Question

Liquid water at 120 kPa enters a 7-kW pump where its pressure is raised to 5.6 MPa. If the elevation difference between the exit and the inlet levels is 10 m, determine the highest mass flow rate of liquid water this pump can handle. Neglect the kinetic energy change of water, and take the specific volume of water to be 0.001 m³/kg.

Answer 1

Answer:

The answer is given below

Explanation:

Things provided in the statement:

Pressure P1 = 120 kPa and P2 = 5.6 MP or 5600 kPa

Power, W = 7 kW

Elevation difference = ∆z = 10 m

Mass of flow = m˙

So potential energy changes may be significant

Specific volume of water V= 0.001 m³/kg

Now putting the values in the formula

Power, W = m˙  x V (P1 - P2) + m˙ x g x ∆z

             7 = m˙  x 0.001 (5600 - 120 ) + m˙ x 9.8 x 10 x (1 kJ/kg/ 1000 m^2/s^2)

             7 = m˙ x 5.48 + m˙ x 0.098

             7 = m ˙x 5.38

            m˙ = 7/5.38

          So mass flow m˙  =  1.301 kJ/s