Explanation Write the formula for velocity of sound at the inlet conditions. c 1 = k R T 1 Here, the specific heat ratio is k , the speed of sound at the inlet condition is c 1 , and the gas constant of air is R . Write formula for the velocity of air at inlet. Ma 1 = V 1 c 1 = V 1 k R T 1 (I) Write the formula of ratio of stagnation temperature to the static temperature at inlet of the nozzle. T 01 T 1 = 1 + k − 1 2 Ma 1 2 T 01 = T 1 ( 1 + k − 1 2 Ma 1 2 ) (II) Here, the inlet static temperature is T 1 , the inlet stagnation temperature is T 01 , the specific heat ratio is k , and Mach number of air at the inlet of nozzle is Ma 1 . Write the formula for stagnation temperature ratio of exit to inlet. T 0 2 T 01 = T 02 / T 0 ∗ T 01 / T 0 ∗ (III) Here, the exit stagnation temperature is T 02 , and the critical stagnation temperature is T 0 ∗ . Write the formula for heat transfer per unit mass ( q ) . q = c p ( T 02 − T 01 ) (IV) Here, the specific heat at constant pressure is c p , the exit stagnation temperature is T 02 , and inlet stagnation temperature is T 01 . Refer Table A-2, “Ideal-gas specific heats of various common gases”. The specific heat ratio ( k ) of air is 1.4 and the specific heat at constant pressure is 1.005 kJ/kg ⋅ K . Refer Table A-34, “Rayleigh flow functions for an ideal gas with k = 1.4 ”. The Rayleigh flow function of inlet stagnation temperature to the critical stagnation temperature corresponding to the inlet Mach number of 0.2494 is 0.2559 . T 01 T 0 ∗ = 0.2559 The Rayleigh flow function of exit stagnation temperature to the critical stagnation temperature corresponding to the exit Mach number of 0.8 is 0.9639 . T 02 T 0 ∗ = 0.9639 Conclusion: Substitute 100 m/s for V 1 , 1.4 for k , 0.2870 kJ/kg ⋅ K for R , and 400 K for T 1 in Equation (I). Ma 1 = 100 m/s ( 1.4 ) ( 0.2870 kJ/kg ⋅ K ) ( 400 K ) = 100 m/s 160.72 kJ/kg × 1000 m 2 / s 2 1 kJ / kg = 100 m/s 400.8989 m/s = 0.2494 Refer Table A-34, “Rayleigh flow functions for an ideal gas with k = 1.4 ”. The Rayleigh flow function of inlet stagnation temperature to the critical stagnation temperature corresponding to the inlet Mach number of 0.2494 is 0.2559 . T 01 T 0 ∗ = 0.2559 Substitute 400 K for T 1 , 1.4 for k , and 0.2494 for Ma 1 in Equation (II). T 01 = 400 K [ 1 + ( 1

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Chapter 17.7, Problem 134RP

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The heat transfer to the air in kJ/kg and the maximum amount of heat transfer without reducing the mass flow rate of air.

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Chapter 17.7, Problem 133RP

Chapter 17.7, Problem 135RP

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180- Dimensions in mm 100 100 D E Steel B Brass 60 kN 40 kN 40-mm diam. 30-mm diam. PROBLEM 2.40 Solve Prob. 2.39, assuming that rod AC is made of brass and rod CE is made of steel. PROBLEM 2.39 Two cylindrical rods, one of steel and the other of brass, are joined at C and restrained by rigid supports at A and E. For the loading shown and knowing that E = 200 GPa and E, 105 GPa, determine (a) the reactions at A and E, (b) the deflection of point C. = R = 45.5 kN ← A

Dimensions in mm 100 100 -180- -120-+- C D Steel B Brass 60 kN 40 kN 40-mm diam. 30-mm diam. E PROBLEM 2.39 Two cylindrical rods, one of steel and the other of brass, are joined at Cand restrained by rigid supports at A and E. For the loading shown and knowing that E = 200 GPa and E₁ = 105 GPa, determine (a) the reactions at A and E, (b) the deflection of point C.

(read image) (Answer: vE = 0.514 m/s)

Chapter 17 Solutions

THERMODYNAMICS LLF W/ CONNECT ACCESS

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 heat transfer to the air in kJ/kg and the maximum amount of heat transfer without reducing the mass flow rate of air.

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The heat transfer to the air in kJ/kg and the maximum amount of heat transfer without reducing the mass flow rate of air.

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