Explanation The figure shown below is a schematic representation of the processes. Figure (1) Write the expression for mass flow rate of saturated vapor using the energy balance. m ˙ W n ˙ g a s = [ y CO 2 ( h ¯ CO 2 ( T 3 ) − h ¯ CO 2 ( T 4 ) ) + y H 2 O ( h ¯ H 2 O ( T 3 ) − h ¯ H 2 O ( T 4 ) ) + y N 2 ( h ¯ N 2 ( T 3 ) − h ¯ N 2 ( T 4 ) ) ] ( h 2 − h 1 ) (I) Here, the mass flow rate of vapor is m ˙ , the total moles of mixture is n ˙ , the mole fraction of carbon dioxide is y CO 2 , the mole fraction of nitrogen is y N 2 , the molar specific enthalpy is h ¯ and the molar specific enthalpy is h . Conclusion: Refer to the Table A-23 “Ideal gas properties of selected gases” to obtain the specific enthalpy of carbon dioxide as 71078 kJ / kmol at 1500 K , the specific enthalpy of carbon dioxide as 22280 kJ / kmol at 600 K the specific enthalpy of water as 57999 kJ / kmol at 1500 K , the specific enthalpy of water as 20402 kJ / kmol at 600 K the specific enthalpy of nitrogen as 47073 kJ / kmol at 1500 K , the specific enthalpy of nitrogen as 17563 kJ / kmol at 600 K . Refer to the Table A-3 “properties of saturated water pressure table” at 25 bar to obtain the value of specific enthalpy of fluid as 962.11 kJ / kg and the value of specific enthalpy of vapor as 2803.1 kJ / kg . Substitute 0.1 for y CO 2 , 0.2 for y H 2 O , 0

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 12.9, Problem 19P

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The mass flow rate of exiting saturated vapor gas mixture in kg / kmol .

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