Question

*Biosystems Engineering - Food Science

A 60% efficient centrifugal pump is being used to transport a liquid food product. If the pump energy Ep is 100 J/kg, the liquid mass flow rate through the pump is 2 kg/s. Determine the power requirement for the pump.

Answer 1

Pump energy Ep = 100 J/Kg (this means, at the OUTPUT the pump uses 100 J of energy for a flow of 1Kg liquid mass.)

Now, the flow rate is 2Kg/s.    (this means, the pump flows 2 Kg of liquid mass per second.)

So, for this flow rate the pump uses (2*100)=200 J of (output) energy per second .

Then, the output power, Pout = (energy needed for the output/time)    # this is the definition of power

   => Pout = 200 J/s = 200 W (1 J/s = 1 Watt) # in short Watt is represented by W

The efficiency is defined as, Eff = (power got at the output / power delivered in the input)

So, the power delivered at the input = (Pout / Eff) = (200 W) / (60%) = (200 W) / (60/100) = (200 * 100 / 60) W = 333.33 W

This is the power required for the pump i.e. 333.33 Watt of power should be delivered in the input.