Question

A pump operating at steady state receives liquid water at 20°C 100 kPa with a mass flow rate of 53 kg/min. The pressure of the water at the pump exit is 5 MPa. The isentropic pump efficiency is 70%. Stray heat transfer and changes in kinetic and potential energy are negligible. What is the power required by the pump?

Answer 1

Answer:

The power required by pump is 6.18 KW.

Explanation:

The ideal output power of the pump is given by:

Pout = ΔP (Volume Flow Rate)

ΔP = Pf - Pi = 5 MPa - 100 KPa = 5 x 10^6 Pa - 0.1 x 10^6 Pa

ΔP = 4.9 x 10^6 Pa

Volume Flow Rate = (Mass Flow Rate)/(Density of Water)

Volume Flow Rate = (53 kg/min)(1 min / 60s)/(1000 kg/m^3)

Volume Flow Rate = 8.83 x 10^(-4) m^3/s

Therefore,

Pout = [4.9 x 10^6 N/m²][8.83 x 10^(-4) m^3/s]

Pout = 4328.33 Watt = 4.33 KW

Now, the required Power (Pin) is related with isentropic efficiency as:

Isentropic Efficiency = Pout/Pin

Pin = [Pout]/[Isentropic Efficiecy]

Pin = (4328.33 Watt)/(0.7)

Pin = 6183.32 Watt = 6.18 KW