presentation on concepts of Hydraulic_Machinery

hydraulic machinery

1. Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Hydraulic Machinery
Pumps, Turbines...


2. Hydraulic Machinery Overview
Types of Pumps
Dimensionless Parameters for Turbomachines
Power requirements
Head-discharge curves
Pump Issues
Cavitation
NPSH
Priming
Pump selection

3. 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)

4. Reciprocating action pumps
Piston pump
can produce very high pressures
hydraulic fluid pump
high pressure water washers
diaphragm pump

5. Peristaltic Pump
Fluid only contacts tubing
Tubing ___ and roller
_______ 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
ID
velocity

6. 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

7. Rotary Pumps
Disadvantages
precise machining
abrasives wear surfaces rapidly
pulsating output
Uses
vacuum pumps
air compressors
hydraulic fluid pumps
food handling

8. 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

9. 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?
0
0.5
1
1.5
2
2.5
0 0.5 1 1.5
revolutions
t
o
t
a
l

f
l
o
w
1st piston
2nd piston
3rd piston
3 pistons
Thirsty Refugees

10. 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

11. (inefficient process)
Turbomachines
impeller
casinghousing
()()
2 1
2 1z t tT Q rV rVré ù= -
ë û
Demour’s 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 _________

12. Pressure Developed by
Centrifugal Pumps
2
2
p
V
h
g
=velocity head
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
equal to the ________ ____ at the periphery of the
impeller
A given pump with a given impeller diameter and
speed will raise a fluid to a certain height regardless
of the fluid density

13. Radial Pumps
Impeller
Vanes
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
2
2
p
V
h
g
=

14. Axial Flow
propeller
also known as
__________ pumps
low head (less than 12
m)
high flows (above 20
L/s)

15. 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
homologous - streamlines are similar
constant ratio of dynamic pressures at corresponding points
also known as kinematic similitude
3
D
Q

same

16. Kinematic Similitude:
Constant Force Ratio

VD
gl
V
2
gl
V

lV
2
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

17. Turbomachinery Parameters
2
2
C
V
p
p



2
C
p
H
h g
V
= impeller
V D
2 2
C
p
H
impeller
h g
D

2 2 3
, , ,
p flow
H
impeller impeller flow flow
h g D Q
C f Re
D D D D

 
 
   
 
 
h e a d
s h
a p
e
d i s c h
a r g
e
r o
u g
h n e s s
impeller (Impeller is better defined)
3
, , , , , ,
flow
p
impeller flow flow
D Q
C f Re F W M
D D D


 
 
 
 
Where is the fluid?

18. 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

19. Impeller Geometry:
Shape Factor
0.18
0.37
1.25
2.33
3.67
axial: high _______, low _______
mixed
radial
mixed
()
3 4
p
Q
S
gh
w
=
pressureflow
pressureflow
S
500
1000
3400
6400
10000
N
()
3 4sp
p
N Q
N
h
=
*
N in rpm, Q in gpm, H in ft
*
Impeller
diameter
N
sp = 2732S
Radial: high _______, low ____

20. Use of Shape Factor:
Specific Speed
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
()
3 4
p
Q
S
gh
w
=
Why multistage?

21. Additional Dimensionless
Parameters
2 2
p
H
h g
C
Dw
=
53
D
P
C
P


3Q
Q
C
D

1 2
3 4
Q
H
C
S
C

impeller
power
Alternate equivalent way
to calculate S.
D is the _______ diameter
P is the _____
(defined at max efficiency)
w pP Qhg=

22. Head-Discharge Curve
circulatory flow
friction
Theoretical head-
discharge curve
Actual head-
discharge curve
s
h
o
c
k
shock
Q
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
V
2
V
2
p
h
2 2
p
H
h g
C
Dw
=
3Q
Q
C
D


23. Pump Power Requirements
w pP Qhg=
s
w
P
P
P
e
m
s
m
P
P
e
p
m
P m
Qh
P
e e
g
=
water
pump
shaft
motor
Water power
Subscripts
w = _______
p = _______
s = _______
m = _______

24. Impeller Shape vs. Power Curves
S
1 - O.33
2 - 0.81
3 - 1.5
4 - 2.1
5 - 3.4
Discharge (% of design)
P
o
w
e
r

(
%
o
f

d
e
s
i
g
n
)
axial
radial
Implicationshttp://www.mcnallyinstitute.com/

25. Affinity Laws
With speed, , held constant:
3
2
1
2
1











P
P
2
11
2
2
p
p
h
h
w
w
æ ö
=
ç ÷
è ø
2
1
2
1



Q
Q
With diameter, D, held constant:
5
1 1
2 2
P D
P D
æ ö
=
ç ÷
è ø
2
11
2
2
p
p
h D
h D
æ ö
=
ç ÷
è ø
3
1 1
2 2
Q D
Q D
æ ö
=
ç ÷
è ø
2 2
p
H
h g
C
Dw
=
3Q
Q
C
D

HQP 
53
D
P
C
P


homologous
Q
Cheld constant

26. 0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0 0.020.040.060.080.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D=0.366 m
Dimensionless Performance
Curves
2 2
p
H
h g
C
Dw
=
3
D
Q
C
Q


Efficiency
(defined at max efficiency)
43
21
H
Q
C
C
S
Efficiency
H
e
a
d
Curves for a particular pump
____________ of the fluid!Independent
 
 
0.5
0.75
0.087
4.57
0.026

shape

27. 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 performance
curves (previous slide).
What will the discharge be?
How large a motor will be needed if motor
efficiency is 95%?
Exercise
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0 0.020.040.060.080.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D=0.366 m

C
H
=
DHg
w
2
D
2
3
D
Q
C
Q
w
=
E
f
f
i
c
i
e
n
c
y
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0 0.020.040.060.080.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D=0.366 m

C
H
=
DHg
w
2
D
2
3
D
Q
C
Q
w
=
E
f
f
i
c
i
e
n
c
y

28. Pumps in Parallel or in Series
Parallel
Flow ________
Head ________
Series
Flow ________
Head ________
Multistage
adds
same
same
adds

29. Cavitation in Water Pumps
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
0
1000
2000
3000
4000
5000
6000
7000
8000
0 10 20 30 40
Temperature (C)
V
a
p
o
r

p
r
e
s
s
u
r
e

(
P
a
)

30. Net Positive Suction Head
NPSH
R - 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
NPSH
A
- pressure in excess of vapor pressure
available at pump inlet
determined by pump installation (elevation above
reservoir, frictional losses, water temperature)
If NPSH
A is less than NPSH
R cavitation will occur

31. Net Positive Suction Head
1
2
2
2
s s v
R
p V p
NPSH
gg g
= + -
s = suction
Total head -p
v!
NPSH
R
increases with Q
2
!
Elevation datum
Absolute pressure
2
2
eye eyev
R
p Vp
NPSH
gg g
= - + At cavitation!
z
How much total head in excess of vapor pressure is available?

32. NPSH
A
2
2
atm v s s v
L
p p p V p
z h
gg g g g
- D - - = + -
2
2
atm s s
reservoir L
p p V
z h
gg g
+ = + +
2
2
atm s s
L
p p V
z h
gg g
- D - = +
atm v
L A
p p
z h NPSH
g g
- D - - =
Subtract vapor pressure
2 2
1 1 2 2
1 2
2 2
L
p V p V
z z h
g gg g
+ + = + + +

33. NPSH
r
Illustrated
P
v
Pressure in excess of
vapor pressure required
to prevent cavitation
NPSH
r
NPSH
r can exceed atmospheric pressure!

34. NPSH problem
Determine the minimum
reservoir level relative to the
pump centerline that will be
acceptable. The NPSH
r 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

35. Pumps in Pipe Systems
60 m
1 km
Pipe diameter is 0.4 m and friction
factor is 0.015. What is the pump
discharge?

h
p
z
2
z
1
h
l

h
p
f(Q)
1 m1 m
2
ph a bQ= -often expressed as

p
1


V
1
2
2g
z
1h
p
p
2


V
2
2
2g
z
2h
l

36. Pumps in Pipe Systems
system curve
0
20
40
60
80
100
120
0 0.2 0.4 0.6 0.8
Discharge (m
3
/s)
H
e
a
d

(
m
)
system operating point
Static head
Head vs. discharge
curve for ________pump
What happens as the static head changes (a tank fills)?
p
h
2 2
p
H
h g
C
Dw
=
Could you solve this
with a dimensionless
performance curve?

37. Priming

C
H
p

2
D
2

pC
H
2
D
2
2 2
p
H
h g
C
Dw
=
density
1.225 kg/m
3
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!

38. 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 pump
priming tank

39. 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

  
                            
 
                                 

40. 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

41. RPM for Pumps
60 cycle
Other options
variable speed
belt drive
number of
poles sync full loadrad/sec
2 3600 3500 367
4 1800 1750 183
6 1200 1167 122
8 900 875 92
10 720 700 73
12 600 583 61
14 514 500 52
16 450 438 46
18 400 389 41
20 360 350 37
22 327 318 33
24 300 292 31
26 277 269 28
28 257 250 26
30 240 233 24

42. 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!
()
3 4
p
Q
S
gh
w
=
()
3 4
pgh
Q
w»
Pump for flume in DeFrees Teaching Lab…
Q = 0.1 m
3
/s, h
p
= 4 m.
Therefore  = 50 rads/s = 470 rpm
Actual maximum rpm is 600!

43. Pump Selection
Material Compatibility
Solids
Flow
Head
NPSH
a
Pump Selection software
A finite number of pumps will come close to
meeting the specifications!

44. Pump Selection Chart
Model X
Model M
http://www.pricepump.com/

45. 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.

46. Gould’s Pump Curves
BEP = 1836 L/s
()
3 4
p
Q
S
gh
w
=
Splitcase double suction
890 rpm = 93.2 rad/s
S=0.787
Check the Power!

47. 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?

48. 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

49. 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.

50. Water in Colomoncagua

51. Waiting for water

52. Water problem:
proposed solutions?
piston pump (80 L/min)
2 km pipeline (2”
galvanized and then 3”
PVC) with rise of 100 m

53. Shape Factor Solution
Create a dimensionless grouping

p


p
Q
23

p
Q
23

43
Eliminate ______
Eliminate _______
Eliminate ______

S
Q
p







34
()
3 4
p
Q
S
gh
w
=
mass
length
time
) (fS  ,,,pQ
2 2 2
2
3
M L
T L L
M T
L
 
 
  

 
   
 
 

2 2 /3
2/ 32 4/ 3
3
1L T
T T
L
 
 
  
 
   
 


54. Pump Curve Solution
2 2
p
H
h g
C
Dw
=

C
Q
Q
D
3
s
revs
rev
/8.62
2
60
min1
min
600






























 
  
037.0
366.0/8.62
/8.92
22
2

ms
smm
C
H
068.0
QC
3
DCQ
Q     smmsQ /21.0366.0/8.62068.0
33

  

3 3
9800 / 0.21 / 2
5.55
0.78 0.95
N m m s m
P kW 
p
m
P m
Qh
P
e e
g
=

55. Pump Curve Solution
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0 0.020.040.060.080.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
D=0.366 m
2 2
p
H
h g
C
Dw
=

C
Q
Q
D
3
Efficiency

56. NPSH solution
18°C
?
A RNPSH NPSH=
atm v
l R
p p
z h NPSH
g
-
D = - - Pap
v
2000
3
/9789mN
101300
atmp Pa=
  
mm
mN
PaPa
z 5.25.0
/9789
2000101300
3



mz14.7
atm v
A L
p p
NPSH z h
g
-
= - D -

57. 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?
What would you do if you were starting an
axial flow pump?
How could reducing the head on a radial
flow pump result in motor failure?
Close the effluent valve
Open the effluent valve
An effluent pipe break would increase the flow and
increase the power requirement

58. Find Q
()()
2 1
2 1z t tT Q rV rVré ù= -
ë û
2
2z tT QV rr=

p
1


V
1
2
2g
z
1
h
p

p
2


V
2
2
2g
z
2
h
l
2
2z tT Q V rw r w=
2
2tz
p
V rT
h
Q g
ww
g
= =
2
2
2 2
2
2
tV rV
z
g g
w
= +
work
Dimensional analysis
Neglect head loss
Datum is reservoir level
Let A = 10 cm
2
2
2tz
V rT
VA g
ww
g
=

59. How could we lift water more
efficiently?
2
2
2 2
2
tV rV
z
g g
w
= +D
22 2
2 2
t
Q A V r g z AVw= - D =
S
h a ft w
o rk a d d e d
K
in e tic e n e rg y
P
o te n tia l e n e rg y
Solve for Q=AV
Decrease V without decreasing Q! (
vv
tt

rr
cs
2
p z
hT
z Q z
w
g
=
D D
222 2
t
p z
A z V r g zz
h T
g w
w
D - DD
=
22z t
T Q V rw r w=

60. Lost energy
2
2
2 2
2
t
V rV
z
g g
w
= +D
2
2
2 2
2
tV rV
z
g g
w
= +D

61. 0.1
1
10
100
1000
0.0001 0.001 0.01 0.1 1 10
Flow (m
3
/s)
P
u
m
p
i
n
g

h
e
a
d

(
m
)
Axial
Mixed
Radial
Positive
displacement
20
40
60
10
200
400
600
100
2000
4000
6000
1000
1246
Selection of Pump Type
P
o
w
e
r

(
k
W
)
p
h