Is this a spiritual type thing or?
Answer:
h1 = 290.16kj/kg
P = 1.2311
Prandil expression at 8
P=p1/p7×pr
=8(1.2311)
=9.85
Enthalpy state at 8 corresponding to 9.85
h1 = 526.13kj/kg
Now prandtl state at 9 that correspond to 1400k.
h9 = 1515.42kj/kg
Pr = 450.5
Prandtl expression at state 10
P= p10/p9×pr
=1/8(450.5)
=56.31
Enthalpy at state 10 corresponding to prandtl 56.31
h10 = 860.39kj/kg
At 520k
h11 = 523.63kj/kg
Answer You ask your coach
Answer:
C. assembly line workers.
Explanation:
Answer:
The pressure upstream and downstream of a shock wave are related as

where,
= Specific Heat ratio of air
M = Mach number upstream
We know that 
Applying values we get

Similarly the temperature downstream is obtained by the relation
![\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B1%7D%7D%7BT_%7Bo%7D%7D%3D%5Cfrac%7B%5B2%5Cgamma%20M%5E%7B2%7D-%28%5Cgamma%20-1%29%5D%5B%28%5Cgamma%20-1%29M%5E%7B2%7D%2B2%5D%7D%7B%28%5Cgamma%20%2B1%29%5E%7B2%7DM%5E%7B2%7D%7D)
Applying values we get
![\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C](https://tex.z-dn.net/?f=%5Cfrac%7BT_%7B1%7D%7D%7B423%7D%3D%5Cfrac%7B%5B2%5Ctimes%201.4%5Ctimes%201.8%5E%7B2%7D-%281.4-1%29%5D%5B%281.4-1%291.8%5E%7B2%7D%2B2%5D%7D%7B%281.4%2B1%29%5E%7B2%7D%5Ctimes%201.8%5E%7B2%7D%7D%5C%5C%5C%5C%5Ctherefore%20%5Cfrac%7BT_%7B1%7D%7D%7B423%7D%3D1.53%5C%5C%5C%5C%5Ctherefore%20T_%7B1%7D%3D647.85K%3D374.85%5E%7Bo%7DC)
The Mach number downstream is obtained by the relation
