Question

Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cross-sectional area of the throat of the converging duct is 0.05 ft2. Determine the mass flowrate in lbm/s through the duct if the receiver pressure is 10 psia.

Answer 1

Answer:

The answer is "0.0728"

Explanation:

Given value:

$P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}$

                                     $= 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia$

if $P$ flow is chocked

if $P>P^{*} \to$ flow is not chocked

When  P= 10 psia < $P^{*}$ $\to$ not chocked

match number:

$\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}$

                       $= \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}$

$M_0=7.625$

$p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}$

  $=0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}$

$\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R$

$\ velocity \ of \ sound \ (C)=\sqrt{KRT}$

                                    $=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3$

R= gas constant=1716

$m=PAV$

    $=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}$