The objective of this exploration is to make you discover the cycle of a gas turbine in the (T, s) entropy thermodynamic chart.
It completes guided exploration (S-M3-V8) where the cycle was presented, with explanations on its settings and its representation in the chart (h, ln (P)) chart.
We will now study the ejector refrigeration cycle.
Click on the following link: Open a file in ThermoptimOpen a file in Thermoptim
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You can also open the diagram using the "Interactive Diagrams" line in the "Special" menu of the simulator screen, which opens an interface that links the simulator and the diagram. Double-click in the field at the top left of this interface to choose the type of diagram desired (here "Ideal gases")..
Once the diagram is open, choose "air" from the protected compound gases substance menu, and select "(T,s)" from the "Chart" menu.
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You can also open this cycle as follows: in the diagram window, choose "Load a cycle" from the Cycle menu, and select "GT_light_25Thin.txt" from the list of available cycles. Then click on the "Connected points" line in the Cycle menu.
The "air inlet" point is located near the abscissa axis, at the intersection with the horizontal isotherm T = 25 ° C and the isobar P = 1 bar. The non-reversible adiabatic compression leads to point 2, on the isobar P = 16 bar.
The heating in the combustion chamber leads to point 3 at the intersection of the isobare P = 16 bar and the isotherm T = 1150 °C.
Evolution (3–4) is a non-reversible adiabatic expansion from 16 bar to 1 bar. The entropy of point 4 is therefore greater than that of point 3.
The Carnot cycle is the one that leads to the best efficiency.
It is given by this formula, T1 and T2 being the temperatures of the hot and cold sources (T1 > T2), expressed in K and not in °C.
We will now examine the temperature differences between the working fluid and the hot and cold sources.
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You can also open this cycle as follows:
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You can also open this cycle as follows:
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You can also open this cycle as follows:
To calculate Carnot efficiency, determine the values in Kelvin of the hot and cold source temperatures, and apply the formula above.
Enter the temperature value of point A in Kelvin
Enter the temperature value of point B in Kelvin
Calculate Carnot's efficiency and enter it
You will now be able to compare the project cycle and the Carnot cycle.
The pressurized air burns at constant pressure with a fuel (light distillate, natural gas).
Given the shape of the isobars in an entropy chart (close to exponentials), the difference is important for this evolution compared to the Carnot cycle which stipulates that the thermal machine exchanges heat at constant temperature with the hot source.
Epansion can be assumed to be adiabatic, but not isentropic. This results in irreversibilities and therefore a difference with the Carnot cycle.
What is the increase in entropy due to expansion?
What would the power of the turbine be if it was isentropic?
Compression can be assumed to be adiabatic, but not isentropic. his results in irreversibilities and therefore a difference with the Carnot cycle.
What would be the compression power if it were isentropic?
What is the increase in entropy due to compression?
In a simple gas turbine, the burnt gases leave the turbine at a temperature close to 500 ° C, and are released into the atmosphere, which constitutes a major source of irreversibility and a notable difference with the Carnot cycle.
This exploration allowed you to familiarize yourself with the representation of the cycle of a gas turbine in the entropy chart and to compare it with the Carnot cycle.
Apparently, the cycle layout presents an anomaly: at the end of the isobaric evolution (2–3) corresponding to the combustion chamber, there is a sudden dropout.
This comes from the fact that the cycle involves several fluids: air, fuel and burnt gases. Strictly speaking, it is impossible to represent it on a single chart.
In this case, points 3 and 4 have the composition of the burnt gases, and their representation on the air chart places them on bad isobars. If we load the chart of the burnt gases, they are on the correct isobars, but this is no longer the case for points 1 and 2.