written 3.6 years ago by | • modified 17 months ago |

**Determine:**

**i) The theoretical discharge of the pump**

**ii) Co-efficient of discharge**

**iii) Slip and the percentage slip of the pump.**

**1 Answer**

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A single acting reciprocating pump, running at 50 rpm, delivers 0.01 m3/s of water. The diameter of the piston is 200 mm and stroke length 400 mm.

written 3.6 years ago by | • modified 17 months ago |

**Determine:**

**i) The theoretical discharge of the pump**

**ii) Co-efficient of discharge**

**iii) Slip and the percentage slip of the pump.**

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written 3.6 years ago by | • modified 3.6 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:$\%slip = (\frac{Q_{th}-Q_{act}}{Q_{th}})\hspace{0.05cm}\times\hspace{0.05cm}100\ \hspace{0.6cm}=\frac{0.01047 - 0.01}{0.01047}\hspace{0.05cm}\times\hspace{0.05cm}100\ \hspace{0.6cm}=4.489\hspace{0.05cm}\%$

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