Respuesta :
Answer:
The flow area at the location where the Mach number is 0.9 is 25.24 cm²
Explanation:
Here we have for isentropic flow;
[tex]\frac{A}{A^*} = \frac{1}{M}(\frac{2}{k+1} (1+\frac{k-1}{2}M^2))^{(\frac{k+1}{2(k-1)} )[/tex]
Where:
A = Area of flow = 36 cm²
M = Mach number at section of = 1.8
k = Specific heat ratio = 1.4
A* = Area at the throat
Therefore, plugging the values we get
[tex]\frac{36}{A^*} = \frac{1}{1.8}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}1.8^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.439[/tex]
Therefore, A* = 36/1.439 = 25.01769 cm²
Where the Mach number is 0.9, we have
[tex]\frac{A}{25.02} = \frac{1}{0.9}(\frac{2}{1.4+1} (1+\frac{1.4-1}{2}0.9^2))^{(\frac{1.4+1}{2(1.4-1)} ) = 1.009[/tex]
Therefore A = 25.020× 1.009 = 25.24 cm²
The flow area at the location where the Mach number is 0.9 = 25.24 cm².
