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Compression power increase 7

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WDWalker

Automotive
Jan 8, 2007
3
Im wondering about the effects on power outbut from raising the compression ratio.

In a 4-stroke automobile engine, of course...

does the power increase raise in perfect(or near perfect) DIRECT PROPORTION with the raising of compression ration??

example-
in a 9.0:1 engine that makes 100hp, would rasing the c/r , say, 10% raise the power 10% also (9.9:1->110hp)?
(20% increase 10.8:1->120hp?)

does it keeps going to extreme amounts? (+100% c/r -- 18:1->200hp)?

i understand that in the extreme amounts as suggested, the engine that was originally designed for 9.0 c/r and 100hp is going to have a very rough time tyring to withstand those stress levels. this is all just in a theoretical mindset.
 
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Compression ratio is only one aspect of a complicated picture. What you are really after is cylinder pressure - which is mainly governed by the comp ration, cam, intake efficiency and exhaust efficiency. And if the cylinder pressure gets too high preignition results. ISZ
 
Short answer

NO

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the relationship is indirect, and according to most books I've read non linear. so that in the example you gave the power increase as a % is lower at the higher compression. "theoretically"
 
The maximum compression ratio that a gasoline engine can handle is (as IceStationZebra said) goverened by many things (check his list) and also the limits imposed on controlled combustion by the onset of detonation. This depends on the fuel in use, the combustion chamber shape and the presence of any 'hot spots' in the design.


Bill
 
Thanks for the replies, guys.

I knew it couldnt have been that mathematically simple.
 
I'm a litte surprised at some of the answers in this thread. At a strictly theoretical level, increasing the compression rato increases the power. At a practical level;

If heads are milled to increase the compression ratio, why is that done?

If you reduced the compression ratio from 8:1 to 6:1, what would happen?
 
the point of my earlier post was to answer op's question as to the shape of a theoretical curve. theoretically if you started at 6:1 increase to 7:1 and keep going, optimizing the whole engine along the way, power continues increasing but in diminishing amounts.
 
FoMoCoMoFo,

Now your response is crystal clear.
 
thanks, FoMoCoMoFo, your explanation makes more sense.

that being decided,

is there such a formula to tell us how the power increase will be by raising the c/r by a given amount?

thanks for all the help guys!
 
There is a formula for the *theoretical* Otto cycle efficiency as a function of its compression ratio. Maybe someone else has it handy; I've forgotten it ... it's in a thermodynamics textbook somewhere buried in the basement in a box that I haven't opened for years.

The theoretical ideal efficiency calculation makes a lot of assumptions that are impossible to achieve in the real world, and it also makes no account of real-world effects like heat losses, pumping losses, the non-zero amount of time that it takes for combustion to take place (ideal Otto-cycle combustion is instantaneous at top-dead-centre, which would destroy a real-world engine), limitations imposed by detonation or self-ignition, etc. The real efficiency of a practical Otto-cycle engine is half or less of what the ideal-cycle efficiency is.
 
Hello BJNX2000,

One has to make assumptions regarding the background knowledge of the person who starts the thread in order to respond effectively. So I apologize if I sound condescending - that is not my desire or intent.

Please first ask yourself - What is compression ratio? If your answer is volume ratio in a closed system, then my following recommendation is: Get a copy of Van Wylen and Sonntag "Classical Thermodynamics" and read chapter 9 on air standard cycles. Then you will have sufficient info to build a simple excel model of the PdV evolution in a standard Otto Cycle. The model will provide you with an "ideal work" answer (no engine friction consideration) which you can use to graph the ideal power response to change in CR, assuming things like constant volume addition of heat and constant ratio of specific heats.

Again, I apologize if my response seems trivial or unhelpful.
 
HI BJNX2000, your original question and my replies dealt with theory, understanding the theory is great, but as most the other posts point out it's too big a leap to a real world rule. way too many variables. but do keep thinking about it, Happy Motoring
 
mimetex.cgi


Code:
REM =======================================================================
REM Compression ratios and engine efficiencies written to a file
REM to partly answer a question posed here
REM [URL unfurl="true"]http://www.eng-tips.com/viewthread.cfm?qid=187729&page=1[/URL]
REM =======================================================================
REM Formulae (see page 8 of 29) and Free Demo BASIC Interpreter from
REM [url=http://sitemaker.umich.edu/mhross/files/fueleff_physicsautossanders.pdf][URL unfurl="true"]http://sitemaker.umich.edu/mhross/files/fueleff_physicsautossanders.pdf[/URL][/url]
REM [url=http://www.cix.co.uk/~rrussell/bbcwin/bbcwdemo.exe][URL unfurl="true"]http://www.cix.co.uk/~rrussell/bbcwin/bbcwdemo.exe[/URL][/url]
REM =======================================================================
      
cv = 7 : REM degrees of freedom x, y, z, like propellor, like spokes & 2*pogo
cp = cv+2
gamma_minus_1 = cp/cv - 1
      
file=OPENOUT("CR_eff.txt")
phi=1                             : REM fuel-air ratio relative to stoichiometric
      
FOR cr = 7 TO 22                  : REM compression ratio
  PRINT# file, STR$(cr)+" "+STR$(FN_FuelAirEfficiency(FN_OttoEfficiency(cr),phi))
  BPUT# file, CHR$(13)
NEXT cr
CLOSE# file
END
      
DEF FN_OttoEfficiency(cr)
= 1 - (1/(cr^gamma_minus_1))
      
DEF FN_FuelAirEfficiency(otto_eff,phi)
= (1 - 0.25*phi)*otto_eff

Code:
7 0.319865101
8 0.335966433
9 0.349667779
10 0.361539399
11 0.371975023
12 0.381257017
13 0.389594231
14 0.397145135
15 0.404032591
16 0.410353626
17 0.416186091
18 0.421593325
19 0.426627495
20 0.431332035
21 0.435743462
22 0.439892746
 
So, I wrote my little program and ran it with phi=1 (stoichiometric air fuel ratio).

Now I just pasted some of those numbers as the following line of text
0.371975023/0.361539399=
into C:\windows\calc.exe and it gave me
1.02886441707007429085204625236432

Now I can finally answer the original question.

Changing a compression ratio from 10:1 to 11:1 on an engine running at stoichiometric would give an expected power increase of about 3%, eg 100 HP to 103 HP.

[ul disc]
[li]My earlier post has a picture generated on the fly by a computer that is usually offline, so if it doesn't show, don't worry. It is just the formulae for engine efficiency from the reference.[/li]
[/ul]
 
I have used an 'online' program to interpret some of my dyno runs on the Mini. It's a 1380cc Austin A series engine. Don't get too excited if the numbers don't match theory, in my experience, they never do. All things remained the same (except for pistons) for all but the last run. That engine was not mine, but the same type and configuration. Fuel was Sunoco 108 for test 1 thru 3, VP 110 for test 4, unknown for test 5 and 6 (most likely Sunoco 116). All tests on same dyno and corrected to std. norms.

10.5:1 Compression ratio---94 hp.
11.5:1 ---97hp
12.5:1---99hp
13.87:1---110hp---107hp (two different runs)
14.5:1---117hp
16.0:1---122(approximate as I did not do this run)

I settled on 13.87:1 (it just worked out that way) at 110.48hp @ 7200rpm.

I have yet to find a computer program that works for my cars, or at least for me (pre computer geek).

Obviously, I guess, I went off track with that "all things remained the same", except...

Rod
 
Thanks.

10.5:1 Compression ratio---94 hp.
11.5:1 ---97hp

We are in the same ball park, 3 HP in 100 HP for adding 1 to the compression ratio at around 10:1, 10.5 in your case.
 
to give you an example of ranges toyota 1zzge, 1.8 liter 10:1 compresion gives 130 HP a 2 2gzze, 1.8 liter 11.5 : 1 is 180 HP. At 12.5 : 1 compression ratio you are approaching the limit for 92 octane gasoline at 100 F.

the problem in converting compression ratio to HP is all kinds of things, the Toyota engines above have ibtirely different storkes, Max RPM, heads/cams/vvt, and ready for this, similar torques...?

On a carburated or fixed injection, or fixed head/cam, the HP cam match equations list above.
 
Back to the OP

I stick by my original answer, however I have 10 min to kill, so in addition to an answer I will explain why, even though that was not asked.

Increasing the compression ratio effects a number of factors.

1) Fuel octane requirement. Higher octane fuel will be a different chemical composition and this will change other properties as well as the octane rating.

2) Increasing compression to the point of detonation increases power, however once exceeding this point, the engine detonates and looses power as well as beating itself to death.

3) To a point, you can retard the spark to avoid detonation, but the retarded spark costs more power than the compression gave.

4) Combustion chamber and or piston top change of shape. This affects flame travel, onset of detonation and combustion speed.

5) Change of combustion chamber surface area which influences energy lost to cooling system, which also effects onset of detonation.

6) Changes to expansion ratio. The greater the expansion of the gas during the power stroke before the exhaust opens, the greater the amount of energy extracted.

7) Effect on manifold pressures. If the chamber volume is smaller at TDC, the engine does not need as many degrees of rotation to double the volume over the piston. This means the pressure drops quicker in the exhaust port, promoting more gas flow. Smaller volume also means a greater amount of the spent gas is sent out the exhaust.

The 10 min is up.


Regards

eng-tips, by professional engineers for professional engineers
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
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