Answer
0.9, 1172.35kPa
Explanation:
Question (in proper order) Â Attached below
Air is flowing inside the throat has following inlet conditions
[tex]P_{0}=1000 kPa[/tex]
[tex]T_{0}=500 K[/tex]
[tex]M=1.8[/tex]
[tex]M=\frac{u}{c}=1.8[/tex]
'u' is the speed of sound in the air
[tex]\Rightarrow u=1.8\times c[/tex]
[tex]=1.8\times 340.29[/tex]
 [tex]=612.522\frac{m}{sec}[/tex]
Therefore volumetric flow rate entering,
[tex]Q=612.522\times 0.0008[/tex]
[tex]=0.4900176\frac{m^{3}}{sec}[/tex]
Using ideal gas equation
PV=nRT
[tex]n=\frac{PV}{RT}[/tex]
[tex]=\frac{1000\times 0.4900176}{8.314\times 500}[/tex]
=0.117878 gmoles/sec
Therefore , mass flow rate
[tex]Mass = 0.117878\times 29[/tex]
[tex]=3.4184 grams/sec[/tex]
Given
[tex]\frac{A}{A_{0}}=2[/tex]
[tex]\Rightarrow A=0.0016.m^{2}[/tex]
Using continuity equation
[tex]A_{1}V_{1}=A_{2}V_{2}[/tex]
[tex]\Rightarrow V_{2}=\frac{A_{1}V_{1}}{A_{2}}[/tex]
[tex]=\frac{0.0008\times 612.522}{0.0016}[/tex]
[tex]=306.261\frac{m}{sec}[/tex]
Hence exit velocity = 306.261 m/sec
Exit Mach number
[tex]M=\frac{u}{c}=\frac{306.261}{340.29}=0.9[/tex]
Temperature will remain same as 500 K
Now
Using Bernoulli's equation
[tex]\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\rho g}+\frac{v_{2}^{2}}{2g}+z_{2}[/tex]
Here
[tex]z_{1} = z_{2}[/tex]
[tex]\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}-\frac{v_{2}^{2}}{2g}=\frac{P_{2}}{\rho g}[/tex]
[tex]\Rightarrow \frac{1000000}{\rho g}+\frac{612.522^{2}}{2g}-\frac{306.261^{2}}{2g}=\frac{P_{2}}{\rho g}[/tex]
[tex]\Rightarrow \frac{1000000}{1.225}+\frac{612.522^{2}}{2}-\frac{306.261^{2}}{2}=\frac{P_{2}}{1.225}[/tex]
[tex]\Rightarrow P_{2}=1172.35kPa[/tex]
