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Question 17.CSGP.80: The coefficient of discharge of a sharp-edged orifice is det......

The coefficient of discharge of a sharp-edged orifice is determined at one set of conditions by use of an accurately calibrated gasometer. The orifice has a diameter of 20 mm and the pipe diameter is 50 mm. The absolute upstream pressure is 200 kPa and the pressure drop across the orifice is 82 mm of mercury.
The temperature of the air entering the orifice is 25°C and the mass flow rate measured with the gasometer is 2.4 kg/min. What is the coefficient of discharge of the orifice at these conditions?

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\begin{aligned}& \Delta P =82 \times 101.325 / 760=10.93\, kPa \\& \Delta T = T _{ i }\left\lgroup\frac{ k -1}{ k } \right\rgroup \Delta P / P _{ i }=298.15 \times \frac{0.4}{1.4} \times 10.93 / 200=4.66 \\& v _{ i }= RT _{ i } / P _{ i }=0.4278 m ^3 / kg , \quad V _{ e }= RT _{ e } / P _{ e }=0.4455 \,m ^3 / kg \\& \mathbf{V} _{ i }= \mathbf{V} _{ e } A _{ e } v _{ i } / A _{ i } v _{ e }=0.1536 \mathbf{V} _{ e } \\& \left( \mathbf{V} _{ e }^2- \mathbf{V} _{ i }^2\right) / 2= \mathbf{V} _{ e }^2\left(1-0.1536^2\right) / 2= h _{ i }- h _{ e }= C _{ p } \Delta T\end{aligned}
\begin{aligned}& \mathbf{V} _{ e }=\sqrt{2 \times 1000 \times 1.004 \times 4.66 /\left(1-0.1536^2\right)}=97.9 \,m / s \\& \dot{ m }= A _{ e } \mathbf{V} _{ e } / v _{ e }=\frac{\pi}{4} \times 0.02^2 \times 97.9 / 0.4455=0.069 \,kg / s \\& C _{ D }=2.4 / 60 \times 0.069= 0 . 5 8\end{aligned}

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