Question AIR IS FLOWING IN A DUCT WITH VELOCITY OF 7.62 M/S, AND A STATIC PRESSURE OF 2.16 CM WATER GAUGE. THE DUCT DIAMETER IS 1.22 M, THE BAROMETRIC PRESSURE OF 99.4 KPA, AND THE GAGE FLUID TEMPERATURE AND AIR TEMPERATURE ARE 30 DEG. C. 1. SOLVE FOR THE TOTAL HEAD AGAINST WHICH FAN WILL OPERATE IN CM OF AIR. 2. COMPUTE FOR POWER OF AIR IN HP 3. COMPUTE FOR POWER USING THE WATER FORMULA IN HP.

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Transcribed Image Text: AIR IS FLOWING IN A DUCT WITH VELOCITY OF 7.62 M/S, AND A STATIC PRESSURE OF 2.16 CM WATER GAUGE. THE DUCT DIAMETER IS 1.22 M, THE BAROMETRIC PRESSURE OF 99.4 KPA, AND THE GAGE FLUID TEMPERATURE AND AIR TEMPERATURE ARE 30 DEG. C. 1. SOLVE FOR THE TOTAL HEAD AGAINST WHICH FAN WILL OPERATE IN CM OF AIR. 2. COMPUTE FOR POWER OF AIR IN HP 3. COMPUTE FOR POWER USING THE WATER FORMULA IN HP.

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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} ...