# Volume Flow and Temperature Rise in Pumps

## Calculate temperature rise in pumps

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No pump is perfect with *100%* efficiency. The energy lost in friction and hydraulic losses are transformed to heat - heating up the fluid transported through the pump.

The temperature rise can be calculated as

dt = P_{s}(1 - μ) / (c_{p}q ρ) (1)

where

dt = temperature rise in the pump (^{o}C)

q = volume flow through the pump (m^{3}/s)

P_{s}= brake power (kW)

c_{p}= specific heat capacity of the fluid (kJ/kg^{o}C)

μ = pump efficiency

ρ = fluid density (kg/m^{3})

Typical relation between the centrifugal pump flow, efficiency and power consumption, is indicated in the figure below:

### Example - Temperature rise in water pump

The temperature rise in a water pump working at **normal conditions** with flow* 6 m ^{3}/h (0.0017 m^{3}/s)*, brake power

*0.11 kW*and pump efficiency of

*28% (0.28)*, can be calculated as

dt = (0.11 kW) (1 - 0.28) / ((4.2 kJ/kg^{o}C) (0.0017 m^{3}/s) (1000 kg/m^{3}))

= 0.011^{ o}C

If the flow of the pump is reduced by throttling the discharge valve, the temperature rise through the pump will increase. If the flow is reduced to *2 m ^{3}/h (0.00056 m^{3}/s)*, the brake power is slightly reduced to

*0.095 kW*and pump efficiency reduced to

*15% (0.15)*, the temperature rise can be calculated as

dt = (0.095 kW) (1 - 0.15) / ((4.2 kJ/kg^{o}C) (0.00056 m^{3}/s) (1000 kg/m^{3}))

= 0.035^{ o}C

With the standard documentation provided by a manufacturer it should be possible to express the temperature rise as a function of volume flow as shown in the figure below:

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