Intake charge temperature v.s. power increase

cbutler

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Jan 31, 2006
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Does anyone know of a relationship between intake charge temperature and hp increase?
Or a relationship between mass flow rate of the intake charge and power would also work.
thanks.
 

Doward

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Jan 11, 2006
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You can't make a generalization like that - colder air makes more power due to an increase in density of the air.

100 degree air at sea level will make more power than 80 degree air up in the Rockies ;)

It's all about making the densest fluid you can!
 

Grimsta

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May 30, 2007
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Exactly, colder air is richer in oxygen engines like oxygen! Thats why my Soup loves it when its cold and rainy around here!
 

GC89

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Jun 13, 2007
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I agree altitude plays a huge role, even on stock boost seattle at 50 degrees was lots of fun compaired to 75 at 2000 feet in spokane. Look at true bar conversion altitude and temperature are actually factored in.
 

Grimsta

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I lived in Colorado for awhile and that just sucked the life out of my little Nissan pickup out there. And here, that little thing hauled ass. Lol it was a faaast little NA KA24E here
 

cbutler

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Jan 31, 2006
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thanks for the replies. I realize that altitude is definately a factor because of the lower atmospheric pressure, but I was wondering if there was a relationship assuming constant pressure.
 

Tun_x

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Apr 1, 2005
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A hotter air charge is also less stable and more prone to detonation!
 

Sawbladz

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Mar 14, 2006
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3p141592654;911803 said:
PV=nRT would be a first order start to this problem.
The Ideal gas equation would get you close but be careful with the assumptions made when using it. This equation neglects both molecular size and intermolecular forces (van der Waals). It is less accurate when using high pressures and lower temperatures.

A better equation would be the van der Waals equation with Avagadro's number.

(P + (n^2*a/V^2))(V-n*b)= nRT
 

cbutler

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Jan 31, 2006
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Sawbladz;911832 said:
The Ideal gas equation would get you close but be careful with the assumptions made when using it. This equation neglects both molecular size and intermolecular forces (van der Waals). It is less accurate when using high pressures and lower temperatures.

A better equation would be the van der Waals equation with Avagadro's number.

(P + (n^2*a/V^2))(V-n*b)= nRT

I have been working on ideal gas law calculations, but as you say there are some assumptions and parameters that are difficult to calculate or measure.
I will take a look at van der waals.
it's for a final engineering project by the way.
 

TurboStreetCar

Formerly Nosechunks
Feb 25, 2006
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I think i read somewhere that every one degree Fahrenheit you change the temperature equals about 1 horsepower. i believe this is also assuming equal barometric pressure and the only variable being temperature.

No idea if its totally true but i know for a fact during the summer, in the day time when its like 80-90 degrees my car definitely feels down on power compared to night time when its like 60-70. the difference was actually quite noticeable.
 

Calculatedrisk

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Feb 2, 2008
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You aren't taking the effect of density on the combustion characteristics into account. It isn't just a matter of more reactants, it's more pressure as well. It will increase the rate of flame travel.
 

Doward

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Allan -
Paragraph said:
Air temperature, in tandem with barometric pressure, has a direct bearing on air density. Warm air is less dense than cool air and consequently contains fewer oxygen molecules in a given volume. That's why we want our engines to breathe cool air through a scoop rather than heated air from beneath the hood. Temperature does matter: Horsepower increases approximately 1 percent for every 10-degree Fahrenheit reduction in temperature. Savvy racers know that their cars run faster in the cooler air of the late rounds even though the barometer doesn't change significantly.

Good find, but it's still a generalization ;)

Calculatedrisk, you may want to double check that. A denser air/fuel mixture burns slower than a less dense air/fuel mixture. Go dense enough, and you are burning basically liquid - go with a low enough density, and you have an explosion ;)
 

Calculatedrisk

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Feb 2, 2008
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No, that is incorrect, a denser mixture will always burn more quickly, and it can be as dense as you want it to be, too low a density won't cause any ill effects except a lack of power.

http://www.innovatemotorsports.com/resources/rich.php

http://www.stealth316.com/2-tmo2.htm

http://www.max-boost.co.uk/max-boost/ignition_deeper.htm

http://books.google.com/books?id=hi...ts=Ne8UTL6cCO&sig=c-_TMqBjyZjzD9fVISbODpXuI4o

If the a/f ratio is incorrect, you will slow down the flame because it has further to travel before it has the chance of encountering fuel/air molecules in the correct ratio to react (either you will get pockets of air with no fuel, or fuel with no air, so no reaction), but not related to density.