The objective of this directed exploration is to make you discover the cycle of a steam power plant in the thermodynamic entropy (T, s) chart.
It completes that exploration(S-M3-V7), where the cycle was presented, with explanations on its settings and its representation in the (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 "Vapors").
Once the diagram is open, choose "water" from the 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 "cycle_lightEnThin.txt" from the list of available cycles. Then click on the "Connected points" line in the Cycle menu.
Points 1 and 2 representing the compression in the liquid state are almost superimposed, and the heating in the liquid state almost coincides with the bubble curve.
The vaporization is done according to a horizontal line segment.
Isobaric overheating corresponds to the maximum peak of the cycle, and irreversible relaxation results in an increase in entropy, point 4 being located in the liquid-vapor equilibrium zone (titer equal to 0.835).
Condensation takes place along the horizontal line segment (4 - 1).
Enter a point in liquid zone
What is the point in two-phase zone?
What is the point with the highest temperature?
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:
What is the temperature difference (water - cold source)?
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You can also open this cycle as follows:
What is the temperature difference (end of superheating - hot source)?
Please note that the hot source temperature must be higher than that of superheating.
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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.
Let us call DeltaThot the difference between the high temperature of the Carnot cycle and the average value of that of water between points 3a and 3
Call DeltaTcold the difference between the low temperature of the Carnot cycle and that of the condenser
Indicate which of these statements is true.
Technologically speaking, with heat exchangers having finite dimensions, the working fluid cannot be at the same temperature as external sources, which constitutes a first difference with the Carnot cycle.
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, as a first approximation, be assumed to be adiabatic, and even isentropic, the irreversibilities taking place in the pump being low. However, we deviate significantly from the Carnot cycle because the temperature at the end of compression remains very close to that of the cold source, instead of that of the hot source
What is the average Delta T between the hot source and the economizer?
This exploration allowed you to familiarize yourself with the representation of the cycle of a steam plant in the entropy chart and to compare it with the Carnot cycle.
Note that water condenses at a temperature higher than that of the cold source