Lecture 7

PUMP PERFORMANCE

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 Pump Performance
• The performance of a pump is determined by the following
efficiencies:
• Volumetric efficiency (ην):It is the ratio of actual flow rate of the
pump to the theoretical flow rate of the pump.
Actual flow rate of the pump QA
Volumetric efficiency    
Theoretical flow rate of the pump QT
• Volumetric efficiency (ην) indicates the amount of leakage that
takes place within the pump. This is due to manufacture tolerances
and flexing of the pump casing under designed pressure operating
conditions.
– For gear pumps, ην = 80%–90%.
– For vane pumps, ην = 92%.
– For piston pumps, ην = 90%–98%.
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 Pump Performance
• Mechanical efficiency (ηm ): It is the ratio of the pump output
power assuming no leakage to actual power delivered to the
pump:
Pump output power assuming no leakages
Mechanical efficiency  m  
Actual power delivered to the pump
• Mechanical efficiency (ηm) indicates the amount of energy
losses that occur for reasons other than leakage. This includes
friction in bearings and between mating parts. This includes
the energy losses due to fluid turbulence. Mechanical pQ
efficiencies are about 90%–95%. We also have the relation m   T

TA N
• Where p is the pump discharge pressure in Pa or N/m2, QT is
the theoretical flow rate of the pump in m3/s, TA is the actual
torque delivered to the pump in Nm and N is the speed of the
pump in rad/s. 3
 Pump Performance
• It (ηm) can also be computed in terms of torque as follows:
Theoretical torque required to operate the pump TT
m  
Actual power delivered to the pump TA

• The theoretical torque (TT) required to operate the pump is

the torque that would be required if there were no leakage.
• The theoretical torque (TT) is determined as follows
VDN  3 N 
TT  N  m   m  2   N m
2  m 
• The actual torque (TA) is determined as follows
P  N m

Actual torqueT A  N  m      N m
  rad s

where   2 N 60. Here N is the speed in RPM.
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 Pump Performance
• Overall efficiency (ηo):It is defined as the ratio of actual
power delivered by the pump to actual power delivered to the
pump.
Actual power delivered by the pump
Overall efficiency o  
Actual power delivered to the pump

• Overall efficiency (ηo) considers all energy losses and can be

represented mathematically as follows:
Overall efficiency o   V m
Q A pQT
 o  
QT T A N

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 Example
• A gear pump has an outside diameter of 82.6 mm, inside
diameter of 57.2 mm and a width of 25.4 mm. If the actual
pump flow is 1800 RPM and the rated pressure is 0.00183
m3/s, what is the volumetric efficiency?

Solution:
 N
We have
QT 
4
 Do2  D i2   d  60
 1800
• Outside diameter Do =82.6mm QT   0.0826 0.0572   0.0254 
2 2

4 60
• Inside diameter Di =57.2 mm
QT  2.125 103
• Width d= 25.4mm Volumetric efficiency is
• Speed of pump N = 1800 RPM 0.00183
V  3
100  86.11%
• Actual flow rate = 0.00183 m3/s 2.125  10

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 Example
• A pump having a volumetric efficiency of 96% delivers 29 LPM
of oil at 1000 RPM. What is the volumetric displacement of
the pump?
Solution:
• Volumetric efficiency of the pump ην = 96%
• Discharge of the pump = 29 LPM
• Speed of pump N=1000 rpm
Actual flow rate of the pump Q
   A
Theoretical flow rate of the pump QT
29
 0.96   QT  30.208 LPM
QT
Volumetric displacement
QT 30.208  103  60
VD    0.0302 L / rev 7
N 60 1000
 Example
• A positive displacement pump has an overall efficiency of 88%
and a volumetric efficiency of 92%. What is the mechanical
efficiency?
• Solution: The overall efficiency is

o  V m
o 88
m    100  95.7%
V 92

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 Example
• Determine the overall efficiency of a pump driven by a 10 HP
prime mover if the pump delivers fluid at 40 LPM at a
pressure of 10 MPa.
• Solution:
Output power  pQ
3
m s 1min
 10  106 N m 2  40 L min  
1000 L s 60s
 6670W
746W
Input power =10 HP   7460W
1HP
Pump output power 6670
o    0.894  89.4%
Pump input power 7460 9
 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..
• 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.
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• 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.

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• The overall and volumetric efficiencies as a function of
speed.

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 • Performance curves of radial piston pump

• 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.

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 Pump 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.

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 Pump Noise
• 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.

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 Pump 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.
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 Pump Cavitation
• Factors Causing Cavitation
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.
• Rules to Eliminate (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.
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 Pump 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.

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