a) p₀ = 1.2484 atm
b) T₀ = 274.5 K
c) p* = 0.6595 atm
d) T* = 228.74 K
e) a* = 303.16 m/s
Explanation:
Given that;
Local Mach number M = 0.7
Pressure P = 0.9 atm
Temperature T = 250 K
First we obtain the following ratios corresponding to Mach Number 0.7 from Appendix(A) table { ISENTROPIC FLOW PROPERTIES }
For M = 0.7; Pressure→ P/P₀ = 0.7209, Temperature; T/T₀ = 0.91075
Now we calculate;
a) value of p₀
we that; P/P₀ = 0.7209
p₀ = p / 0.7209
we substitute
p₀ = 0.9 / 0.7209
p₀ = 1.2484 atm
b) value of T₀
we know that;
T/T₀ = 0.91075
T₀ = T / 0.91075
we substitute
T₀ = 250 / 0.91075
T₀ = 274.5 K
c) value of p*
To obtain value of p*, we say;
p* = 0.52828 × 1.2484
p* = 0.6595 atm
d) value of T*
To obtain value of T*, we say;
T* = 0.8333 × 274.5
T* = 228.74 K
e) value of a*
we know that;
a* = √(γRT*)
so
a* = √(1.4 × 287 × 228.74)
a* = √(91907.732)
a* = 303.16 m/s