(a)
The equilibrium composition of the product gases from combustion of liquid propane.
(a)
Answer to Problem 38P
The equilibrium composition of the product gases from combustion of liquid propane is
Explanation of Solution
Write the stoichiometric equation for combustion of 1 kmol of liquid propane
Write the chemical reaction equation for dissociation of carbon dioxide.
Write the formula to calculate the equilibrium constant
Here, number of moles of
Conclusion:
The stoichiometric coefficients from the Equation (II),
Balance for
Re-write the actual combustion equation for the combustion of 1 kmol of liquid propane
Calculate the total number of moles of products
From the Table A-28 of “Natural logarithms of the equilibrium constant
Substitute
Substitute 3 for
Thus, the equilibrium composition of the product gases from combustion of liquid propane is
(b)
The amount of heat transfer for the combustion of liquid propane.
(b)
Answer to Problem 38P
The amount of heat transfer for the combustion of liquid propane is
Explanation of Solution
Write the energy balance equation to determine the heat transfer
Here, number of moles of products is
Write the expression to calculate the molar flow rate of liquid propane
Here, mass flowrate of propane is
Write the expression to calculate the rate of heat transfer
Conclusion:
Rewrite the Equation (V) for the combustion of liquid propane.
Here, enthalpy of formation at
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 |
Enthalpy of formation at 285 K |
Enthalpy of formation at 298 K |
Enthalpy of formation at 1200 K |
- | - | - | ||
0 | 8696.5 | 8682 | 38,447 | |
0 | 8286.5 | 8669 | 36,777 | |
- | 9904 | 44,380 | ||
- | 9364 | 53,848 |
Substitute
Substitute
Substitute
Thus, the amount of heat transfer for the combustion of liquid propane is
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Chapter 16 Solutions
Thermodynamics: An Engineering Approach
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