Explanation Express the combustion equation. { ( 0 .65CH 4 + 0.08 H 2 + 0.18 N 2 + 0.03 O 2 + 0.06 CO 2 ) + a th ( O 2 + 3.76 N 2 ) → x CO 2 + y H 2 O + z N 2 } (I) Here, methane is CH 4 , hydrogen is H 2 , oxygen is O 2 , nitrogen is N 2 , carbon dioxide is CO 2 , water is H 2 O and stoichiometric coefficient of air is a th . Perform the mass balance to calculate the coefficients in Equation (I) Carbon balance: x = 0.65 + 0.06 = 0.71 Hydrogen balance: 4 ( 0.65 ) + 2 ( 0.08 ) = 2 y y = 1.38 Oxygen balance: 0.3 + 0.06 + a th = x + y 2 0.3 + 0.06 + a th = 0.71 + 1.38 2 a th = 1.31 Nitrogen balance: 0.18 + 3.76 a th = z 0.18 + 3.76 ( 1.31 ) = z z = 5.106 Substitute 0.71 for x , 1.38 for y , 1.31 for a th and 5.106 for z in Equation (I). { ( 0 .65CH 4 + 0.08 H 2 + 0.18 N 2 + 0.03 O 2 + 0.06 CO 2 ) + 1.31 ( O 2 + 3.76 N 2 ) → 0.71 CO 2 + 1.38 H 2 O + 5.106 N 2 } (II) Express the partial pressure of the moisture air. P v , in = ϕ air P sat@25°C (III) Here, relative humidity of air is ϕ air and saturation pressure at temperature of 25°C is P sat@25°C . Express the amount of moisture that accompanies dry air. 4.76 a th = 4.76 ( 1.31 ) = 6.24 kmol of dry air Express the number of moles of the moisture in the air. N v , in = P v , in P total N total = P v , in P total ( 6.24 + N v , in ) (IV) Here, total pressure is P total and number of total moles is N total . Rewrite Equation (II) as, { ( 0 .65CH 4 + 0.08 H 2 + 0.18 N 2 + 0.03 O 2 + 0.06 CO 2 ) + 1.31 ( O 2 + 3.76 N 2 ) + N v , in H 2 O → 0.71 CO 2 + ( 1.38 + N v , in ) H 2 O + 5.106 N 2 } (V) Express mass of air. m air = 1.31 N air M air + N H 2 O M H 2 O (VI) Here, number of moles of air is N air and molar mass of air and water is M air and M H 2 O respectively. Express mass of fuel. m fuel = N CH 4 M CH 4 + N H 2 M H 2 + N N 2 M N 2 + N O 2 M O 2 + N CO 2 M CO 2 (VII) Here, number of moles of methane, hydrogen, nitrogen, oxygen, and carbon dioxide is N CH 4 , N H 2 , N N 2 , N O 2 and N CO 2 respectively, and molar masses of methane, hydrogen, nitrogen, oxygen, and carbon dioxide is M CH 4 , M H 2 , M N 2 , M O 2 and M CO 2 respectively. Express the theoretical air fuel ratio. AF th = m air m fuel (VIII) Conclusion: Since each kmol of O 2 in air is accompanied by 3 .76 kmol of N 2 , thus N air = 1 kmol O 2 + 3.76 kmol N 2 = 4.76 kmol air Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses. M C = 12 kg / kmol M H 2 = 2 kg / kmol M air = 29 kg / kmol M N 2 = 28 kg / kmol M O 2 = 32 kg / kmol Express molar mass of water. M H 2 O = 2 M H + M O = 2 ( 1 ) + 16 = 18 kg / kmol Express molar mass of carbon dioxide. M CO 2 = M C + 2 M O = 12 + 2 ( 16 ) = 44 kg / kmol Express molar mass of methane

9th Edition

ISBN: 9781260917055

Author: CENGEL

Publisher: MCG

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Chapter 15.7, Problem 30P

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