(a) Explanation Write the expression for the combustion reaction. C + ( O 2 + 3.76 N 2 ) → C O 2 + 3.76 N 2 (I) Here C is carbon, N is nitrogen, O is oxygen and C O 2 is carbon dioxide Write the expression for mole fraction of carbon dioxide after combustion. y C O 2 = n C O 2 n t (II) Here, n C O 2 is number of mole of carbon dioxide and n t is total number of mole in equation (I) Write the expression for mole fraction of nitrogen after combustion. y N 2 = n N 2 n t (III) Here, n N 2 is number of mole of nitrogen and n t is total number of mole in equation (I) Write the expression for the entropy of carbon dioxide. s ¯ C O 2 = s ¯ o C O 2 − R ¯ ln y C O 2 (IV) Here, s ¯ o C O 2 is the absolute enthalpy of formation of carbon dioxide, R ¯ is the universal gas constant and s ¯ C O 2 is the entropy of carbon dioxide. Write the expression for the entropy of nitrogen. s ¯ N 2 p = s ¯ o N 2 − R ¯ ln y N 2 (V) Here, s ¯ o N 2 p is the absolute enthalpy of formation of nitrogen for product side, R ¯ is the universal gas constant and s ¯ N 2 is the entropy of nitrogen. Write the expression for the mole fraction of oxygen before combustion. y O 2 = n O 2 n t (VI) Here, n O 2 is number of mole of oxygen and n t is total number of mole in equation (I) Write the expression for mole fraction of nitrogen before combustion. y N 2 = n N 2 n t (VII) Here, n N 2 is number of mole of nitrogen and n t is total number of mole in equation (I) Write the expression for the entropy of nitrogen. s ¯ N 2 r = s ¯ o N 2 − R ¯ ln y N 2 (VIII) Here, s ¯ o N 2 r is the absolute enthalpy of formation of nitrogen for reactant side, R ¯ is the universal gas constant and s ¯ N 2 is the entropy of nitrogen. Write the expression for the entropy of oxygen. s ¯ O 2 = s ¯ o O 2 − R ¯ ln y O 2 (IX) Here, s ¯ o O 2 is the absolute enthalpy of formation of oxygen for reactant side, R ¯ is the universal gas constant and s ¯ O 2 is the entropy of nitrogen. Write the expression for the entropy produced from combustion. σ c v n F = [ ( s ¯ C O 2 + 3.76 s ¯ o N 2 p ) − s ¯ f u e l − ( s ¯ O 2 + 3.76 s ¯ o N 2 r ) ] (X) Write the expression for the exergy destruction . E ˙ d n ˙ F = T o σ c v n F (XI) Here, E ˙ d is the exergy destruction, n ˙ F is the mass flow rate of fuel, T o is the temperature and σ c v is the calorific value of fuel. Conclusion: Substitute 1 mol for n C O 2 and 4.76 mol for n t in equation (II). y C O 2 = 1 mol 4.76 mol = 0.210084 Substitute 3.76 mol for n N 2 and 4.76 mol for n t in equation (III). y N 2 = 3.76 mol 4.76 mol = 0.789916 Refer Table A-25 “Thermo chemical properties of selected substance” for the value of s ¯ o C O 2 as 321.8144 kJ / kmol ⋅ K with respect to temperature at T = 500 K and pressure at P = 1 atm . Substitute, 8.314 kJ / kmol ⋅ K for R ¯ , 0.210084 for y C O 2 and 321.8144 kJ / kmol ⋅ K for s ¯ o C O 2 in equation (IV). s ¯ C O 2 = 321.8144 kJ / kmol ⋅ K − 8.314 kJ / kmol ⋅ K ln 0.210084 = 321.8144 kJ / kmol ⋅ K − 8.314 kJ / kmol ⋅ K × ( − 1.56 ) = 321.8144 kJ / kmol ⋅ K − ( − 12.97 kJ / kmol ⋅ K ) = 334.7863 kJ / kmol ⋅ K Refer Table A-25 “Thermo chemical properties of selected substance” for the value of s ¯ o N 2 as 259 (b) To determine The exegetic efficiency of the reactor.

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ISBN: 2818440116926

Author: MORAN

Publisher: WILEY CONS

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Chapter 13.8, Problem 104P

(a)

To determine

The rate of exergy destruction per kmol of carbon.

(b)

To determine

The exegetic efficiency of the reactor.

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Chapter 13.8, Problem 103P

Chapter 13.8, Problem 105P

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