# 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 of the fluid (kJ/kg^{o}C)

μ = pump efficiency

ρ = fluid density (kg/m^{3})

A typical relation between flow, efficiency and power consumption for a centrifugal pump:

### Example - Temperature rise in water pump

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

Specific heat of water *c _{p} = 4.2 kJ/kg^{o}C*.

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

*0.095 kW*and the 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 manufacturing documentation the temperature rise versus throttling can be expressed as:

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