Question

A pump steadily delivers 8.94 kg/s of water at the conditions given below. Calculate the pump power (hp).

The rate of heat transfer from the pump to the surroundings is Q = 1.61 kW.

There are no changes in kinetic or potential energy.

Pump Inlet Temperature = 50^oC

Pump Inlet Pressure = 1.75 MPa

Pump Exit Temperature = 50^oC

Pump Exit Pressure = 4.17 MPa

Answer 1

Lets see whats happening here.

It is isothermal process but, it will be only valid in steady conditions which we have.

Exit temperature is staying same to inlet temperature only because pump is rejecting 1.61 kW of heat to surrounding maintaining 50 degree celcius at outlet. If there had no heat transfer of 1.61 kW. Then exit temperature might have been more than 50 degree celcius.

So, it means whatever power we will calculate by pressure difference , we will have to add heat loss 1.61 kW in it.

First we need discharge (volume flow rate ) V

[ 1000 kg/m³ is the density of water ]

V = 8.94 x 10^-3 m³/sec

Pump power using pressure difference

W = 21634.8 watt = 21.634 kW

Total Pump power

W_T = W + Q

= 21.634 + 1.61

W_T = 23.2448 kW = 23244.8 watt

We know

746 watt = 1hp

therefore,

Pump Power = 31.16 HP

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