Question 1 |

A pump with an efficiency of 80% is used to draw groundwater from a well for
irrigating a flat field of area 108 hectares. The base period and delta for paddy
crop on this field are 120 days and 144 cm, respectively. Water application
efficiency in the field is 80%. The lowest level of water in the well is 10 m
below the ground. The minimum required horse power (h.p.) of the pump is
________. (round off to two decimal places)

(Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m^3)

(Consider 1 h.p. = 746 W; unit weight of water = 9810 N/m^3)

25.64 | |

36.25 | |

30.82 | |

48.32 |

Question 1 Explanation:

Base period (B) = 120 days

\begin{aligned} Delta (\Delta )&=144 cm=1.44m\\ Duty(D)&=\frac{8.64B}{\Delta }=\frac{8.64 \times 120}{1.44}\\ D&=720\frac{hec}{cumec}\\ Q&=\frac{Area}{D}=\frac{108}{720}=\frac{3}{20}m^3/s\\ Q_{applied}&=\frac{Q}{\eta _a}=\frac{3}{20 \times 0.8}=\frac{3}{16}\\ \text{Power required}&=\frac{mgh}{t}\\ &=\rho _w \times \left ( \frac{volume}{t} \right )g \times h\\ &=\rho _w Qgh=\gamma Qh\\ &=9810 \times \frac{3}{16} \times 10\\ &=\frac{73575}{4}watt \end{aligned}

Horse power of pump = \frac{\text{Power required}}{746 \times \text{efficiency}}=\frac{73575}{4 \times 746 \times 0.8}=30.82hp

\begin{aligned} Delta (\Delta )&=144 cm=1.44m\\ Duty(D)&=\frac{8.64B}{\Delta }=\frac{8.64 \times 120}{1.44}\\ D&=720\frac{hec}{cumec}\\ Q&=\frac{Area}{D}=\frac{108}{720}=\frac{3}{20}m^3/s\\ Q_{applied}&=\frac{Q}{\eta _a}=\frac{3}{20 \times 0.8}=\frac{3}{16}\\ \text{Power required}&=\frac{mgh}{t}\\ &=\rho _w \times \left ( \frac{volume}{t} \right )g \times h\\ &=\rho _w Qgh=\gamma Qh\\ &=9810 \times \frac{3}{16} \times 10\\ &=\frac{73575}{4}watt \end{aligned}

Horse power of pump = \frac{\text{Power required}}{746 \times \text{efficiency}}=\frac{73575}{4 \times 746 \times 0.8}=30.82hp

Question 2 |

If a centrifugal pump has an impeller speed of N (in rpm), discharge Q (in m^{3}/s ) and the total head H (in m), the expression for the specific speed N_{s} of the pump is given by

N_{s}=\frac{NQ^{0.5}}{H^{0.5}} | |

N_{s}=\frac{NQ^{0.5}}{H} | |

N_{s}=\frac{NQ^{0.5}}{H^{0.75}} | |

N_{s}=\frac{NQ}{H^{0.75}} |

Question 2 Explanation:

Specific speed of pump.

N_{\mathrm{s}}=\frac{N \sqrt{Q}}{H^{3 / 4}}=\frac{N Q^{0.5}}{H^{0.75} }

N_{\mathrm{s}}=\frac{N \sqrt{Q}}{H^{3 / 4}}=\frac{N Q^{0.5}}{H^{0.75} }

Question 3 |

Identify the FALSE statement from the following:

The specific speed of the pump increases with

The specific speed of the pump increases with

increase in shaft speed | |

increase in discharge | |

decrease in gravitational acceleration | |

increase in head |

Question 3 Explanation:

The specific speed of a pump is given by,

N_{S}=\frac{N \sqrt{Q}}{\left(H_{m}\right)^{3 / 4}}

N_{S}=\frac{N \sqrt{Q}}{\left(H_{m}\right)^{3 / 4}}

Question 4 |

The allowable Net Positive Sustion Head (NPSH) for a pump provided by the
manufacturer for a flow of 0.05 m^{3}/s is 3.3 m. The temperature of water is 30^{\circ}C (vapour pressure head absolute = 0.44 m), atmosphere pressure is 100 kPa absolute and the head loss from the reservoir to pump is 0.3 N-m/N. The maximum height of the pump above the sustion reservoir is

10.19 m | |

6.89 m | |

6.15 m | |

2.86 m |

Question 4 Explanation:

The maximum suction lift,

\begin{aligned} h_{s}&=\left(\frac{p_{a}-p_{v}}{\gamma_{w}}\right)-N P S H-h_{ts} \\ p_{a} &=100 k P a_{i} \\ \gamma_{w} &=1000 \times 9.81=9810 \mathrm{N} / \mathrm{m}^{3} \\ \frac{p_{v}}{\gamma_{w}} &=0.44 \mathrm{m} \\ \frac{p_{a}}{\gamma_{w}} &=\frac{100 \times 10^{3}}{9810}=10.19 \mathrm{m} \\ \therefore \quad h_{s} &=10.19-0.44-3.3-0.3 \\ &=6.15 \mathrm{m} \end{aligned}

\begin{aligned} h_{s}&=\left(\frac{p_{a}-p_{v}}{\gamma_{w}}\right)-N P S H-h_{ts} \\ p_{a} &=100 k P a_{i} \\ \gamma_{w} &=1000 \times 9.81=9810 \mathrm{N} / \mathrm{m}^{3} \\ \frac{p_{v}}{\gamma_{w}} &=0.44 \mathrm{m} \\ \frac{p_{a}}{\gamma_{w}} &=\frac{100 \times 10^{3}}{9810}=10.19 \mathrm{m} \\ \therefore \quad h_{s} &=10.19-0.44-3.3-0.3 \\ &=6.15 \mathrm{m} \end{aligned}

Question 5 |

A pump can lift water at a discharge of 0.15m^{3}/s to a head of 25 m. The critical cavitation number (\sigma _{c}) for the pump is found to be 0.144. The pump is to be installed at a location where the barometric pressure is 9.8 m of water and the
vapour pressure of water is 0.30 of water. The intake pipe friction loss is 0.40 m.
Using the minimum value of NPSH (Net Positive Suction Head), the maximum
allowable elevation above the sump water surface at which the pump can be
located is

9.8m | |

6.2m | |

5.5m | |

none of these |

Question 5 Explanation:

\begin{aligned} \sigma_{c} &=\text { NPSH/H } \\ \therefore \quad 0.144 &=\frac{\text { NPSH }}{25} \\ \therefore \quad \text { NPSH } &=3.6 \mathrm{m} \\ \text { NPSH } &=H_{\text {atm }}-H_{v}-h_{s}-h_{L} \\ 3.6 &=9.8-0.3-h_{s}-0.4 \\ \therefore \quad h_{s} &=5.5 \mathrm{m} \end{aligned}

There are 5 questions to complete.