parts
 self priming
 handles solids easily
 Disadvantage
 inefficient
 Uses
 deep well pumping
 pumping water mixed with solids
http://spaceflight.nasa.gov/shuttle/upgrades/ojp.html
Turbomachines
 Demour’s centrifugal pump - 1730
 Theory
 conservation of angular momentum
 conversion of kinetic energy to potential energy in flow
expansion ___________
(inefficient ________
process)
ù
Tz = r Q é
 Pump components
1 t1 û
ë r2Vt2 - rV
 rotating element - ___________
impeller
 encloses the rotating element and seals the pressurized
casing or _________
housing
liquid inside - ________
(
) (
)
Pressure Developed by
Centrifugal Pumps
 Centrifugal pumps accelerate a liquid
 The maximum velocity reached is the velocity of the
periphery of the impeller
 The kinetic energy is converted into potential energy
as the fluid leaves the pump
 The potential energy developed is approximately
V2
velocity head
equal to the ________
____ at the periphery of the hp =
2g
impeller
 A given pump with a given impeller diameter and
speed will raise a fluid to a certain height regardless
of the fluid density
Radial Pumps
 also called _________
centrifugal pumps
 broad range of applicable flows and heads
2
V
 higher heads can be achieved by increasing the
hp =
2g
_______
diameter or the ________
rotational ______
speed of the impeller
Flow Expansion
Discharge
Casing
Suction Eye
Impeller
Impeller
Vanes
Axial Flow
also known as
__________
pumps
propeller
low head (less than 12
m)
high flows (above 20
L/s)
Dimensionless Parameters for
Turbomachines
 We would like to be able to compare pumps
with similar geometry. Dimensional analysis to
the rescue...
 To use the laws of similitude to compare
performance of two pumps we need
 exact geometric similitude
 all linear dimensions must be scaled identically
 roughness must scale
same
 homologous - streamlines are similar
 constant ratio of dynamic pressures at
corresponding points
 also known as kinematic similitude
Q
D 3
Kinematic Similitude:
Constant Force Ratio
 Reynolds
 ratio of inertial to _______
viscous forces
 Froude
 ratio of inertial to ________
gravity force
 Weber
VD

V2
gl
2
V
l
 ratio of inertial to _______
______ forces
surface-tension
 Mach
 ratio of inertial to _______
elastic forces

V
c
V
gl
Turbomachinery Parameters

D flow

Q 
C p  f  Re, F ,W , M ,
,
,
 Where is the fluid?
3

Dimpeller D flow  D flow 

 2p
Cp 
V 2
hp g
2
 2 Dimpeller
hp g
CH = 2
V
V   Dimpeller
CH 
hp g
2
 2 Dimpeller

D flow

Q 
 CH  f  Re,
,
,

3

Dimpeller D flow  D flow 

impeller
(Impeller is better defined)
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?
S  f ( , Q, p,  )
Exercise
N sp =
N Q
*
34
(h )
p
Impeller Geometry: w Q
S=
Shape Factor
(gh )
34
p
Impeller
diameter
N
500
S
0.18
1000
0.37
radial
3400
1.25
mixed
6400
2.33
mixed
10000
*N
3.67
in rpm, Q in gpm, H in ft
pressure low ____
flow
Radial: high _______,
pressure
flow
axial: high _______,
low _______
Nsp = 2732S
Use of Shape Factor:
Specific Speed
S=
w Q
(gh )
 The maximum efficiencies for all pumps occurs
when 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 a
discharge and a head.
 Shape factor defined for specific cases
 Double suction
 Treat like two pumps in parallel
 Multistage (pumps in series)
 Use Q and H for each stage
Why multistage?
34
p
Additional Dimensionless
Parameters
hp g
CH = 2 2
w D
Q
 D3
CQ 
CP 
P
 3 D 5
S
CQ1 2
CH3 4
D is the _______
impeller diameter
Pw = g Qhp
P is the _____
power
Alternate equivalent way
to calculate S.
(defined at max efficiency)
Head-Discharge Curve
 circulatory flow inability of finite
number of blades to
guide flow
V2
 friction - ____
hp g
CH = 2 2
w D
hp
 shock - incorrect angle
V2
of blade inlet ___
 other losses




bearing friction
packing friction
disk friction
internal leakage
Theoretical headdischarge curve
Actual headdischarge curve
Q
Q
CQ 
 D3
Pump Power Requirements
Pw = g Qhp
eP 
Pw
Ps
em 
Ps
Pm
Pm =
g Qhp
eP em
Water power
Subscripts
w = _______
water
p = _______
pump
s = _______
shaft
m = motor
_______
Power (% of design)
Impeller Shape vs. Power Curves
radial
axial
S
1 - O.33
2 - 0.81
3 - 1.5
4 - 2.1
5 - 3.4
Discharge (% of design)
http://www.mcnallyinstitute.com/
Implications
Affinity Laws

With diameter, D, held constant:
hp g
CH = 2 2
w D
Q
CQ 
 D3
Q1
Q2

1
hp1
2
hp 2
homologous
CQ  held constant
P  QH
CP 
P
 3 D 5
 1 
  
 
P2
 2
2
æw1 ö
=ç ÷
èw2 ø
3
P1
With speed, , held constant:
3
Q1 æD1 ö
=ç ÷
Q2 èD2 ø
hp1
hp 2
2
æD1 ö
=ç ÷
èD2 ø
5
P1 æD1 ö
=ç ÷
P2 èD2 ø
Dimensionless Performance
Curves
0.08
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.07
hp g
CH = 2 2
w D
0.06
0.05
0.04
0.03
0.02
D=0.366 m
0.01
0
0
 0.087 
0.75
0.026


0.02
0.5
S
12
Q
C
C
34
H
 4.57
0.04
0.06
CQ 
(defined at max efficiency)
Q
D 3


0.08
Efficiency
0.1
shape
Curves for a particular pump
Independent of the fluid!
____________
Pump Example
0.08
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.07
CH 
Hg
 2D 2
0.06
0.05
0.04
0.03
0.02
D=0.366 m
0.01
0
0
0.02
0.04
0.06
CQ 
Q
0.08
Efficiency
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 performance
curves (previous slide).
What will the discharge be?
How large a motor will be needed if motor
efficiency is 95%?
0.1
D 3
Exercise
Pumps in Parallel or in Series
Parallel
adds
Flow ________
same
Head ________
Series
Flow ________
same
Head ________
adds
Multistage
Cavitation in Water Pumps
8000
7000
Vapor pressure (Pa)
 water vapor bubbles
form when the pressure
is less than the vapor
pressure of water
 very high pressures
(800 MPa or 115,000
psi) develop when the
vapor bubbles collapse
6000
5000
4000
3000
2000
1000
0
0
10
20
Temperature (C)
30
40
Net Positive Suction Head
 NPSHR - absolute pressure in excess of vapor
pressure required at pump inlet to prevent
cavitation
 given by pump manufacturer
 determined by the water velocity at the entrance to the
pump impeller
 NPSHA - pressure in excess of vapor pressure
available at pump inlet
 determined by pump installation (elevation above
reservoir, frictional losses, water temperature)
 If NPSHA is less than NPSHR cavitation will occur
Net Positive Suction Head
Elevation datum
2
Absolute pressure
z
1
ps Vs2 pv
NPSH R = +
g 2g g
s = suction
Total head -pv!
2
pv Veye
At cavitation!
NPSH R =
+
g
g 2g
peye
NPSHR increases with Q2!
How much total head in excess of vapor pressure is available?
NPSHA
p1 V12
p2 V22
+
+ z1 = +
+ z2 + hL
g 2g
g 2g
patm
ps Vs2
+ zreservoir = +
+ hL
g
g 2g
patm
ps Vs2
- Dz - hL = +
g
g 2g
patm
pv ps Vs2 pv
- Dz - hL = +
g
g
g 2g g
patm
p
- Dz - hL - v = NPSH A
g
g
Subtract vapor pressure
NPSHr Illustrated
Pv
NPSHr
Pressure in excess of
vapor pressure required
to prevent cavitation
NPSHr can exceed atmospheric pressure!
NPSH problem
Determine the minimum
reservoir level relative to the
pump centerline that will be
acceptable. The NPSHr for
the pump is 2.5 m. Assume
you have applied the energy
equation and found a head
loss of 0.5 m.
?
18°C
Exercise
Pumps in Pipe Systems
Pipe diameter is 0.4 m
and friction factor is
0.015. What is the pump
discharge?
60 m
1 km
V12
p2 V22

 z1  hp  
 z 2  hl 11m
m
 2g
 2g
p1
hp  z 2  z 1  hl
hp  f(Q)
often expressed as
hp = a - bQ 2
Pumps in Pipe Systems
system operating point
120
hp
Head (m)
100
Head vs. discharge
pump
curve for ________
80
60
40
Could you solve this
with a dimensionless
performance curve?
Static head
20
0
0
0.2
0.4
0.6
3
Discharge (m /s)
0.8
hp g
CH = 2 2
w D
What happens as the static head changes (a tank fills)?
Priming
hp g
CH = 2 2
w D
 The pressure increase created is
proportional to the _______
density of the fluid
p
CH 
being pumped.
 2 D 2
 A pump designed for water will be
2 2
p

C

D
unable to produce much pressure
H
increase when pumping air
1.225 kg/m3
 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!
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
priming tank
to vacuum pump
 Solutions
 foot valve
 priming tank
 vacuum source
 self priming
foot valve
Self-Priming Centrifugal Pumps
Require a small volume of liquid in the
pump
Recirculate this liquid and entrain air from
the suction side of the pump
The entrained air is separated from the
liquid and discharged in the pressure side of
the pump
Variable Flows?
How can you obtain a wide range of flows?
Valve
__________________________
Multiple pumps (same size)
__________________________
Multiple pumps (different sizes)
__________________________
Variable speed motor
__________________________
Storage tank
__________________________
Why is the flow from two identical pumps
usually less than the 2x the flow from one
pump?
RPM for Pumps
60 cycle
Other options
variable speed
belt drive
number of
poles
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
sync
full load
3600
3500
1800
1750
1200
1167
900
875
720
700
600
583
514
500
450
438
400
389
360
350
327
318
300
292
277
269
257
250
240
233
rad/sec
367
183
122
92
73
61
52
46
41
37
33
31
28
26
24
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=
w Q
34
(gh )
p
34
gh )
(
w»
p
Q
Pump for flume in DeFrees Teaching Lab…
Q = 0.1 m3/s, hp = 4 m.
Therefore  = 50 rads/s = 470 rpm
Actual maximum rpm is 600!
Pump Selection
 Material Compatibility
 Solids
 Flow
 Head
 NPSHa
 Pump Selection software
 A finite number of pumps will come close to
meeting the specifications!
Pump Selection Chart
http://www.pricepump.com/
Model M
Model X
End of Curve Operation
 Right of the BEP (Best Efficiency Point)
 is sufficient NPSH available for the pump to operate
properly?
 fluid velocities through the suction and discharge
nozzles of the pump could be extremely high, resulting
in increased pump and system noise (and wear)
 Left of BEP operation
 high thrust loads on the pump bearings and mechanical
face seals result in premature failure.
 The pump is oversized, resulting in lower efficiency
and higher operating and capital costs.
Gould’s Pump Curves
S=
w Q
34
(gh )
p
890 rpm = 93.2 rad/s
Splitcase double suction
BEP = 1836 L/s
S=0.787
Check the Power!
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?
Pump Summary
 Positive displacement vs. turbomachines
 Dimensional analysis
 Useful for scaling
 Useful for characterizing full range of pump
performance from relatively few data points
 Turbomachines convert shaft work into increased
pressure (or vice versa for turbines)
 The operating point is determined by where the
pump and system curves intersect
 NPSH
Water problem?
Early in my college days I took a break and spent 17 months in Salvadoran refugee
camps in Honduras. The refugee camps were located high in the mountains and for
several of the camps the only sources of water large enough to 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 pipes were
bursting frequently. Piston pumps were used. The refugees were complaining because
they needed water. The Honduran army battalion was nervous because they didn’t want
any refugees leaving the camp. There was only one set of spare parts (valve springs and
valves) for the pump and the last set of parts only lasted a few days. The pump repair
crew didn’t want to start using the pump until the real cause of the problem was fixed
because spare parts have to be flown in from Miami.
Water in Colomoncagua
Waiting for water
Water problem:
proposed solutions?
2 km pipeline (2”
galvanized and then 3”
PVC) with rise of 100 m
piston pump (80 L/min)
Shape Factor Solution
Create a dimensionless grouping
S  f ( , Q, p,  )
p

p
Q 2 3
p
Q 2 3 4 3
mass
Eliminate ______
length
Eliminate _______
time
Eliminate ______
M L
 T 2 L2   2 

 L
T2 
M 
 
 L3 
 2
2/3 
L
T

 1 
 T 2  L3 2 / 3   T 4 / 3 


S
S=
 Q
34
p 
 
  
w Q
34
(gh )
p
Pump Curve Solution
 600rev   1 min   2 

  62.8 / s

  
 
 

 min   60s   rev 

2m 9.8m / s 2 
CH 
 0.037
62.8 / s 2 0.366m 2
hp g
CH = 2 2
w D
Q
CQ 
D 3
Q  CQD
Pm =
CQ  0.068
3
g Qhp
eP em
Q  0.06862.8 / s 0.366m   0.21m3 / s
3
9800 N / m  0.21m / s   2m 

P
 5.55kW
3
3
 0.78 0.95
Pump Curve Solution
0.08
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.07
hp g
CH = 2 2
w D
0.06
0.05
0.04
0.03
0.02
D=0.366 m
0.01
0
0
0.02
0.04
0.06
Q
CQ 
D 3
0.08
0.1
Efficiency
NPSH solution
NPSH A = NPSH R
?
patm - pv
NPSH A =
- Dz - hL
g
Dz =
z 
18°C
patm - pv
- hl - NPSH R
g
101300 Pa   2000 Pa 
9789 N / m3
z  7.14m
 0.5m  2.5m
pv  2000 Pa
patm = 101300 Pa
  9789 N / m3
Implications of Power Curves
You are going to start a radial flow pump
powered by an electric motor. You want to
reduce the starting load on the motor. What
can you do? Close the effluent valve
What would you do if you were starting an
axial flow pump? Open the effluent valve
How could reducing the head on a radial
flow pump result in motor failure?
An effluent pipe break would increase the flow and
increase the power requirement
Find Q
(
) (
)
ù
Tz = r Q é
r
V
rV
2
t
1
t
2
1
ë
û
Tz = r QVt2 r2
Tzw = r QwVt2 r2
Tzw wVt2 r2
hp =
=
gQ
g
work
wVt2 r2
Tzw
=
gVA
g
V12
p2 V22

 z1  hp  
 z 2  hl
 2g
 2g
p1
wVt2 r2
g
V22
=
+ z2
2g
Let A = 10 cm2
Dimensional analysis
Datum is reservoir level
Neglect head loss
How could we lift water more
efficiently?
vt
wVt2 r2
g
V22
=
+ Dz Solve for Q=AV
2g

r
Tzw = r QwVt2 r2
Q = A 2wVt2 r2 - 2 g Dz = AV2
Decrease V without decreasing Q! (
hp
Tzw
=
Dz g Q Dz
Dz g ADz 2wVt2 r2 - 2 g Dz
=
hp
Tzw
cs2
Lost energy
wVt2 r2
g
wVt2 r2
g
V22
=
+ Dz
2g
V22
=
+ Dz
2g
Selection of Pump Type
Positive
displacement
Radial
hp
Pumping head (m)
100
Mixed
10
Axial
1
1
0.1
0.0001
0.001
0.01
0.1
Flow (m3/s)
2 4 6
1
6000
4000
2000
1000
600
400
200
100
60
40
20
10
10
Power (kW)
1000