Answer
0.9, 1172.35kPa
Step-by-step explanation:
Question (in proper order) Attached below
Air is flowing inside the throat has following inlet conditions
![T_(0)=500 K](https://img.qammunity.org/2021/formulas/engineering/college/vlecxwou9wjw60dcu20ch9154xx6v693om.png)
![M=1.8](https://img.qammunity.org/2021/formulas/engineering/college/gy6joyo9buvdefoyd4zdazi8mgagy4xpvy.png)
![M=(u)/(c)=1.8](https://img.qammunity.org/2021/formulas/engineering/college/p3391trdipj0hn982jniqk90j1y0oiqmqg.png)
'u' is the speed of sound in the air
![\Rightarrow u=1.8* c](https://img.qammunity.org/2021/formulas/engineering/college/ryih1i3wxu72ceywtcnxyrocqp8qcmaiml.png)
![=1.8* 340.29](https://img.qammunity.org/2021/formulas/engineering/college/6ms84vzut28p3b77wlj58075z22xfr0op5.png)
![=612.522(m)/(sec)](https://img.qammunity.org/2021/formulas/engineering/college/d0hsijrmhomj5yqj0c1ugun7915k6wicme.png)
Therefore volumetric flow rate entering,
![=0.4900176(m^(3))/(sec)](https://img.qammunity.org/2021/formulas/engineering/college/pdtd3cwvxhy9fuli2ycp35xvqd39j7k5ma.png)
Using ideal gas equation
PV=nRT
![n=(PV)/(RT)](https://img.qammunity.org/2021/formulas/engineering/college/ki1yx8lpgcfmh4d9rnxysugq50to01dgw7.png)
=0.117878 gmoles/sec
Therefore , mass flow rate
![Mass = 0.117878* 29](https://img.qammunity.org/2021/formulas/engineering/college/2webuxe39alv2m63ecq2zsv0d4l5t9yl83.png)
![=3.4184 grams/sec](https://img.qammunity.org/2021/formulas/engineering/college/a22tejyzjzha8hlwaxarrjl6ukppw5b92t.png)
Given
![(A)/(A_(0))=2](https://img.qammunity.org/2021/formulas/engineering/college/om3x98vu75skzoludplt96n2pw27a9jmnd.png)
![\Rightarrow A=0.0016.m^(2)](https://img.qammunity.org/2021/formulas/engineering/college/bbm6hzfzyv74nna5tvbpqnzwnqynkomjqr.png)
Using continuity equation
![A_(1)V_(1)=A_(2)V_(2)](https://img.qammunity.org/2021/formulas/engineering/college/6agippsbodgar4jgcm53uu775sp9v8d7bc.png)
![\Rightarrow V_(2)=(A_(1)V_(1))/(A_(2))](https://img.qammunity.org/2021/formulas/engineering/college/b7q7i94gpv96f792amwu3ajsdphyxcz0te.png)
![=(0.0008* 612.522)/(0.0016)](https://img.qammunity.org/2021/formulas/engineering/college/d9dct5zotdnjfbvauzw8visp7uo1kb68st.png)
![=306.261(m)/(sec)](https://img.qammunity.org/2021/formulas/engineering/college/qrcup3ec3xq4ypohaitl2wefhqv2rb9t1z.png)
Hence exit velocity = 306.261 m/sec
Exit Mach number
![M=(u)/(c)=(306.261)/(340.29)=0.9](https://img.qammunity.org/2021/formulas/engineering/college/zx5lh61mi9x4jhz6yg18bwq178f106pw6d.png)
Temperature will remain same as 500 K
Now
Using Bernoulli's equation
![(P_(1))/(\rho g)+(v_(1)^(2))/(2g)+z_(1)=(P_(2))/(\rho g)+(v_(2)^(2))/(2g)+z_(2)](https://img.qammunity.org/2021/formulas/engineering/college/9xbfhuhyhes7qgu9fkrtg80vabjret311v.png)
Here
![z_(1) = z_(2)](https://img.qammunity.org/2021/formulas/engineering/college/vazmunn4zqf7zwv50c8k0brs1lqdze4ftf.png)
![(P_(1))/(\rho g)+(v_(1)^(2))/(2g)-(v_(2)^(2))/(2g)=(P_(2))/(\rho g)](https://img.qammunity.org/2021/formulas/engineering/college/nxa2sfxe2dw6n7555au6mosnin5v3mhy8l.png)
![\Rightarrow (1000000)/(\rho g)+(612.522^(2))/(2g)-(306.261^(2))/(2g)=(P_(2))/(\rho g)](https://img.qammunity.org/2021/formulas/engineering/college/x3k7jdjlxe260c1d491x8xg4ljarykvdcz.png)
![\Rightarrow (1000000)/(1.225)+(612.522^(2))/(2)-(306.261^(2))/(2)=(P_(2))/(1.225)](https://img.qammunity.org/2021/formulas/engineering/college/2t2x4riy3kdo4skmb1rdjjh9z7yhjvj0kf.png)
![\Rightarrow P_(2)=1172.35kPa](https://img.qammunity.org/2021/formulas/engineering/college/rcu4xrk6tcbb0rc583p0wm5ezdln1orrbu.png)