Explanation Write the expression for the gas constant in Btu / lb ⋅ °R . R 1 = R ¯ M (I) Here, universal gas constant is R ¯ , gas constant is R 1 and the molecular mass is M . Write the expression for the specific capacity. c p = k R 1 ( k − 1 ) (II) Here, specific heat constant is c p and the other constant is k . Write the expression for the specific heat at constant volume. c v = c p − R 1 (III) Here, specific heat at constant volume is c v . Write the expression for the exit temperature. T e = T 1 ( p e p 1 ) k − 1 k (IV) Here, exit temperature is T e , exit pressure is p e , inlet pressure is p 1 and inlet temperature is T 1 . Write the expression for the final state temperature. T 2 = T 1 ( p 2 p 1 ) k − 1 k (V) Here, final state temperature is T 2 , final state pressure is p 2 , inlet pressure is p 1 and inlet temperature is T 1 . Write the expression for the gas constant in ft ⋅ lbf / lb ⋅ ° R . R 2 = R ¯ M (VI) Here, universal gas constant is R ¯ , gas constant is R 2 and the molecular mass is M . Write the expression for the volume of the tank. V = R 2 W ˙ c v c v ( p 1 − p 2 ) + c p T e ( p 2 T 2 − p 1 T 1 ) (VII) Conclusion: Substitute 1.986 Btu for R ¯ and 28.97 lb ⋅ °R for M in Equation (I). R 1 = 1.986 Btu 28.97 lb ⋅ °R = 0.068553676 Btu / lb ⋅ °R Substitute 0.0685536 Btu / lb ⋅ °R for R 1 and 1.4 for k in Equation (II). c p = ( 0.0685536 Btu / lb ⋅ °R ) 1.4 ( 1.4 − 1 ) = 0.09597514 Btu / lb ⋅ °R 0.4 = 0.2399378 Btu / lb ⋅ °R ≈ 0.24 Btu / lb ⋅ °R Substitute 0.24 Btu / lb ⋅ °R for c p and 0.0685536 Btu / lb ⋅ °R I for R 1 in Equation (III). c v = 0.24 Btu / lb ⋅ °R − 0.0685536 Btu / lb ⋅ °R = ( 0.24 − 0.0685536 ) Btu / lb ⋅ °R = 0.1714 Btu / lb ⋅ °R Substitute 1000 °R for T 1 , 1 atm for p e , 30 atm for p 1 and 1.4 for k in Equation (IV)

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ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

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Chapter 6.13, Problem 167P

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The volume of the tank.

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Chapter 6.13, Problem 166P

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