Explanation Write equation of the molar mass of the mixture of oxygen and nitrogen. M m = y O 2 M O 2 + y N 2 M N 2 (I) Here, the molar mass of the mixture is M m , the molar mass of the oxygen is M O 2 , molar mass of nitrogen is M N 2 , number of moles of oxygen is y O 2 , and number of moles of nitrogen is y N 2 . Refer Table A-1, “Molar mass, gas constant, and critical-point properties”, obtain the following properties. The molar mass of oxygen ( M O 2 ) is 31.999 kg/kmol ≃ 32 kg/kmol . The molar mass of nitrogen ( M N 2 ) is 28.013 kg/kmol ≃ 28 kg/kmol . Refer Table A-2, “Ideal-gas specific heats of various common gases”, obtain the following properties. The specific heat ratio ( k ) of oxygen is 1.395 ≃ 1.4 . The specific heat ratio ( k ) of nitrogen is 1.4 . Here, the specific heat ratios of oxygen and nitrogen are equal. Hence, the ratios of critical temperature, pressure, density to stagnation temperature, pressure, and density are expressed as follows. Write the formula for the critical temperature of the mixture. T ∗ = T 0 ( 2 k + 1 ) (II) Here, the critical temperature of mixture is T ∗ , the ratio of specific heats is k , and stagnation temperature of mixture is T 0 . Write the formula for the critical pressure of the mixture. P ∗ = P 0 ( 2 k + 1 ) k / ( k − 1 ) (III) Here, the critical pressure of mixture is P ∗ , and the stagnation pressure of mixture is P 0 . Write the formula for the critical density. ρ ∗ = P ∗ R m T ∗ (IV) Here, the critical density of mixture is ρ ∗ , and the gas constant of mixture is R m . The gas constant of mixture ( R m ) is expressed as follows. R m = R u M m (V) Here, the universal gas constant is R u . Refer Table A-1, “Molar mass, gas constant, and critical-point properties”. The universal gas constant ( R u ) is 8.314 kJ/kmol ⋅ K . Conclusion: It is given that the molar mixture is equivalent. Hence, the number of moles of oxygen and nitrogen are as follows. y O 2 = 0.5 y N 2 = 0.5 Substitute 0.5 for y O 2 , 0.5 for y N 2 , 32 kg/kmol for M O 2 , and 28 kg/kmol for M N 2 in Equation (I)

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 129RP

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The critical temperature, pressure, and density for stagnation and pressure of 550 K and 350 kPa for the equimolar mixture of oxygen and nitrogen.

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Chapter 17.7, Problem 128RP

Chapter 17.7, Problem 130RP

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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 critical temperature, pressure, and density for stagnation and pressure of 550 K and 350 kPa for the equimolar mixture of oxygen and nitrogen.

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The critical temperature, pressure, and density for stagnation and pressure of 550 K and 350 kPa for the equimolar mixture of oxygen and nitrogen.

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