Take the pump as a control volume, shown on the left. Denote the
inlet by subscript 1 and exit by subscript 2. The definition of isentropic efficiency of pump is:
η _{P} v(P_{2} -
P_{1})/(h_{2} - h_{1})
It is assumed that the pump operates at ambient temperature. Hence the specific volume of water is
v = 0.001 m^{3}/kg
The pump increases the pressure of the water up to 300
kPa, which means the pressure difference of water between the exit and inlet is 300 kPa. That is
P_{2} - P_{1} = 300 kPa
Substituting the given efficiency, which is 85%, and the pressure difference above to the definition of isentropic efficiency gives,
h_{2} - h_{1} = v(P_{2} - P_{1})/η _{P}
=
0.001(300,000)/0.85 = 352.9 J/kg
With the assumptions that the pump is adiabatic and the kinetic and potential
energies are negligible, the energy balance of the pump reduces to
The mass flow rate is given as 20 kg/min. Substituting the enthalpy difference
and the mass flow rate to the energy balance yields,
=
-20/60(352.9) = -117.6 W
The negative sign means the pump needs power input. |