Question 4. Air is flowing in a duct with a velocity of 6.62 m/s and static pressure of 4.16 cm of water gauge. The duct diameter is 1.42 m, the barometric pressure 109.4 kPa and the gauge Fluid temperature and air temperature are 30°C. What is the total pressure against which the fan will operate in cm of water?

\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. ...