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Efficiency in Pumps or Fans

For a fluid flow process involving a pump or fan the overall efficiency is related to the

• hydraulic
• mechanical
• volumetric

losses in the pump or fan.

Hydraulic Loss and Hydraulic Efficiency

Hydraulic losses relates to the construction of the pump or fan and is caused by the friction between the fluid and the walls, the acceleration and retardation of the fluid and the change of the fluid flow direction.

The hydraulic efficiency can be expressed as:

ηh = w / (w + wl)                            (1)

where

w = specific work from the pump or fan (J/kg)

wl = specific work lost due to hydraulic effects (J/kg)

Mechanical Loss and Mechanical Efficiency

Mechanical components - like transmission gear and bearings - creates mechanical losses that reduces the power transferred from the motor shaft to the pump or fan impeller.

The mechanical efficiency can be expressed as:

ηm = (P - Pl) / P                              (2)

where

ηm = mechanical efficiency

P = power transferred from the motor to the shaft (W)

Pl = power lost in the transmission (W)

.

Volumetric Loss and Volumetric Efficiency

Due to leakage of fluid between the back surface of the impeller hub plate and the casing, or through other pump components - there is a volumetric loss reducing the pump efficiency.

The volumetric efficiency can be expressed as:

ηv = q / (q + ql)                               (3)

where

ηv = volumetric efficiency

q = volume flow out of the pump or fan (m3/s)

ql = leakage volume flow (m3/s)

Total Loss and Overall Efficiency

The overall efficiency is the ratio of power actually gained by the fluid to power supplied to the shaft. The overall efficiency can be expressed as:

η = ηhηmηv                               (4)

where

η = overall efficiency

The losses in a pump or fan converts to heat that is transferred to the fluid and the surroundings. As a rule of thumb - the temperature increase in a fan transporting air is approximately 1oC.

Example - Hydraulic Efficiency for a Pump

An inline water pump works between pressure 1 bar (1 105 N/m2) and 10 bar (10 105 N/m2). The density of water is 1000 kg/m3. The hydraulic efficiency is ηh = 0.91.

The actual water head (water column) can be calculated as:

h = (p2 - p1) /γ

= (p2 - p1) /ρ g

= ((10 105 N/m2) - (1 105 N/m2)) / (1,000 kg/m3) (9.81 m/s2)

= 91.7 m - water column

The pump must be constructed for the specific work:

wc = g h /ηh

= (9.81 m/s2) (91.7 m) / 0.91

= 988.6 (J/kg, m2/s2)

The construction or design head is:

h =wc / g

= (988.6 m2/s2) / (9.81 m/s2)

= 100.8 m - water column

Related Topics

• Pumps

Design of pumping systems and pipelines. With centrifugal pumps, displacement pumps, cavitation, fluid viscosity, head and pressure, power consumption and more.

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