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Question for turbo gurus

Joined
8 July 2002
Messages
2,493
I am not knowledgeable about turbocharging.
Is there a rule of thumb equating a turbocharged engine power to a larger NA engine?

I believe a 3 litre NSX NA engine gets it's cylinder fill from atmospheric pressure of 14.7 lbs/sq.in. or 1 atmosphere
Does the same 3 litre engine if turbocharged with a 9 lb boost have about 1.6 atmospheres of pressure for the cylinder fill?

Does that mean the power output would be about the same as an NA engine with 60% more capacity?
i.e. 1.6 x 3 litres = 4.8 litre equivalent?
 
No, because there isn't a direct relationship. Mainly because it's doesn't quite scale like that. An N/A engine's cylinder is created by a vacuum on the intake stroke of the piston, while a boosted car is filled with inlet pressure; neither have much relation to atmospheric pressures. The other thing to remember is that the air is compressible so the relationship with volume and density is not linear. Combustion is not linear in scale either. Also, remember that it's not the air that produces power, but the fuel (gasoline) that has the energy potential, so it's more dependent on how much fuel you can utilize. Too many factors related to power production that will determine what you can develop and what it is comparable to; air volume only being one of them.
 
I am not knowledgeable about turbocharging.
Is there a rule of thumb equating a turbocharged engine power to a larger NA engine?

I believe a 3 litre NSX NA engine gets it's cylinder fill from atmospheric pressure of 14.7 lbs/sq.in. or 1 atmosphere
Does the same 3 litre engine if turbocharged with a 9 lb boost have about 1.6 atmospheres of pressure for the cylinder fill?

Does that mean the power output would be about the same as an NA engine with 60% more capacity?
i.e. 1.6 x 3 litres = 4.8 litre equivalent?

if you are looking to do the math, here is a handy equation I use,

cid x rpm / 3456 = CFM * 0.086 = lb/min

For example
355x6000/3456 = 616~ * 0.069 = 42. lb/min, or 420 flywheel horsepower

To factor in boost now, multiple lb/min by pressure ratio, so for 14.5psi of boost you multiple by 2.0 so the total would be 840 flywheel horsepower. That is what a 355 cubic inch motor with 100% volumetric efficiency at 6000rpm with 14.5psi of boost should make at the flywheel. You can multiply by .85 to reflect rear wheel horsepower (take out 15% or so).

The 'trick' here is in the turbocharger constant, or 0.086, which takes into account the air density vs temperature (you can change this number around to reflect colder air if you know what you are doing). The equation assumes the turbo is also in an efficient range and of course the 100% VE of the engine (which many engines do not have when they are that large at that rpm).

Another 'quick' way to do the dirty math is simply double the output for every 14.5psi of boost, then subtract roughly 10-20% for 'losses' due to air temp and so forth. So if you had a 2.0L engine which made 150 horsepower, and you push 14.5psi boost at it, you could just double it to 300 then take out a little bit (say 15%) for air temp losses. And walaa- my 2.0L engine with 15psi makes about 250rwhp, so figure 280 to the flywheel, so our initial estimate of 300 minus 15% or so was nearly exactly right.
 
if you are looking to do the math, here is a handy equation I use,
cid x rpm / 3456 = CFM * 0.086 = lb/min
For example
355x6000/3456 = 616~ * 0.069 = 42. lb/min, or 420 flywheel horsepower
To factor in boost now, multiple lb/min by pressure ratio, so for 14.5psi of boost you multiple by 2.0 so the total would be 840 flywheel horsepower. That is what a 355 cubic inch motor with 100% volumetric efficiency at 6000rpm with 14.5psi of boost should make at the flywheel. You can multiply by .85 to reflect rear wheel horsepower (take out 15% or so).
The 'trick' here is in the turbocharger constant, or 0.086, which takes into account the air density vs temperature (you can change this number around to reflect colder air if you know what you are doing). The equation assumes the turbo is also in an efficient range and of course the 100% VE of the engine (which many engines do not have when they are that large at that rpm).
Another 'quick' way to do the dirty math is simply double the output for every 14.5psi of boost, then subtract roughly 10-20% for 'losses' due to air temp and so forth. So if you had a 2.0L engine which made 150 horsepower, and you push 14.5psi boost at it, you could just double it to 300 then take out a little bit (say 15%) for air temp losses. And walaa- my 2.0L engine with 15psi makes about 250rwhp, so figure 280 to the flywheel, so our initial estimate of 300 minus 15% or so was nearly exactly right.

Okay so a 2 liter engine naturally aspirated has 1 atmosphere (14.7 lbs/sq in) of fill pressure and makes 150 hp.
With boost we have 14.7 lbs/sq in of natural pressure plus boost of 14.5 lbs/sq in so we have 150 hp base plus 150 hp less about 15 % boost loss for a total of about 280 hp.

Looking at our 3 litre C30 engine with a 270 base hp, adding 14.5 of boost would add another 270 x .85 for a total of about 500 hp.
So far so good?

Now let's look at the new 3.5 litre NSX engine.
It's rated at 143 hp/litre for a total of about 500 hp.
The current Porsche 4 litre GT3 naturally aspirated engine puts out about 125 hp/litre or about 500 hp.
The Ferrari 488 3.9 turbocharged litre engine puts out about 169 hp/litre or about 661 hp.

So if we use the Porsche NA output as a base for a modern direct injection engine and apply that output on the NSX 3.5 litre engine we'd get about 437 hp unboosted.
So the apparent gain on the NSX engine with boost is about 64 hp in total or 18 extra hp per litre.
Using your rule of thumb formula with 14.5 lbs of boost the new NSX engine should yield about 437 x .85 or about 370 extra hp.
As we're only seeing a 64 hp gain does that mean Honda is only applying about 2.5 lbs/sq in of boost to the new NSX engine?

Using your formula it would look like the Ferrari engine is using about 7 lbs of boost.

- - - Updated - - -

An N/A engine's cylinder is created by a vacuum on the intake stroke of the piston, while a boosted car is filled with inlet pressure; neither have much relation to atmospheric pressures. The other thing to remember is that the air is compressible so the relationship with volume and density is not linear. Combustion is not linear in scale either. Also, remember that it's not the air that produces power, but the fuel (gasoline) that has the energy potential, so it's more dependent on how much fuel you can utilize. Too many factors related to power production that will determine what you can develop and what it is comparable to; air volume only being one of them.

My basic understanding is the descending piston in a cylinder on the intake stroke creates a low pressure center.
Atmospheric pressure of 14.7 lbs/sq/in forces an air/fuel mix into this low pressure area and so we have a cylinder fill of some efficiency depending on intake design etc.

I'm reading that in a forced induction system we have the usual atmospheric pressure plus the extra pressure created by the boost impeller and this forces more air fuel mix into the cylinder then you would get unboosted.
So we get a larger volume of air fuel mix in the chamber and when it ignites there is more gas volume to expand and this creates more force on the piston on the power stroke hence more hp.
And he more volume of air/fuel mix that can be crammed into the cylinder on the intake cycle the more power you can make.

I realize there are a great number of variables like air temperature, intake design and so on affecting how efficient the cylinder fill is under boost.

If a modern Porsche GT3 engine make 125 hp per litre naturally aspirated, then if Porsche wanted 690 hp like the Ferrari 488 it would seem they would need about a 5.5 litre NA engine.

In the case of new new NSX engine at 3.5 litres if it made the same NA power as the Porsche GT3 engine it would make 3.5 l x 125 hp/l or about 430 hp.
It's boosted hp is rated at 143 hp/l so about an 18 hp/l gain from the boost and would equate to about a 3.8 l Porsche NA engine.

Would it seem to you that the new NSX engine is a conservative build or am I missing a salient point?
 
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I don't see any reason to assume that the i boosted NSX engine (or any engine designed for boost) would have the same specific output as the GT3 engine.
 
I don't see any reason to assume that the i boosted NSX engine (or any engine designed for boost) would have the same specific output as the GT3 engine.

I'm not sure I understand what you're saying.
Are you saying that the NSX engine built for boost and has a lower compression ratio would mean it's output in NA form would be lower than the GT3 engine?
If so, what would you say the NSX engine with a 10:1 compression ratio would produce as a base power output?
 
Are you saying that the NSX engine built for boost and has a lower compression ratio would mean it's output in NA form would be lower than the GT3 engine? If so, what would you say the NSX engine with a 10:1 compression ratio would produce as a base power output?
Yes, the compression ratio seems to be one of the first-order effects for specific output. I believe the 991 GT3 has a compression ratio of 12.9:1, which an engine designer should be able to use to raise the specific output as compared with an engine with a 10:1 compression ratio. Of course, there must be factors affecting specific output other than compression ratio (e.g, scavenging efficiency, inlet tuning, flow losses) but I am not knowledgeable about the range or relative significance of those.
 
Yes, the compression ratio seems to be one of the first-order effects for specific output. I believe the 991 GT3 has a compression ratio of 12.9:1, which an engine designer should be able to use to raise the specific output as compared with an engine with a 10:1 compression ratio. Of course, there must be factors affecting specific output other than compression ratio (e.g, scavenging efficiency, inlet tuning, flow losses) but I am not knowledgeable about the range or relative significance of those.

I think I get the drift now.
Porsche is running a 12.9:1 ratio and is getting 125 hp/l
The new NSX is at 10:1
The Ferrari 488 is 9.4:1 but I read will get boost as high as 1.8 bar, about 26 lb/sq in at peak boost.

So The 488 engine in NA from would likely have the lowest power output per l followed by the NSX assuming all other things constant.
But as you say determining the base output of the 488 and NSX is not easily determined.
Modern engines are very complex no?
 
My basic understanding is the descending piston in a cylinder on the intake stroke creates a low pressure center.
Atmospheric pressure of 14.7 lbs/sq/in forces an air/fuel mix into this low pressure area and so we have a cylinder fill of some efficiency depending on intake design etc.

You are going the right way. Here are some more ideas for you to consider.

1. The weight of the engine and its drivetrain plays a role. Two 4L engines or 2L engines can have dramatically different RWHP numbers because one of them could be 4WD or have a heavier rotating assembly.
2. the compression ratio of the engine plays an enormous role. boost pressure raises compression and when you start off with a high static ratio and add boost the power output can increase dramatically beyond what the basic math formula will predict, but these situations also typically call for better fuel ("race fuels")
3. The max RPM of the engine is very important, as in the math above the final RPM will tell us the max flow in mass/time of a given engine before boost is applied.
4. IF the engine needs to draw air using it's piston, this is a sacrifice to engine output, aka parasitic loss. A turbocharger is able to improve fuel economy because it will help push air into the cylinder, lowering the amount of energy required to pull in air. This is not a "power consideration" i.e. you will not make significant more power on a dyno by negating this loss; but it does add up over time while cruising in the form of MPG.
5. The more output you make per displacement, typically the less efficient the engine will become while using higher and higher boost pressures. In other words, a small engine, say 2.0L, making 300hp/liter, will be throwing more fuel/cooling at the combustion reaction than the same engine only making 200hp/liter with the same fuel. You probably already know that boosted engines run with richer a/f ratios. This is not to say that they would not make more power with a higher(leaner) a/f ratio; The cooling effects of say methanol for example, reduce the temp of the combustion reaction to make it safer/slower and this results with a loss of energy, kind of like throwing away $$ to keep the engine happy. Temperature rise = $$ from fuel, you use the fuel to generate the temp rise so this is in essence the $$ you spent on the fuel going up in flames.
 
You are going the right way. Here are some more ideas for you to consider.

1. The weight of the engine and its drivetrain plays a role. Two 4L engines or 2L engines can have dramatically different RWHP numbers because one of them could be 4WD or have a heavier rotating assembly.
2. the compression ratio of the engine plays an enormous role. boost pressure raises compression and when you start off with a high static ratio and add boost the power output can increase dramatically beyond what the basic math formula will predict, but these situations also typically call for better fuel ("race fuels")
3. The max RPM of the engine is very important, as in the math above the final RPM will tell us the max flow in mass/time of a given engine before boost is applied.
4. IF the engine needs to draw air using it's piston, this is a sacrifice to engine output, aka parasitic loss. A turbocharger is able to improve fuel economy because it will help push air into the cylinder, lowering the amount of energy required to pull in air. This is not a "power consideration" i.e. you will not make significant more power on a dyno by negating this loss; but it does add up over time while cruising in the form of MPG.
5. The more output you make per displacement, typically the less efficient the engine will become while using higher and higher boost pressures. In other words, a small engine, say 2.0L, making 300hp/liter, will be throwing more fuel/cooling at the combustion reaction than the same engine only making 200hp/liter with the same fuel. You probably already know that boosted engines run with richer a/f ratios. This is not to say that they would not make more power with a higher(leaner) a/f ratio; The cooling effects of say methanol for example, reduce the temp of the combustion reaction to make it safer/slower and this results with a loss of energy, kind of like throwing away $$ to keep the engine happy. Temperature rise = $$ from fuel, you use the fuel to generate the temp rise so this is in essence the $$ you spent on the fuel going up in flames.

I can see designing, building and tuning a boosted engine is very complex.
With so many variables it's hard use simple rules of thumb to come up with estimated power output.
Thank you for taking the time to explain all this.
 
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