Volume Flow and Temperature Rise in Pumps

Calculate temperature rise in pumps

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.

pump power si imperial units

The temperature rise can be calculated as

dt = Ps (1 - μ) / (cp q ρ)          (1)


dt = temperature rise in the pump (oC)

q = volume flow through the pump (m3/s)

Ps = brake power (kW)

cp = specific heat of the fluid (kJ/kgoC)

μ = pump efficiency

ρ = fluid density (kg/m3)

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

pump flow power and efficiency

Example - Temperature rise in water pump

Temperature rise in a water pump working at normal conditions with flow 6 m3/h (0.0017 m3/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/kgoC) (0.0017 m3/s) (1000 kg/m3))

    = 0.011 oC

Specific heat of water cp = 4.2 kJ/kgoC.

If the flow through the pump is reduced by throttling the discharge valve the temperature rise increase. If the flow is reduced to 2 m3/h (0.00056 m3/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/kgoC) (0.00056 m3/s) (1000 kg/m3))

    = 0.035 oC

With manufacturing documentation the temperature rise versus throttling can be expressed as:

pump temperature increase

Related Topics

  • Pumps - Piping systems and pumps - centrifugal pumps, displacement pumps - cavitation, viscosity, head and pressure, power consumption and more

Related Documents

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