Drivetrain (Cooper S) MINI Cooper S (R53) intakes, exhausts, pulleys, headers, throttle bodies, and any other modifications to the Cooper S drivetrain.

Drivetrain 10 deg reduction in intake temps = 2% hp increase...

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Old Apr 5, 2004 | 06:31 AM
  #1  
Super_MINI's Avatar
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I read something yesterday that I thought I might share. It said that for every 10 degrees of intake temperature reduction on a forced induction (turbo or SC) engine expect a 2% increase in HP. For a normally aspirated engine it was 1%. Does this sound like a good generalization for the MCS?
 
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Old Apr 5, 2004 | 06:43 AM
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sdanaher's Avatar
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i'd think that might be subject to diminishing returns... else we'd have some people thinking about that 0 kelvin air producer..

hmm... i wonder how absolute zero effects my engine's life.. I'd assume making my engine so cold that particles cease to move is BAD?

lol
 
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Old Apr 5, 2004 | 07:08 AM
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Cool!!! (Literally)

Now, understand that even though our engines follow the rules of Physics, that this one is a small part of the total efficiency, and this is posted as kind of a Physics irony.

For every 10 degrees F. you lower the intake temperature, the carnot efficiency drops 2%

Plug in 75 degrees F TsubH (Air temperature in) and
150 degrees F TsubC (Air temperature out) and see -14% efficiency
Then use 65 and see -16 55=-18 etc etc.
Physics site

Cool huh? Irony is great!

(Now, there will be a HP gain lowering the intake air temperature. I don't know how much)


 
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Old Apr 5, 2004 | 07:09 AM
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Yeah, air particles not moving CANT be good...but theres only one way to find out! I say we have a Mini run on the moon.
 
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Old Apr 5, 2004 | 08:11 AM
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So are we trying to increase the horse power of your MCS or
just talking physics? :???:
 
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Old Apr 6, 2004 | 10:09 PM
  #6  
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More physics:

Auto air/gas ratio, or the so-called stochiometric ratio is precisely measured and engineered for optimum performance (ideally 14.7:1, air to fuel). Since warmer air contains fewer molecules of oxygen than colder air for a given volume of air, the stochiometric ratio becomes less than ideal based on a set standard temperature, which is why a car with a CAI theoretically will perform better than one without.

But then, the oxygen sensor is supposed to regulate this fuel/air ratio, so this puzzles me.

Just my 02 cents worth.
 
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Old Apr 7, 2004 | 07:41 AM
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in doing this analysis, we need to distinguish the difference between the operating temperature of the combustion process and the temperature its surroundings (upon which the thermodynamic efficiency depends) from the temperature of the air entering the process.

lowering the temperature of the incoming air will increase its density linearly with temperature, giving you a power increase. this is well documented in other threads and on other sites.

the temperature of the incoming air doesn't have as big an effect on the operating temperature of the engine, since that depends on the energy released by the combustion.

ok,ok, off the engineering lecturn again,
flyboy2160


 
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Old Apr 7, 2004 | 07:56 AM
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From: PA/NJ
>>More physics:
>>
>>Auto air/gas ratio, or the so-called stochiometric ratio is precisely measured and engineered for optimum performance (ideally 14.7:1, air to fuel). Since warmer air contains fewer molecules of oxygen than colder air for a given volume of air, the stochiometric ratio becomes less than ideal based on a set standard temperature, which is why a car with a CAI theoretically will perform better than one without.
>>
>>But then, the oxygen sensor is supposed to regulate this fuel/air ratio, so this puzzles me.
>>
>>Just my 02 cents worth.

02 more cents:

The oxygen sensor works in conjunction with the Mass Air Flow sensor in a closed loop. If the air is colder, there is more of it (more mass) and the computer responds by adding more fuel to achieve the 14.7:1 ratio. More fuel = more power.

Along these lines, this is why fuel injected cars run fine at higher altitudes (allbeit with less power) while carburated cars need re-jetting or they'll run rich. The computer recalculates the fuel needs automatically.
 
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Old Apr 7, 2004 | 07:57 AM
  #9  
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Something to consider is the difference between temperature at the air filter and temperature in the intake manifold. Let's say the supercharger increases air temperature by 200F at 6500 rpm in 4th gear (I measured a similar worst-case rise a couple days ago). Now, let's say the stock intercooler is 60% efficient (again, a worst-case scenario).

Two conditions:

1) Air temperature is 70F at the filter, ambient temp is 70F
Intercooler inlet = 270 F
Intercooler outlet = 150 F

1) Air temperature is 80F at the filter, ambient temp is 70F
Intercooler inlet = 280 F
Intercooler outlet = 154 F

The difference between 150F and 154F is less than 0.7% difference in absolute temperature, or less than 0.7% difference in the number of air molecules in the manifold. Airflow is proportional to horsepower (all other things being equal), so a 200 hp MINI might be 201 hp with a 10 F reduction in temp at the air filter.
 
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Old Apr 7, 2004 | 08:02 AM
  #10  
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>>The difference between 150F and 154F is less than 0.7% difference in absolute temperature, or less than 0.7% difference in the number of air molecules in the manifold. Airflow is proportional to horsepower (all other things being equal), so a 200 hp MINI might be 201 hp with a 10 F reduction in temp at the air filter.

What if you increase the efficiency of the intercooler? Looking under the hood of a MINI from an air-in/air-out perspective, it seems a lot could be done here...
 
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Old Apr 7, 2004 | 08:31 AM
  #11  
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From: Lansdale, PA
Increasing the efficiency of the intercooler will bring the air closer to ambient temp.

E = (i-o)/(i-a)

o = i - (i-a)*E

So:

A) Air temperature is 80F at the filter, ambient temp is 70F, intercooler efficiency is 70%
Intercooler inlet = 280 F
Intercooler outlet = i - (i-a)*E = 280 - (280 - 70) * 0.7 = 133 F

B) Air temperature is 80F at the filter, ambient temp is 70F, intercooler efficiency is 80%
Intercooler inlet = 280 F
Intercooler outlet = i - (i-a)*E = 280 - (280 - 70) * 0.8 = 112 F

C) Air temperature is 80F at the filter, ambient temp is 70F, intercooler efficiency is 90%
Intercooler inlet = 280 F
Intercooler outlet = i - (i-a)*E = 280 - (280 - 70) * 0.9 = 91 F

Making the intercooler more efficient has a WAY bigger effect than lowering intake temps.

Using a 90 % efficient intercooler, we could get away with having inlet temps several hundred degrees hotter
 
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Old Apr 7, 2004 | 08:39 AM
  #12  
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I think that all of this theory is great. But what I am looking for is to keep summer IAT temps below the threshold at which the ECU retards timing and dumps fuel. Does anyone happen to know that number?

Not making hp but preventing its decline.
 
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Old Apr 7, 2004 | 09:21 AM
  #13  
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I logged my IAT and timing on the dyno and didn't really see the timing get retarded even though I got around 200F for IAT.


 
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Old Apr 7, 2004 | 09:25 AM
  #14  
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>>I think that all of this theory is great. But what I am looking for is to keep summer IAT temps below the threshold at which the ECU retards timing and dumps fuel. Does anyone happen to know that number?
>>
>>Not making hp but preventing its decline.

176 degrees F...can anyone confirm or refute this? Also wondering and want to be sure.

M
 
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