Calculating Power of a Water Pump: Enthalpy vs Pressure Difference

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SUMMARY

The discussion centers on calculating the power of a water pump using two methods: enthalpy values and pressure difference. Participants highlight that using enthalpy involves multiplying flow rate (in kg/s) by enthalpy values to derive power in kW, while the pressure difference method requires converting flow rate to m³/s and multiplying by the pressure difference for an energy balance. The discrepancy between the two methods is attributed to factors such as pump type and efficiency, particularly for centrifugal and positive displacement pumps.

PREREQUISITES
  • Understanding of fluid dynamics and pump operation principles
  • Knowledge of thermodynamic properties, specifically enthalpy
  • Familiarity with flow rate conversions (kg/s to m³/s)
  • Basic concepts of pump efficiency and types (centrifugal vs. positive displacement)
NEXT STEPS
  • Research the thermodynamic properties of water, focusing on enthalpy values
  • Learn about calculating pump efficiency for centrifugal and positive displacement pumps
  • Explore energy balance equations in fluid mechanics
  • Utilize online calculators for pump power calculations and efficiency comparisons
USEFUL FOR

Engineers, mechanical designers, and anyone involved in pump selection and performance analysis will benefit from this discussion.

Benjamin Crump
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Trying to calculate the power of a pump, it's pumping water.

It has a flowrate in and out in kg/s. Surely I should look up the enthalpy values and just mutiply with the flow rate to get power in kW.
However I know that if I convert flowrate to m3/s by dividing 1000 and multiplying by the pressure difference would I not be doing an energy balance? I get a different answer in both methods, which one is the valid method here?
 
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Hello Ben, :welcome:

Is the difference you get really that shocking ?

There's a calculator here to compare your results
 
What kind of pump? If it's a centrifugal then you must know the efficiency at the particular pumping rate and differential pressure. If it's positive displacement of some sort then friction of the mechanical elements enters into the equation. There are too many unknowns in your question as stated.
 

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