# Pump and Fan Efficiency

## Overall pump and fan efficiency is the ratio - power actually gained by the fluid - to shaft power supplied

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

- hydraulic
- mechanical
- volumetric

loss in the pump or fan.

### Hydraulic Loss and Hydraulic Efficiency

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

The hydraulic efficiency can be expressed as:

η_{h}= w / (w + w_{l})(1)

where

η_{h}= hydraulic efficiency

w= specific work from the pump or fan

w_{l}= specific work lost due to hydraulic effects

### Mechanical Loss and Mechanical Efficiency

Mechanical components - as transmission gear and bearings - generates a mechanical loss that reduces the power transferred from the motor shaft to the pump or fan impeller.

The mechanical efficiency can be expressed as:

η_{m}= (P - P_{l}) / P(2)

where

η_{m}= mechanical efficiency

P= power transferred from the motor to the shaft

P_{l}= power lost in the transmission

### 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:

η_{v}= q / (q + q_{l})(3)

where

η_{v}= volumetric efficiency

q= volume flow out of the pump or fan

q_{l}= leakage volume flow

### Total Loss and Overall Efficiency

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

η=η_{h}η_{m}η_{v}(4)

where

η= overall efficiency

The losses in the pump or fan converts to heat transferred to the fluid and the surroundings. As a rule of thumb the temperature increase in a fan transporting air is approximately 1 ^{o}C.

### Example - Hydraulic Efficiency for a Pump

An inline water pump works between pressure* 1 bar (1 10 ^{5} N/m^{2}) *and

*10 bar (10 10*Density of water is

^{5}N/m^{2}).*1000 kg/m*The hydraulic efficiency

^{3}.*η*

_{h}*= 0.91*.

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

h = (p_{2}- p_{1}) /γ

= (p_{2}- p_{1}) /ρ g

=((10 10^{5}N/m^{2}) - (1 10^{5}N/m^{2})) / (1,000 kg/m^{3}) (9.81 m/s^{2})

=91.7m - water column

The pump must be constructed for the specific work:

w_{c}= g h /η_{h}

= (9.81 m/s^{2}) (91.7 m) / 0.91

=988.6m^{2}/s^{2}

The construction or design head is:

h =w_{c}/ g

= (988.6 m^{2}/s^{2}) / (9.81 m/s^{2})

=100.8m - water column

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