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:
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