(a) Explanation The Figure-(1) shows the process of dual cycle on the P - v diagram. Figure-(1) Write the expression for the first law of thermodynamics for the heat addition process from state 2 to state 3. q 23 = ( u 3 − u 2 ) + w 23 (I) Here, the heat addition from the process 2-3 is q 23 , the internal energy at point 3 is u 3 , the internal energy at point 2 is u 2 and the work addition from 2-3 is W 23 . Since the work done from process 2-3 is zero so, w 23 = 0 Substitute 0 for W 23 , 1400 kJ / kg for q 23 in Equation (I). q 23 = m ( u 3 − u 2 ) (II) Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (III) Here, the variables denote by x and y are specific internal energy and specific enthalpy at given condition. Formula for the isentropic compression is, V r 2 = 1 r V r 1 (IV) Here, the compression ratio is r and the relative volume at state 2 is V r 2 . Write the expression for the heat added in the process 2-3. q 23 = 2 3 q (V) Here, the total heat added is q and the heat added in the process 2-3 is q 23 . Write the expression for the heat addition process from 3-4. q 34 = h 4 − h 3 (VI) Here, the heat added is q 34 . Calculate the heat added in the process 3-4. q 34 = 1 3 q (VII) Write the expression for the isentropic expansion at state 4-5. V r 5 = r × T 3 T 4 × V r 4 (VIII) Conclusion: Refer to Table A-22 “Ideal Gas Properties of Air” to obtain the specific internal energy at temperature 300 K as, u 1 = 214.07 kJ / kg Here, the specific internal energy at state 1 is u 1 . Refer to Table A-22 “Ideal Gas Properties of Air” to obtain the critical volume at temperature 300 K as, V r 1 = 621.2 Here, the critical volume is V r 1 Substitute the given values in the Equation (III). V r 2 = 1 9 ( 621.2 ) = 69.02 Refer to Table A-22 “Ideal Gas Properties of Air” to obtain the specific internal energy at critical volume 69.02 as, u 2 = 514.49 kJ / kg Refer to Table A-22 “Ideal Gas Properties of Air” to obtain the temperature at critical volume 69.02 as, T 2 = 720.7 K Substitute 1400 kJ / kg for q in Equation (IV). q 2 − 3 = 2 3 ( 1400 kJ / kg ) = 933.33 kJ / kg Substitute 933.33 kJ / kg for Q 23 and 514.49 kJ / kg for u 2 in Equation (II). 933.33 kJ / kg = ( u 3 − 514.49 kJ / kg ) u 3 = 933.33 kJ / kg + 514.49 kJ / kg u 3 = 1447.82 kJ / kg Refer to the Table A22 “Ideal Gas Properties of Air” to obtain the Temperature at specific enthalpy 1447.82 kJ / kg as, T 3 = 1758.5 K Thus, the temperature at state 3 is 1758.5 K . Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (IX) Here, the variables denote by x and y are specific internal energy and specific enthalpy at given condition. Show the specific internal energy in Table (1). Specific internal energy, ( kJ / kg ) Specific Enthalpy ( kJ / kg ) 1439.8 1941.6 1447.82 ? 1487.2 2003.3 Calculate the specific enthalpy at the given condition. Substitute 1439.8 kJ / kg for x 1 , 1447.82 kJ / kg for x 2 , 1487.2 kJ / kg for x 3 , 1941.6 kJ / kg for y 1 , and 2003.3 kJ / kg for y 3 in Equation (IX). y 2 = ( 1447.82 kJ / kg − 1439.8 kJ / kg ) ( 2003.3 kJ / kg − 1941.6 kJ / kg ) ( 1487.2 kJ / kg − 1439.8 kJ / kg ) + 1941.6 kJ / kg = 10.44 kJ / kg + 1941.6 kJ / kg = 1952.04 kJ / kg Write the specific enthalpy at state 3. h 3 = 1952 (b) To determine The work done by the cycle per unit mass of the air. (c) To determine The thermal efficiency. (d) To determine The mean effective pressure.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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Question

Chapter 9.14, Problem 34P

(a)

To determine

The temperature at the end of each heat addition process.

(b)

To determine

The work done by the cycle per unit mass of the air.

(c)

To determine

The thermal efficiency.

(d)

To determine

The mean effective pressure.

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Chapter 9.14, Problem 30P

Chapter 9.14, Problem 35P

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