Timeline for Why does adiabatic expansion occur in the carnot process?
Current License: CC BY-SA 4.0
12 events
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| Jun 26, 2022 at 22:15 | comment | added | Chemomechanics | Great! I added some more dialogue to address these points. | |
| Jun 26, 2022 at 22:14 | history | edited | Chemomechanics | CC BY-SA 4.0 |
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| Jun 26, 2022 at 22:00 | comment | added | iwab | thank you!! If you could add that point to your answer I mark it as solved :-) have a nice evening (here it's midnight 😴) | |
| Jun 26, 2022 at 21:57 | comment | added | Chemomechanics | Yes. For this and other reasons, the Carnot cycle proceeds infinitely slowly; it provides maximum efficiency and maximum work but zero power (work collected per unit time). In real engines, we sacrifice efficiency (not that we have any choice) to push processes to run conveniently fast, wasting some heat but providing useful power. | |
| Jun 26, 2022 at 21:53 | comment | added | iwab | Thank you! So I can imagine that I move objects with a mass m on the frictionless surface of the piston sideways away, and since mv^2/2 applies I must make the whole theoretically infinitely slow, so that v-->0 and therefore E=0? | |
| Jun 26, 2022 at 21:46 | comment | added | Chemomechanics | Work is transferred to and from the system during the adiabatic steps, if that's what you're asking. (Over a full cycle, they cancel out.) Any additional energy to move the weights can be reduced to an arbitrary low level (e.g., by sliding them in and out of place over low-friction surfaces). Thus, this additional energy is ignored for idealized cycles. | |
| Jun 26, 2022 at 20:23 | comment | added | iwab | This is the way, ok keep the surrounding pressure lower than your system's pressure, you will get the work from the internal energy of the system. But doesnt it require some energy to lower the pressure somehow? Wouldn't adding or removing weights from a piston require energy? | |
| Jun 26, 2022 at 20:08 | comment | added | Chemomechanics | And yes, you must constantly adjust the applied pressure to nearly match the system pressure to maintain quasistatic conditions and minimize (in the ideal case, eliminate) entropy generation. An example is adding or removing weights from a piston pressing on the system. | |
| Jun 26, 2022 at 7:02 | comment | added | Chemomechanics | The adiabatic step is needed to avoid a temperature difference between the system and the hot and cold reservoirs. Heat transfer over a temperature difference would generate entropy, which we can't afford in this perfectly efficient (Carnot) heat engine cycle. | |
| Jun 26, 2022 at 3:32 | comment | added | iwab | I wonder, wouldn't there have to be some kind of pressure change in the environment to allow the gas to expand/compress adiabatically in the first place? Wouldn't that need some other work, so to say the energy to set up the right conditions to allow adiabatic expansion/compression? | |
| Jun 26, 2022 at 3:29 | comment | added | iwab | Thank you for the detailed answers, but still I don't understand this passage: "I'll wrap up the isothermal expansion at the higher temperature with some adiabatic expansion to cool the fluid down so that there's no entropy generation when I put it next to the cool body." Why does this process happen at all then? So why does the gas continue to expand by itself after insulation to cool down? Couldn't it have reached the same volume already with the isothermal expansion? | |
| Jun 25, 2022 at 22:14 | history | answered | Chemomechanics | CC BY-SA 4.0 |