Answer
a) $V_{1}=0.06297\frac{m^3}{s}$
$\rho_{1}=31.76\frac{kg}{m^3}$
$V_{2}=0.05667\frac{m^3}{s}$
b) $V_{1}=0.06164\frac{m^3}{s}$
$\rho_{1}=32.44\frac{kg}{m^3}$
$V_{2}=0.05474\frac{m^3}{s}$
c) $V_{1}$ $(2.16\% error)$
$\rho_{1}$ $(2.10\% error)$
$V_{2}$ $(3.53\% error)$
Work Step by Step
a) Based on the ideal gas equation:
$V_{1}=\frac{mRT_{1}}{P_{1}}=\frac{2\frac{kg}{s}*0.1889\frac{kPam^3}{kgK}*500K}{3000kPa}=0.06297\frac{m^3}{s}$
$\rho_{1}=\frac{P_{1}}{RT_{1}}=\frac{3000kPa}{0.1889\frac{kPam^3}{kgK}*500K}=31.76\frac{kg}{m^3}$
$V_{2}=\frac{mRT_{2}}{P_{2}}=\frac{2\frac{kg}{s}*0.1889\frac{kPam^3}{kgK}*450K}{3000kPa}=0.05667\frac{m^3}{s}$
b) Using the generalized compressibility chart:
$P_{R,1}=\frac{P_{1}}{P_{cr}}=\frac{3MPa}{7.39MPa}=0.407$
$T_{R,1}=\frac{T_{1}}{T_{cr}}=\frac{500K}{304.2K}=1.64$
$P_{R,2}=0.407$
$T_{R,2}=\frac{T_{2}}{T_{cr}}=\frac{450K}{304.2K}=1.48$
From Fig A-15
$Z_{1}=0.979$
$Z_{2}=0.966$
$V_{1}=\frac{Z_{1}mRT_{1}}{P_{1}}=\frac{0.979*2\frac{kg}{s}*0.1889\frac{kPam^3}{kgK}*500K}{3000kPa}=0.06164\frac{m^3}{s}$
$\rho_{1}=\frac{P_{1}}{Z_{1}RT_{1}}=\frac{3000kPa}{0.979*0.1889\frac{kPam^3}{kgK}*500K}=32.44\frac{kg}{m^3}$
$V_{2}=\frac{Z_{2}mRT_{2}}{P_{2}}=\frac{0.966*2\frac{kg}{s}*0.1889\frac{kPam^3}{kgK}*450K}{3000kPa}=0.05474\frac{m^3}{s}$
c) Errors:
$V_{1}$ $(2.16\% error)$
$\rho_{1}$ $(2.10\% error)$
$V_{2}$ $(3.53\% error)$
