Liquid octane (C8H18) enters a steady-flow combustion chamber at 25°C and 8 atm at a rate of 0.8 kg/min. It is burned with 200 percent excess air that is compressed and preheated to 500 K and 8 atm before entering the combustion chamber. After combustion, the products enter an adiabatic turbine at 1300 K and 8 atm and leave at 950 K and 2 atm. Assuming complete combustion and T0 = 25°C, determine (a) the heat transfer rate from the combustion chamber, (b) the power output of the turbine, and (c) the reversible work and exergy destruction for the entire process.
(a)
The rate of heat transfer from the combustion chamber.
Answer to Problem 114RP
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:
Perform unit conversion of temperature at state 1 from degree Celsius to Kelvin.
For air temperature enter in the combustion chamber,
Write the combustion equation of 1 kmol for
Here, liquid octane is
Express the stoichiometric coefficient of air by
Substitute
Refer Appendix Table A-18, A-19, A-20, and A-23, obtain the enthalpy of formation, at 298 K, 500 K, 950 K, and 1500 K for
Substance | |||||
-249,950 | --- | --- | --- | --- | |
0 | 14,770 | 8682 | 42,033 | 26,652 | |
0 | 14,581 | 8669 | 40,170 | 28,501 | |
-241,820 | --- | 9904 | 48,807 | 33,841 | |
-393,520 | --- | 9364 | 59,552 | 40,070 |
Refer Equation (X), and write the number of moles of reactants.
Here, number of moles of reactant octane, oxygen and nitrogen is
Refer Equation (X), and write the number of moles of products.
Here, number of moles of product carbon dioxide, water, oxygen and nitrogen is
Substitute the value from 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 rate of power output of the turbine.
Answer to Problem 114RP
The rate of power output of the turbine is
Explanation of Solution
Determine the power output of the adiabatic turbine from the steady-flow energy balance equation for non-reacting gas mixture.
Determine the rate of work done of the adiabatic turbine.
Conclusion:
Substitute the value from table (I) of substance in Equation (XI).
Substitute
Thus, the rate of power output of the turbine is
(c)
The exergy destruction rate from the combustion chamber.
The rate of reversible work done in the combustion chamber.
Answer to Problem 114RP
The exergy destruction rate from the combustion chamber is
The rate of reversible work done in 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.
Write the expression for exergy destruction during this process.
Here, the thermodynamic temperature of the surrounding is
Determine the rate of the reversible work done of the combustion chamber.
Conclusion:
Refer Equation (XV) for reactant and product to calculation the entropy in tabular form as:
For reactant entropy,
Substance |
(T, 1 atm) | ||||
1 | 1.00 | 360.79 | 17.288 | 360.79 | |
37.5 | 0.21 | 220.589 | 4.313 | 8,110.34 | |
141 | 0.79 | 206.630 | 15.329 | 26,973.44 | |
For product entropy,
Substance |
(T, 1 atm) | ||||
8 | 0.0437 | 266.444 | -20.260 | 2,293.63 | |
9 | 0.0490 | 230.499 | -19.281 | 2,248.02 | |
25 | 0.1366 | 241.689 | -10.787 | 6,311.90 | |
141 | 0.7705 | 226.389 | 3.595 | 31,413.93 | |
Substitute
Substitute
Substitute
Thus, the exergy destruction rate from the combustion chamber is
Substitute
Thus, the rate of reversible work done in the combustion chamber is
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Chapter 15 Solutions
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