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Intuitively to me it seems likes increasing compression ratio would require more work to compress the gasses before ignition, so you'd just end up getting back what you put in - like a spring. What am I missing?

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To close voters this asks why a given system is more efficient, not something like "how do I make this engine more efficient". Seems on topic to me. – Kyle Oman Jul 29 '14 at 21:48
And on the question: pure guessing, but perhaps you get more energy out of a compressed burn than an uncompressed burn (i.e. higher fraction of fuel undergoes combustion reaction). Without a difference like this, you're right, compression would be fairly pointless energy-wise. – Kyle Oman Jul 29 '14 at 21:49
@Kyle: No, compression is a good thing even if combustion is complete in both cases. As Floris says, the heating makes much higher pressure after compression, which energy can be extracted. – Ross Millikan Jul 29 '14 at 22:20
Shouldn't "more efficient" be "higher in torque?" – Ben Crowell Jul 30 '14 at 0:16

3 Answers 3

You are missing the fact that when I have a lot of air in a tight space, and I then heat it up, I get really high pressure. You have to draw yourself a diagram of pressure vs volume - compressing the cold gas requires a certain $\int P \cdot dV$ of work, but then I heat the gas and the subsequent expansion takes me along a different curve where the work extracted is higher.

The more I move left to right in that diagram, the larger the (difference in) swept area, and the more work extracted per cycle for the same heat in. Basic thermodynamics.

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Note also that under the adiabatic assumption (no heat flows in or out by conduction during the compression / expansion) the work you do to compress the gas further is returned to you during expansion. In reality, the gas heats and will give up some of that heat to the environment, and there is friction between the piston and the cylinder which results in dissipation of energy, but from the intuitive point of view, the work you do to compress the gas comes back at the end of the cycle - even before you add combustion.

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I think you could rephrase your question much better, please take the time to detail things if you wish for concise and appropriate answers from the community.

If I understand correctly you wish to understand in an intuitive manner, why compressing at cold is energetically less costly?

Conceptually speaking, when you want to compress gas which is already at a high temperature $T_1$, you're faced with a stronger resistance from the gas molecules as they are already highly agile and try to expand to reach their equilibrium state again. So by the same line of logic, when the gas is colder before compression, let's say at $T_2 < T_1$, then clearly they have much lower kinetik energies and hence easier to be pushed into a compressed state.

To further complete this view, same logic applies when expanding a gas, meaning it is easier for a gas to undergo an expansion when it is already hot, the molecules are already activated and able to move towards equilibrium. These two ideas, i.e. compressing at cold and expanding at warm temperatures, are at the heart of motor engines. A good and neat example would be Stirling's engine, where the "displacer" is used to create these exact scenarios (for the gas) that I just described.

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Imagine two cylinders of the same size when expanded, one that compresses $5:1$ and the other compresses $10:1$. As you fill them with the same amount of fuel/air mixture, the energy loss in the second is compressing from the $5:1$ point to the $10:1$ point. At this point the $10:1$ has something more than twice the pressure of the $5:1$ (Due to the heating of compression). Now when you burn the fuel, the pressure is still twice as high as the other, but much higher. The energy extracted when expanding from the $10:1$ point to the $5:1$ point is much higher than the cost of compression, so we are winning.

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