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Expert Answer

Step 1

Given that: 

                EN=62Discharge through the pump, Q=0.05 m3/sTotal head, HT=110 mSpeed of pump impeller, N=2062 rpmSpecific speed, Ns=20Speed ratio, ∅=0.9Efficiency of pump, η=0.88Diameter of similar pump, D2=0.25m and N2=1250 rpm$$\begin{gathered} EN=62 \\ Discharge through the pump, Q=0.05 m^3/s \\ Total head, H_T=110 m \\ Speed of pump impeller, N=2062 rpm \\ Specific speed, N_s=20 \\ Speed ratio, \varnothing =0.9 \\ Efficiency of pump, \eta =0.88 \\ Diameter of similar pump, D_2=0.25m and N_2=1250 rpm \end{gathered}$$

Step 2

Let HP=head upto which a pump can lift water.Now,            Ns=N1Q√Hp3/4Or,             20 = 20620.05√Hp3/4Or,           Hp=65.613 mSo, Number of pumps required = HTHP=11065.613=1.7≈2 pumps.Now,            v=∅2gHp⎯⎯⎯⎯⎯⎯⎯⎯⎯√=πD1N60Or,                       0.92x9.81x65.613⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√=πD1x206260Or,                     D1 = 0.30 mSo, diameter of pump, D1=0.30mNow, Power of pump, P=ρHpQ75η=1000x65.613x0.0575x0.88=49.71 hpWhen Diameter of pump, D2=0.25m and speed, N2=1250 rpm: Using, H2H1=(N2N1)2(D2D1)2Or,     H265.613=(12502062)2(0.250.30)2Or,   H2=16.744 mSo, new pumping head, H2=16.744 mAgain,         Q2Q1=(N2N1)(D2D1)3Or,              Q20.05=(12502062)(0.250.30)3Or,               Q2=0.0175 m3/sSo, New pump discharge, Q2=0.0175 m3/s&          P2P1=(N2N1)3(D2D1)5Or,       P249.71=(12502062)3(0.250.30)5Or,    P2=4.450 hp$$\begin{gathered} \mathrm{Let} {\mathrm{H}}_{\mathrm{P}}=\mathrm{head} \mathrm{upto} \mathrm{which} \mathrm{a} \mathrm{pump} \mathrm{can} \mathrm{lift} \mathrm{water}. \\ \mathrm{Now}, {\mathrm{N}}_{\mathrm{s}}=\frac{{\mathrm{N}}_1\sqrt{\mathrm{Q}}}{{{\mathrm{H}}_{\mathrm{p}}}^{3/4}} \\ \mathrm{Or}, 20 = \frac{2062\sqrt{0.05}}{{{\mathrm{H}}_{\mathrm{p}}}^{3/4}} \\ \mathrm{Or}, {\mathrm{H}}_{\mathrm{p}}=65.613 \mathrm{m} \\ \mathrm{So}, \mathrm{Number} \mathrm{of} \mathrm{pumps} \mathrm{required} = \frac{{\mathrm{H}}_{\mathrm{T}}}{{\mathrm{H}}_{\mathrm{P}}}=\frac{110}{65.613}=1.7\approx \mathbf{2}\mathbf{ }\mathbf{pumps}\mathbf{.} \\ \mathrm{Now}, \mathrm{v}=\varnothing \sqrt{2{\mathrm{gH}}_{\mathrm{p}}}=\frac{{\mathrm{\pi D}}_1\mathrm{N}}{60} \\ \mathrm{Or}, 0.9\sqrt{2\mathrm{x}9.81\mathrm{x}65.613}=\frac{{\mathrm{\pi D}}_1\mathrm{x}2062}{60} \\ \mathrm{Or}, {\mathrm{D}}_1 = 0.30 \mathrm{m} \\ \mathrm{So}, \mathrm{diameter} \mathrm{of} \mathrm{pump},\mathbf{ }{\mathbf{D}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{m} \\ \mathrm{Now}, \mathrm{Power} \mathrm{of} \mathrm{pump}, \mathrm{P}=\frac{{\mathrm{\rho H}}_{\mathrm{p}}\mathrm{Q}}{75\mathrm{\eta }}=\frac{1000\mathrm{x}65.613\mathrm{x}0.05}{75\mathrm{x}0.88}=\mathbf{49}\mathbf{.}\mathbf{71}\mathbf{ }\mathbf{hp} \\ \mathrm{When} \mathrm{Diameter} \mathrm{of} \mathrm{pump}, {\mathrm{D}}_2=0.25\mathrm{m} \mathrm{and} \mathrm{speed}, {\mathrm{N}}_2=1250 \mathrm{rpm}: \\ \mathrm{Using}, \frac{{\mathrm{H}}_2}{{\mathrm{H}}_1}={\left(\frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}\right)}^2{\left(\frac{{\mathrm{D}}_2}{{\mathrm{D}}_1}\right)}^2 \\ \mathrm{Or}, \frac{{\mathrm{H}}_2}{65.613}={\left(\frac{1250}{2062}\right)}^2{\left(\frac{0.25}{0.30}\right)}^2 \\ \mathrm{Or}, {\mathrm{H}}_2=16.744 \mathrm{m} \\ \mathrm{So}, \mathrm{new} \mathrm{pumping} \mathrm{head}, {\mathbf{H}}_{\mathbf{2}}\mathbf{=}\mathbf{16}\mathbf{.}\mathbf{744}\mathbf{ }\mathbf{m} \\ \mathrm{Again}, \frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}=\left(\frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}\right){\left(\frac{{\mathrm{D}}_2}{{\mathrm{D}}_1}\right)}^3 \\ \mathrm{Or}, \frac{{\mathrm{Q}}_2}{0.05}=\left(\frac{1250}{2062}\right){\left(\frac{0.25}{0.30}\right)}^3 \\ \mathrm{Or}, {\mathrm{Q}}_2=0.0175 {\mathrm{m}}^3/\mathrm{s} \\ \mathrm{So}, \mathrm{New} \mathrm{pump} \mathrm{discharge}, {\mathbf{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0175}\mathbf{ }{\mathbf{m}}^{\mathbf{3}}\mathbf{/}\mathbf{s} \\ \& \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}={\left(\frac{{\mathrm{N}}_2}{{\mathrm{N}}_1}\right)}^3{\left(\frac{{\mathrm{D}}_2}{{\mathrm{D}}_1}\right)}^5 \\ \mathrm{Or}, \frac{{\mathrm{P}}_2}{49.71}={\left(\frac{1250}{2062}\right)}^3{\left(\frac{0.25}{0.30}\right)}^5 \\ \mathrm{Or}, \mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{P}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{450}\mathbf{ }\mathbf{hp} \end{gathered}$$

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