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Are the Heat Engine (HE) and Internal Combustion Engine (ICE) not the same because ICE only exchanges the heat to the surrounding[only one reservoir]? And I have a question what if it's an SI engine which has a spark plug to be a hot reservoir, and surrounding would be a cold reservoir like the following picture.

enter image description here

I showed this picture to my Prof. He said it was wrong, and he will answer it next week, but I just want to know it now. I try to look up in the book or some websites but still can't find the answer.

Since it has 2 reservoirs why it's not the Heat Engine (I know that they are not the same, but according to the reasons above isn't it should be the same?)

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The internal combustion engine Otto Cycle is modeled thermodynamically as shown in the PV and TS diagrams below. (Note: intake and exhaust strokes not shown).The heat addition (spark ignition) is modeled as a constant volume heat addition. It does not occur at constant temperature as you depicted. The heat rejection is modeled as a constant volume process. Again it does not occur at constant temperature.

Hope this helps.

enter image description here

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  • $\begingroup$ Thank you so so much. Your explanation is very clear and on point. Thank youu T^T $\endgroup$ – Chayutpong Leartrattanaporn Jan 16 at 16:14
  • $\begingroup$ @ChayutpongLeartrattanaporn So you accept the answer? $\endgroup$ – Bob D Jan 16 at 16:50
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The spark doesn't act as a reservoir at all, that is why.

The driving force of the internal combustion engine is not coming from the heat of the spark. It is coming from the energy released in the combustion process. This energy is not stored thermally in the spark, as a reservoir would do. Instead, the energy is stored chemically in the fuel being burned. Once the spark plug ignites the fuel, it would be more accurate to consider this heated fuel as the hot reservoir in your thermodynamic system. An ICE is still technically a heat engine, BTW.

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  • $\begingroup$ Thank you so so much. Your explanation is very clear and on point. Thank you T^T $\endgroup$ – Chayutpong Leartrattanaporn Jan 16 at 16:14

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