Explanation The fluid properties before the shock is denoted by suffix 1: Write the expression to calculate the stagnation temperature before shock T 01 . T 01 = T 1 + V 1 2 2 c p (I) Here, temperature before shock is T 1 , and specific heat at constant pressure is c p . Write the thermodynamic relation between stagnation pressure and temperature. P 01 P 1 = ( T 01 T 1 ) k / ( k − 1 ) (II) Here, stagnation pressure before shock is P 01 , pressure before shock is P 1 , and ratio of specific heats is k . Write the expression to calculate the velocity of sound in fluid before the shock c 1 . c 1 = k R T 1 (III) Here, and gas constant of air is R . Write the relation between Mach number and fluid velocity before the shock. Ma 1 = V 1 c 1 (IV) Here, Mach number before shock is Ma 1 . The fluid properties after the shock is denoted by suffix 2: Write the expression to calculate the velocity of sound in fluid after the shock. c 2 = k R T 2 (V) Here, velocity of sound before the shock is c 2 , and gas constant of air is R. Write the relation between Mach number and fluid velocity after the normal shock. Ma 2 = V 2 c 2 (VI) Here, velocity of air after the normal shock is V 2 . Conclusion: The properties before the shock: Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the following properties for air at room temperature. c p = 1.005 kJ / kg ⋅ K R = 0.287 kJ / kg ⋅ K k = 1.4 Substitute 230 K for T 1 , 815 m / s for V 1 , and 1.005 kJ / kg ⋅ K for c p in Equation (I). T 01 = 230 K + ( 815 m / s ) 2 2 × 1.005 kJ / kg ⋅ K = 230 K + ( 815 m / s ) 2 2 × 1.005 kJ / kg ⋅ K ( 1 kJ / kg 1000 m 2 / s 2 ) = 230 K + 330 .46 K = 560.46 K Substitute 26 kPa for P 1 , 560.46 K for T 01 , 1.4 for k and 230 K for T 1 in Equation (II). P 01 26 kPa = ( 560.46 K 230 K ) 1.4 / ( 1.4 − 1 ) P 01 = 26 kPa ( 560.46 K 230 K ) 1.4 / ( 0.4 ) P 01 = 587.3 kPa Thus, the stagnation pressure of air before the shock is 587.3 kPa . Substitute 1.4 for k , 0.287 kJ / kg ⋅ K for R , and 230 K for T 1 in Equation (III)

Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684

9th Edition

ISBN: 9781260048667

Author: Yunus A. Cengel Dr.; Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 17.7, Problem 76P

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The stagnation pressure and Mach number before of the shock, and pressure, temperature, velocity, Mach number, and stagnation pressure after of the shock.

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Chapter 17.7, Problem 75P

Chapter 17.7, Problem 77P

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Thermodynamics: An Engineering Approach ( 9th International Edition ) ISBN:9781260092684

Ch. 17.7 - A high-speed aircraft is cruising in still air....Ch. 17.7 - What is dynamic temperature?Ch. 17.7 - Prob. 3P Ch. 17.7 - Prob. 4P Ch. 17.7 - Prob. 5P Ch. 17.7 - Prob. 6P Ch. 17.7 - Calculate the stagnation temperature and pressure...Ch. 17.7 - Prob. 8P Ch. 17.7 - Prob. 9P Ch. 17.7 - Prob. 10P

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The stagnation pressure and Mach number before of the shock, and pressure, temperature, velocity, Mach number, and stagnation pressure after of the shock.

Solution Summary: The author explains the stagnation pressure and Mach number before and after the shock, and the velocity of sound in air after.

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The stagnation pressure and Mach number before of the shock, and pressure, temperature, velocity, Mach number, and stagnation pressure after of the shock.

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