Answer:
155.38424 K
2.2721 kg/m³
Explanation:
[tex]P_1[/tex] = Pressure at reservoir = 10 atm
[tex]T_1[/tex] = Temperature at reservoir = 300 K
[tex]P_2[/tex] = Pressure at exit = 1 atm
[tex]T_2[/tex] = Temperature at exit
[tex]R_s[/tex] = Mass-specific gas constant = 287 J/kgK
[tex]\gamma[/tex] = Specific heat ratio = 1.4 for air
For isentropic flow
[tex]\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K[/tex]
The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by
[tex]\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3[/tex]
The density of the flow at the exit is 2.2721 kg/m³