Explanation Write the expression for energy rate balance at the heat exchanger. m ˙ h ( h 4 − h 3 ) = m ˙ c ( h 1 − h 2 ) (I) Here, the mass flow rate of hot stream is m ˙ h , mass flow rate of cold stream is m ˙ c , specific enthalpy of combustion gases at inlet of heat exchanger is h 3 , specific enthalpy of combustion gases at outlet of heat exchanger is h 4 , specific enthalpy at the exit of first turbine is h 1 and specific enthalpy at the inlet of combustor is h 2 . Write the expression for change in exergy from exit to inlet of cold stream. m ˙ c ( e f 4 − e f 3 ) = m ˙ c { ( h 4 − h 3 ) − T o ( s 4 o − s 3 o − R ln p 4 p 3 ) } (II) Here, the specific flow exergy at 3 is e f 3 , the specific flow exergy at 4 is e f 4 , the standard entropy at 3 is s 3 o , the standard entropy at 4 is s 4 o , the pressure at 4 is p 4 , the pressure at 3 is p 3 and the ambient temperature is T o . Write the expression for change in exergy from inlet to exit of combustion gas. m ˙ h ( e f 1 − e f 2 ) = m ˙ h { ( h 1 − h 2 ) − T o ( s 1 o − s 2 o − R ln p 1 p 2 ) } (III) Here, the specific flow exergy at 1 is e f 1 , the specific flow exergy at 2 is e f 2 , the standard entropy at 1 is s 1 o , the standard entropy at 2 is s 2 o , the pressure at 1 is p 1 , the pressure at 2 is p 2 and the ambient temperature is T o . Write the expression for exergetic efficiency of the heat exchanger. ε = m ˙ c ( e f 4 − e f 3 ) m ˙ h ( e f 1 − e f 2 ) (IV) Here, the exergetic efficiency is ε . Conclusion: Refer to Table A-22 “Ideal Gas Properties of Air” to obtain values of specific enthalpy as 1068.89 kJ / kg and value of standard entropy as 2.99034 kJ / kg ⋅ K for temperature of 1020 K . Refer to Table A-22 “Ideal Gas Properties of Air” to obtain values of specific enthalpy as 617.53 kJ / kg and value of standard entropy as 2.42644 kJ / kg ⋅ K for temperature of 610 K . Refer to Table A-22 “Ideal Gas Properties of Air” to obtain values of specific enthalpy as 888.27 kJ / kg and value of standard entropy as 2.79783 kJ / kg ⋅ K for temperature of 860 K . Substitute 1068.89 kJ / kg for h 1 , 617.53 kJ / kg for h 3 , 888.27 kJ / kg for h 4 , 90 kg / s for m ˙ c and 90 kg / s for m ˙ h in Equation (I). 90 kg / s ( 888.27 kJ / kg − 617.53 kJ / kg ) = 90 kg / s ( 1068.89 kJ / kg − h 2 ) ( 1068.89 kJ / kg − h 2 ) = 270.74 kJ / kg h 2 = 798.15 kJ / kg Refer to Table A-22 “Ideal Gas Properties of Air” to obtain values of standard entropy at exit of heat exchanger by using interpolation method of two variables. 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 (V) Here, the variables denote by x and y are specific enthalpy and standard entropy. Show the standard entropy with respect to the specific enthalpy as in below table: Specific enthalpy, kJ / kg Standard entropy, kJ / kg ⋅ K 789.11 kJ / kg 2.67595 kJ / kg ⋅ K 798.15 kJ / kg ? 800.03 kJ / kg 2.69013 kJ / kg ⋅ K Substitute 789.11 kJ / kg for x 1 , 798.15 kJ / kg for x 2 , 800.03 kJ / kg for x 3 , 2.67595 kJ / kg ⋅ K for y 1 , and 2.69013 kJ / kg ⋅ K for y 3 in Equation (V). y 2 = [ ( 798.15 kJ / kg − 789.11 kJ / kg ) ( 2.69013 kJ / kg ⋅ K − 2.67595 kJ / kg ⋅ K ) ( 800.03 kJ / kg − 789.11 kJ / kg ) + 2

Fundamentals of Engineering Thermodynamics, Binder Ready Version

8th Edition

ISBN: 9781118820445

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

Publisher: WILEY

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Chapter 7.7, Problem 100P

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The exergetic efficiency of heat exchanger.

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Chapter 7.7, Problem 99P

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Fundamentals of Engineering Thermodynamics, Binder Ready Version

Ch. 7.7 - Prob. 1E Ch. 7.7 - Prob. 2E Ch. 7.7 - Prob. 3E Ch. 7.7 - 4. A pail of water initially at 20°C freezes when...Ch. 7.7 - Prob. 5E Ch. 7.7 - Prob. 6E Ch. 7.7 - 7. After a vehicle receives an oil change and lube...Ch. 7.7 - Prob. 8E Ch. 7.7 - Prob. 9E Ch. 7.7 - Prob. 10E

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