Answer:
The answer is "0.0728"
Explanation:
Given value:
[tex]P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\[/tex]
[tex]= 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia\\\\[/tex]
if [tex]P<P^{*} \to[/tex] flow is chocked
if [tex]P>P^{*} \to[/tex] flow is not chocked
When P= 10 psia < [tex]P^{*}[/tex] [tex]\to[/tex] not chocked
match number:
[tex]\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}[/tex]
[tex]= \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}[/tex]
[tex]M_0=7.625[/tex]
[tex]p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}[/tex]
[tex]=0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}[/tex]
[tex]\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\[/tex]
[tex]\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\[/tex]
[tex]=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\[/tex]
R= gas constant=1716
[tex]m=PAV\\\\[/tex]
[tex]=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}[/tex]