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.
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.
Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
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 head = static + dynamic head\text { dynamic head } \begin{aligned}& =\frac{y^{2}}{2 g}=\frac{7.62^{2}}{2 \times 9.81} \mathrm{~m} \\h_{d} & =2.96 \mathrm{~m}\end{aligned}\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 prers }}{\text { eairg }}density of air at 30^{\circ} \mathrm{C} at 99.419 \mathrm{a}\begin{array}{l}\text { bair }=\frac{P}{R_{T}}=\frac{99.4}{0.287 \times(30+273)} \\\text { Cair }=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} ...