Answer:
The answer is "[tex]112.97 \ \frac{ft}{s}[/tex]"
Explanation:
Air flowing into the[tex]p_1 = 20 \ \frac{lbf}{in^2}[/tex]
Flow rate of the mass [tex]m = 230.556 \frac{lbm}{s}[/tex]
inlet temperature [tex]T_1 = 700^{\circ} F[/tex]
Pipeline[tex]A= 5 \times 4 \ ft[/tex]
Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:
[tex]\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}[/tex]
[tex]= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}[/tex]
[tex]V= \frac{mv}{A}[/tex]
[tex]= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}[/tex]