Community Answer

Honor CodeSolved 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

Get Started

\therefore ANSWER:-Given data:-Velocity (V)=6.62Pressure \left(P_{1}\right)=4.16 \mathrm{~cm}Diameter (D)=I \cdot UeP_{\text {baro }}= Atmospheric pressure =109.4 \mathrm{kPa}Temperature of air \left(T_{1}\right)=30^{\circ} \mathrm{C}=30+273.15 \mathrm{~K}=303.15 \mathrm{~K}Absolute static pressure:-\begin{aligned}P_{a b s} & \left.=P_{1} \text { (gauge }\right)+P_{b a r o} \\P_{a b s} & =e . g \cdot h+109.4 \times 10^{3} \mathrm{~Pa} \\P_{a b s} & =1000 \times 9.81 \times\left(\frac{4.16}{100}\right)+\left(109.4 \times 10^{3}\right) \\P_{a b s} & =408.096+109.4 \times 10^{3} \\& =109405.096 \mathrm{~Pa} \\P_{a b s} & =109.405 \mathrm{kpa}\end{aligned}velocity of sound (C)=\sqrt{1 \cdot R \cdot T_{1}}\begin{array}{l}=\sqrt{1.4 \times 287 \times 303.15} \\c=349 \mathrm{~m} / \mathrm{s}\end{array} \quad\left\{\begin{array}{l}k=1.4 \\R=287 \\\text { constant }\end{array}\right\}Mach Number\begin{array}{l}M_{a}=\frac{V}{c} \\M_{a}=\frac{6.62}{349} \\M_{a}=0.018\end{array}now calculate total pressure ( \left.P_{01}\right)\begin{aligned}\frac{P_{O I}}{P_{a b s}} & =\left[1+\frac{k-1}{2}(\mathrm{Ma})^{2}\right]^{\frac{k}{k-I}} \\\frac{P_{0 I}}{109.40} & =\left[1+\frac{1.4-1}{2} \times(0.018)^{2}\right]^{\frac{1.4}{1.4-1}} \\\frac{P_{01}}{109.40} & =\left[1+0.2 \times\left(3.24 \times 10^{-4}\right)\right]^{3.5} \\P_{01} & =[1.0000648]^{3.5} \times 109.40 \\& =1.00022 \times 109.40 \\& =109.424 \mathrm{KPa} \\& =109424.8 \mathrm{~Pa}\end{aligned}Now convert \mathrm{Pa} in to \mathrm{cm} of waterSo we have to multiply with 0.0101972\begin{array}{l}P_{01}=109424.8 \times 0.0101972 \\P_{O 1}=1115.8224 \mathrm{~cm} \text { of water }\end{array}Dear student i hope you understand. If you have any doubt then comment it. Thank you. ...