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.

(a)

Flow

Power

(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. Undersized plumbing.

2. Clogged lines or suction filters.

3. High fluid viscosity.

4. 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. Keep suction line velocities below 1.2 m/s.

2. Keep the pump inlet lines as short as possible.

3. Minimize the number of fittings in the inlet line.

4. Mount the pump as close as possible to the reservoir.

5. 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

Area of pipe =Flow through pipe

Velocity of flow

Now

Flow = 40 LPM = 40/60 LPS = 40/60 × 310 m3/s

Area of pipe =340  10

60 1.2

= 0.555 × 310 m

2

Let the bore of the pipe be of diameter D. Then

Area of pipe 2

3 20.555  10 m4

πD

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 through pipe

Flow velocity=Area of pipebore

Now

Area of pipe bore = 2 2 2 6 215 mm   177 mm 177  1  0 m4

π

So

Flow velocity = 3

6

60  103.77 m / s

60 177 10

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.

1.14Pump Selection

The main parameters affecting the selection of a particular type of pump are as follows:

1. Maximum operating pressure.

2. Maximum delivery.

3. Type of control.

4. Pump drive speed.

5. Type of fluid.

6. 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 200

Mechanical

handling

250

Mobile 300

Press work 800

Table 1.4Operating pressure and size ranges for hydraulic pump types

Pump Type

Maximum

Pressure

Maximum

Delivery

(L/min)

Speed(RPM) Min.

Filtratio

n(µm)

Pulsation

Noise

Level(

dB)

η (%)

From To From To From To

External gear 40 300 0.25 760 500 3000 100 High 90 70–90

Internal gear 100 210 0.6 740 3000 4000 100 Low 85 75–90

Vane 50 140 6 360 500 3000 50 Low 80 65–80

Balanced vane 140 175 2 620 500 300 50 Low 85 70–90

Axial piston

(swash plate) 200 350 1 1450 200 2000 25 High 90

80–90

Axial piston

(bent-axis) 250 350 17 3500 200 2000 25 High 90

50–90

Radial piston 350 1720 0.3 1000 200 2000 50 High 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:

10

(final)dB increase 10  log 10 log 10 dB

(initial)

I

I

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 Overall Efficiency

Piston

Plunger in line 99 95

Radial 95  90

Axial 95  90

Precision gear pumps 95 90

Vane pump 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 m

3/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, DV = 120 3cm , 3

A  0.0015 mQ /s, 1  440 rpmN ,

   60 barP , input torque A 1  30 N mT .

Total number of working hours available  250  1  2   3000 h

Volumetric displacement in m3/rev is

DV =

33

3120 cm 1 m0.000120 m /rev

rev 100 cm

Theoretical dischargecan be calculated as

3

T D

14400.000120  rev/s 0.00288 m /s

60Q V N

Now we can calculate the volumetric efficiency as

Av

T

0.001552.08%

0.00288

Q

Q

Mechanical efficiency is given by 5

Tm

A

60 10 0.00288 1728088.2%

2  19603130 1440

60

pQ

T

Note the product AT gives power in units of Nm/s (W) where AT has a unit of Nm and shaft

speed has units of rad/s.

The overall efficiency is

o m v  0.882 0.5208 0.459 45.9%η

Alternativelyoverall efficiency can also be calculated as

5

Ao

A

60 10 0.0015 9000  45.9%

2  19603130 1440

60

pQη

T

Now since the mechanical efficiency is known, we can calculate the theoretical torque as

T A m 130 .882 114.7 N mT T

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

A

2 130 1440 19603 W 19.603 kW

60T

Next we calculate the electrical input power.

Electric motor input power is given by

Electric motor output power 19.60322.28 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

Rs/ye13.28

ar4

 534627 318760 Rs / year22.28

Since

13.28459.2

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 0.203 m

Cylinder rod diameter 0.102 m

Extending speed of cylinder 0.0762 m/s

External load on a cylinder 178000 N

Pump volumetric efficiency 92%

Pump mechanical efficiency 90%

Pump speed 1800 RPM

Pump inlet pressure −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.

Figure 1.22</figure caption>

Solution:

(a) Volumetric displacement of pump.

pump-actual piston piston ext

2 3

   

0.203 0.0762 0.00247 m /s4

Q A V

= 2.47 LPS

3

pump-actual

pump-theoretical

Vol

0.00247 m  0.00268

0.92 s

QQ

Now

pump-theoretical D  Q V N

D

18000.00268    

60V

D V 0.0000893 3m 0.0893 L

(b) Input power required to drive the pump

Pump output power = actual( )p Q

blank-end piston rod-end piston rod external load on cylinder( )p A p A A F

2 2 2

blank-end 0.203 345000 (0.203 0.102 ) 1780004 4

p

blank-end 5758000 Pa 5758 kPap

Pump output power = {(5758 517 27.6)0.00247} 15.6 kW

Pump input power = v m

Pump output power 15.618.8 kW

0.92 0.90

(c) Input torque required to drive the pump

Pump input power is

Cylinder

Pump

Motor

Directional control Valve

Breather

Reservoir

External

load

Check valve

F

rev 1 min 2  rad1800 188 rad/s

min 60 sT T

Now 18800 W 188 rad/sT

So torque required to drive pump is T = 100 Nm.

(d) Power delivered to load is

external load on cylinder cyl.extF V

= 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 pressure-

compensated 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

blank-end 5758000 Pa 5758 kPap

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

Electric motor input power = 6871

22.16247.4

24.3 kW

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

16.9724.1

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

0.440.63 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