(a) Explanation Calculate heat addition in the cycle. Q add = m a ( u 3 − u 2 ) (I) Here, heat addition in the cycle is Q add , mass of air is m a , internal energy at state 3 is u 3 and internal energy at state 2 is u 2 . Write the expression for compression ratio. r = V 1 V 2 (II) Here, compression ratio is r , volume at state 1 is V 1 and volume at state 2 is V 2 . Write the expression for relative volume at state 2. v r 2 = ( V 2 V 1 ) ( v r 1 ) (III) Here, relative volume at state 2 is v r 2 and relative volume at state 1 is v r 1 . Write the expression for volume ratio. V 3 V 4 = 1 ( V 1 V 2 ) (IV) Here, volume at state 3 is V 3 and volume at state 4 is V 4 . Write the expression for relative volume at state 4. v r 4 = v r 3 ( V 3 V 4 ) (V) Here, relative volume at state 4 is v r 4 and relative volume at state 3 is v r 3 . Write the expression for mass of air. m a = P 1 V 1 R T 1 (VI) Here, mass of air is m a , gas constant for air is R . Conclusion: Refer to Table A-22 “Ideal gas properties of air” to obtain internal energy ( u 1 ) as 206.91 kJ/kg and relative volume v r 1 at state 1 as 676.1 Substitute 8 for V 1 V 2 in Equation (II). r = 8 Substitute 1 8 for V 2 V 1 and 676.1 for v r 1 in Equation (III). v r 2 = ( 1 8 ) ( 676.1 ) = 84.5125 ≈ 84.513 Refer to Table A-22 “Ideal gas properties of air” to obtain relative volume ( v r 21 ) as 85.34 , relative volume ( v r 22 ) as 81.89 , internal energy ( u 21 ) as 473.25 kJ/kg and internal energy ( u 22 ) as 481.01 kJ/kg corresponding to relative volume 84.513 . Write the expression for interpolation to obtain internal energy at state 2. u 2 = ( u 21 ) + ( v r 2 − v r 21 v r 22 − v r 21 ) ( u 22 − u 21 ) (VII) Here, internal energy at state 2 is u 2 . Substitute 473.25 kJ/kg for u 21 , 84.513 for v r 2 , 85.34 for v r 21 , 81.89 for v r 22 and 481.01 kJ/kg for u 22 in Equation (VII). u 2 = ( 473.25 kJ/kg ) + ( 84.513 − 85.34 81.89 − 85.34 ) ( 481.01 kJ/kg − 473.25 kJ/kg ) = ( 473.25 kJ/kg ) + ( 0.23971 ) ( 7.76 kJ/kg ) ≈ 475.11 kJ/kg Refer to Table A-22 “Ideal gas properties of air” to obtain relative volume ( v r 21 ) as 85.34 , relative volume ( v r 22 ) as 81.89 , temperature ( T 21 ) as 650 K and temperature T 22 as 660 K corresponding to relative volume 84.513 . Write the expression for interpolation to obtain temperature at state 2. T 2 = ( T 21 ) + ( v r 2 − v r 21 v r 22 − v r 21 ) ( T 22 − T 21 ) (VIII) Here, temperature at state 2 is T 2 . Substitute 84.513 for v r 2 , 85.34 for v r 21 , 81.89 for v r 22 , 650 K for T 21 and 660 K for T 22 in Equation (VIII). T 2 = ( 650 K ) + ( 84.513 − 85.34 81.89 − 85.34 ) ( 660 K − 650 K ) = ( 650 K ) + ( 0.2397 ) ( 10 K ) ≈ 652.397 K Refer to Table A-22 “Ideal gas properties of air” to obtain internal energy ( u 3 ) as 1872.4 kJ/kg and relative volume ( v r 3 ) at state 3 as 2.012 . Substitute 8 for V 1 V 2 in Equation (IV). V 3 V 4 = 1 8 Substitute 2.012 for v r 3 and 1 8 for V 3 V 4 in Equation (V). v r 4 = 2.012 ( 1 8 ) = ( 2 (b) To determine The net work of the cycle. (c) To determine The thermal efficiency of the cycle. (d) To determine The mean effective pressure of the cycle.

8th Edition

ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

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Question

Chapter 9.14, Problem 3P

(a)

To determine

The heat addition in the cycle.

(b)

To determine

The net work of the cycle.

(c)

To determine

The thermal efficiency of the cycle.

(d)

To determine

The mean effective pressure of the cycle.

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FUNDAMENTALS OF ENGINEERING THERMODYNAM

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