Explanation Write the expression ratio of static temperature at exit to the temperature stagnation T e T 0 = 2 2 + ( k − 1 ) Ma 2 T e = T 0 ( 2 2 + ( k − 1 ) Ma 2 ) (I). Here, the exit temperature is T e , stagnation temperature is T 0 , ratio of specific heats of nitrogen is k , and Mach number is Ma . Write the expression ratio of static pressure at exit to the pressure stagnation P e P 0 = ( T e T 0 ) k / ( k − 1 ) P e = P 0 ( T e T 0 ) k / ( k − 1 ) (II). Here, the exit pressure is P e , and stagnation pressure is P 0 . Write the expression for the stagnation density of the flow ( ρ 0 ) . ρ 0 = P 0 R T 0 (III). Write the expression for the exit velocity of nitrogen through the nozzle after the normal shock. V 2 = Ma 2 k R T 2 (IV). Here, exit velocity of nitrogen after the normal shock is V 2 . Write the expression for the mass flow rate of air at the inlet. m ˙ = ρ e A e V e (V). Here, mass flow rate at the exit is m ˙ , density of air at the exit is ρ e , area of the combustion chamber at the exit is A e , and velocity of air at the exit is V e . Conclusion: From Table A-1E, “Molar mass, gas constant, and critical-point properties”, obtain the following properties for air. k = 1.4 R = 0.06855 Btu / lbm ⋅ R or 0.3704 psia ⋅ ft 3 / lbm ⋅ R Substitute 1.4 for k , 2 for Ma , and 100 ° F for T 0 in Equation (I). T e = 100 ° F ( 2 2 + ( 1.4 − 1 ) ( 2 ) 2 ) = ( 100 + 460 ) R ( 2 2 + ( 1.4 − 1 ) ( 2 ) 2 ) = 311 R Thus, the pressure for a Mach number Ma = 2 is 311 R . Substitute 1.4 for k , 100 ° F for T 0 , 311 R for T e and 150 psia for P 0 in Equation (II). P e = 150 psia ( 311 R 100 ° F ) 1

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Chapter 17.7, Problem 65P

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The pressure, temperature, velocity, and mass flow rate in the test section for a Mach number Ma = 2.

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Chapter 17.7, Problem 64P

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