Answer:
Explanation:
Given that:
The pressure differential between the outlet and the inlet of the pump at full load is measured to be ΔP is 211 kPa.
the flow rate through the pump is 18 L/s
The Formula of flow rate.
[tex]{Q_f} = \frac{{{P_o}}}{{\Delta P}}[/tex]
Rearrange
[tex]{P_o} = \Delta P\left( {{Q_f}} \right)[/tex]
Substitute [tex]211{\rm{ kPa}}\, for\, \Delta P \,and\, 18{\rm{ L/s}}\, for\, {Q_f}[/tex]
[tex]\begin{array}{c}\\{P_o} = \left( {211{\rm{ kPa}}} \right)\left( {\left( {18{\rm{ L/s}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ }}{{\rm{m}}^{\rm{3}}}{\rm{/s}}}}{{1\;{\rm{L/s}}}}} \right)} \right)\\\\ = 3.8{\rm{ kW}}\\\end{array}[/tex]
The formula for the overall efficiency of the pump.
[tex]\eta = \frac{{{P_o}}}{{{{\rm{P}}_{in}}}} \times 100[/tex]
input the values 3.8kW for [tex]{P_o}[/tex] and 5kW for [tex]{P_i}[/tex]
[tex]\begin{array}{c}\\\eta = \left( {\frac{{3.8{\rm{ kW}}}}{{5{\rm{ kW}}}}} \right) \times 100\\\\ = 76\% \\\end{array}[/tex]
The overall efficiency of the glycerin pump is 76% .
The overall efficiency of the pump is 76%.
Q = 18 L/s
[tex]= 18 \times 10^{-3}[/tex]
The pump power should be
[tex]= 211 \times 18 \times 10^{-3}\\\\ [/tex]
= 3.798 kW
Now the overall efficiency should be
[tex]= 3.798 \div 5[/tex]
= 76%
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