A pump operating at steady state receives saturated liquid water at 50°C with a mass flow rate of 30 kg/s. The pressure of the water at the pump exit is 1.5 MPa. If the pump operates with negligible internal irreversibilities and negligible changes in kinetic and potential energy, determine power required in kW. (Moran, 01/2018, p. P-67) Moran, M. J., Shapiro, H. N., Boettner, D. D., Bailey, M. B. (01/2018). Fundamentals of Engineering Thermodynamics, Enhanced eText, 9th Edition [VitalSource Bookshelf version]. Retrieved from vbk://9781119391388 Always check citation for accuracy before use.

Given Information:

Temperature = T₁ = 50 °C

Mass flow rate = m = 30 kg/s

Exit Pressure = P₂ = 1.5 MPa = 1500 kPa

Required Information:

Power = P = ?

Power = 45.16 kW

Explanation:

The power of the pump can be found using,

P = m*W

Where m is the mass flow rate and W is the work done by pump.

Work done is given by

W = vf*(P₂ - P₁)

Where vf is the specific volume and its value is found from the saturated water temperature table.

at T =  50 °C

vf = 0.001012 m³/kg

P₁ = 12.352 kPa

P = m*W

P = m*vf*(P₂ - P₁)

P = 30*0.001012*(1500 - 12.352)

P = 45.16 kW

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