Question

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

Answer 1

Answer:

The answer is "$112.97 \ \frac{ft}{s}$"

Explanation:

Air flowing into the$p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline$A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

      $= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

   $= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$