Question

Water is pumped from a lake to a storage tank 18 m above at a rate of 70 L/s while consuming 20.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of 232 BERNOULLI AND ENERGY EQUATIONS the pump–motor unit and (b) the pressure difference between the inlet and the exit of the pump.

Answer 1

(a) 60.59 %

(b) 176.58 kPa

Step-by-step explanation:

`1 m^(3)=1000 liters`

The volume flow rate, \bar V is given as 70 l/s which can be written as
`70/1000=0.07 m^(3)/s`

The free body diagram of the storage tank is as attached figure

The change in energy between two points is given by

`\triangle E=\frac {P_1-P_2}{\rho}+ \frac {V_1^(2)-V_2^(2)}{2} + g(z_1-z_2)`

The velocity at both points are zero hence kinetic energy is zero. Also, pressure at both points is the same since both points only experience atmospheric pressure. To imply also that the flow energy is zero.

Therefore,

`\triangle E=0+0+9.81 m/s^(2)(0-18 m)=-176.58 J/kg`

`\triangle E= 176.58 J/kg`

Mass flow rate is given by

`\bar m= \rho \bar V= 1000 Kg/m^(3) * 0.07 m^(3)/s=70 kg/s`

Change in energy for
`\bar m= 70 Kg/s` will be

`\triangle\bar E=\bar m* \triangle E= 70 kg/s * 176.58 J/kg=12360.6 W`

Power consumption, W= 20.4 kW= 20400 W

The overall efficiency of the pump motor system will be

`Efficiency=\frac {\triangle\bar E}{W}* 100=\frac {12360.6}{20400}* 100=60.59118\approx 60.59`

Therefore, the efficiency is 60.59%

(b)

Assuming that the height difference between inlet and exit of the pump are negligible and also kinetic energy difference between the inlet and outlet is negligible then

`\triangle E=\frac {P_(inlet)-P_(outlet)}{\rho}+ \frac {V_(inlet)^(2)-V_(outlet)^(2)}{2} + g(z_(inlet)-z_(inlet))`

Therefore,

`\triangle E=\frac {P_(inlet)-P_(outlet)}{\rho}+0+0`

`\triangle E=\frac {P_(inlet)-P_(outlet)}{\rho}`

Already, we have

`\triangle \bar E= \bar m (\triangle E)`

`12360.6 W= 70 kg/s (\frac {P_(inlet)-P_(outlet)}{\rho})`

The pressure difference will then appear as

`P_(inlet)- P_(outlet)=\rho * \frac {12360.6 W}{70 kg/s}= 1000 Kg/m^(3) * \frac {12360.6 W}{70 kg/s}=176.58* 10^(3) Pa= 176.58 kPa`

Therefore, pressure difference between inlet and outlet is 176.58 kPa