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

Answer:

W = - 523.425 W = -0.5234 kW

Negative sign show power input to the pump

Explanation:

By using energy balanced at state q and state 2

$\dot m ( h_1 +\frac{v_1^2}{2} + gz_1) + Q = \dot m ( h_2 +\frac{v_2^2}{2} + gz_2) + w$

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

$v_1 =v_f = 1.009 \times 10^{-3} m^3/kg$

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

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

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

$W = 1.5\times 1.009\times 10^[-3} (1-3) \times 10^5 + 1.5\times 9.81\times(0-15)$

W = - 523.425 W

Negative sign show power input to the pump