PH-LecNotes-HH2 (AY2013-V1.0)
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Power Hydraul i cs Module PH-LecNotes -HH2 (AY2013-V1.0)
A pump when connected to a hydraulic circuit causes the fluid in the reservoir to flow to the circuit and isregarded as the source of hydraulic power in the circuit. However, the pump itself is driven by a prime
mover like an electric motor (as in most industrial applications) or from an internal combustion engine(especially in mobile or vehicular applications).
1. FUNCTION OF A PUMP
A pump converts input mechanical power from a prime mover such as an electric motor or an internalcombustion engine, into hydraulic power.
It must be remembered that a pump creates flow and NOT pressure. Pressure results from theresistance to the flow. A pump thus creates the flow necessary for the development of pressure.
2. TWO BROAD CLASSIFICATION OF PUMPS
There are two broad classes of pumps:
non-positive displacement pumps, and
positive displacement pumps
CHAPTERH2
HYDRAULIC PUMPS
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2.1 Non-positive Displacement Pumps
Non-positive displacement pumps do not separate the inlet flow from the outlet flow. An example isthe centrifugal pump as shown in Figure 1(a)). Hence an increase in pressure at the outlet affectsthe inlet flowrate. Thus flowrate decreases with increase in pressure (Figure 1(b)). Such a pump isnot good for hydraulics application because its flow is pressure dependent and their use isrestricted to fluid transfer.
2.2 Positive Displacement Pumps
Positive displacement pumps are pumps where the outlet flow is separated from the inlet. When
driven by a prime mover, it creates a partial vacuum at the inlet port which enables atmosphericpressure to force fluid from the reservoir into the pump. The mechanical action of the pump thentraps this fluid within the pump cavity and transports it through the pump and forces it out into thehydraulic system. Theoretically, all the fluid that enters the inlet must be discharged. However inreality, as the pressure increases, there will be more leakage due to clearance gaps (Figure 2).
Figure 1(a) (b)
Outlet Inlet
Impeller imparts
centrifugal force
To cause pumping
action
Figure 2
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3. TYPES OF POSITIVE-DISPLACEMENT PUMPS
The three main types of positive displacement pumps are:
gear pumps,
vane pumps and
piston pumps.
Gear Pump
One typical gear pump is the external gear pump. It consists of a pump housing and two meshinggears (Figure 3). One gear is driven by the drive shaft attached to the prime mover. The other gearis driven by the meshing gear teeth. In this way, fluid is displaced from the inlet port to the outletport when the gears are driven. This is a fixed displacement pump.
Vane Pump
One typical vane pump is the fixed displacement balanced vane pump (Figure 4). It consists of anelliptical track ring, a rotor and vanes placed inside equally spaced slots on the rotor. Duringrotation, centrifugal force drives the vanes outward against the elliptical track where they make firmcontact and follow the contour of the track. The elliptical track permits two intake positions and twooutlet positions (see Figure 4) which are opposite each other and hence the hydrostatic forces onthe rotor are always balanced. The pump housing is specially constructed to merge the two inletpositions into one inlet port, and the two outlet positions into one outlet port.
The vane pump is also a fixed displacement.
Figure 3
Figure 4
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4. FIXED DISPLACEMENT VERSUS VARIABLE DISPLACEMENT PUMPS
Displacement is the volume of fluid pushed out by the pump for every revolution that the pumpshaft turns.
Fixed displacement pumps will produce a
fixed flowrate at a certain speed of operation. Forvariable displacement pumps, the flowrate may be varied even when operating at a fixed
speed.
Most pumps are driven by electric motors running at fixed speed. It is not common to usevariable speed motor to drive fixed displacement pumps in order to obtain variable flowrate. Theflowrate can also be controlled by flow control valves. This method wastes energy because theexcess flow must return to the reservoir through the pressure relief valve.
5. ANALYSIS OF A SIMPLE FIXED POSITIVE DISPLACEMENT PUMP
A simple fixed displacement pump is shown in Figure 6. Point (1) and (2) are the pump inlet and
outlet respectively.
Since the pump converts MECHANICAL WORK to HYDRAULIC ENERGY, the input power tothe pump will be mechanical and the output power will be hydraulic.
Let
p : the pump volumetric displacement [cm3/rev]
n p : the pump speed or the speed of the prime mover [rev/min, rpm]Q1 : the pump suction flowrate or theoretical delivery at (1) [litre/min ]Q2 : the actual pump delivery at (2) [litre/min ]T p : the actual input torque at the pump shaft required to drive the pump [ Nm]T p,th : the theoretical input torque to drive the pump [ Nm]
P p : the pressure rise across the pump [bar]
Pp= P2 - P1
Figure 6
(1)
(2)n p, T p n p, T p,th
(a) (b)
where p = pump displacement (cm3/rev)
Q = flowrate (cm3/min)
n p = pump rotational speed (rev/min)
Fixed Displacement
Pump
Variable Displacement
Pump
p pnQ
1
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5.1 Flowrate
Theoretical pump delivery, Q1 = pump displacement x number of rev per min
p p
nQ 1
In practice, the actual pump delivery Q2 , will be less than the theoretical delivery Q1 as a result of
leakage and slippage.
Hence, volumetric efficiency is defined as
1
2
Q
Qv p
Loss of flow as a result of leakages or leakage flow through case drains of pump is computed as,
21 QQQ L
(1)
(2)Q2
Q1
Q L
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EXAMPLE 1
A hydraulic gear pump that has a volumetric efficiency of 95 % is operating at a speed of 1200 rpm.
a) Determine the pump’s displacement in cm3/rev if it delivers an output flowrate of 45 /min.
b) Calculate the new output flowrate in /min if the prime mover is rotating at 800 rpm.
Solution:
(a) The pump theoretical flowrate Q1 will be computed
min368.4795.0
1
21
12
Q
QQ v p
But
p pnQ 1 and
the displacement is calculated as
min47.39min1200
min10368.47 333
1 cmrev
cm
n
Q
p
p
(b) The pump speed is 1200 rpm giving an output flowrate of 45 /min.
If the pump speed is 800 rpm, an output flowrate will be min301200
45800
5.2 Power Required to Drive the Pump
Because of friction, slippage and windage losses across the couplings, the actual torque T p from
the prime mover at (a) is therefore greater than the ideal torque T p,th required at (b) to drive pumpshaft,
ie T p T p,th
Q1
Q2 n p
(1)
(2)
(a) (b)
T pn p
T p,th
n p
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Now, Mechanical Input Power at (a) is
p p T n 2
and Mechanical Power required at (b) is
th p p T n ,2
Therefore, mechanical efficiency of a pump p m , which is defined as ratio of the Mechanical Power
At (b) to the Mechanical Power At (a) is given as,
p
th p
m p
p p
th p p
m p
T T
T n
T n
aat Power Input Mechanical
bat Power Mechanical
,
,
2
2
)(
)(
Now, let us consider the pump, the Hydraulic Power Available At (1) for theoretical flowrate
1QP p
and the Output Hydraulic Power Available At (2)
2QP p
Logically, the Mechanical Power from the ideal torque at (b) should be converted to the Hydraulic
Power at theoretical flowrate at (1).
p
p p
m p
p p
p p p
p p
p
m p
T
P
T n
nP
T n
QP
aat Power Mechanical
at Power Hydraulic
aat Power Mechanical
bat Power Mechanical
2
2
2
)(
)1(
)(
)(
1
Now, overall efficiency of a pump p o , is defined as the ratio of Output Hydraulic Power at (2) to
the Input Mechanical Power at (a).
Therefore,
p p
p
o p
o p
T n
QP
aat Power Input Mechanical
at Power Output Hydraulic
2
)(
)2(
2
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Alternatively.
)(
)(
)1(
)2(
)(
)2(
aat Power Mechanical Input
bat Power Mechanical
at Power Hydraulic
at Power HydraulicOutput
aat Power Mechanical Input
at Power HydraulicOutput
o p
o p
m pv po p
There are some important consideration for pumps :
(a) Cavitation In a well-designed system, suction flow is the theoretical flow. In practice, the suction conditionmay become less than ideal, e.g. a suction strainer may be choked with dirt and no provision ismade to allow for by-pass flow (bad design!). Such condition will cause cavitation that will leadto physical damage to the pump in a very short time.
(b) Pressure The term "pressure" in this module shall be referred to as gauge pressure. Since hydraulicsystems work at high pressure (say 50 to 300 bars), the stated gauge pressures are useddirectly in the calculation with negligible error.
(c) Outlet Pressure, P2 The outlet pressure is proportional to the load up to the limiting pressure of the system set bythe pressure relief valve.
(d) Inlet Pressure, P1 In system design, the inlet or suction pressure always refer to the pressure just outside thepump's inlet port, and it is due to the height of the liquid of the fluid level above the inlet port (positive suction pressure) or below (negative suction pressure). Unless otherwise stated, theinlet pressure is usually assumed to be zero in the system calculation, since its magnitude issmall compare to the system's working pressure range.
(e) Pressure Rise, Pp= P2 - P1
The energy from the prime mover is used to convert low pressure inlet flow to higher pressureoutlet flow. In cases where non-zero inlet pressure is higher, the pressure rise which is Pp= P2 - P1 should be used instead of P2 , for pump's calculation.
(f) Case drain The pumping mechanism of a pump leaks continuously while in operation. This leakage flowmust be allowed to drain back to the reservoir to keep the pressure in the pump's housing low.
Volumetric Efficiency Mechanical Efficiency
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EXAMPLE 2
Figure 7 shows a hydraulic piston pump driven by an ac motor. The pump is able to deliver 60litres/min when the ac motor is rotating at 1200 rpm. The operating pressure is 120 bar and thepump’s volumetric and mechanical efficiencies are 90 % and 85% respectively. Calculate,
a) the pump’s displacement in cm3/rev
b) the mechanical power in kW provided by the ac motor
c) the torque in Nm generated by the ac motor to operate the pump
Solution:
a) The pump theoretical flowrate Q1 will be computed
min67.66
90.0
1
21
12
Q
QQ v p
Apply p pnQ 1 , and
the displacement is calculated as
min56.55min1200
min1067.66 333
1 cmrev
cm
n
Q
p
p
b) The overall efficiency is computed as 765.085.09.0 m pv po p
kW
or Watts
sm
m N
QPPower Input Mecanical
Power Input Mechanical
Power Output Hydraulic
o p
p
o p
6863.15
3.15686
765.0
601060
1012033
25
2
c) Torque can be computed from Mechanical Input Power
m N
srev
sm N
n
Power Input MecanicalT
T nPower Input Mechanical
p
p
p p
83.124
6012002
3.15686
2
2
Q1
Q2
n p
Figure 7
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6. APPENDIX : PUMP RATING AND SELECTION
6.1 Pump rating
Pumps are rated according to :-
Maximum allowable operating pressure (kPa or Bar)
Flowrate (/min, m3/min) at maximum drive speed of the pump (rev/min)
Displacement (cm3/rev)
Mechanical efficiency, volumetric efficiency, and overall efficiency, presented graphically, asthey are not constant throughout the operating range
6.2 Pump selection
Table 1 summarises the characteristics of the main types of pumps.
Table 1 : Pump Characteristics
Pumptype
Output Max. operatingpressure
Other characteristics
Gear Fixed displacement Low pressure
(P 150 bar)
Durable, dirt tolerant,noisy
VaneFixed / variabledisplacement
(Low to mediumvolume)
Quite high
( 150 P 200 bar)
Reliable, efficient, easy to
maintain, less dirt tolerant.
Suggest to use oil withanti
-wear additives
PistonFixed / variable,reversibledisplacement
(Low to high volume)
High pressure
(P 200 bar)
High efficiency, not dirttolerant
The main factors affecting the selection of pumps are listed below :
1. Maximum operating pressure
2. Maximum delivery
3. Pump efficiency4. Pump speed
5. Types of control
6. Noise
7. Size and weight
8. Cost
9. Maintenance and spares
10. Type of fluid and contamination
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Maximum Operating Pressure
The rated pressure of the pump must be higher than the maximum pressure required to producethe necessary force or torque at the actuator, i.e.,
for linear actuator, eg cylinder, force is computed as
AreaessureForce Pr and
for rotary actuator, eg hydraulic motor, torque is computed as
pn
Power T
2
60
The main advantage of higher system pressure:
Smaller actuator can be used for the same force transmitted.
Disadvantages of higher system pressure:1. Expensive
2. Limited choice of equipment
3. Requires good maintenance
Maximum Delivery
The delivery of the pump Q must be sufficient for the required speed of the actuator, i.e.
for a cylinder, linear velocity of piston Area
QFlowv E , and
for a hydraulic motor, motor speednt Displaceme Motor
QFlown
Pump delivery can be stated in terms of:
1) volume delivered at a given speed and pressure, eg: 5 litres/min at 200 bar and 1000 rpm.
2) displacement (swept volume per revolution of the pump) eg: 10 cm3 /rev
3) characteristic curves showing delivery versus pressure at various speeds.
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An example is shown in Figure 8.
Pump Speed
The speed of the pump is the speed of its prime mover. The flowrate of a pump varies directly withthe speed. The manufacturer's catalogue will show the minimum and the maximum operating
speeds. The lower the speed, the longer the pump life.
Type of fluid and fluid contamination
Pumps are designed to operate with a particular viscosity of fluid. Other types of fluids may affectthe lubrication of the pump.
Certain pumps are more tolerant of fluid contamination than others.
Figure 8