Answer:
The Mach number M₂ is 0.351636 and the velocity, V₂ is 445.15 ft/s
Explanation:
Here we have
Temperature = 212 °F = 373.15 K
Pressure = 115 psia
Velocity = 250 ft/s = 72.6 m/s
Area, A₁ = 1 ft² = 0.09290304 m²
Mach number, M = V/c
c = √(γRT)
γ for air at 212 °F and 115 psia is 1.03/0.7412 =1.39
R for air = 287.05 J/kg K
Therefore, c = √(1.39·287.05·373.15) = 385.86 m/s
Mach number = V/c = 72.6/385.86 = 0.197
Therefore,
![\frac{A_1}{A^*} = \frac{1}{M_1} [\frac{2}{(\gamma +1)} (1+\frac{(\gamma -1)}{2} M_1^2)]^{\frac{(\gamma +1)}{2(\gamma -1)} }](https://tex.z-dn.net/?f=%5Cfrac%7BA_1%7D%7BA%5E%2A%7D%20%3D%20%5Cfrac%7B1%7D%7BM_1%7D%20%5B%5Cfrac%7B2%7D%7B%28%5Cgamma%20%2B1%29%7D%20%281%2B%5Cfrac%7B%28%5Cgamma%20-1%29%7D%7B2%7D%20M_1%5E2%29%5D%5E%7B%5Cfrac%7B%28%5Cgamma%20%2B1%29%7D%7B2%28%5Cgamma%20-1%29%7D%20%7D)
Where
Which gives the area at the throat as
= 3.09×10⁻² m²
For the section with Area = 85 in² = 0.0548386 m²
We have
From
![\frac{A_1}{A^*} = \frac{1}{M_1} [\frac{2}{(\gamma +1)} (1+\frac{(\gamma -1)}{2} M_1^2)]^{\frac{(\gamma +1)}{2(\gamma -1)} }](https://tex.z-dn.net/?f=%5Cfrac%7BA_1%7D%7BA%5E%2A%7D%20%3D%20%5Cfrac%7B1%7D%7BM_1%7D%20%5B%5Cfrac%7B2%7D%7B%28%5Cgamma%20%2B1%29%7D%20%281%2B%5Cfrac%7B%28%5Cgamma%20-1%29%7D%7B2%7D%20M_1%5E2%29%5D%5E%7B%5Cfrac%7B%28%5Cgamma%20%2B1%29%7D%7B2%28%5Cgamma%20-1%29%7D%20%7D)
Where:
A₂ = 0.0548386 m²
γ = 1.39
= 3.09×10⁻² m² Plugging in the values, we obtain M as
M₂ = 0.351636
Where the speed of sound is given by
c = V₂/M₂ = 385.86 m/s
V₂ = M₂ × 385.86 m/s = 0.351636 × 385.86 m/s = 135.682 m/s
V₂ = 135.682 m/s or 445.15 ft/s
The Mach number M₂ = 0.351636 and the velocity, V₂ = 445.15 ft/s.