Respuesta :
Answer:
At the point of minimum pressure, [tex]\frac{P}{P_{\infty}} = 0.7095[/tex]
Solution:
As per the question;
Critical Mach no. for the air foil, [tex]M_{critical} = 0.8[/tex]
[tex]M_{\infty}[/tex] = 0.8
Now, for the value of the minimum pressure point:
Pressure of the air foil is critical, [tex]P_{critical}[/tex] when [tex]M_{critical} = M_{\infty}[/tex]
Now,
[tex]\frac{P_{critical}}{P_{\infty}} = \frac{P_{critical}}{P}\times \frac{P}{P_{\infty}}[/tex] (1)
When M = 1:
[tex]\frac{P_{critical}}{P} = (\frac{5 - 1}{2}M^{2} + 1)^{- \frac{5}{5 - 1}})[/tex]
[tex]\frac{P_{critical}}{P} = (\frac{5 - 1}{2}1^{2} + 1)^{- \frac{5}{5 - 1}} = 0.2532[/tex] (2)
When [tex]M_{\infty} = 0.8[/tex]:
[tex]\frac{P_{critical}}{P} = (\frac{5 - 1}{2}M_{\infty}^{2} + 1)^{- \frac{5}{5 - 1}})[/tex]
[tex]\frac{P_{critical}}{P} = (\frac{5 - 1}{2}\times 0.8^{2} + 1)^{\frac{5}{5 - 1}} = 2.8016[/tex] (3)
Now, from the eqns (1), (2) and (3):
[tex]\frac{P}{P_{\infty}} = 0.2532\times 2.8016 = 0.7095[/tex]
