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An academic turbo question

B
Jan 11, 2013
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I have a 18 850 putting out 165 hp at sea level, losing 3% per 1000 feet of elevation so at 10000 feet it is putting out 115 hp. I have a turbo running at 5 pounds of boost and if I get 10 hp per pound for 50 extra hp(I know these numbers aren’t exact but they will do for a discussion). My question is, if I’m at 10000 feet with the turbo am I now at 165 hp or am I at 215 hp? This is what happens when you work with your riding buddies and get into these discussions. By the way I call this academic because I already have the turbo just have to put it on so I don’t really care what the answer is just curious.
 
N

nuttyn01

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Dec 3, 2007
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turbo hp

If you are running 5lbs at all elevations you can figure a 1% loss for every 1000' of elevation.

stock 165HP sea level
5lbs of boost 215HP sea level

at 10,000 feet you have lost 21.5 hp

If you run with Electronic boost control your boost increases or decreases relative to atmospheric pressure/elevation to maintain hp
 
B
Jan 11, 2013
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I don’t disagree with anything you said but the other question I have is that if sea level atmospheric pressure is 14.7 and 10000ft atmospheric pressure is 10.1 doesn’t the turbo have to get past the 4.6 loss before it can give more hp over stock? I just can’t wrap my head around where the math falls down.
 

kanedog

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If you are running 5lbs at all elevations you can figure a 1% loss for every 1000' of elevation.

stock 165HP sea level
5lbs of boost 215HP sea level

at 10,000 feet you have lost 21.5 hp

If you run with Electronic boost control your boost increases or decreases relative to atmospheric pressure/elevation to maintain hp
3% per 1000ft, not 1% so 215hp x 3%=65hp loss. Net hp @10,000ft=150hp.

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justinkredible56

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I just researched this a couple weeks back, this is what I found:

NA engines typically lose 3% power per 1k feet of elevation gained.

Turbo engines (unless they have active altitude compensation) will typically lose around 1% power per 1k feet of elevation gained.
 

kanedog

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I just researched this a couple weeks back, this is what I found:

NA engines typically lose 3% power per 1k feet of elevation gained.

Turbo engines (unless they have active altitude compensation) will typically lose around 1% power per 1k feet of elevation gained.
I dont agree as there is 3% less oxgen available to all parts of the engine. 150hp na will lose 45hp at 10,000 ft. Just cuz there's a turbo on it doesn't make it lose less hp. Now if you feed it more oxygen by cranking up the boost, it will recover that hp to get back to 150hp or more.

Sent from my SM-G920W8 using Tapatalk
 
B
Jan 11, 2013
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I dont agree as there is 3% less oxgen available to all parts of the engine. 150hp na will lose 45hp at 10,000 ft. Just cuz there's a turbo on it doesn't make it lose less hp. Now if you feed it more oxygen by cranking up the boost, it will recover that hp to get back to 150hp or more.

Sent from my SM-G920W8 using Tapatalk

I’ve heard this too but at sea level the atmosphere is 20.9% oxygen and at 10000ft it is 14.3 for a difference of 6.6% so I don’t think that the O2 plays into it that much because no matter how much boost I’m running it is still only 14.3% O2. But I don’t know, this is usually the point where my head starts to hurt.:)
 
S
Oct 15, 2008
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All other factors remaining constant, a turbo impeller shaft will turn 1-2% faster for every 1000ft of elevation increase due to the incoming cold air not being as dense. This is where the 1% loss per 1k originates when turbocharged. I would assume due to active air/fuel ratio changes as altitude increases/decreases that the exhaust temp/velocity doesn't change appreciably therefore creating enough push on the exhaust impeller to generate faster shaft speed due to a loss of air density at higher elevation. Have to remember that denser air takes more pressure to flow through the intake tract than less dense air. Kind of backwards to think about but it's true.

Essentially the same as trying to flow the same amount of any liquid through a smaller or larger hose, if you lower the density of the liquid, the same quantity of liquid will flow with less pressure required. if you then apply the same pressure as previously applied to the intake tract, air speed will increase significantly and more oxygen will be delivered by the air speed required to generate the same intake pressure as the lower elevation had. Less oxygen per cubic ft of air but significantly higher air speed compared to sea level.

I think it makes sense anyway but my wife tells me I'm wrong a lot so who knows.
 
M

mtn-doo

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Another way to visualize the O2 content and understanding what altitude and pressure play on it is;.

Visualize a balloon, 1 cubic foot of air in it at sea level. It contains a little over 20% oxygen.

Take that same balloon up to 10,000 ft. The "size" of the balloon becomes much larger because of the lack of atmospheric pressure... there is still one cubic foot of air in it. The lack of "pressurization" allows the cubic foot of air to "spread out". The balloon at 10,000 ft is much larger in size but still contains the same amount of oxygen. The oxygen molecules are simply spread out in a larger area. Spread out, farther apart.

You can easily see now that "pressurizing" the air , condenses the O2 content in it. A turbocharger simply compresses , "pressurizes" the air being fed into the combustion chamber. You can control this benefit by simply increasing or decreasing the boost. You can pressurize, "Boost", slightly to off set and maintain stock power at all elevations, or over pressurize to increase HP way over stock.

Of course you know increasing O2 requires an increase in fuel to maintain correct ratios.
 
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S
Oct 15, 2008
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I would assume that the air being less dense at altitude will allow/require a larger volume to move at any given time in order to maintain a given boost level above ambient pressure. This being the reason when the same boost level is maintained at both elevations, that the loss is less than the NA rate of 3% per 1K. More cubic feet of ambient air will flow through the engine at elevation when the same charge pressure is realized.

The ambient air in this case just so happens to be more compressible at elevation than at sea level therefore requiring more volume to achieve the same pressure differential(boost). Turbos are semi self compensating due to this principle.
 

Wheel House Motorsports

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Food for thought.

Ideal Gas law. PV=nRT

Boost psi is truely unrelated to horsepower. Pressure is just a measure of restriction in flow. All of this talk is under the assumption the turbo system is of 'average' efficiency and using a properly sized turbo charger. You can pump air through a really inefficient system and poorly sized turbo resulting in Boost PSI all generate from higher temperature. No actual flow gains (HP) were made, just increased pressure.

14.7 at sea level. 10.1 @10k. one would assume 4.6psi of boost would recover you to sea level HP. the problem is that a turbo is not 100% efficient. Yes you get back to 14.7psi of intake pressure. BUT due to compressing inefficiency, your going to have higher intake temperatures, given the same starting air temp. This means although its the same pressure, its slightly less dense air, so not full HP recovery. Most turbos when operating in an area of proper efficiency compress air at about 6-70% efficiency. Call it 66.6% (2/3). This is why a turbo recovers 2 of the 3% loss due to elevation (1% remaining loss vs the full 3% NA)
 
J

JJ_0909

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Nov 16, 2009
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Food for thought.

Ideal Gas law. PV=nRT

Boost psi is truely unrelated to horsepower. Pressure is just a measure of restriction in flow. All of this talk is under the assumption the turbo system is of 'average' efficiency and using a properly sized turbo charger. You can pump air through a really inefficient system and poorly sized turbo resulting in Boost PSI all generate from higher temperature. No actual flow gains (HP) were made, just increased pressure.

14.7 at sea level. 10.1 @10k. one would assume 4.6psi of boost would recover you to sea level HP. the problem is that a turbo is not 100% efficient. Yes you get back to 14.7psi of intake pressure. BUT due to compressing inefficiency, your going to have higher intake temperatures, given the same starting air temp. This means although its the same pressure, its slightly less dense air, so not full HP recovery. Most turbos when operating in an area of proper efficiency compress air at about 6-70% efficiency. Call it 66.6% (2/3). This is why a turbo recovers 2 of the 3% loss due to elevation (1% remaining loss vs the full 3% NA)

This is pretty good.

But really MAP (manifold absolute pressure) is manifold absolute pressure. The motor doesn't know what elevation its at, and pressure is pressure. So if you are achieving a MAP of 14.7 (sea level), its the same be it "natural" or "forced".

The big exception here is charge temps, which is a variable. We could get into the math behind this, but I'd strongly argue for sleds, especially the way we generally ride them these days, its not as big of a factor (until it is ;) ) as WHM suggests. It becomes a much bigger factor on those huge pulls in deep snow. Not so much of a factor with the in-and-out of the throttle tree riding with an occassional WOT burst we usually are after these days.

This is what I've seen from our data logging nayway.
 
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jim

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Nov 26, 2007
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Losing HP at elevation is based on pressure loss...and less density...which means less combustion power.

Simply, if you have 165HP at 14.7PSI, I would argue that you have 11.2 HP per PSI. So, going up high to 10 PSI ATM, means you have 112HP. Add 5 lbs and you are back to stock. Add more (most PG kits add 8 to 9 PSI that high, and you get closer to that 200HP mark.

That said, you have losses and heat from pressurizing the air. I don't think the 200HP PG thing is really that valid...but that is just me. Seat of the pants, I'd say most PG kits are more like 180 to 190...but the difference is the torque and that the motor just bulldozes through what would normally drop track speed...holds consistent Rs and track speed during the climbs...which all adds up.

Some might argue...but, yes, with turbo efficiency losses (heat and flow) and equating to the pressure lost at elevation, I don't see the 200HP claim either. And don't feel it.

Now, you want to feel HP...ride a 4 stroke at 20lbs on race gas. That is big HP.
 

tdbaugha

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Mass flow rate is what dictates hp. PV=nRT. n is what you're calculating. P is a small part of the puzzle
 
J

jim

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PV=nRT is based on static, Newtonian fluid characteristics.

A turbo, as many have pointed out, is a dynamic monster where the most accomplished PHds using the most sophisticated flow analysis tools can only begin to scratch the surface when it comes to full comprehension.

That said, because this is dynamic and includes a motor (you model the combustion cycle with heat in/heat out and pressure/volume differentials the turbo simply modifies), the 2nd law of thermodynamics kicks in and you have Volume in, volume out, Pressure in, Pressure out, volume, and heat (Q) and disorder/entropy losses.

More like the below (and, no, I'm not a smarty pants...I barely passed the test for this stuff...it is incredibly complex even when watered down like this and the math is ridiculous):
attachment.php


Regardless...it is fun to talk about.
 

Schanny

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Does anyone on here Fly a Turbo'd Airplane that goes from sea level to 16 or 17k feet?
 
X
Oct 8, 2009
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Everyone is noting some great thoughts here. And, it looks like someone is putting his thermodynamics to work. But, it is best to address the issue in the simplest terms possible, so those who are not engineers do not pass out.

Mass flow dictates HP. All of the reasons stated speak to sampling mass flow. The only thing left to say is turbos create restriction to exhaust flow, requiring more boost and increasing piston temp. Turbine sizing is a critical component to evaluate power output. There is not a all size fits all rule of thumb for HP loss with elevation since changing back pressure (turbine sizes) greatly influences power output. In the end, you have to do the math, data log, or dyno it. Unfortunately, too many people are focused on compressor wheel flow rates, which is why companies market HP ranges for turbos on compressor wheel size. In reality, the turbine wheel and area radius dictate power level. Only then should the compressor wheel be matched to the mass flow rate demanded by the motor.

In short, think about this in rough terms. If I lose 1 psi MAP, I need roughly 1 psi to get back where I started. Boosted engines, like NA engines, lose power with elevation if boost is static. Electronic Boost Control just ensures MAP remains constant; the gauge only changes. Finally, who cares what the HP is if it feels good to you. Most companies inflate dyno numbers anyway by manipulating dyno inputs and reporting peak numbers.
 
X
Oct 8, 2009
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I also wanted to say great post to Jim. You are right about your assumptions. Most pump gas setups fall short because they have to over come flow restrictions from small turbos and pumping losses in addition to the loss in atmospheric pressure. At 9000 ft, 200 hp is more around 9 psi; a little higher or lower depending on turbine efficiency (turbo size).

To your point, if you want a point of reference, try running next to a properly setup turbo yamaha or doo 1200 turbo up a steep hill with lots of powder. The big sleds will out pull a turbo 800; especially at 20 psi. Haha! At lower boost, the 850 is more fun to ride. At high boost (20+), RIP turbo 850.
 
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