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Given:-V=7.62 \mathrm{~m} / \mathrm{sec}static pressure =2.16 \mathrm{~cm} waterduct dianeter d=1.22 \mathrm{~m}Barometric presure =99.4 \mathrm{kp}.Temperature =30^{\circ} \mathrm{C}(1) Total hend = static + dynamic head\begin{array}{l}\text { dynamic head }=\frac{V^{2}}{L g}=\frac{7.6 L^{2}}{2 \times 9.81} \mathrm{M} \\h_{d}=2.96 \mathrm{~m} \\\end{array}\text { Static head }=\frac{p}{e g}Cair = density of air2e= density of water in \mathrm{kg} / \mathrm{m}^{3}p= Pressure in pascalh_{s}=\frac{e g \times \text { static press }}{\text { lairg }}density of air at 30^{\circ} \mathrm{C} at 99.419 \mathrm{a}\begin{array}{l}S_{\text {air }} \frac{P}{R_{T}}=\frac{99.4}{0.287 \times(30+273)} \\P_{\text {air }}=1.143 \mathrm{~kg} / \mathrm{m}^{3}\end{array}\begin{array}{l}h_{s}=\frac{1000 \times 9.61 \times\left(\frac{2.16}{100}\right) \mathrm{m}}{1.143 \times 9.81} \\h_{s}=18.89 \mathrm{~m} \text { of air }\end{array}\begin{aligned}\text { Total head } & =18.89+2.96 \\& =21.85 \mathrm{~m} \\& =2185 \mathrm{~cm}-\text { Answer }\end{aligned}(2) Power of air =\dot{n} g \times total head\begin{array}{l}\dot{m}=\rho A \times \text { xelocity } \\\dot{m}=1.143 \times \frac{\pi}{4} \times\left(1.22^{2}\right) \times 7.62 \\\dot{m}=10.1814 \mathrm{~kg} / \mathrm{sec}\end{array}\begin{aligned}\text { Power } & =10.1814 \times 9.81 \times 21.85 \\& =2182.384 \text { watt Anvw }\end{aligned}(3) by using water fromula total head in terms of water\begin{aligned}& =2.96 \mathrm{~m}+\frac{2.16}{100} \mathrm{~m} \\h_{T} & =2.9816 \mathrm{~m} \\\text { Hower } & =m g h_{T} \\& =10.1814 \times 9.81 \times 2.9816 \\& =297.8 \text { watt } \\& =0.4 \mathrm{HP}\end{aligned} ...