This document discusses different types of hydraulic pumps, including their basic operating principles and comparisons. It provides equations to calculate pump parameters such as theoretical flow rate, volumetric displacement, efficiency, and torque. For example, it defines that positive displacement pumps capture and transfer fixed amounts of fluid, while centrifugal pumps impart velocity to fluid to create pressure. Gear pump displacement can be calculated based on gear dimensions. Pump efficiency is affected by factors like leakage and viscosity.

2. Hydraulic Pump
Hydrodynamic
pump
Positive displacement
pump
Gear
pump
Vane
pump
Piston
pump
Centrifugal
pump
Axial
pump

3.  Generally used for low pressure, high volume
flow applications.
 Not capable of withstanding high pressure
 Normally the maximum pressure capacity is
limited to 250-300 psi.
 Used for transporting fluids from one location
to another.
 E.g. – Centrifugal pump
 - Axial pump

5.  Eject a fixed amount of fluid into the system
per revolution.
 Capable of overcoming the pressure from
mechanical loads and friction.
 Advantages:
◦ High pressure capability (up to 12,000 psi)
◦ Small, compact size
◦ High volumetric efficiency
◦ Small changes in efficiency throughout the design
pressure range
◦ Great flexibility in term of pressure and speed
ranges

7. Gear pump always produce fixed
volume displacement. Thus the
volumetric displacement of a gear
pump can be represented by:
 LDDV ioD
22
4


where Do, Di, and L is
referred to outside diameter,
inside diameter and the width
of the gear teeth.

8.  A gear pump teeth has a 25 mm width, 75
mm outside diameter and 50 mm inside
diameter. What is the volumetric
displacement of the gear pump in liter per
revolution?

11. Basic Comparison - Dynamic Pumps
Vs. Positive Displacement Pumps
Dynamics Positive Displacement
Mechanics Imparts velocity to the liquid
resulting in a pressure at the
outlet (pressure is created
and flow results).
Captures confined amounts
of liquid and transfers it
from the suction to the
discharge port (flow is
created and pressure
results).
Performance Flow varies with changing
pressure.
Flow is constant with
changing pressure.
Viscosity Efficiency decreases with
increasing viscosity due to
frictional losses inside the
pump (typically not used on
viscosities abov e850 cSt).
Efficiency increases with
increasing viscosity.

12. Basic Comparison - Dynamic Pumps
Vs. Positive Displacement Pumps
Centrifugal Positive Displacement
Efficiency Efficiency peaks at best-
efficiency-point. At higher or
lower pressures, efficiency
decreases.
Efficiency increases with
increasing pressure.
Inlet
Conditions
Liquid must be in the pump
to create a pressure
differential. A dry pump will
not prime on its own.
Negative pressure is created
at the inlet port. A dry pump
will prime on its own.

13.  Assuming an ideal pump, with no internal
leakage, no friction, and no pressure losses,
the pump flow rate is given by the following
expression:

14.  the input mechanical power is equal to the
increase in the fluid power

15.  A gear pump of 12.5 cm3 geometric volume
operated at 1800 rev/min delivers the oil at
16 MPa pressure. Assuming an ideal pump,
calculate the pump flow rate, Qt, the increase
in the oil power, ΔN, the hydraulic power at
the pump exit line, Nout, and the driving
torque, Tt, if the inlet pressure is 200 kPa.

17.  Hydraulic power delivered to the fluid by the
real pumps is less than the input mechanical
power due to the volumetric, friction, and
hydraulic losses.
 The actual pump flow rate, Q, is less than the
theoretical flow, Qt, mainly due to:
◦ Internal leakage
◦ Pump cavitation and aeration
◦ Fluid compressibility
◦ Partial filling of the pump due to fluid inertia

18.  The effect of leakage is expressed by the
volumetric efficiency, ηv, defined as follows
◦ Q - actual pump flow rate
◦ Qt - theoretical flow rate
 ηv indicates amount of leakage that takes place in the
pump

19.  Mechanical efficiency (ηm): Indicates amount of
energy losses due to reasons other than leakage.
ηm = pQt/ωTA
= (pump output power, no leakage)/(actual power
delivered to pump)
where p : pump discharge pressure [Pa]
Qt: pump theoretical flow rate [m3/s]
TA : theoretical torque delivered to pump [Nm]
ω : radial pump speed [rad/s]
ω = 2πN / 60

20.  Or
ηm = TT/TA
= (theoretical torque to operate pump)/(actual
torque delivered to pump)
where
TT [Nm] = (V [m3] × P [Pa])/2π
TA = (actual power delivered to pump [W])/(2πN/60
[rpm])

21.  Total efficiency: ηtot = ηvol × ηm
where ηtot : total efficiency
ηvol : volumetric efficiency
ηm : mechanical/motor efficiency

22.  A leakage of oil from a pump is 6% at 230 bar. Calculate
the total efficiency if the flow rate at 0 bar is 10 dm3min-1
and the motor efficiency is 75%.
 Solution:
Q (P = 0 bar) = 10 dm3min-1
Q (P = 230 bar) = 10 × 0.94 = 9.4 dm3min-1
ηmotor = 0.75, ηvol = 9.4/10 = 0.94
Therefore
ηtot = ηmotor × ηvol = 0.705 (= 70.5 %)

23.  A pump has a displacement volume of 100 cm3. It
delivers 0.0015 m3/s at 1000 rpm and 70 bars. The
prime mover input torque is 120 Nm.
a) What is the overall efficiency of the pump?
b) What is the theoretical torque required to operate
the pump?

24.  a) From QT = V × n,
Given V = 100 cm3/rev
= 0.0001 m3/rev
QT = V × n
= 0.0001 m3/rev × (1000/60 revs-1)
= 0.00167 m3/s

25.  Solve volumetric efficiency
ηvol = QA/QT
= 0.0015/0.00167 = 0.898 = 89.8%
Solve mechanical efficiency
ηm = PQT/TAN
= (70 × 105)(0.00167)/(120)(1000 × (2π/60))
= 0.93 = 93%
Therefore, ηtot = 0.93 × 0.898 = 0.835 = 83.5%

26.  b) ηm = TT/TA
TT = ηm × TA = 0.93 × 120 = 112 Nm

27.  The pump in Example 2 is driven by an electric motor
having an overall efficiency of 85%. The hydraulic
system operates 12 hours per day for 259 days per
year. The cost of electricity is RM0.11 per kWh.
Determine:
a) The yearly cost of electric to operate the hydraulics
system.
b) Amount of yearly cost of electricity that is due to the
inefficiencies of the electric motor and pump.

28. Fluid power can be calculated from the pressure and the flow rate. It is also the
output power of a hydraulic pump. The following equation applies:
P = p x Q
Where P is referred as Pump output power [W], p is referred as Pressure [Pa]
and Q is referred to Flow rate [m3/s].
Flow, Q
Pressure, p
Fluid
Power
Power (Fluid Power/Output Power)

29.  A hydraulic pump produced a flow rate of 4.2 l/min
of oil. The resistance in hydraulic system produced a
working pressure of 60 bar for the pump. What is the
output power produced by the pump?

30.  Pump torque is calculated as force (F) time the
distance from the force to the pivoted point (d).
T=F x d
 Pump torque can also be calculated as the relation of
pressure and pump delivery.
 222
Vp
N
Qp
N
P
T





31.  A hydraulic pump produced a flow rate of 4.2 l/min of
oil. The resistance in hydraulic system produced a
working pressure of 60 bar for the pump. It is powered
by an electric motor which rotates at 1000 rpm. If there
is no loss of energy from the electric motor to the
pump, calculate the theoretical torque produced by the
electric motor to drive the pump.