Explanation Write the expression to calculate the mass fractions of the constituent ( mf i ) in a mixture. mf i = m i m m (I) Here, mass of constituent is m i and mass of the mixture is m m . Write the expression to calculate the number of moles of constituent ( N i ) in a mixture. N i = m i M i = mf i × 100 lbm M i (I) Here, the total mass of the mixture is 100 lbm , mass of the gas constituent is m i and the molecular mass of the constitute is M i . Write the expression to calculate the total number of moles of gases ( N m ) in products. N m = N A + N B + .............. + N Z (II) Here, number of moles of compound A is N A , number of moles of compound B is N B , and the number of moles of compound Z is N Z . Write the expression to calculate the mole fraction of a constituent ( y i ) in a mixture. y i = N i N m (III) Write the expression to calculate the partial pressure of the water vapor ( P v ) in the product gases. P v = ( N v N prod ) P prod (IV) Here, number of moles water vapor in products is N v , number of moles of products is N prod , and the partial pressure of the product is P prod . Conclusion: From the Table A-1 of “Molar mass, gas constants, and critical-point properties”, select the molar masses of carbon, hydrogen, oxygen, Sulphur, and nitrogen as, M C = 12 kg / kmol M H 2 = 2 kg / kmol M O 2 = 32 kg / kmol M S = 32 kg / kmol M N 2 = 28 kg / kmol Here -5 lbm is the mass of ask Substitute 61 .40 lbm for m C , and 95 lbm for m m in Equation (I) . mf C = 61 .40 lbm 95 lbm = 0.6463 Substitute 5 .79 lbm for m H 2 , and 95 lbm for m m in Equation (I) . mf C = 5 .79 lbm 95 lbm = 0.6463 Substitute 5 .79 lbm for m O 2 , and 95 lbm for m m in Equation (I) . mf O 2 = 25 .31 lbm 95 lbm = 0.2664 Substitute 5 .79 lbm for m N 2 , and 95 lbm for m m in Equation (I) . mf N 2 = 1 .09 lbm 95 lbm = 0.01147 Substitute 1 .41 lbm for m S , and 95 lbm for m m in Equation (I) . mf S = 1 .41 lbm 95 lbm = 0.01484 Consider for 100 kg of ash. Substitute 0.6463 for mf C and 12 kg / kmol for M C in Equation (II). N C = 0.6463 × 100lbm 12 lbm / lbmol = 5.386 lbmol Substitute 0.06095 for mf H 2 and 2 kg / kmol for M H 2 in Equation (II). N H 2 = 0.06095 × 100lbm 2 kg / kmol = 3.048 lbmol Substitute 0.2664 for mf O 2 and 32 kg / kmol for M O 2 in Equation (II). N O 2 = 0.2664 × 100lbm 32 kg / kmol = 0.8325 lbmol Substitute 0.01147 for mf N 2 and 28 kg / kmol for M N 2 in Equation (II). N N 2 = 0.01147 × 100lbm 28 kg / kmol = 0.04096 lbmol Substitute 0.01484 for mf S and 32 kg / kmol for M S in Equation (II). N S = 0.01484 × 100lbm 32 lbm / lbmol = 0.04638 lbmol Substitute 5.386 lbmol for N C , 3

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ISBN: 9781259822674

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 15.7, Problem 96RP

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The temperature at the dew point for the gas-vapor mixture.

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Chapter 15.7, Problem 95RP

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The temperature at the dew point for the gas-vapor mixture.

Solution Summary: The author calculates the mass tions of the constituent (mf_i) in a mixture.

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The temperature at the dew point for the gas-vapor mixture.

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