Lecture 9
HYDRAULIC PUMPS [CONTINUED]
1.11 Pump Performance Curve
Pump performance characteristics are first analyzed independently of the rest of hydraulic
system and then as a part of the system. Both sets of data are valuable to the designer.
Analyzing the pump by itself gives an indication of its capabilities and performance based on
the speed of rotation, internal geometry, cost factors, etc., whereas analyzing pump
performance in system essentially determines pump system compatibility. In the first case,
the system designer may observe performance curves to see if a specific pump has the
pressure and volume flow rate to operate a given set of actuators. In a second instance, the
system designer may be computing the noise, vibration, cavitation and flow characteristics of
a specific pump before or after installation to determine if the pump and existing system are
compatible. Where the two are necessarily complimentary, in practice much of hands-on
work is completed independently. Pump performance characteristics are interpreted from data
in tabular form and then graphed.
Figure 1.20 shows a graphical representation of a typical positive displacement pump. Figure
1.20(a) represents the relationship between input power and pump output flow of a variable
displacement piston pump as a function of pump speed.Observe the linear relationship
between the discharge flow and pump speed. Figure 1.20(b) gives curves of overall and
volumetric efficiencies as a function of speed. Performance curves of radial piston pump are
given in Fig.1.20(c). Discharge flow of these pumps is nearly constant over a broad pressure
range.Discharge flow can be varied infinitely between the point of inflection on the constant
discharge portion of the curve and zero flow.
Flow
Power
(a)
(b)
(c)
Figure 1.20Pump performance curves
1.12Pump Noise
Pump noise is an important parameter used to determine the performance. Any increase in
noise indicates increased wear and eventually pump failure. Pumps are good generators but
poor radiators of noise. Noise is not just the sound coming directly from the pump, but also
from the vibration and fluid pulsation produced by the pump. Pumps are small in size and
hence, they are poor radiators of noise. Reservoirs, electric motors and piping being largerin
size are better radiators. Hence,a pump-induced vibration can cause audible noise greater than
that coming from the pump. Fixed displacement pumps are less noisy than variable
displacement pumps because of their rigid construction.
Figure 1.21 Pumpnoise characteristics
As can be seen from Fig.1.21, the pump speed has a strong effect on noise compared to
displacement and pressure. To reduce the noise levels, electric motors are used and the most
advantageous combination of size and pressure is selected to produce the needed power.
1.13Pump Cavitation
During the working of a positive displacement pump, vacuum is created at the inlet of the
pump. This allows atmospheric pressure to push the fluid in. In some situations, the vacuum
may become excessive, and a phenomenon known as cavitation occurs. When the pressure of
the liquid reaches a low enough level, it vaporizes or boils. Cavitation is the formation of oil
vapor bubbles due to a very low pressure (high vacuum) on the inside of the pump. The low
pressure also causes air, which is dissolved in the oil to come out of the solution and form
bubbles. These air and oil vapor bubbles collapse when they reach the outlet side of the
pump, which is under a high pressure. The collapsing of these vapor bubbles causes
extremely high localized pressure and fluid velocity. These pressures are so high that they
cause pitting of metal and consequently decrease the life and efficiency of the pump.
1.13.1 Factors Causing Cavitation
Cavitation is caused by the following factors:
1.
2.
3.
4.
Undersized plumbing.
Clogged lines or suction filters.
High fluid viscosity.
Too much elevation head between the reservoir and the pump inlet.
1.13.2 Rules to Eliminate (Control) Cavitation
Following are the rules to control cavitation:
1.
2.
3.
4.
5.
Keep suction line velocities below 1.2 m/s.
Keep the pump inlet lines as short as possible.
Minimize the number of fittings in the inlet line.
Mount the pump as close as possible to the reservoir.
Use low-pressure drop inlet filters.
6. Use proper oil as recommended by the pump manufacturer.
Example 1.14
Calculate the pipe bores required for the suction and pressure lines of a pump delivering 40
L/min using a maximum flow velocity in the suction line of 1.2 m/s and a maximum flow
velocity in the pressure line of 3.5 m/s.
Solution:Consider the suction line
Flow = Average velocity × Flow area
Flow through pipe
Area of pipe =
Velocity of flow
Now
Flow = 40 LPM = 40/60 LPS = 40/60 × 103 m3/s
40  103
Area of pipe =
= 0.555 × 103 m2
60  1.2
Let the bore of the pipe be of diameter D. Then
πD 2
3
2
Area of pipe  0.555 10 m 
4
 D = 0.0266 m
Minimum bore suction pipe = 26.6 mm.
Note: in all calculations great care must be taken to ensure that units are correct.
Alternatively, if a flow velocity of 1m/s is used then suction pipe bore can be of diameter 29
mm. The required diameter of the pressure line can be calculated in a similar manner taking
the flow velocity as 3.5 m/s. Here the minimum bore of pressure pipe is equal to 15.6 mm.
It is unlikely that a pipe having the exact bore is available, in which case select a standard
pipe having a larger bore. Alternatively, a smaller bore pipe may be chosen but it will be
necessary to recheck the calculation to ensure that the flow velocity falls within the
recommended range. That is, a standard pipe with an outside diameter of 20 mm and a wall
thickness of 2.5 mm is available. This gives an internal diameter of 15 mm.
Flow velocity =
Flow through pipe
Area of pipe bore
Now
Area of pipe bore =
π
152 mm2  177 mm2  177  106 m2
4
So
60  103
 3.77 m / s
60  177  106
This is satisfactory. It is also important to ensure that the wall thickness of pipe is sufficient
to withstand the working pressure of the fluid.
Flow velocity =
1.14Pump Selection
The main parameters affecting the selection of a particular type of pump are as follows:
1.
2.
3.
4.
5.
6.
Maximum operating pressure.
Maximum delivery.
Type of control.
Pump drive speed.
Type of fluid.
Pump contamination tolerance.
7. Pump noise.
8. Size and weight of a pump.
9. Pump efficiency.
10. Cost.
11. Availability and interchangeability.
12. Maintenance and spares.
1.14.1 Maximum Operating Pressure
This is determined by the power requirement of the circuit, the particular
application,availability of components, type of fluid and to some extent the environment and
level of labor both using and maintaining the equipment.
In general, the higher the operating pressure, the higher the component cost and the lower the
choice of components. The main advantage of higher working pressures is the reduction in
fluid flow rates for a given system power, resulting in smaller pumps, smaller bore pipes and
smaller components. The disadvantage is that at higher working pressures, the
compressibility of the fluid used can have considerable adverse effects where precision
control is required over a wide range of loads.
The general tendency is toward increased operating pressures. Typical maximum pressures
for fewapplications are given in Table 1.3. The operating pressures of pumps depend to some
extent on the fluid used. A fire-resistant fluid is generally not as good lubricant as a mineral
oil. So to give a reasonable pump life expectancy when using a fire-resistant fluid, the
maximum operating pressure must be reduced and it is advisable to consult the pump
manufacturer.
The maximum operating pressure and range of flow rates for different types of currently
available hydraulic pumps are shown in Table 1.4.The figures given cover a range of sizes
and makes; maximum valuesof delivery and pressure are not applicable to one pump.
Table 1.3System maximum pressure in relation to application
Application
Pressure (bar)
Machine tool
Mechanical
handling
Mobile
Press work
200
250
300
800
Table 1.4Operating pressure and size ranges for hydraulic pump types
Pump Type
Maximum
Pressure
Maximum
Delivery
(L/min)
From
To
0.25
760
0.6
740
External gear
Internal gear
From
40
100
To
300
210
Vane
50
140
6
Balanced vane
Axial piston
(swash plate)
Axial piston
(bent-axis)
Radial piston
140
175
200
Speed(RPM)
Min.
Filtratio
n(µm)
Pulsation
Noise
Level(
dB)
100
100
High
Low
90
85
From
500
3000
To
3000
4000
360
500
3000
50
Low
80
2
620
500
300
50
Low
85
350
1
1450
200
2000
25
High
90
250
350
17
3500
200
2000
25
High
90
350
1720
0.3
1000
200
2000
50
High
90
η (%)
70–90
75–90
65–80
70–90
80–90
50–90
80–90
1.14.2 Maximum Delivery
The pump system selected must be capable of delivering the maximum flow rate demanded
by the circuit. If the circuit demand is reasonably constant, a fixed displacement pump is
chosen. When the demand is at a series of fixed levels, a multi-pump system is used. For
demands which vary within a relatively narrow band, a variable displacement pump is used.
If there is a wide variance in system demand, an accumulator circuit may best satisfy the
requirements.
Pump capacities are stated by manufactures for a particular viscosity fluid at given operating
temperatures and pressures. Any increase in temperature and hence a reduction in viscosity or
an increase in operating pressure causes more leakage across the pump and consequently
reduces the pump delivery. As the pump wears the leakage will increase.It is usual to select a
pump with a capacity about 10% higher than that required to make an allowance for the
reduction in volumetric efficiency with wear. Pumps are available with flows from a fraction
of 1 LPM to–1000 LPM and above.
1.14.3 Type of Control
Various types of pump controls are available such as manual servo control, pressure
compensated control, constant power control and constant flow control. The choice of control
is dependent upon the circuit requirement such as complexity, accuracy of control, cost, type
of machining operation,etc.The designer has to choose carefully the type of control after a
detailed study of system characteristics.
1.14.4Pump Drive Speed
Amajority of pumps are driven directly from the prime mover – electric motor or internal
combustion engine–so the proposed drive speed is known. The fluid delivery rate is
proportional to the speed of rotation. Each design has a minimum and maximum operating
speed: the faster the pump runs, the shorter its life.
1.14.5Types of Fluid
Pumps are designed to operate within a particular range of fluid viscosity. Mineral oils of the
correct viscosity work satisfactory with most pumps provided the oil is clean. Operating with
synthetic or water-based fluids reduces the working life of a pump that relies on the hydraulic
fluid to lubricate the bearings and moving parts. When any fluid other than a mineral oil is to
be used, it is advisable to seek the pump manufacturer’s advice.
1.14.6Fluid Contamination
Any fluid contamination causes pump damage. Precision pumps with very fine clearances are
more susceptible to damage. If a contaminated fluid has to be pumped, such as in a cleanup
loop, particular attention must be paid to pump selection. Non-precision gear pumps, lobe
pumps and gerotor pumps are the most dirt tolerant.Whichever type is used, a strainer must
be fitted in the suction line. In the case of precision pumps, the manufacturer’s
recommendation on filtration must be followed; otherwise the life of pump will be drastically
reduced and the maker’s warranty voided.
1.14.7Pump Noise
Noise has become increasingly important environmentally. Operating levels vary
considerably between the pumps of the same type but of different makes.The manufacturers
are working on those aspects which most affect the emission of noise– port plate design,
bearings, flow passages, pressure controls, materials and methods of mounting. Generally, the
sound generated increases with speed and pressure. Certain kinds do, however, propagate
lower noise levels, in particular, those with internal gears. A multi-stage internal gear pump is
marketed by one manufacturer under the name Q pump, with Q signifying quiet.
Example 1.15
The intensity (in units of W/m2) of the noise of a pump increases by a factor of 10 due to
cavitation. What is the corresponding increase in noise level in decibels?
Solution:
dB increase  10  log
I (final)
 10  log10  10 dB
I (initial)
1.14.8Size and Weight of a Pump
Generally, not only the overall size and weight of a hydraulic system is important in mobile
installations, but also the whole system is important, as the size and weight of a pump is only
part of the whole system. In a mobile hydraulic field, the trend is to reduce the weight of the
hydraulic system by increasing the operating pressure,reducing the size of the reservoir and
using efficient oil coolers.
The best power-to-weight ratios can usually be achieved in the 200–300 bar operating
pressure range. The actual sizeand weight of a pump depend upon the particular
manufacture’s design. Very light compact units have been developed for use in the aerospace
industry but these tend to be extremely expensive.
1.14.9Efficiency
Reciprocating pumps tend to have higher efficiencies than rotary pumps. The actual
efficiency depends on design, operating pressure, speed and fluid viscosity.Table1.5gives an
indication of the range of efficiencies of various types of pumps.
Table 1.5Efficiency ranges of pumps
Pump Type
Volumetric Efficiency
Piston
Plunger in line
 99
Radial
 95
Axial
 95
Precision gear pumps
 95
Vane pump
 90
Overall Efficiency
 95
 90
 90
 90
 80
1.14.10Cost
The initial cost of a pump is usually of secondary importance to running and maintenance
costs.Gear pumps are cheaper, vane and piston pumps are expensive.
1.14.11Availability and Interchangeability
A number of gear pump manufacturers produce units to CETOP and SAE standards so far as
the external dimensions are concerned. This gives direct interchangeability between gear
pumps of different manufacturers. The shafts, mounting flanges and port connections of most
of the other types also comply with various international standards allowing a degree of
interchangeability.
1.14.12Maintenance and Spares
In every type of pump, the components involved in pumping worn out after a time and need
replacing. In gear pumps, it is usual to replace the entire pump. With some types of vane
pumps, the wear parts are grouped together as a cartridge that can easily be replaced without
dismantling the pump drive. In the case of piston pumps, it may be advisable to ensure that
the manufacturer offers a fast overall service for critical applications to carry a spare pump in
stock.
Example 1.16
A pump has a displacement volume of 120 cm3. It delivers 0.0015 m3/s at 1440 RPM and 60
bar. If the prime mover input torque is 130 Nm. What is the overall efficiency of the pump?
What is the theoretical torque required to operate the pump. The pump is driven by an electric
motor having an overall efficiency of 88%. The hydraulic system operates 12 h/d for 250
days per year. The cost of electricity is Rs 8 per kWh. Determine the yearly cost of electricity
to operate the hydraulic system. The amount of the yearly cost of electricity that is due to the
inefficiencies of the electric motor and pump.
Solution: Given volumetric displacement, VD = 120 cm3 , QA  0.0015 m3 /s, N  1440 rpm ,
P  60 bar , input torque TA  130 N m .
Total number of working hours available  250  12  3000 h
Volumetric displacement in m3/rev is
3
120 cm3  1 m 
3
VD =

  0.000120 m /rev
rev
100
cm


Theoretical dischargecan be calculated as
QT  VD N  0.000120
1440
rev/s  0.00288 m3 /s
60
Now we can calculate the volumetric efficiency as
Q
0.0015
v  A 
 52.08%
QT 0.00288
Mechanical efficiency is given by
pQT 60 105  0.00288 17280
m 


 88.2%
2
TA
19603
130 1440 
60
Note the product TA gives power in units of Nm/s (W) where TA has a unit of Nm and shaft
speed has units of rad/s.
The overall efficiency is
ηo  m v  0.882  0.5208  0.459  45.9%
Alternativelyoverall efficiency can also be calculated as
pQA 60 105  0.0015 9000
ηo 


 45.9%
TA 130 1440  2  19603
60
Now since the mechanical efficiency is known, we can calculate the theoretical torque as
TT  TA m  130  .882  114.7 N m
Thus, due to mechanical losses within the pump, 130 Nm of torque are required to drive the
pump instead of 114.7 Nm.
First we calculate the mechanical input power the electric motor delivers to the pump.
Pump input power (kW) isgiven by
TA  130 1440 
2
 19603 W  19.603 kW
60
Next we calculate the electrical input power.
Electric motor input power is given by
Electric motor output power
19.603

 22.28 kW
Electric motor overall efficiency
0.88
So
Yearly cost = Power rate  Time per year  Unit cost of electricity
= 22.28  12 h/d  250 d/y  8 Rs/kWh
= Rs 534627
The total loss equals kW loss due to electric motor plus the kW loss due to pump. Thus, we
have
Total loss  (1  0.88)  22.28  (1  0.459) 19.603
 2.674  10.61  13.284 kW
Yearly cost due to inefficiencies is
13.284
 534627 Rs/year  318760 Rs / year
22.28
Since
13.284
 59.2
22.28
we conclude that 59.2% of the total cost of electricity is due to inefficiency of the electric
motor and pump. This also means that only 49.8% of the electrical power entering the electric
motor is transferred into hydraulic power at the pump outlet port.
Example 1.17
For the fluid power system of Fig. 1.22, the following data are given:
Cylinder piston diameter
Cylinder rod diameter
Extending speed of cylinder
External load on a cylinder
Pump volumetric efficiency
Pump mechanical efficiency
Pump speed
Pump inlet pressure
0.203 m
0.102 m
0.0762 m/s
178000 N
92%
90%
1800 RPM
−27600 Pa
(i)The total pressure drop in the line from the pump discharge port to the blank end of the
cylinder is 517000 Pa.
(ii) The total pressure drop in the return line from the rod end of the cylinder = 345000 Pa.
Determine the
(a) Volumetric displacement of the pump.
(b) Input power required to drive the pump.
(c) Input torque required to drive the pump.
(d) Percentage of pump input power delivered to the load.
Motor
Check valve
Cylinder
F
Breather
External
load
Pump
Reservoir
Directional control Valve
Figure 1.22</figure caption>
Solution:
(a) Volumetric displacement of pump.
Qpump-actual  Apiston Vpiston ext


4
 0.2032  0.0762  0.00247 m3 /s
= 2.47 LPS
Qpump-actual 0.00247
m3
Qpump-theoretical 

 0.00268
Vol
0.92
s
Now
Qpump-theoretical  VD N
1800
60
3
 VD  0.0000893 m  0.0893 L
 0.00268 VD 
(b) Input power required to drive the pump
Pump output power = (p)Qactual
 pblank-end  Apiston  prod-end  ( Apiston  Arod )  Fexternal load on cylinder
 pblank-end 

4
 0.2032  345000 

4
 (0.2032  0.1022 )  178000
 pblank-end  5758000 Pa  5758 kPa
Pump output power = {(5758  517  27.6)0.00247}  15.6 kW
Pump output power
15.6
Pump input power =

 18.8 kW
 v  m
0.92  0.90
(c) Input torque required to drive the pump
Pump input power is
T   T 1800
rev 1min 2 rad

 188 rad/s
min 60
s
Now
18800 W  T 188 rad/s
So torque required to drive pump is T = 100 Nm.
(d) Power delivered to load is
Fexternal load on cylinder Vcyl.ext
= 178000 0.0762 = 13600 W = 13.6 kW
Percent of pump input power delivered to load = 13.6/18.8 100 = 72.3%
Example 1.19
The system of in Example 1.17 contains a fixed displacement pump with a pressure relief
valve set at 6871 kPa. The system operates 20 h/d for 250 days in a year. The cylinder is
stalled in its fully extended position 70% of the time. When the cylinder is fully extended,
0.0633 LPS leaks past its piston.
(a) If the electric motor driving the pump has an efficiency of 85% and the cost of electricity
is Rs 10 per kWh, find the annual cost of electricity for powering the system
(b) It is being considered to replace fixed displacement pump with a pressurecompensated pump (compensator set at 6871 kPa) that cost Rs 250000
more. How long will it take for the pressure-compensated pump to pay for
itself if its overall efficiency is same as fixed displacement pump?
Solution
(a)Annual cost of electricity for powering the system
pblank-end  5758000 Pa  5758 kPa
The total pressure drop in the line from the pump discharge port to the blank end of the
cylinder is 517000 Pa.
Pump inlet pressure = −27600 Pa
Pump discharge pressure = 5758000 Pa  517000 Pa  27600 Pa  6247.4 kPa
Pump input power = 18.8 kW
Electric motor input power = 18.8/0.85 = 22.1 kW
Thus with the cylinder fully extended (pressure relief valve set at 6871 kPa) we have
6871
Electric motor input power =
 22.1  24.3 kW
6247.4
Thus, the yearly cost of electricity is
Yearly cost = Power rate × Time per year × Unit cost of electricity
= 0.30 × 22.1 (kW) 20 h/d  250 d/yr  10/kWh + 0.70 24.3 20 250 10
= Rs 1182000/year
(b) The fixed displacement pump produces 2.47 LPS at 6871 kPa when the cylinder is fully
extended. Leakis 0.0633 LPS through the cylinder plus 2.407 LPS through the relief valve.
Thus, when the cylinder is fully extended, we have power lost with a fixed displacement
pump
pQ = 6871  0.00247 = 16.97 kW
Hence, the electric motor input power is
16.97
 24.1
0.828  0.85
The overall efficiency of the pump 82%.The pressure-compensated pump would produce
only 0.0633 LPS at 6871 kPa when the cylinder is fully extended. For this case we have the
power lost with pressure-compensated pump is
pQ = 6871 × 0.0000633 = 0.44 kW
Hence, the electric motor input power is
0.44
 0.63 kW
0.828  0.85
Thus, the kW power saved while cylinder is fully extended = 24.1−0.63 = 23.47 kW
Savings per year = 23.47 0.70  20 250 10 = Rs 821450 per year
Time to pay for pump = Rs 250000/821450 = 0.3 years
Objective-Type Questions
Fill in the Blanks
1. Non-positive displacement pumps are primarily ______type units that have a
______clearance between rotating and stationary parts.
2. Flow rate of pump does ______ with head in the case of a positive displacement pump,
while it ______ with head in the case of a non-positive displacement pump.
3. In a lobe pump, the output may be slightly ______ because of the smaller number of
______ elements.
4. In a screw pump, there are no pulsations at ______ speed; it is a very quiet operating
pump.
5. In a balanced vane pump, the rotor and vanes are contained within a ______ eccentric cam
ring and there are ______ inlet segments and two outlet segments during each revolution.
State True or False
1. The volumetric capacity of a positive displacement pump is less than that of a non-positive
displacement pump.
2. Too low elevation head between the reservoir and the pump inlet causes cavitation.
3. Efficiency is almost constant with the head in the case of non-positive displacement
pumps.
4. The sole purpose of pumps is to create pressure.
5. Mechanical efficiency indicates the amount of energy losses that occur for reasons other
than leakage.
Review Questions
1. What is a positive displacement pump? In what ways does it differ from a centrifugal
pump?
2. Define the source of hydraulic power (pump).
3. Explain the working principle of a pump.
4. Pumps do not pump pressure. Justify this statement.
5. What is the function of a pump in a hydraulic system?
6. How is the pumping action in positive displacement pumps accomplished?
7. How the volumetric efficiency of a positive displacement pump is determined?
8. List the advantages of hydrostatic pumps over hydrodynamic pumps.
9. Give the classification of hydrostatic pumps used in a fluid power system.
10. What is the difference between a fixed displacement pump and a variable displacement
pump?
11. What types of pumps are available in variable displacement designs?
12. How can the vane pump/piston pumps be made as variable displacement pumps?
13. Name three designs of external gear pumps.
14. Name two designs of internal gear pumps.
15. What are the advantages of screw pumps over other gear pumps?
16. Why is the operation of screw pump quiet?
17. How can the unbalanced vane pump be used as a variable displacement pump?
18. What is a pressure-compensated vane pump and how does it work?
19. What is meant by a balanced design vane pump?
20. Name the important considerations when selecting a pump for a particular application.
21. Why a gear pump cannot be used as a variable displacement pump?
22. How can the displacement of an axial piston pump be varied?
23. What is pump cavitation and what is its cause?
24. How is pressure developed in hydraulics systems?
25. Why centrifugal pumps are rarely used in fluid power systems?
26. Draw the graphical symbols for the fixed displacement and variable displacement pumps.
27. Which parameters affect the noise level of a positive displacement pump?
28. What is meant by the pressure rating of a positive displacement pump?
29. Name the four rules that control or eliminate cavitation of a pump.
30. Comment on the relative comparison in performance among gear, vane and piston pumps.
31. What are the two ways of expressing a pump size?
32. What are pump characteristic curves? Draw the same for the positive displacement
pumps.
33. How is the capability of a variable displacement pump affected by the addition of
pressure compensation?
34. Name the three principal ways in which noise reduction can be accomplished.
35. What are the most common things apart from pressure or speed that can cause a pump to
fail? Explain each.
36. Where are external gear pumps used?
Answers
Fill in the Blanks
1.Velocity, large
2.Not change, decreases
3.Greater pulsation,meshing
4.Higher;
5.Double, two
State True or False
1.True
2.False
3.False
4.False
5.True