A 5 kilowatt pump operating at steady state draws in liquid

A 5 - kilowatt pump operating at steady state draws in liquid water at 1.0 bara 15 degree C and delivers it at 5.0 bara at an elevation 6 m above the inlet. There is no significant change in velocity between the inlet and exit, and the local acceleration of gravity is 9.8 m/s^2. Would it be possible to pump 7.5 m^3 in 10 min or less? Explain.

Solution

P_h(kW) =q g h / (3.6 *10⁶)            (1)

where

P_h(kW) = hydraulic power (kW)

q = flow capacity (m³/h)

= density of fluid (kg/m³)

g = gravity (9.81 m/s²)

h = differential head (m)

q=(p*3.6 *10⁶)/( g h)

q=(5*3.6 *10⁶)/(1000*9.8*6)

q=306 m³/h

q=306/60 m³/min

q=5.1 m³/min

so in 10 min q=5.1*10=51 m³/min

for 7.5 liter time = 7.5/5.1=1.47 min

answer=it is possible to pump the water in less than 10 min ie. in 1.47 min.