Answer:
[tex]\dot W_{pump} = 16264.922\,W\,(16.265\,kW)[/tex]
Explanation:
The pump is modelled after applying Principle of Energy Conservation, whose form is:
[tex]\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}[/tex]
The head associated with the pump is cleared:
[tex]h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})[/tex]
Inlet and outlet velocities are found:
[tex]v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }[/tex]
[tex]v_{1} \approx 4.278\,\frac{m}{s}[/tex]
[tex]v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }[/tex]
[tex]v_{2} \approx 11.573\,\frac{m}{s}[/tex]
Now, the head associated with the pump is finally computed:
[tex]h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m[/tex]
[tex]h_{pump} = 7.705\,m[/tex]
The power that pump adds to the fluid is:
[tex]\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}[/tex]
[tex]\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)[/tex]
[tex]\dot W_{pump} = 16264.922\,W\,(16.265\,kW)[/tex]
