Liquid octane (C8H18) enters a steady-flow combustion chamber at 25°C and 1 atm at a rate of 0.25 kg/min. It is burned with 50 percent excess air that also enters at 25°C and 1 atm. After combustion, the products are allowed to cool to 25°C. Assuming complete combustion and that all the H2O in the products is in liquid form, determine (a) the heat transfer rate from the combustion chamber, (b) the entropy generation rate, and (c) the exergy destruction rate. Assume that T0 = 298 K and the products leave the combustion chamber at 1 atm pressure.
FIGURE P15–87
(a)
The rate of heat transfer from the combustion chamber.
Answer to Problem 83P
The rate of heat transfer from the combustion chamber is
Explanation of Solution
Write the energy balance equation using steady-flow equation.
Here, the total energy entering the system is
Substitute
Here, the enthalpy of formation for product is
Calculate the molar mass of the
Here, the number of carbon atoms is
Determine the rate of mole flow rates of the product.
Here, the mass flow rate is
Determine the heat transfer rate from the combustion chamber.
Conclusion:
Write the theoretical combustion equation of for
Here, liquid octane is
Calculate the stoichiometric coefficient of air by
Substitute
From the Table-26, “Enthalpy of formation, Gibbs function of formation, and absolute entropy at
Substance | |
-249,950 | |
0 | |
0 | |
-285,830 | |
-393,520 |
Refer Equation (VII), and write the number of moles of reactants.
Here, number of moles of reactant octane, oxygen and nitrogen is
Refer Equation (VII), and write the number of moles of products.
Here, number of moles of product carbon dioxide, water, oxygen and nitrogen is
Substitute the value table (I) of substance in Equation (II).
Therefore the heat transfer for
Substitute 8 for
Substitute
Substitute
Thus, the rate of heat transfer from the combustion chamber is
(b)
The entropy generation rate from the combustion chamber.
Answer to Problem 83P
The entropy generation rate from the combustion chamber is
Explanation of Solution
Write the expression for entropy generation during this process.
Write the combustion equation of Equation (VI)
Here, the entropy of the product is
Determine the entropy at the partial pressure of the components.
Here, the partial pressure is
Determine the entropy generation rate from the combustion chamber.
Conclusion:
Refer Equation (X) for reactant and product to calculation the entropy in tabular form as:
For reactant entropy,
Substance |
(T, 1 atm) | ||||
1 | 1.00 | 360.79 | --- | 360.79 | |
18.75 | 0.21 | 205.14 | -12.98 | 4089.75 | |
70.50 | 0.79 | 191.61 | -1.96 | 13646.69 | |
For product entropy,
Substance |
(T, 1 atm) | ||||
8 | 0.0944 | 213.80 | -19.62 | 1867.3 | |
9 | --- | 69.92 | --- | 629.3 | |
6.25 | 0.0737 | 205.04 | -21.68 | 1417.6 | |
70.50 | 0.8319 | 191.61 | -1.53 | 13616.3 | |
Substitute
Substitute
Thus, the entropy generation rate from the combustion chamber is
(c)
The exergy destruction rate from the combustion chamber.
Answer to Problem 83P
The exergy destruction rate from the combustion chamber is
Explanation of Solution
Write the expression for exergy destruction during this process.
Here, the thermodynamic temperature of the surrounding is
Conclusion:
Substitute
Thus, the exergy destruction rate from the combustion chamber is
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Chapter 15 Solutions
Thermodynamics: An Engineering Approach
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