Dyno correction factor vs air density

[fullcircle69]

I posted the question on the thermodynamics forum because I did not feel it was really an automotive based question. I have still yet to confirm this so I am posting here geared more towards an automotive question. I am not trying to cross post so I hope the moderator understands, just looking for another point of view.


Basically I would like to know why the cf will indicate more power when calculated from two atmoshpere that have the SAME air density, but one is made from a higher pressure?

Atmophere A: 70F 29inHg density 1.163kg/m^3

Atmoshere B: 80F 29.547inHg density 1.163kg/m^3


My assumption based on the ideal gas law is that both atmoshperes will appear identical to the engine and therefore make the same power.

using the cf equation:
atmosphere A = 1.002
atmoshpere B = 0.990

So is the cf equation correct that atmosphere B will create more power or is the equation just an approximation and really should make the same power?

 

[JSteve2]

The correction factor uses pressure to the first power, and temperature to the 0.5 power, because those were the best fit. It's just an empirical fit, so it shouldn't be taken as anything exact. It's really not any more complicated than that.

There are various theoretical reasons why pressure and temperature have different effects on the combustion and realized power even where the same density is achived. However, that discussion is not really needed here, although it is somewhat baked into the equation used by nature of those powers being the best fit through the ranges the correction factor equation is relevant for.

 

[fullcircle69]

This was exactly my concern. I really can't reason why this should happen. I would think if the cylinder has the same amount of O2 it could make the same power.

So do you feel it is plausible that you may make the same power with the same density eventhough one air sample comprises a higher temperature or pressure?

 

[JSteve2]

You will not make exactly the same power if you vary the temperature and pressure, even though you have the same density of air going in. Among the things that change: heat transfer rate from the cylinder during combustion and power generation, rate of combustion (and crank angle degrees over which the power is generated), and exhuast gas temperature. Even volumetric efficiency probably varies but I'm not sure how signficantly.

For small variations in temperature or pressure, I would expect the difference in power generated to be insignificant for the same air density. However, for large variations in pressure/temperature, even where you net conditions that achieve the same density, I would expect the differences to be noticeable.

E.g. If we take the CF coefficients of pressure and square root of temperature, we should expect 100 hp at sea level and 100 deg F to correlate to roughly 93 hp at 4,000 feet and 22 deg F (which should be similar air density if my math was right). That's not earth shattering, but it's noticable, and it's on the order of magnitude of what we should expect based on the factors above that cause the differences.

 

[fullcircle69]

Just so that I am clear.

You do feel that there will be an independent temperature and/or pressure effect. And you feel that the cf equation, altough it may not be calcualting the horsepower exactly, it is correct that it is showing an increase in power when the temperature and pressure are higher at the same density.

In other words, the cf equation does NOT have enough error that it could have just as easily shown the power being less.

 

[JSteve2]

I'll try my best.

1) There will be independent temperature and pressure effects beyond just the effect on density.
2) The CF equation is matching a trend in a narrowly defined window of pressures and temperatures. Within that window, I think it is correct that, generally, a higher temperature/pressure at the same density will give you higher horsepower.

To respond to your last statement:
The CF equation matches a real-world situation with a smooth curve. Generally, it will trend in the right direction. There may be times, in a real engine, where a trend may briefly reverse (e.g. when a mode behavior occurs, such as a new fueling table being used at certain temperatures or pressures, an actuator saturates, etc.), but generally the CF equation is correct if used within the data ranges it is designed for.

 

[gruntguru]

Have you considered humidity? If the two air states you are considering have the same Relative Humidity, the water content (and therefore the oxygen content) will be different.

Previous posts are correct however. Elevated air temperature has less effect than reduced pressure (density being equal). For one thing, hotter air will absorb less heat (and therefore suffer less temp rise and density fall) during the intake process.

 

[fullcircle69]

Based on some other information I have come across:

If the value is opened for a long enough period of time to let the cylinders equalize with the atmosphere then the density in both cylinders will be the same. But if the valve is opened only for a short time then the rate at which air flows into the cylinder will be dependent on the pressure differential, thereby the atmosphere with the higher pressure will fill the cylinder with more air while the valve is opened and be able to make more power.

I am sure the higher temperature will also allow for easier combustion, but I think this is a much smaller effect then the pressure pushing the air in.

Does all this sound reasonable?

 

[JSteve2]

That does sound like one of the factors at play.

 

[black2003cobra]

The correction factor with the resulting square root of (absolute) temperature dependence is actually derived from the expression for 1-D compressible flow through a constriction. It isn't empirical.

 

[black2003cobra]

P.S. Regarding the duplicate thread, it has been recommended by other long timers here, (patprimer) that one red flags those themselves. Personally, I think this is the more appropriate forum for your thread. Cheers!

 

[Lou Scannon]

If the value is opened for a long enough period of time to let the cylinders equalize with the atmosphere then the density in both cylinders will be the same. But if the valve is opened only for a short time then the rate at which air flows into the cylinder will be dependent on the pressure differential, thereby the atmosphere with the higher pressure will fill the cylinder with more air while the valve is opened and be able to make more power.

Uhmm, my initial reaction to that is that it is fallacious to consider only the ambient pressure effect on the intake side, since the same ambient pressure is presumably also acting on the exhaust side.

 

[JSteve2]

Black2003cobra, that is just the theoretical underpinning for the curve fit. The coefficients for the equation are not determined on first principles, and they have been tweaked. The equation also only works for a narrow range of temperatures and pressures, since it is not a model.

Hemi, that depends upon the timing of valve events. You are not worried about energy to fill and empty the cylinder, you are primarily looking at how much air ends up in the cylinder at the end of the intake stroke. When your exhaust valve is closed, the differential pressure at the intake valve is dominated by the dynamic volume and intake manifold pressure, and it doesn't much matter what exhaust pressure you were working against. It will slightly change your dead volume, which is ignored, introducing some error.

 

[black2003cobra]

Hi Steve – What the OP is referring to is the correction factor as described in SAE J1394. (Please refer to his other thread in the Thermo forum). Actually, the coefficient and constant that appear in that expression are not determined empirically, either. They come about by assuming a mechanical efficiency of 85%, which is the default value used by the standard. The coefficient out front is simply 1/?m = 1/0.85 = 1.18, (rounded off), and the constant of -0.18 is simply from –(1/?m – 1) = -(1/0.85 – 1). The derivation follows from the expression for steady-state compressible flow by scaling only iHP, while keeping fHP constant. Humidity is taken into account by using only the partial pressure of dry air…vapor pressure is taken out. There are other assumptions. I think the original confusion on the OP's part is that downstream density is not equal to the upstream stagnation density.

 

[JSteve2]

black2003cobra,
Good info - thank you. Is it only good for a narrow range because that's where the assumptions are good?