Q2GATE 2017MCQ

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

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📖 Explanation

Specific speed of pump. $N_{\mathrm{s}}=\frac{N \sqrt{Q}}{H^{3 / 4}}=\frac{N Q^{0.5}}{H^{0.75} }$

Q3GATE 2006MCQ

Identify the FALSE statement from the following: The specific speed of the pump increases with

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📖 Explanation

The specific speed of a pump is given by, $N_{S}=\frac{N \sqrt{Q}}{\left(H_{m}\right)^{3 / 4}}$

Q4GATE 2004MCQ

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

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📖 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}

Q5GATE 2002MCQ

A pump can lift water at a discharge of 0.15 $m^{3}/s$ 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

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📖 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}

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