Explanation Write the expression for the energy balance equation. E in − E out = Δ E system (I) Here, the total energy entering the system is E in , the total energy leaving the system is E out , and the change in the total energy of the system is Δ E system . Substitute Q in for E in , W b , o u t for E out , and Δ U for Δ E system in Equation (I) Q i n − W b , o u t = Δ U Q i n = m ( u 3 − u 1 ) + W b , o u t (II) Here, the mass of the piston cylinder device is m , the final specific internal energy is u 3 , the initial specific internal energy is u 1 , and the work done during the process is W b , o u t . Determine the initial volume of the air. ν 1 = m R T 1 P 1 (III) Here, the mass of a piston-cylinder device is m , the initial pressure of a piston-cylinder device is P 1 , the initial temperature is T 1 , and the universal gas constant for air is R . Calculate the final volume which is doubles of initial volume. ν 3 = 2 × ν 1 (IV) Determine the final temperature of an ideal gas for polytrophic process. P 1 ν 1 T 1 = P 3 ν 3 T 3 T 3 = P 3 ν 3 P 1 ν 1 T 1 (V) Here, the initial pressure of air is P 1 , the final pressure of air is P 3 , the initial volume of air is ν 1 , and the final volume of air is ν 3 . Determine the work done during the constant pressure process. W b , o u t = ∫ 1 2 P d ν = P ( ν 3 − ν 1 ) (VI) Conclusion: From the Table A-1, “Molar mass, gas constant, and critical-point properties” to obtain the value of gas constant for air as 0.2870 kJ / kg ⋅ K . Substitute 3 kg for m , 0.2870 kJ / kg ⋅ K for R, 27 ° C for T 1 , and 200 kPa for P 1 in Equation (I). ν 1 = ( 3 kg ) ( 0.2870 kJ / kg ⋅ K ) ( 27 ° C ) 200 kPa = ( 0.861 kJ / kg ⋅ K × ( 1 kPa ⋅ m 3 1 kJ ) ) ( 27 ° C+273 ) 200 kPa = 258

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ISBN: 9781260917055

Author: CENGEL

Publisher: MCG

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Chapter 4.5, Problem 79P

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The work done by the air.

The heat transfer to the air during this process.

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Chapter 4.5, Problem 78P

Chapter 4.5, Problem 80P

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The work done by the air. The heat transfer to the air during this process.