The effects of higher CR

[Lothar]

I've seen a lot of discussion about going to higher compression pistons in order to enhance performance. Most of the discussion revolves around horsepower, but wouldn't there be a more significant increase in torque. I know that diesel engines run at much higher cr than gasoline engines and typically have a different relationship between hp and torque.

Could someone explain in a real world example, like a 3.0 L engine, how torque is increased when increasing from 8.5:1 CR to say 10.5:1?

[beepbeep]

horsepower is nothing but RPM x torque, so higher horsepower usually means higher torque. Unfortunately, max engine HP means peak power , which isn't saying so much (as peak torque usually occurs at lower RMP's).

[island911]

Higher compression means higher thermal efficiencey (you get more out of the "bang"). . . push the piston with more force -> more torque.
Those emmission devices like catalytic converters and air pumps are/were there to help finish of whats "left on the table" thermo-chemically speaking.
Higher compression also means faster combustion. . .which can lead to detonation. . .which our piston engines dont handle so well. So it's a balance game.

[jluetjen]

horsepower is nothing but RPM x torque, so higher horsepower usually means higher torque.

BeepBeep - I think that you have your conclusion backwards. Higher HP does not necessarily mean higher torque, it could just be higher rev's (like in an F1 engine). But higher torque at a given rev will mean a higher HP at that engine speed.

Looking at some comparative data for the different "early 911" configurations, I've come to a couple of conclusions:

* Torque is generally a function of engine capacity. You'll be hard pressed to find a very wide distribution of torque numbers for a given engine capacity (take note of the VE discussion below)

* Once the CR gets above 9.0:1, the increase in HP or Torque is pretty marginal. If the CR is really low; like 7.0:1, then yes increasing the CR will most likely increase torque -- with the following exception:

*When using cams with a lot of overlap (such as the early S cam), a higher static CR can help fill in the bottom of the rev range when the engine is "off cam". Basically my suspicion is, when the engine is off-cam, the intake tuning is not having an affect and in fact some of the charge is going out the exhaust pipe (thus causing poor emissions). It also decreases the volumetric efficiency of the motor so that it acts like a smaller motor (and develops the torque of a smaller motor). All of this changes when the engine gets "up on the cam" in which case now the charge is being packed into he chamber which can often deliver a VE of more then 1:1 (and as a result the torque of a larger motor).

* As you mentioned, HP is torque * RPM, but the torque peak is generally at a lower RPM then the HP peak. I've come to the conclusion that the torque peak will start to drop off as gas speed goes over a critical speed (~80 m/s in a 911). The trick is that often the torque drops off slower then the rev's increase, so the HP keeps climbing. But at some point the torque drops off a cliff at which point the HP will start to drop. The end result is that at that point the torque numbers are already well below the peak, so the change in CR isn't going to have much of an impact.

Note that I'm sort of self-taught on this, so I'm welcome to correction.

[island911]

Originally posted by jluetjen
* Once the CR gets above 9.0:1, the increase in HP or Torque is pretty marginal. If the CR is really low; like 7.0:1, then yes increasing the CR will most likely increase torque --

This should help with the "How Much" question;

[tbitz]

I'm not a thermal Eng, but I'd guess an approximate gain in power could be calculated by:

percent increase = (C.R.2/C.R.1)^0.286

For the case stated above:

percent increase = (10.5/8.5)^0.286 = 1.062 or 6.2%

On a 180Hp SC engine you'd get 11Hp more.

EDIT: My above equation is incorrect. Look below for an explanation...

[jluetjen]

Tbitz; wouldn't that equation be accurate for torque rather then HP?

BTW - I've seen a number of times that high CR is critical for success in an air restricted motor such as F3, ALMS/FIA GT or Nascar Restricter plate engines. In those cases I guess they've gone up to 15:1 or beyond. I guess this makes sense since the engine will be limited in it's ability to fill the cylinder at higher rev's by the restrictor. Since the chamber has less air in it, it's possible to go to a higher CR to try to make up for the loss in VE. I guess the risk of detonation is lessoned by the reduction in chamber pressures. Just thinking on the fly...

[ChrisBennet]

Originally posted by island911
...
Higher compression also means faster combustion. . .which can lead to detonation. . .which our piston engines dont handle so well. So it's a balance game.

island911,
I was under the impression that the flame front travelled at the same speed regardless of compression.
I can think of a couple of theories why detonation is more likely with higher compression. 1. The increased heat caused by the increased compression (think diesel). 2. The shape of the high compression pistons.
Educate me,
Chris

[tbitz]

Originally posted by jluetjen
Tbitz; wouldn't that equation be accurate for torque rather then HP?
...

If your only changing CR (nothing else ie: cams), then the profile of the torque curve should not change. If this is the case then,
peak torque and peak Hp occur at the same rpm before and after the CR changes, then both would increase by the same percentage after the CR changes.

[69911e]

quote:If your only changing CR (nothing else ie: cams), then the profile of the torque curve should not change. If this is the case then,
peak torque and peak Hp occur at the same rpm before and after the CR changes, then both would increase by the same percentage after the CR changes.

-----------------------------------------------

This is not exactly true because effective CR changes with RPM due to changing volumetric efficiency. Because the thermal efficiency is not linear with respect to CR, this would change the shape of the torque and HP curves slightly (but it would change both).

ED

[island911]

Originally posted by ChrisBennet

island911,
I was under the impression that the flame front travelled at the same speed regardless of compression.
I can think of a couple of theories why detonation is more likely with higher compression. 1. The increased heat caused by the increased compression (think diesel). 2. The shape of the high compression pistons.
Educate me,
Chris

Heat and pressure both make the chemical reaction happen faster. Firecrackers for instance; Light the fuse, and the powder in the fuse burns at a fairly constant rate (you hope) Then, once the flame front gets inside the firecracker, it goes bang! THis is because the paper around the powder keeps the pressure on long enough to make the burn rate go way up, really fast.

(Simply) The piston in your motor is doing it’s best to remove the increasing pressure (make HP).
As you increase the starting pressure, you had better match it with increasing piston speed. (or other sol’n) If you don’t, the burn rate will snowball with the pressure rise and give detonation.

Make sense?

[Tinker]

Tony,

I am trying to comprehend your equation. What is ^.286??

Tinker

[island911]

I'm not sure what Tony has going there either. but . . .

theoretical thermal efficiency n(rc) = 1-(1/rc)^0.4 (See curve above)

Then:
8.5:1 gives n=57.5%
10.5:1 gives n=61.0%

making for a theoretical jump of ~6% (180HP -> 191HP with 6%)

[tbitz]

Originally posted by Tinker
Tony,

I am trying to comprehend your equation. What is ^.286??

Tinker

This means "to the exponent of 0.286".

Originally posted by island911


I'm not sure what Tony has going there either. but . . .

I'm not sure either

When I made that statement I was thinking of difference in enthalpy, where:

Enthalpy = Cp * Temperature * (CR ^0.286)

However CR is not compression ratio, but rather pressure ratio, so the equation does not apply here since compression ratio and pressure ratio are not the same, sorry

I'll stick with Electrical Eng....

[island911]

Originally posted by tbitz
I'm not sure either
. . ..
I'll stick with Electrical Eng....

EE? You had ME fooled.

[emcon5]

Originally posted by tbitz
I'm not a thermal Eng, but I'd guess an approximate gain in power could be calculated by:

percent increase = (C.R.2/C.R.1)^.286

For the case stated above:

percent increase = (10.5/8.5)^0.286 = 1.062 or 6.2%

On a 180Hp SC engine you'd get 11Hp more.

I'm not an engineer of any sort, but I don't think the numbers add up.

Between the US and ROW 80's SC's, the 9.3:1 had 180hp, and the 9.8:1 had 204hp with the same cam/heads. The fuel system was slightly different, but still CIS, and the exhaust was a same except for a pre-muffler instead of a cat.

If half a point more compression = 24HP, I can''t see how 2 full points (i.e. 4 times more) will only give 11hp.

Can someone 'splain this?

Tom

[Matt Holcomb]

Originally posted by emcon5
If half a point more compression = 24HP, I can't see how 2 full points (i.e. 4 times more) will only give 11hp.

Well, if half a point yields to an increase of 24HP, then my 11.3:1 Carrera RS motor is developing close to 360HP!


Matt Holcomb
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[jluetjen]

If half a point more compression = 24HP, I can''t see how 2 full points (i.e. 4 times more) will only give 11hp.

Simple: It's not a linear function. Note the exponential value of r. In addition to the decreasing return from increasing the CR; note that when using Tony's comparison, percentage change between a CR of 6:1 and a CR of 7:1 (~16%) is noticably greater then the difference between a CR of 10:1 and 11:1 (~10%). And so it goes as the CR's go up.

But I suspect that there is more in play then just the CR which accounts for that 24 HP delta. Specifically the omission of the catalytic converter -- "early" converters were notorious for being restrictions in the exhaust system. And a restricted exhaust is a great way to kill HP ( especially at high RPM's when a lot of air is moving) without making a big impact on the torque.

[ChrisBennet]

Before you go comparing Euro and US compression ratios be aware that the stated factory compression ratios are not the actual (measured) ratios. Here are some numbers courtesy of an old post by Kevin Roush:
Compression Ratios
Factory (Real) Motor
9.8:1 (9.2) 3.2L [This doesn't make sense - Chris]
7.0:1 (6.5) 3.3 Turbo
11.3:1 (10.4) 3.6L

"A USA 3.2 is billed as 9.8:1 compression I believe , but worse,
once CC'd actually it comes out to a true measured compression ratio of around 9.2 :1 similarly go all the other Porsche motors (3.3 turbos stated 7:1- actual 6.5.1, 11.3:1 3.6 actual 10.4:1) for reasons I have never had clearly explained to me."
Kevin Roush

-Chris

[jluetjen]

To add fuel to the fire, here is a chart I made up that compares the BMEP's of the early 911 "Normal" with the 3 versions of 911E. If CR really made a big difference, you'd most likely see some indication here between the '70 and the '72E.



Note that the cars had the following CR's:
Normal: 9.0:1
'69E: 9.1:1
'70E: 9.1:1
'72E: 8.0:1

If the 1.1:1 drop in CR from '70 to '72 really made a difference it should jump out here, especially since I've graphed the BMEP rather then torque. (Note that both the Torque and BMEP curves have the same shape, but BMEP is normalized so that changes in capacity won't change the data.)

A few of observations...
1) Comparing the '67 Normal with the '69E; note the big difference that the MFI make at the top and the bottom of the rev range without really affecting the peak torque in the center.
2) The '72E actually had a slightly different cam design. The profile was the same but the lobe center was increased which reduced the overlap. I suspect that this accounts for some of the recovery in performance since it might have increased the effective CR.
3) Most significantly, note that the peak torque didn't change by much if at all. You can check the graphs for the S's and the T's here, but I'm not seeing a huge change in the performance of the 911 as the CR's were reduced. I guess it goes back to the diminishing returns discussion.