Engine Efficiency Cycle: 500K-300K, Isothermal Compression & Expansion

[nokia8650]

process A: isothermal compression at low temperature with an input of work of 83 J
process B: constant volume increase in pressure with an energy input by heating of 200 J
process C: isothermal expansion at high temperature with work output of 139 J
process D: constant volume cooling to the original pressure, volume and temperature

In this cycle, the energy input in process B is the same as the energy rejected in
process*D.

Complete the table by applying the first law of thermodynamics to each process and to the whole cycle. Which I have done:

http://img12.imageshack.us/img12/8733/21144137.jpg
http://g.imageshack.us/img12/21144137.jpg/1/

The highest and lowest temperatures of the air during the cycle are 500 K and 300 K. Show that the thermal efficiency of the ideal cycle is equal to the maximum possible efficiency for any heat engine working between these temperature limits.


For this part, I am fine with calculating the maximum possible efficiency using the temperatures given. However, the markscheme for finding the actual thermal efficiency of the ideal cycle says:

http://img10.imageshack.us/img10/3898/15977913.jpg
http://g.imageshack.us/img10/15977913.jpg/1/


Can someone please explain the reasons for using 56 and 139 - I would have thought that 139J of work is done, since this much work is done during the isothermal expansion, with the input of 200 + 139 = 339J, since this is the total energy input. Can someone please explain what is wrong with this reasoning.

Thank you very much.

 

[anathema]

That is a good question. In order to understand the reasoning behind using 56 and 139, it is important to understand the concept of work and energy in thermodynamics. In thermodynamics, work is defined as the transfer of energy from one system to another. This transfer of energy can occur in various forms, such as mechanical, electrical, or thermal energy. In the case of a thermodynamic cycle, work is done on the system during the compression process, and work is done by the system during the expansion process.

In the given cycle, process A and process C are isothermal processes, meaning that the temperature of the system remains constant during these processes. During an isothermal process, the work done is equal to the change in internal energy of the system. Therefore, in process A, the work done on the system is 83 J, and in process C, the work done by the system is 139 J.

On the other hand, process B and process D are constant volume processes, meaning that the volume of the system remains constant during these processes. In a constant volume process, no work is done by the system or on the system. However, in process B, energy is input into the system through heating, and in process D, energy is rejected from the system through cooling. Therefore, the energy input in process B is equal to the energy output in process D, which is 200 J.

Now, to calculate the thermal efficiency of the ideal cycle, we need to consider the total energy input and output of the system. In this case, the total energy input is 200 J (from process B) and 139 J (from process C), and the total energy output is 83 J (from process A) and 200 J (from process D). Therefore, the net energy input is 339 J (200 J + 139 J), and the net energy output is 283 J (83 J + 200 J). Using these values in the formula for thermal efficiency, we get:

Thermal efficiency = (net energy output/net energy input) * 100% = (283 J/339 J) * 100% = 83.5%

This value is different from the value obtained in the markscheme, which is 56%. The reason for this difference is that the markscheme is calculating the thermal efficiency for the ideal cycle, while you were calculating the thermal efficiency for the actual cycle. In the ideal cycle, the

 

[nlightner]

The reason for using 56 J and 139 J in the calculation of the thermal efficiency is because these values represent the net work done during the isothermal expansion and compression processes respectively. This means that the values take into account the work done by the system (positive values) and the work done on the system (negative values).

In the isothermal compression process, the input of 83 J of work is used to compress the gas, resulting in a decrease in the internal energy of the system (since work is being done on the system). This means that the net work done in this process is -83 J, which is why this value is used in the calculation of the thermal efficiency.

Similarly, in the isothermal expansion process, the output of 139 J of work is the net work done by the system, since the gas is expanding and doing work on its surroundings. This is why this value is used in the calculation of the thermal efficiency.

In summary, the values used in the calculation of the thermal efficiency take into account the direction of work (positive or negative) and represent the net work done by the system.