Concept explainers
A steady-flow combustion chamber is supplied with CO gas at 37°C and 110 kPa at a rate of 0.4 m3/min and air at 25°C and 110 kPa at a rate of 1.5 kg/min. Heat is transferred to a medium at 800 K, and the combustion products leave the combustion chamber at 900 K. Assuming the combustion is complete and T0 = 25°C, determine (a) the rate of heat transfer from the combustion chamber and (b) the rate of exergy destruction.
(a)
The rate of heat transfer from the combustion chamber.
Answer to Problem 93P
The rate of heat transfer from the combustion chamber is
Explanation of Solution
Determine the volume of the CO in the combustion chamber.
Here, the universal gas constant is
Determine the mass flow rate of the CO in the combustion chamber.
Here, the volume flow rate of the CO in the combustion chamber is
Determine the molar air-fuel ratio.
Here, the mass flow rate of the air is
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
Determine the heat transfer per kg of CO.
Here, the molar mass of the CO is
Determine the rate of heat transfer.
Conclusion:
Perform unit conversion of temperature at state 1 from degree Celsius to Kelvin.
For air temperature enter in the combustion chamber,
For CO temperature enter in the combustion chamber,
From the Table A-1 “Molar mass, gas-constant, and critical-point properties”, obtain the value of molar mass for air and carbon monoxide and universal gas constant of CO as:
Substitute
Substitute
Substitute
Here, the number of mole of oxygen used per mole of carbon monoxide is
Write the combustion equation of 1 kmol for
Here, stoichiometric coefficient of air is
From the Table-21, 19, 18, and 20, obtain the enthalpy of formation, at 298 K, 310 K, and 900 K for
Substance | ||||
-110,530 | 8669 | 9014 | --- | |
0 | 8682 | --- | 27,928 | |
0 | 8669 | --- | 26,890 | |
-393,520 | 9364 | --- | 37,405 |
Substitute the value of substance in Equation (V).
Therefore the heat transfer for
Substitute
Substitute
Thus, the rate of heat transfer from the combustion chamber is
(b)
The exergy destruction from the combustion chamber.
Answer to Problem 93P
The exergy destruction from the combustion chamber is
Explanation of Solution
Write the expression for the relation of reversible work using the exergy balance on the combustion chamber.
In the ideal gas table the value of entropy for 1 atm is equal to 101.325 kPa of pressure. Each reactant of the entropy and the product is to be calculated at the partial pressure of the components which is equal to:
Here, the mole fraction of component
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
Write the expression for rate of exergy destruction during this process.
Here, the thermodynamic temperature of the surrounding is
Conclusion:
The entropy calculation can be presented in tabular form as:
For reactant entropy,
Substance |
(T, 1 atm) | ||||
1 | 1.00 | 198.678 | 0.68 | 198.00 | |
0.637 | 0.21 | 205.04 | -12.29 | 138.44 | |
2.400 | 0.79 | 191.61 | -1.28 | 462.94 | |
For product entropy,
Substance |
(T, 1 atm) | ||||
1 | 0.2827 | 263.559 | -9.821 | 273.38 | |
0.137 | 0.0387 | 239.823 | -26.353 | 36.47 | |
2.400 | 0.6785 | 224.467 | -2.543 | 544.82 | |
Substitute
Substitute
Thus, the exergy destruction from the combustion chamber is
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Chapter 15 Solutions
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