Explanation Write the chemical reaction equation for the combustion of 1 lbmol methane. CH 4 + 2 ( O 2 + 3.76 N 2 ) → CO 2 + 2 H 2 O + 7.52 N 2 (I) Write the energy balance equation for the combustion of methane. − Q out = ∑ N P ( h ¯ f o + h ¯ − h ¯ o − P v ¯ ) P − ∑ N R ( h ¯ f o + h ¯ − h ¯ o − P v ¯ ) R − Q out = ∑ N P ( h ¯ f o + h ¯ 1520 R − h ¯ 537 R − R u T ) P − ∑ N R ( h ¯ f o − R u T ) R (II) Here, the amount of heat transfer for the combustion of liquid propane is Q out , number of moles of product is N P , number of moles of reactants is N R , enthalpy of vaporization is h ¯ f o , enthalpy of formation at a temperature is h ¯ , the enthalpy of formation at the standard reference state is h ¯ o , pressure is P , specific volume is v ¯ , individual gas constant is R u , and the temperature is T . Conclusion: From the Table A-18 through Table A-26, obtain the enthalpies of vaporization and enthalpies of formation for different substances as in Table 1. Substance Enthalpy of vaporization h ¯ f o ( Btu / lbmol ) Enthalpy of formation at 537 R h ¯ 537 R ( Btu / lbmol ) Enthalpy of formation at 1260 R h ¯ 1260 R ( Btu / lbmol ) CH 4 − 32 , 210 - - O 2 0 3725.1 9096.7 N 2 0 3729.5 8859.3 CO 2 − 169 , 300 4027.5 11,661.0 H 2 O ( g ) − 104 , 040 4258.0 10,354.9 Rewrite the Equation (II) for the heat transfer from combustion process. − Q out = { [ N CO 2 ( h ¯ f o + h ¯ 1260 R − h ¯ 537 R − R u T ) CO 2 + N H 2 O ( h ¯ f o + h ¯ 1260 R − h ¯ 537 R − R u T ) H 2 O + N N 2 ( h ¯ f o + h ¯ 1260 R − h ¯ 537 R − R u T ) N 2 ] P − [ N CH 4 ( h ¯ f o − R u T ) CH 4 + N O 2 ( h ¯ f o − R u T ) O 2 + N N 2 ( h ¯ f o − R u T ) N 2 ] R } (III) Here, number of moles of CO 2 is N CO 2 , number of moles of H 2 O is N H 2 O , number of moles of O 2 is N O 2 , number of moles of N 2 is N N 2 , and number of moles of CH 4 is N CH 4

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ISBN: 9780073398174

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

Publisher: McGraw-Hill Education

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Chapter 15.7, Problem 67P

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Chapter 15.7, Problem 65P

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