# Efficiency in Pumps or Fans

## The overall pump and fan efficiency is the ratio power gained by the fluid to the shaft power supplied.

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:

η_{h}= w / (w + w_{l})(1)

where

η_{h}= hydraulic efficiency

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

w_{l}= 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:

η_{m}= (P - P_{l}) / P(2)

where

η_{m}= mechanical efficiency

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

P_{l}=powerlost 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:

η_{v}= q / (q + q_{l})(3)

where

η_{v}= volumetric efficiency

q= volume flow out of the pump or fan (m^{3}/s)

q_{l}= leakage volume flow(m^{3}/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=η_{h}η_{m}η_{v}(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 *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*The density of water is

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

^{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})

= 91.7 m - 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

= 988.6 (J/kg, m^{2}/s^{2})

The construction or design head is:

h =w_{c}/ g

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

= 100.8 m - water column