Answer
$A_{1} = 0.1333ft^{2}$
Work Step by Step
#1
The mass flow rate determined from
$m = \frac{A_{2} V_{2}}{v_{2}} = \frac{A_{2}V_{2}P_{2}}{RT_{2}} $
We replace the values in the equation
$m=\frac{(0.01ft^{2})(100\frac{ft}{s})(200psia)}{(2.6809psia \times ft^{3}\frac{psia\times ft^{3}}{lbm\times R})(1060R)} $
$m = 0.07038\frac{lbm}{s}$
therefore
$A_{1} = \frac{mv1}{V_{1}} = \frac{mRT_{1}}{V_{1}P_{1}}$
We replace the values in the equation
$A_{1} = \frac{(0.07038\frac{lbm}{s})(2.6809\frac{psia\times ft^{3}}{lbm\times R})(530R)}{(50\frac{ft}{s})(15psia)}$
$A_{1} = 0.1333ft^{2}$
