Explanation Given: The initial volume of the system ( V 1 ) is 3 kg. The mass of the air ( m ) is 3 kg. The initial temperature at state 1 ( T 1 ) is 27 ° C . The initial pressure at state 1 ( P 1 ) is 200 kPa . The ratio of volume ( V 3 V 1 ) is 2. The final pressure at state 2 ( P 2 ) is 400 kPa. Calculation: Refer the Table (A-4 through A-6), at initial pressure of 250 kPa , The value of initial specific volume ( ν 1 = ν g @ 250 kPa ) is 0.71873 m 3 / kg The specific internal energy ( u 1 = u g @ 250 kPa ) is 2536.8 kJ / kg . Calculate the initial volume of the air. V 1 = m R T 1 P 1 V 1 = 3 kg × ( 0.287 kPa ⋅ m 3 / kg ⋅ K ) × 27 ° C 200 kPa = 3 kg × ( 0.287 kPa ⋅ m 3 / kg ⋅ K ) × ( 27 + 273 ) K 200 kPa = 1.29 m 3 Write the relation between the initial and final volume. V 3 = 2 V 1 V 3 = 2 × 1.29 m 3 = 2.58 m 3 Calculate the final Temperature ( T 3 ) of hydrogen in the tank. P 1 V 1 T 1 = P 3 V 3 T 3 T 3 = P 3 V 3 P 1 V 1 T 1 T 3 = 400 kPa 200 kPa × 2 × ( 27 ° C ) = 400 kPa 200 kPa × 2 × ( 27 + 273 ) K =1200 K Determine the work done during the constant pressure process ( W b , o u t ) . W b , o u t = ∫ 2 3 P d ν = P ( ν 3 − ν 2 ) W b , o u t = ( 200 kPa ) × ( 2

5th Edition

ISBN: 9780078027680

Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala

Publisher: McGraw-Hill Education

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Properties of Pure Substances

A substance with fixed chemical composition throughout is called a pure substance like water, air, and nitrogen. A pure substance does not have to be a single element or compound. A mixture of two or more phases of a pure substance is also a pure substan…

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Chapter 5, Problem 73P

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The work done by the air and the amount the amount of heat transfer in the process.

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Chapter 5, Problem 72P

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Chapter 5 Solutions

Fundamentals of Thermal-Fluid Sciences

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