Talk:Compression ratio

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Untitled

Should this information be in a seperate article, or in an article on internal combustion engines? I ask this for two reasons:

The term is also used in data compression, but more importantly...
It doesn't seem a topic that is self-contained enough to warrant a full-blown article.

Repy: I found it to be very helpful the way it is. 192.75.238.205 21:53, 18 May 2006 (UTC) JT[reply]

Also think about indexability for search engines. (No pun intended.) Someone typing in "compression ratio" on Google will now get linked to this page, because it has it's own article. I think is helpful if this information is easily accessible like that. J. Geerink (talk) 00:00, 12 December 2024 (UTC)[reply]

First sentence

I replaced this:

"The mixture of fuel and air is a single number that can be used to predict the performance of any internal-combustion engine."

with this:

"The compression ratio is a single number that can be used to predict the performance of any internal-combustion engine."

Since that is what the sentence is trying to say, and, also, wikipedia style says that the title should be the subject of the first sentence of an article.

-- human 16:03, 26 June 2006 (UTC)[reply]

Natural gas

"I am a manufacture of high pressure cng (compressed natural gas) compressors and cng engines and vehicles and there is an error in your statement concerning the compression ratio natural gas engines can operate on. They in fact can operate in the diesel range above 14 to 1 without pre-detonation (knock)."

This comment was in the article itself, I am moving it here so someone can verify, etc., and perhaps turn it into a useful statement somewhere in the article.

-- human 16:49, 26 June 2006 (UTC)[reply]

Clean up

I have removed the 'clean up' tag. Has anyone any objections to this? - Ballista 07:34, 1 August 2006 (UTC)[reply]

Error in formula

The formula should be either

{\mbox{CR}}={\frac {(\pi b^{2}s)/4+V_{c}}{V_{c}}}

OR

{\mbox{CR}}={\frac {\pi (.5b)^{2}s+V_{c}}{V_{c}}}

and not

{\mbox{CR}}={\frac {(\pi (.5b)^{2}s)/4+V_{c}}{V_{c}}}

-- ElroyH 18:25, 4 October 2006 (UTC)[reply]

Error in formula... still.

The volume of a cylinder is the area of the top of the cylinder times the height of the cylinder if the circle on top is the same area as the circle on the bottom. That formula for the volume of a circle is (pi)*(radius^2). For a cylinder you just multiply the area by the height. (pi)*(radius^2)*(h). The bore size is the diameter, not the radius. The radius is half of the bore. Square it! To apply this to a internal combustion cylinder you must change the height to stroke length add the additional volume of the head's combustion area. So a formula should look like so:

{(pi)*[(.5*bore)^2]*stroke}+(volume of head)

Basically the only thing I am changing is that the bore is in diameter and not in radius. If you would half the bore and then square it you would get the correct answer you are looking for.

Yeah, I see this as wrong too. Why the confusion in the formula is baffling. It's PI * r2 * h or PI * bore * stroke_length. Edit: I think the OP got confused with the formula for displacement https://en.wikipedia.org/wiki/Bore_(engine)

Dear Mr/Ms IP. Which formula in the article do you think is incorrect? The closest one to what you describe is this one, but to me it looks correct.

V_{d}={\tfrac {\pi }{4}}b^{2}s

Cheers, 1292simon (talk) 22:38, 4 August 2019 (UTC)[reply]

Gas

Which is gas with the best compression ration at room temperature ? (this is, uses less space to be stored).--Mac (talk) 17:16, 27 January 2008 (UTC)[reply]

relationship of CR vs power / torque developed

Is there any way of estimating how much of a power boost an increased compression ratio would give? In looking to buy a motorcycle I've noticed some engines have what appear to be distinctly lower CRs than I'm used to with cars (presumably built both for reliability where maintenance may be patchy and flexibility to use low-grade fuel, neither of which are necessary where I live) as well as low power outputs compared to what has been demonstrated as possible in harder-tuned models of the same displacement. Be nice to know how far it needs to be wound up to, say, reach the legal maximum power output for that class of bike / license, whether it would then be into the zone of potentially needing high-grade fuel (going up from ~7.5 to 10.5 or more). Or on the other hand, how much more power would be available by winding it up to just below where it would start to pink on regular gasoline (so as to get the best possible economy!). 193.63.174.10 (talk) 12:27, 22 June 2009 (UTC)[reply]

compression ratio vs overall pressure ratio

the entry comparing compression ratio to pressure ratio needs some help. the math relates CR = PR * T1 / T2, which is true, but doesn't help calculate PR, since T2 is unknown. the more relevant math would seem to be Adiabatic_process#Ideal_gas_.28reversible_case_only.29, which shows that P*V^gamma is constant, gamma is 1.4 for air (stated more explicitly in other spots, eg Heat_capacity_ratio). based on this,

P_{2}V_{2}^{\gamma }=P_{1}V_{1}^{\gamma }

PR={P_{2} \over P_{1}}=({V_{1} \over V_{2}})^{\gamma }=(CR)^{\gamma }

and then there's the table of CR vs PR. running this equation thru octave, i get different results than those in the table. here's the composite:

Compression ratio and overall pressure ratio are interrelated as follows:

Compression ratio	1:1	2:1	5:1	10:1	15:1	20:1	25:1	35:1
Pressure ratio	1:1	3:1	10:1	22:1	40:1	56:1	75:1	110:1
PR by math/octave	1.00:1	2.64:1	9.52:1	25.12:1	44.31:1	66.29:1	90.60:1	145.11:1

i'll call the provided numbers the 22:1 PR. not claiming that i'm right, just that this seems like a pretty straight-forward application of the theory, and that the numbers don't match the numbers given. where did the 22:1 PR numbers come from ? are they based on theory, in which case what was the gamma value used ? and if they're experimental, where's the reference ? going to mark this section as not verifiable ... i've googled for a few hours and haven't found anything matching the 22:1 PR

Nqzero (talk) 01:04, 14 September 2009 (UTC)[reply]

if the person who made that table was using any gamma value at all, it was not constant. it goes from log₂3 (approximately 1.58) at CR=2:1 to log₃₅110 (approximately 1.32) at CR=35:1. it could be because the fluid in question is a mixture of gasoline and air, rather than just plain air; the gasoline may evaporate, which would absorb heat and cause the value to change. i don't know enough about that to say for sure though.
Qbert203 (talk) 05:50, 5 February 2010 (UTC)[reply]

Yes, agreed, the previous version was wrong. I've fixed it in the article. 173.168.102.141 (talk) 15:44, 12 June 2010 (UTC)[reply]

adding a fluid (water or fuel ) will reduce final pressure by evaporation .Wdl1961 (talk) 01:32, 6 February 2010 (UTC)[reply]

Gasohol

Someone randomly put in that users of gasohol may notice higher knock resistance. Not only was this out of place, 10% ethanol gasoline, the type mentioned in the sentence comes in the same octane ratings as normal gasoline, typically 87,89,91. The ethanol is just added to gasoline that is not as high octane to start with. The presence of ethanol doesn't mean higher octane unless its lots of ethanol, or high grade gasoline to begin with.174.3.107.124 (talk) 05:56, 17 October 2009 (UTC)[reply]

External combustion engine

What sort of External combustion engine has a compression ratio? Biscuittin (talk) 23:50, 3 November 2012 (UTC)[reply]

Qestion

Compression and pressure ratios are different on the talk page and main page? Anyway it doesn't make sense to me: if you reduce a volume by a factor of two, the pressure increases by a factor of two?--Pawlin (talk) 08:21, 22 April 2014 (UTC)[reply]

Only if the temperature also remains constant. Generally the compression ratio is calculated from geometry beforehand (this is easy to do for either piston engines or for turbines), but the pressure ratio is the result of this process. As we assume an adiabatic process, it's quite hard to predict the pressure change (i.e. how much of the work done on the fluid during compression becomes a pressure rise, and how much goes to a temperature rise), so this is usually measured instead. Ratio of specific heats allows a theoretical prediction of this that's only a little more complicated than the ideal gas law, but this is such an approximation that (especially for turbines) it's not terribly useful. Andy Dingley (talk) 12:46, 22 April 2014 (UTC)[reply]

Tanks Andy: but why are the differnet values on Overall pressure ratio and Compression ratio?--Pawlin (talk) 08:21, 22 April 2014 (UTC)[reply]

There are several differences, but the main one in practice is that compression ratio is applied to piston engines, pressure ratio to gas turbines. Compression ratio is the cause of the compression, pressure ratio is the result of this compression and increase in pressure. Compression ratio can be calculated easily from the engine's geometry. Pressure ratio is measured. Andy Dingley (talk) 21:36, 23 April 2014 (UTC)[reply]

No. I meant the tables on Overall pressure ratio and Compression ratio differnet, I don't know which one to use? --Pawlin (talk) 11:24, 30 April 2014 (UTC)[reply]

Oh, use Compression_ratio#Compression_ratio_versus_overall_pressure_ratio

The one at Overall_pressure_ratio#Compression_ratio_versus_overall_pressure_ratio is using a lower value for

\gamma

(about 1.3 rather than 1.4), hence the numerical differences, and more importantly its explanation is unreadable nonsense. Feel free to replace it with the better one! Andy Dingley (talk) 12:53, 30 April 2014 (UTC)[reply]

Carnot cycle

In internal combustion engines, it is generally true that the higher the compression ratio, the higher the thermal efficiency. According to the Carnot cycle, the thermal efficiency depends on temperature difference, not pressure difference. How are these two related? Biscuittin (talk) 07:54, 11 August 2014 (UTC)[reply]

They aren't much related at all. The Carnot cycle is a highly idealized cycle in which heat is added at constant temperature during one quarter of the cycle, and then heat is lost at constant temperature in another quarter of the cycle. The notable feature of the Carnot cycle is that it is the most thermally efficient possible for any heat engine operating between the maximum and minimum temperatures nominated for the particular cycle. No other cycle operating between the same maximum and minimum temperatures can match the thermal efficiency of the Carnot cycle. The cycles that are relevant to internal combustion, reciprocating engines are the Otto cycle and the Diesel cycle. Dolphin (t) 11:59, 11 August 2014 (UTC)[reply]

Thank you. Biscuittin (talk) 18:24, 18 August 2014 (UTC)[reply]

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non-combustion compression ratios

This page seems to give a lot of weight to internal combustion engines - there are plenty of processes where the compression ratio is important, that are entirely unrelated to ICEs. Examples include vacuum components, where the compression ratio of e.g. turbomolecular pumps, is a key design variable that tells you how much of a pressure gradient can be maintained across the device, and is usually limited by backflow. Similarly, the compression ratio for high-pressure industrial processes is important, even when combustion is not invovled. This can tell you about how high a pressure you can obtain by inserting a compressor into your process, and where you expect to find the equilibrium between backflow from the upstream processs, and the downstream.

Would be nice if this article could include more detail here 79.76.186.174 (talk) 10:36, 30 September 2020 (UTC)[reply]

non-cars

Is someone able to give the compression ratio for non-cars, as Spitfires, U-boat, pocket battleships ...?thanks 151.29.133.9 (talk) 14:30, 1 May 2023 (UTC)[reply]