Explanation Express the total mass of the coal when the ash is substituted. m total = 100 − m ash (I) Here, mass of ash is m ash . Express the mass fraction of carbon. m f C = m C m total (II) Here, mass of carbon is m C . Express the mass fraction of hydrogen. m f H 2 = m H 2 m total (III) Here, mass of hydrogen is m H 2 . Express the mass fraction of oxygen. m f O 2 = m O 2 m total (IV) Here, mass of oxygen is m O 2 . Express the mass fraction of nitrogen. m f N 2 = m N 2 m total (V) Here, mass of nitrogen is m N 2 . Express the mass fraction of sulphur. m f S = m S m total (VI) Here, mass of sulphur is m S . Express the number of moles of carbon. N C = m f C M C (VII) Here, molar mass of carbon is M C . Express the number of moles of hydrogen. N H 2 = m f H 2 M H 2 (VIII) Here, molar mass of hydrogen is M H 2 . Express the number of moles of oxygen. N O 2 = m f O 2 M O 2 (IX) Here, molar mass of oxygen is M O 2 . Express the number of moles of nitrogen. N N 2 = m f N 2 M N 2 (X) Here, molar mass of nitrogen is M N 2 . Express the number of moles of sulphur. N S = m f S M S (XI) Here, molar mass of sulphur is M S . Express the total number of moles. N m = N C + N H 2 + N O 2 + N N 2 + N S (XII) Express the mole fraction of carbon. y C = N C N m (XIII) Express the mole fraction of hydrogen. y H 2 = N H 2 N m (XIV) Express the mole fraction of oxygen. y O 2 = N O 2 N m (XV) Express the mole fraction of nitrogen. y N 2 = N N 2 N m (XVI) Express the mole fraction of sulphur. y S = N S N m (XVII) Express the partial pressure of water in the product mixture. P v = P sat@50°C (XVIII) Here, saturation pressure at temperature of 50°C . Express the water mole fraction. y v = P v P (XIX) Here, pressure is P . Express actual mole fraction of water in the products. y v = N H 2 O N CO 2 + N H 2 O + N SO 2 + N N 2 + N O 2 (XX) Here, number of moles of carbon dioxide, water, sulfur dioxide, nitrogen and oxygen is N CO 2 , N H 2 O , N SO 2 , N N 2 and N O 2 respectively. Conclusion: 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 O 2 = 32 kg / kmol M S = 32 kg / kmol M air = 29 kg / kmol M N 2 = 28 kg / kmol Here, molar mass of air is M air . Substitute 8.62 for m ash in Equation (I). m total = 100 − 8.62 = 91.38 kg Substitute 79 .61 kg for m C and 91 .38 kg for m total in Equation (II). m f C = 79 .61 kg 91 .38 kg = 0.8712 = 87.12 kg Substitute 4 .66 kg for m H 2 and 91 .38 kg for m total in Equation (III). m f H 2 = 4 .66 kg 91 .38 kg = 0.05100 = 5.10 kg Substitute 4 .76 kg for m O 2 and 91 .38 kg for m total in Equation (IV). m f O 2 = 4 .76 kg 91 .38 kg = 0.05209 = 5.209 kg Substitute 1 .83 kg for m N 2 and 91 .38 kg for m total in Equation (V). m f N 2 = 1 .83 kg 91 .38 kg = 0.02003 = 2.003 kg Substitute 0 .52 kg for m S and 91 .38 kg for m total in Equation (VI). m f S = 0 .52 kg 91 .38 kg = 0.00569 = 0.569 kg Substitute 87 .12 kg for m C and 12 kg / kmol for M C in Equation (VII). N C = 87 .12 kg 12 kg / kmol = 7.26 kmol Substitute 5 .10 kg for m H 2 and 2 kg / kmol for M H 2 in Equation (VIII). N H 2 = 5 .10 kg 2 kg / kmol = 2.55 kmol Substitute 5 .209 kg for m O 2 and 32 kg / kmol for M O 2 in Equation (IX)

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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ISBN: 9781260048636

Author: CENGEL

Publisher: MCG

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Chapter 15.7, Problem 97RP

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The fraction of the water in the combustion products that is liquid and the fraction that is vapor.

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Chapter 15.7, Problem 96RP

Chapter 15.7, Problem 98RP

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CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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The fraction of the water in the combustion products that is liquid and the fraction that is vapor.

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The fraction of the water in the combustion products that is liquid and the fraction that is vapor.

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