written 6.0 years ago by | • modified 3.8 years ago |
Determine:
i) The theoretical discharge of the pump
ii) Co-efficient of discharge
iii) Slip and the percentage slip of the pump.
written 6.0 years ago by | • modified 3.8 years ago |
Determine:
i) The theoretical discharge of the pump
ii) Co-efficient of discharge
iii) Slip and the percentage slip of the pump.
written 6.0 years ago by | • modified 6.0 years ago |
Given:
$N = 50\hspace{0.05cm}rpm, \hspace{0.25cm}Q_{act} =0.01\hspace{0.05cm}m^3/s\\ D = 200\hspace{0.05cm}mm = 0.2\hspace{0.05cm}m\\ L = 400\hspace{0.05cm}mm = 0.4\hspace{0.05cm}m$
To Find: $Q_{th} = \hspace{0.05cm}?,\hspace{0.25cm}C_d = \hspace{0.05cm}?,\hspace{0.25cm}\% slip\hspace{0.05cm} = \hspace{0.05cm}?$
Solution
Theoretical discharge:
$Q_{th} = \frac{A\hspace{0.05cm}\times\hspace{0.05cm}L\hspace{0.05cm}\times\hspace{0.05cm}N}{60} = \frac{\pi\hspace{0.05cm}\times\hspace{0.05cm}D^2\hspace{0.05cm}\times\hspace{0.05cm}L\hspace{0.05cm}\times\hspace{0.05cm}N}{4\hspace{0.05cm}\times\hspace{0.05cm}60}\\ \hspace{0.5cm}=\frac{\pi\hspace{0.05cm}\times\hspace{0.05cm}0.2^2\hspace{0.05cm}\times\hspace{0.05cm}0.4\hspace{0.05cm}\times\hspace{0.05cm}50}{4\hspace{0.05cm}\times\hspace{0.05cm}60}\\ \hspace{0.5cm}= 0.01047 m^3/s$
Coefficient of discharge: $C_d = \frac{Q_{act}}{Q_{th}} = \frac{0.01}{0.01047} =0.955\\$ Percentage slip:$\%slip = (\frac{Q_{th}-Q_{act}}{Q_{th}})\hspace{0.05cm}\times\hspace{0.05cm}100\ \hspace{0.6cm}=\frac{0.01047 - 0.01}{0.01047}\hspace{0.05cm}\times\hspace{0.05cm}100\ \hspace{0.6cm}=4.489\hspace{0.05cm}\%$