Explanation Write the expression for the velocity at nozzle exit. V 2 = V 1 2 + ( h 1 − h 2 ) (I) Here, the velocity at the nozzle inlet is V 1 , the specific enthalpy at the nozzle inlet is h 1 and the specific enthalpy at the nozzle exit is h 2 . Write the expression for the reduced temperature at inlet. T R 1 = T 1 T C (II) Here, the critical temperature of the gas is T C and temperature of the gas at initial state is T 1 . Write the expression for the reduced temperature at exit. T R 2 = T 2 T C (III) Here, temperature of the gas at final state is T 2 . Write the expression for the reduced pressure at initial state. p r 1 = p 1 p C (IV) Here, the initial pressure is p 1 and the critical pressure is p C . Write the expression for the reduced pressure at final state. p r 2 = p 2 p C (V) Here, the final pressure is p 2 . Write the expression for the entropy departure value at final state. s ¯ o ( T 2 ) − R ¯ [ ( s ¯ * − s ¯ ) R ¯ ] 2 = s ¯ o ( T 1 ) + R ¯ ln p 2 p 1 − R ¯ [ ( s ¯ * − s ¯ ) R ¯ ] 1 (VI) Here, the characteristic gas constant is R ¯ , the entropy departure values corresponds to final temperature is s ¯ o ( T 2 ) , the entropy departure values corresponds to initial temperature is s ¯ o ( T 1 ) , entropy departure at reduced pressure and temperature at initial state is [ ( s ¯ * − s ¯ ) R ¯ ] 1 and entropy departure at reduced pressure and temperature at final state is [ ( s ¯ * − s ¯ ) R ¯ ] 2 . Write the expression for the specific enthalpy between two states. h ¯ 1 − h ¯ 2 = h ¯ * 1 − h ¯ 2 − R ¯ T C [ ( h ¯ * − h ¯ R ¯ T C ) 1 − ( h ¯ * − h ¯ R ¯ T C ) 2 ] (VII) Here, the enthalpy departure value corresponds to initial temperature is h ¯ * 1 , the enthalpy departure value corresponds to final temperature is h ¯ 2 , enthalpy departure at reduced pressure and temperature at initial state is ( h ¯ * − h ¯ R ¯ T C ) 1 and enthalpy departure at reduced pressure and temperature at final state is ( h ¯ * − h ¯ R ¯ T C ) 2 . Conclusion: Refer to table A-1 “Atomic molecular weight and critical properties of selected elements and compounds” to obtain the critical properties of the oxygen as molecular weight M = 32 kg / k ⋅ mol , critical temperature T C = 154 K and critical pressure p C = 50.5 bar . Substitute 300 K for T 1 and 154 K for T C in Equation (II). T R 1 = 300 K 154 K = 1.9480 ≈ 1.95 Substitute 60 bar for P 1 and 50.5 bar for p C in equation (IV). p r 1 = 60 bar 50.5 bar = 1.1881 ≈ 1.19 Substitute 30 bar for P 2 and 50.5 bar for p C in Equation (V). p r 2 = 30 bar 50.5 bar = 0.59 Refer to table A-23 “Ideal gas properties of the selected gases” to obtain the enthalpy departure at reduced pressure p r 1 = 1.19 and the reduced temperature T R 1 = 1.95 as 0.35 . Refer to figure A-5 “Generalized entropy correction chart” to obtain the entropy departure at reduced pressure p r 1 = 1.19 and the reduced temperature T R 1 = 1.95 as 0.14 . Here, specific entropy at initial state is equal to the final state. Refer to table A-23 “Ideal gas properties of the selected gases” to obtain the entropy departure value at temperature T 1 = 300 K as 205

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 11.9, Problem 91P

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Chapter 11.9, Problem 90P

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Chapter 11 Solutions

Fundamentals of Engineering Thermodynamics

Ch. 11.9 - Prob. 1E Ch. 11.9 - Prob. 2E Ch. 11.9 - 3. What is an advantage of using the Redlich–Kwong...Ch. 11.9 - To determine the specific volume of superheated...Ch. 11.9 - Prob. 5E Ch. 11.9 - Prob. 6E Ch. 11.9 - Prob. 7E Ch. 11.9 - Prob. 8E Ch. 11.9 - Prob. 9E Ch. 11.9 - Prob. 10E

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