Simple gas turbine, compression ratio 16, turbine inlet temperature 1065 °C
In the Thermoptim-UNIT portal, you will find a summary presentation of the combustion chambers and a Diapason session on combustion.
The complete combustion of a CHa formula fuel with pure oxygen would be governed by the equation below: it takes 1 oxygen molecule to form CO2, and a/4 oxygen molecules to form a/2 H2O.
However, in energy systems, combustion is almost always made with air as an oxidizer.
Since dry air is approximately 79% nitrogen and 21% oxygen, the nitrogen/oxygen ratio is 79/21 = 3.762.
That's why the air can be represented by O2 + 3.762 N2.
The complete combustion with air of a CHa formula fuel is therefore governed by the equation below:
It is the one that leads to the highest end-of-combustion temperature. We will consider it as a reference combustion.
These equations mean that the stoichiometric combustion of a CHa fuel mole requires (1 + a/4) oxygen moles and produces 1 mole of carbon dioxide and a/2 moles of water.
If the oxidizer is air, 3.762 (1 + a/4) moles of nitrogen are also involved, but because they do not react with the fuel, they end up in the burnt gases. Nitrogen is said to remain inert.
When the combustion is not stoichiometric, it can be defined in several ways:
Either by excess air e, which as the name suggests, represents the amount of excess air
Or by the air factor lambda, which is equal to 1 + excess air
Either by the R richness, ratio of the number of moles (or mass) of fuel contained in a specified amount of mixture, to the number of moles (or mass) of fuel in the stoichiometric mixture.
We will use lambda preferentially in the following, because that is the air multiplier factor in the combustion equation.
This small calculator allows you to study the influence of a and lambda on this equation. The table below shows you the number of moles of the products and their total, how the molar fractions are calculated, and their values for the chosen setting.You can test the two oxidizers used previously, oxygen or atmospheric air.
If this calculator doesn't open in your Thermoptim browser, you can access it via this link: https://thopt.s4e2.com:8443/CalcCombInitEn/index.jsp
It suffices to replace lambda by its value.
Loading the model is done by opening the diagram file and a properly
configured project file.
Start by loading the model, then complete the other suggested
activities.
Click on the following link: Open a file in Thermoptim
You can also:
The model appears in the diagram editor.
We find the air inlet, the compressor, the combustion chamber, to
which is connected a process-point entitled "fuel", and the expansion
turbine.
The three substances put into play in the model are air,
methane (CH4 'methane), and fumes (burnt gases).
Double-click on the "combustion chamber" process.
If you are unable to do so, click this button
Here is the screen of the combustion chamber.
The options "set pressure" and "by the upstream point" are selected,
which means that the downstream point "3" pressure will be equal to
that of the upstream point "2", i.e. 16 bar.
The efficiency of the combustion chamber is equal to 1, which means
that it is not cooled: all the energy released by the combustion is
used to heat the burnt gases.
Finally, the "dissociation" option is not checked, which means that
the combustion is complete: there will be no unburnt gas in the fumes.
Once the process is calculated, the results obtained are as
follows:
It is easy to verify that this composition is that given by the methane combustion equation, using the small calculator on the previous page.
It is enough to replace lambda with its value, the decimal separator being the point.
The values provided by Thermoptim are slightly different due to the presence of argon in the air, whereas the previous equation did not include any, this gas being counted in nitrogen.
The figure below shows the synoptic view of the initial model where the combustion chamber was replaced by an "exchange" process.
We note that the efficiencies are roughly the same: 32% for the
initial cycle, and 32.57% for the new one, i.e. a difference of less
than 2%.
The useful power output was 230 kW while it is 255 kW with the
combustion chamber, which represents an increase of almost 11%.
How can we explain such a difference ?
Click this button
You can also open the chart using the line “Interactive charts” of
the “Special” menu of the simulator screen, which opens an interface
that links the simulator and the charts. Double-click in the field
located at the top left of this interface to choose the type of
chart desired (here “Ideal gases”).
Once the chart is open, select “(h, p)” in the “Chart” menu, and
click on the line “Load a gas from the data base” in the Gas Editor menu,
and choose “burnt gases” among the Protected compound gases.
If you do not have access to the gas turbine (h, P) charts, open
the Help/Global settings menu in the Thermoptim simulator screen,
check the "(h, P) chart for gas" at the bottom on the left, then
close the window by clicking on "OK".
Restart Thermoptim. (h, P) charts should be displayable.
If the window is displayed correctly, but without showing the
iso-value lines of the chart, it is undoubtedly that the display
interval of the values is not the right one, because of the
previous settings.
It is possible to modify the calculation and display values by
adjusting the parameters in the chart menus:
Click this button
The two cycles are superimposed for the phases of compression (1-2) and heating (2-3), but the two points 3 are shifted, which is surprising at first glance.
This is because their enthalpies are different, even though their pressure and temperature are the same.
We have represented the two cycles in a single chart, but, strictly speaking, we are not allowed to do so for the cycle with combustion chamber, since points 1 and 2 do not correspond to the same fluid as points 3 and 4.
You can, if you wish, see what becomes of the plot of these two
cycles in the entropy chart.
All you have to do is activate line (T, s) in the Chart menu.
The chart that appears corresponds to the image below.
Points 3 are at the same temperature, but their entropies are
different, which explains why they do not have the same abscissa.
The curve (2-3) of the heating in the combustion chamber has no
physical meaning, due to the change in the composition of the working
fluid.
This is why it does not follow the isobar P = 16 bar.
This disturbed plot comes, as we have said, from the fact that we
should not strictly represent this cycle in a chart relating to a
single fluid.
If we had displayed the cycles in the air chart, the curves (2-3) and
(3-4) would be even further removed from reality.
Let us recall the screen of the combustion chamber, whose specific
settings appear in the lower right part.
If we consider that the combustion reaction is not complete, it is possible to take into account the dissociation of part of the CO2 into CO and O2.
Here is what you get.
But above all, the combustion efficiency is no longer equal to 1. It
has dropped a little because of the presence of unburnt gas.
If you display the new smoke composition, here is the result.
If you look at the overall efficiency of the gas turbine set in this
way, you will see that it has increased very slightly, which may seem
paradoxical because there are unburnt gases.
The reason is that the balance calculated by Thermoptim does not take
into account the LHV lost by the fumes.
It is determined by considering only the energy released by
combustion, excluding that which remains in the flue gases.
Refer to volume 2 of the software package reference manual for all the possible settings of a combustion chamber.
It is that which is taken into account in the equation corresponding to the law of mass action.
We will assume that it is 950 ° C.
We must also enter the value of the dissociation rate, which we will assume to be 0.02, that is to say 2%.
Then recalculate the "combustion chamber" process.