Explanation Determine the inlet stagnation temperature. T 01 = T 1 ( 1 + k − 1 2 Ma 1 2 ) (I) Here, the Mach number is Ma and the specific heat ratio is k . Determine the inlet stagnation pressure. P 01 = P 1 ( 1 + k − 1 2 Ma 1 2 ) k / ( k − 1 ) (II) Determine the exit stagnation temperature from energy equation. Q ˙ = m ˙ air c p ( T 02 − T 01 ) T 01 = T 02 + Q ˙ m ˙ air c p (III) Here, the rate of heat transfer to the duct is Q ˙ and the mass flow rate of the air is m ˙ air . Determine the maximum value of stagnation temperature is T 0 ∗ . T 0 ∗ = T 01 T 01 / T 0 ∗ (IV) Determine the stagnation temperature ratio at the exit. Ratio stagnation = T 02 T 0 ∗ (V) Here, the maximum value of stagnation temperature is T 0 ∗ . Determine the stagnation pressure in the duct. P 02 P 01 = P 02 / P 0 ∗ P 01 / P 0 ∗ (VI) Here, the ratio of Rayleigh flow for inlet pressure is P 01 / P 0 ∗ and the ratio of Rayleigh flow for outlet pressure is P 02 / P 0 ∗ . Determine the pressure drop in the duct. Δ P 0 = P 01 − P 02 (VII) Conclusion: From the Table A-2, “Ideal-gas specific heats of various common gases” to obtain value of universal gas constant, specific heat of pressure, and the specific heat ratio of air at 300 K temperature as 0.287 kJ / kg ⋅ K , 1.005 kJ / kg ⋅ K and 1.400 . Substitute 700 K for T 1 , 1.4 for k , and 0.2 for Ma in Equation (I). T 01 = ( 700 K ) ( 1 + ( 1.4 − 1 ) 2 ( 0.2 ) 2 ) = ( 700 K ) ( 1 + 0.008 ) = 705.6 K Substitute 600 kPa for P 1 , 1.4 for k , and 0.2 for Ma in Equation (II). P 01 = ( 600 kPa ) ( 1 + ( 1.4 − 1 ) 2 ( 0.2 ) 2 ) 1.4 / ( 1.4 − 1 ) = ( 600 kPa ) ( 1 + 0.008 ) 1.4 / ( 0.4 ) = 616.96 kPa ≅ 617.0 kPa Substitute 0.3 kg / s for m ˙ air , 705.6 K for T 01 , 300 kJ / s for Q ˙ and 1.005 kJ / kg ⋅ K for c p in Equation (III). T 02 = 705.6 K + 300 kJ / s ( 0

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 102P

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The Mach number at the duct exit.

The drop in stagnation pressure P01−P02 during this process.

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Chapter 17.7, Problem 101P

Chapter 17.7, Problem 103P

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Chapter 17 Solutions

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 Mach number at the duct exit. The drop in stagnation pressure P 01 − P 02 during this process.

Solution Summary: The author analyzes the Mach number at the duct exit and the drop in stagnation pressure during this process.

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The Mach number at the duct exit.

The drop in stagnation pressure P01−P02 during this process.

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Chapter 17 Solutions

The Mach number at the duct exit. The drop in stagnation pressure P 01 − P 02 during this process.

Solution Summary: The author analyzes the Mach number at the duct exit and the drop in stagnation pressure during this process.

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The Mach number at the duct exit. The drop in stagnation pressure P01−P02 during this process.

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Chapter 17 Solutions

The Mach number at the duct exit.

The drop in stagnation pressure P01−P02 during this process.