Question

Liquid water flows steadily through a work producing device with a mass flow rate of 100 kg/s. At the inlet the conditions are: p1=1000 kPa, T1=25 deg-C, Vel1 = 30 m/s. At the exit, p2=100 kPa, u2= u1 (specific internal energy remains constant), and Vel2 = 20 m/s. Assume no heat transfer, no change in elevation between the inlet and exit, and density of water to be 1000 kg/m3.

(a). Determine the power produced (w/o neglecting the effect of change in kinetic energy)

(b). Determine the power produced (neglecting the effect of change in kinetic energy).

Answer 1

since there is no change in internal energy, there will be no change in temperature ;

pressure at 1 = 1000 kPa and flow rate is 100 Kg/s and velocity is 30m/s

100 kg per second is also 0.1 m³/s velocity is 30m/s, so cross section area would be 1/300 m²

pressure is energy/volume so power is pressure*volume/time so power = 1000*1000*0.1 = 100000 watts

or 100 Kilowatts.

now if we consider loss in kinetic energy, initial liquid kinetic energy is 100*30*30/2

final kinetic energy is 100*20*20/2

so the loss is : 250*100 = 25000 joules/second or 25 Kilowatts

so power produced would be 125 kilowatts