What is the difference between compression ratio and cut off ratio?

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Discussion Overview

The discussion centers on the differences between compression ratio and cutoff ratio in the context of internal combustion engines, exploring their definitions, implications for efficiency, and underlying physics. Participants seek to clarify these concepts and their relationships to engine performance.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants define the cutoff ratio as the ratio of the volume after combustion to the volume before combustion, while the compression ratio is described as the ratio of maximum volume to minimum volume.
  • There is a claim that efficiency increases with compression ratio and decreases with cutoff ratio, although the reasons for this relationship are questioned.
  • One participant explains that in internal combustion engines, more fuel/air mixture can be compressed into the cylinder, potentially increasing power output, which is linked to volumetric efficiency.
  • Another participant expresses confusion about the relationship between compression ratio and efficiency, seeking a simpler explanation of the underlying physics.
  • A later reply attempts to clarify that the efficiency increase with higher compression ratios is related to the ability to recover more energy during the expansion phase after combustion.
  • It is suggested that the cutoff ratio affects the average expansion ratio of the air particles, impacting overall efficiency.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and agreement on the definitions and implications of compression and cutoff ratios. Some concepts remain contested, particularly regarding the physics behind efficiency changes.

Contextual Notes

Participants highlight limitations in their understanding of the thermodynamics involved, indicating a need for clearer explanations of how these ratios affect engine performance.

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I know the cut off ratio is of the specific volumes, but is that it? And what exactly do they tell you
 
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The cutoff ratio is the ratio of the volume after combustion to the volume before combustion. The compression ratio is the ratio of the maximum volume to the minimum volume.

Efficiency goes up with compression ratio and down with cutoff ratio.
 
Why does the efficiency go up with compression ratio and the cutoff ratio down?
(I don't really understand the physics/thermodynamics behind it

Thank you
 
a simple way to look at this is ...in internal combustion engine- power comes from very fast burning of fuel / air mixture. the more of this mixture you can cram into the cylinder before ignition, the more power..and ( this is a very simplified explanation) the for volumetric efficiency. If you opened up a cylinder and let atmospheric pressure fill it up , and then closed the cylinder, you would have a compression ration of 1 to 1. Since we have crank shaft and pistons and valves moving around the IC sucks in air /fuel mixture and compresses it when the valve train closes the intake port. This mixture is squeezed when the piston moves toward the top of the cylinder and if you measured the amount the volume was squeezed or compressed at the end of this action your now have the ratio of the fuel / air mixture before and after the compression. Most lame passenger cars today have about 8:t or 8 to 1 compression ration. When you light off this mixture your get a measureable amount of work that can be done with this mixture. The piston is moved toward the bottom of the cylinder bu the fast burning action of the ignited mix and we have force being applied to the top of the piston. F x D = work...since the whole mechanism is rotating via crank shaft this action happens many times per minute ( RPM = revolutions per minute) so we have F x D / Time = Power

If all things are equal during this process, and we can cram twice as much fuel / air mixture into this same cylinder and light it we ( theoretically) should produce twice the power so the IC with a 16:t compression ration would make a lot more power and hence the volumetric efficiency thing..
 
Is it possible to explain it much simpler please? ( I am very weak in physics.)

So basically,

What I know so far is (in simple steps)

1) In a box, air is compressed.

2) Fuel is added

so

3) ignites

4) and produce work


if compression ratio goes up, efficiency increases. But why? because we can put more Air in the box ? I am sorry I am totally confused.
 
student23 said:
Is it possible to explain it much simpler please? ( I am very weak in physics.)

So basically,

What I know so far is (in simple steps)

1) In a box, air is compressed.

2) Fuel is added

so

3) ignites

4) and produce work


if compression ratio goes up, efficiency increases. But why? because we can put more Air in the box ? I am sorry I am totally confused.

The reason lies mostly in your 4th point: «and produce work». To answer your question, you need to ask the real question: How does it produce work?

When you ignite the fuel mixture, the heat produced transfer energy to the air inside the cylinder. Now you need to recover that energy to produce mechanical work. At this point, you need to look at air as if it was a compressed spring. If a compressed spring expands, it releases its stored energy producing mechanical work (creating some motion, somehow). The more it expands, the more energy you can recover. When the piston goes down in a cylinder, it gives space to the burned fuel mixture to expand. The compression ratio that was, say, 10:1 became an expansion ratio of 10:1. The greater the expansion ratio, the greater the energy recovered (or, in other words, work produced) for the same amount of fuel burned. Thus, the «higher compression ratio = higher efficiency» statement.

So, the important thing is not how much you compress the air before the ignition, but how much you expand it after the fuel is completely burned.

Now you can also see the importance of the cutoff ratio. The first particles of air that were heated up when the piston was at top dead center (TDC) will have a full expansion ratio of 10:1. But the ones that were heated up at the end the combustion process may have an expansion ratio of only 8:1 (because the piston is not at TDC anymore). Hence a higher cutoff ratio will give less efficiency because it means that the «average» expansion ratio for all air particles is lower.
 
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