Hydraulic Pump

Hydrodynamic pump

Positive displacement pump

Gear pump

Vane pump

Piston pump

Centrifugal pump

Axial pump

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

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

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.

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?

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.

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.

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

the input mechanical power is equal to the increase in the fluid power

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.

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

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

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

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])

Total efficiency: ηtot = ηvol × ηm

where ηtot : total efficiency

ηvol : volumetric efficiency

ηm : mechanical/motor efficiency

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 %)

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?

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

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%

b) ηm = TT/TA

TT = ηm × TA = 0.93 × 120 = 112 Nm

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.

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)

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?

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

PT

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.