Question

A pump, operating at steady state, is drawing water from a reservoir at T.-15°C and p1= 1 bar and the mass flow rate is 1.5 kg/s. The exit pressure when the water enters a storage tank located 15 m above the pump inlet is 3 bar. The water temperature remains constant from inlet to outlet at T=15°C. Neglect both kinetic energy Changes and heat transfer between the pump and the surroundings. Assume g=9.81 m/s2. Determine the power needed by the pump, in kW.

Answer 1

W = - 523.425 W = -0.5234 kW

Negative sign show power input to the pump

Step-by-step explanation:

By using energy balanced at state q and state 2

`\dot m ( h_1 +(v_1^2)/(2) + gz_1) + Q = \dot m ( h_2 +(v_2^2)/(2) + gz_2) + w`

As it is given neglect kinetic energy and heat transfer therefore above equation rduece to

`\dot m ( h_1 + gz_1) = \dot m ( h_2 + gz_2) + W`

`W = \dot m ( h_1-h_2) + \dot m g (z_1 - Z_2)`

As temp remain cosntant , so enthalapy difference is givena s

`h_1 -h_2 = v_f (p_1 - p_2)`

from saturated water tables, for temperature 15 degree celcius specific volume of water is

`v_1 =v_f = 1.009 * 10^(-3) m^3/kg`

`W = \dot m ( h_1-h_2) + \dot m g (z_1 - Z_2)`

`W =  \dot m v_f (p_1 - p_2)+ \dot m g (z_1 - Z_2)`

putting zi =0, z2 = 15, m= 1.5 kg/s

`W = 1.5* 1.009* 10^[-3} (1-3) * 10^5 + 1.5* 9.81*(0-15)`

W = - 523.425 W

Negative sign show power input to the pump