Altitude effects on NA vs. FI

[flyboy86]

So being here at 7,248 feet above sea level, I notice a significant drop in HP from sea level. I'm starting to consider a blower, and I got to thinking, which setup notices a bigger decrease in power when it is brought to altitude, NA or FI?

It makes sense to think of it mathematically, clearly, an NA car brings in X amount of air, and combusts X amount of air. The super/turbo-charger is what i don't understand as great. Does it bring in 2X air? or Xsquared?

My first guess is that the FI car would see a bigger drop in power, because it would amplify the change in air density. (i.e., an exponential drop with FI compared to linear with NA). I clearly don't know enough about how turbos or superchargers work. Can anybody help with this question? I plan on asking my propulsion professor tomorrow in class.

 

[Dusstbuster]

....Depending on what you're asking...I only know 1 thing. Put a blower/turbo on the car and the drop in power due to altitude won't matter compared to naturally aspirated.

 

[thomas91169]

it wont be as much of a loss as compared to n/a. i would elaborate but i dont have the mental capacity right now to write a novel on forced induction theory 201.

 

[droid#83853]

The charger is forcing more air in, making up for the thinner air at altitude. You still would make more power at sea level though. Your neighbors must be mountain goats at that height.

You are in the gotdamn AF Academy. This is so sad. You should know about this sort of stuff. Go ask one of your instructors about turbocharged and non-turbocharged piston aircraft engines. What are they teaching you? If you want to be a pilot, you must know this stuff. It is in the FAA private pilot exam. End comments.

 

[flyboy86]

The charger is forcing more air in, making up for the thinner air at altitude. You still would make more power at sea level though. Your neighbors must be mountain goats at that height.

You are in the gotdamn AF Academy. This is so sad. You should know about this sort of stuff. Go ask one of your instructors about turbocharged and non-turbocharged piston aircraft engines. What are they teaching you? If you want to be a pilot, you must know this stuff. It is in the FAA private pilot exam. End comments.

Man, I'm sorry. However, I don't think you understand what i'm asking. Yes, I fully understand that a turbocharged plane will perform better at high altitude compared to a non turbocharged plane. Anybody can understand that. Thats not my question.

Take a 300 HP NA V8 at sea level, and a 300 HP turbocharged 4 banger. Then bring both of them up to say 15,000 feet or so. Which one will loose more power? Thats my question. This is what I was thinking...

Lets say at sea level, atmospheric pressure is 2 psi. The NA car then combusts that same 2 psi. But lets say the turbo boosts it to 10 psi. Now, lets take those two cars up on top of a mountain. Now the atmospheric pressure is 1 psi. In this situation, the NA engine would be seeing 1 psi, for a total drop of 1 psi. However, if the turbo works like I think it does, it would be turning that 1 psi into 5 psi, for a drop of 5 psi from sea level. So if those assumptions are correct, the FI car saw a decrease of 5 psi, where the NA car saw a drop of 1 psi.

I'm sorry if I disappointed you with my question. Maybe I should think harder about my questions first.

 

[droid#83853]

I shouldn't have come off that hard. You have a lot to live up to in a service academy. I see what you are asking. No, the power loss is not quite like that. At 1 ATA the n/a puts out 1 ATA, and the turbo would put out say 3 ATA. At 0.5 ATA the n/a sees around 0.5-0.6 ATA, whereas the turbo would put out something like 2.7 ATA. Let me find the formula. I have it somewhere. The curve is dynamic of course. There are a lot of variables, but I am sure you get the general idea.

 

[LaserRed01GT]

I can appreciate your question, but I think you are focusing on the wrong side of this. I doesn't matter which application loses the most net power, it's which application makes the most net power.

 

[04YELLOWGT]

I can appreciate your question, but I think you are focusing on the wrong side of this. I doesn't matter which application loses the most net power, it's which application makes the most net power.

ding, ding, ding. We have a winner!

 

[flyboy86]

I shouldn't have come off that hard. You have a lot to live up to in a service academy. I see what you are asking. No, the power loss is not quite like that. At 1 ATA the n/a puts out 1 ATA, and the turbo would put out say 3 ATA. At 0.5 ATA the n/a sees around 0.5-0.6 ATA, whereas the turbo would put out something like 2.7 ATA. Let me find the formula. I have it somewhere. The curve is dynamic of course. There are a lot of variables, but I am sure you get the general idea.

I understand what you're saying, but I don't buy it yet. So far, nobody can explain WHY the FI engine can maintian its power. What you just described means that the turbo is getting more efficient with a decreased pressure. If someone can prove me wrong, i'll eat my words, but the more I think about it, the more I convince myself that with equally powered cars at sea level, the NA car will retain more of its power at altitude. Here is another way to think about it......

An N/A V8 derives its power from combusting large volumes of lower pressure air, wheras a turbo'd 4 banger gets the same power from combusting small volumes of higher pressure air. So from that, you can say that the power is a function of two things, (assuming the same octane gas, etc) , which are displacment and density of the air. As you increase the altitude, the density changes, but the displacement does not.

The turbo'd 4 banger makes its money based on a high density of air. The V8 does it with a large displacment. Now, on top of a mountain, the V8 still has its displacment, but the turbo lost its advantage with a decreased density.
So with increased altitude, one thing remains constant, displacment, and that happens to be the advantage of the V8 over the 4 banger.

Like I said, I could be completely wrong, but I would like somebody to be able to explain why I am wrong. Any thoughts?

 

[Dusstbuster]

Think of it like this...

As both cars got higher and higher, eventually the naturally aspirated car wouldn't have enough air to run off and would die.

The turbo/supercharged car would still be able to go higher because the boosting apparatus would cram more air in and allow it to go higher.

Thinking of it that way, you can see the NA car is losing its power where the FI car isn't. If it works this way to the extreme end I don't see why it wouldn't work this way as the altitude increases closer to earth.


here is an excerpt from Borg Warners website too:

The high-altitude performance of a turbocharged engine is significantly better. Because of the lower air pressure at high altitudes, the power loss of a naturally aspirated engine is considerable. In contrast, the performance of the turbine improves at altitude as a result of the greater pressure difference between the virtually constant pressure upstream of the turbine and the lower ambient pressure at outlet. The lower air density at the compressor inlet is largely equalized. Hence, the engine has barely any power loss.

Also read this. I don't want to but after a quick skim it looks like it might have some answers for your question.
http://web.usna.navy.mil/~dfr/flying/turbo.pdf

 

[007]

Hmm. Got me thinking. I may be completely off base but this seem to make sense to me...

NA sucks in a certain voulme of air. Higher alt = lower air pressure = less oxygen = less power.
Exactly the same is true for FI.

But it gets interesting because you can adjust the boost of your blower by changing the pulley...

Where you live you have atm press of about 11PSI, vs about 15PSI at sea level. When we talk about boost we are talking about differential pressure between the atm and the intake. I.e. 9PSI boost at sea level makes the absolute intake air pressure about 24 PSIa. The same boost at your altitude is going to get you 20 PSIa, thus the power difference. But surely you could run a 13lb pulley, and at your altitude you'd make 11+13=24PSI in your intake... exactly the same as a 9PSI kit at sea level.

In that case you just work the blower harder to make up for the low atmospheric pressure. You don't work the motor any harder than the 9 PSI kit at sea level (since the pressure in the motor is identical), and you should make about the same power.

Seems to me that running boost at high altitudes allows you to negate the effects of the thin air by running a more aggresive pulley. Of course youll have too much boost if you drive down to sea level, and you'll blow your motor up!

Thoughts?

 

[LaserRed01GT]

Well, there is a big difference between the dynamics of pressurizing the intake manifold with a blower and pressurizing the intake manifold with a turbo. With a turbo setup, you have the luxury of making more boost by means of controlling the waste gate with a manifold vacuum reference, pneumatics, or electrical. With a turbo setup, vacuum is highest when the throttle blades are closed, then as you open the throttle, the absolute pressure within the manifold begins to increase(vacuum decreases). The more the vacuum within the intake manifold, the tighter the diaphragm is within the wastegate valve and consequently the more exhaust gas is bypassed from the turbine. The more exhaust gas that is bypassed, the slower the turbine speed and the less boost you will make. So at idle or relatively low throttle position, the wastegate keeps the turbos from making boost that would be unnecessary at that particular engine load/throttle position.

Just the opposite applies when you step on the loud pedal. The vacuum is immediately decreased, and the wastegate begins to close and allow more exhaust gas to go by the turbine making steadily increased boost.

So, as you go higher and higher into elevation, the atmospheric absolute pressure will obviously become less. The wastegate can “sense” this, and appropriately control the boost so as to keep the boost up. Now granted, the boost gauge observed pressure and the specific density of the air are not a linear relationship as compared to a third axis being elevation, but you will not be losing that much power at 7,000+ ft with a turbo and properly setup wastegate. Now a blower is a different animal. The only way to get more boost from a blower is to spin it faster. That either means increasing the crank speed to blower speed pulley ratio, or changing the internal step up ratio with the reduction gearing inside the blower.

Let’s say you take a car that makes 300 RWHP N/A at sea level up to 7,500 ft, it will make roughly 232.5 RWHP or sustain a 22.5% loss.
Take a turbo charged car that makes 500 RWHP at sea level up to 7,500 ft, it will make roughly 460 RWHP or sustain a 8% loss. The mathematics involved in finitely modeling the compressor maps at various elevations, or more specifically different atmospheric absolute pressures, can get very in depth and very complex. The majority of the relatively small loss from the turbo motor is not directly because of the less dense air, but because of the slightly decreased volumetric efficiency of the motor due to the higher turbine speeds required to maintain the boost.

But, suffice it to say, the turbo is much better as your net (overall) power will still be higher than N/A no matter what the elevation (atmospheric absolute pressure

 

[flyboy86]

That was the kind of answer I was looking for, thanks! Kind of funny, I heard the exact same explanation twice today. I just got back from Aero 361, which is a propulsion class. Today was a lab, and we fired up a J69 turbojet. Wow, that was really cool. Also, they had in the lab a 454 with a blower on it that makes 1200 hp, the best part is that they use two of them just to start an F100 turbofan. Imagine, 2400 hp just to start the damn thing! Anyways, there is a PhD down there, who runs the engine lab, and of course I asked him the same question I posted on here. I got the same answer that LaserRed01GT just posted.

I think the reason I had it backwards is because I didn't understand how a turbo works. I did not know that they make more pressure than the engine can use. I was asuming that a blower or turbo took atmospheric air in, and compresses it a certain ratio, and I thought that ratio stayed constant.

Thanks for the help guys, I learned a lot from this question I had.

 

[thomas91169]

close.

a na car will only be able to suck in so much air and lean out the fuel mixture so much.

when going to FI you are adding fuel in to the mix whenever you add more air. now when you get to higher altitudes, you still lose oxygen at the same rate, but since more air is getting forced in, you can still maintain proper a/f ratios, whereas the n/a car would just plain run out of enough oxygen to create proper a/f ratios.

the most ultimate thing would be a sensor that can detect oxygen ratios as the altitude gets higher, and re-map the fuel and timing curves to adjust and maintain the proper a/f ratio. since most cars use the amount of air coming in to run their fuel and timing maps, the ecu has no way of knowing the amount of oxygen coming in.

 

[LaserRed01GT]

a na car will only be able to suck in so much air and lean out the fuel mixture so much.

when going to FI you are adding fuel in to the mix whenever you add more air. now when you get to higher altitudes, you still lose oxygen at the same rate, but since more air is getting forced in, you can still maintain proper a/f ratios, whereas the n/a car would just plain run out of enough oxygen to create proper a/f ratios.

the most ultimate thing would be a sensor that can detect oxygen ratios as the altitude gets higher, and re-map the fuel and timing curves to adjust and maintain the proper a/f ratio. since most cars use the amount of air coming in to run their fuel and timing maps, the ecu has no way of knowing the amount of oxygen coming in.

It has absolutely nothing to do with air/fuel ratios.

All fuel injection cars that are equipped with an electronically monitored powertrain control module can closely estimate the altitude in which the vehicle is being operated. They all do it in various ways using some complex thermodynamic algorithms, but by using a combination of input from the oxygen sensors, TPS, IAT, MAF, MAP, and BARO sensors, they can closely estimate the altitude and adjust the fuel trims respectively.

Now we don't have BARO or MAP sensors, but Ford Mustangs have some unique thermodynamic algorithms within the EEC that use data from the DPFS(Differential Pressure Feedback Sensor) sensor to also make some calculations to closely estimate the current operating altitude. The DPFS is primarily used by the EEC to report EGR exhaust gas flow observations, but it is also used to estimate altitude. Fuel injection computers have to know what the air density and absolute pressure/altitude are in order to correctly estimate the volumetric efficiency tables and variables.

In automobile engines, you would have to be on top of Mt. Everest for a N/A motor to stop running from lack of Oxygen.

 

[flyboy86]

You seem to know alot about the subject. How does altitude effect the octane? I live and do most of my driving at 7,000 feet, should I be using a higher or lower octane fuel?

 

[thomas91169]

It has absolutely nothing to do with air/fuel ratios.

but by using a combination of input from the oxygen sensors, TPS, IAT, MAF, MAP, and BARO sensors, they can closely estimate the altitude and adjust the fuel trims respectively.

 

[LaserRed01GT]

You seem to know alot about the subject. How does altitude effect the octane? I live and do most of my driving at 7,000 feet, should I be using a higher or lower octane fuel?

Theoretically you should be able to use the lowest pump octane available at higher altitudes like yourself.

Our octane ratings here in the US are advertised as the average of the RON and MON. The RON, or F1, is basically a measure of how resistant gasoline is to detonation (spark knock). It is measured relative to a mixture of 2,2,4-trimethylpentane (iso-octane) and n-heptane. So a gasoline at the pump rated at 87 has the same knock resistance as a mixture of 87% isooctane and 13% n-heptane. Iso-octane by itself is relatively resistant to detonation, whereas n-heptane by itself is not, but the combination of both, and more importantly the mixture ratio, is what’s important. Race gas, or octane ratings higher than 100, are found by measuring the amount of tetraethyl lead that must be added to pure iso-octane to exude the same resistance to spark knock as a sample fuel without tetraethyl lead.

Lots of people get wrapped up in assuming that their static compression ratio is the most important determining factor in detonation resistance….it’s not. The most important effect on detonation is the absolute or effective compression. Static compression and even dynamic compression calculations are just theoretical whereas the effective or measured compression pressures are absolute and are what’s important. Given that, and being that your car is naturally aspirated, your motor should have a lower effective compression at your altitude as compared to sea level. Since the incoming air charge on a n/a motor is essentially governed by its pumping efficiency, your actual cylinder pressures will be less than at sea level. This means the PCM will add less fuel and theoretically there should be much less chance of detonation.

I have only driven a car once at that altitude when I drove from Seattle to Yakima to purchase a race car, but I could feel the difference in power when I was going through the mountains. But, you should be fine on 87 octane.