|Thermodynamics and Propulsion
This is given in terms of the moles of the different constituents, and it reduces to the more familiar form for a single fluid (say air) with no reactions occurring. We need to specify one parameter as the basis of the solution; 1 kmole of fuel, 1 kmole of air, 1 kmole total, etc. We use 1 kmole of fuel as the basic unit and examine the burning of hydrogen.
The reactants and the products are both taken to be at and , so the inlet and exit and are specified. The control volume is the combustion chamber. There is no shaft work done and the SFEE is in the form of Equation (15.2). The enthalpy of the entering gas is zero for both the hydrogen and the oxygen (elements have enthalpies defined as zero at the reference state). If the exit products are in the gaseous state, the exit enthalpy is therefore related to the enthalpy of formation of the product by:
If the water is in a liquid state at the exit of the process:
There is more heat given up if the products emerge as liquid. The difference between the two values is the enthalpy needed to turn the liquid into gas at : .
A more complex example is provided by the burning of methane (natural gas) in oxygen, producing
The components in this reaction equation are three ideal gases (methane, oxygen, and ) and liquid water. We again specify that the inlet and exit states are at the reference conditions so that:
Suppose the substances which comprise the reactants and the products are not at and . If so, the expression that connects the reactants and products is
Equation (15.4) shows that we must compute the enthalpy difference between the reference conditions and the given state if the inlet or exit conditions are not the reference pressure and temperature.
There are different levels of approximation for the computation:
When doing cycle analysis, do we have to consider combustion products and their effect on specific heat ratio ( is not 1.4)? (MP 15.6)