Answer:
(a) 60.59 %
(b) 176.58 kPa
Explanation:
[tex]1 m^{3}=1000 liters[/tex]
The volume flow rate, \bar V is given as 70 l/s which can be written as [tex]70/1000=0.07 m^{3}/s[/tex]
The free body diagram of the storage tank is as attached figure
The change in energy between two points is given by
[tex]\triangle E=\frac {P_1-P_2}{\rho}+ \frac {V_1^{2}-V_2^{2}}{2} + g(z_1-z_2)[/tex]
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,
[tex]\triangle E=0+0+9.81 m/s^{2}(0-18 m)=-176.58 J/kg[/tex]
[tex]\triangle E= 176.58 J/kg[/tex]
Mass flow rate is given by
[tex]\bar m= \rho \bar V= 1000 Kg/m^{3} \times 0.07 m^{3}/s=70 kg/s[/tex]
Change in energy for [tex]\bar m= 70 Kg/s[/tex] will be
[tex]\triangle\bar E=\bar m\times \triangle E= 70 kg/s \times 176.58 J/kg=12360.6 W[/tex]
Power consumption, W= 20.4 kW= 20400 W
The overall efficiency of the pump motor system will be
[tex]Efficiency=\frac {\triangle\bar E}{W}\times 100=\frac {12360.6}{20400}\times 100=60.59118\approx 60.59[/tex]
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
[tex]\triangle E=\frac {P_{inlet}-P_{outlet}}{\rho}+ \frac {V_{inlet}^{2}-V_{outlet}^{2}}{2} + g(z_{inlet}-z_{inlet})[/tex]
Therefore,
[tex]\triangle E=\frac {P_{inlet}-P_{outlet}}{\rho}+0+0[/tex]
[tex]\triangle E=\frac {P_{inlet}-P_{outlet}}{\rho}[/tex]
Already, we have
[tex]\triangle \bar E= \bar m (\triangle E) [/tex]
[tex]12360.6 W= 70 kg/s (\frac {P_{inlet}-P_{outlet}}{\rho})[/tex]
The pressure difference will then appear as
[tex]P_{inlet}- P_{outlet}=\rho \times \frac {12360.6 W}{70 kg/s}= 1000 Kg/m^{3} \times \frac {12360.6 W}{70 kg/s}=176.58\times 10^{3} Pa= 176.58 kPa[/tex]
Therefore, pressure difference between inlet and outlet is 176.58 kPa
