Turbo Rule of thumb

[Vortex347]

Hi all, anyone able to give me a basic rule of thumb power and torque increase for various boost pressures on a turbo
e.g. 20lbs- 10% power and 7% torque increase <<<Just an example (not meant to be correct)
Boost pressures between 8 and 86lbs is the goal (8,14,18,26,32,40) - whatever you want to do be awesome if you could do every one between 8 and 86lb or give me a derivable formula (rule of thumb) for torque and horsepower based off of boost pressure.

Cheers guys and Thanks in advance

[Brian Coat]

To a first approximation it will scale linearly to ABSOLUTE manifold pressure.

This assumes the engine is not limited by knock or fuel flow (both highly relevant in F1).

[flynfrog]

look at the HP equation


H. P.= PLAN/33,000 in which:

P. Mean effective pressure in pounds per square inch.

L. Stroke in feet.

A. Piston area in square inches.

N. Number of power strokes per minute.

You are increasing the P. So for every 1 BAR or ~14.7 PSI you add your NA hp. Example a 100Hp NA engine will make 200 at 14.7 PSI This is ignoring all losses but gets you in the ballpark.


http://chestofbooks.com/crafts/popular- ... Power.html

[gruntguru]

To the above replies, I would like to add "temperature". To be more precise CAT (charge air temperature) the temperature of the air as it enters the cylinder head.

So MAP (Manifold Absolute Pressure) and CAT are equally importantant - why? Because what supercharging does, is increase the density of the air entering the engine. Denser air means more molecules in each litre and that means more oxygen. Supercharging does not change the VE of the engine. (By "VE" I mean the volumetric airflow at the cylinder head. The massflow increases, but only due to the increased mass in each litre of air.)

It is easy to calculate density change using the ideal gas law: PV = nRT.

Rearrange it to n/V = P/RT. Think of n/V as density. R is a constant. So the equation is saying "density is proportional to pressure divided by temperature". (P and T must both be in absolute units.)

Example. NA engine with intake air at 1 atm and 20*C. Same engine with turbo - 2 atm and 80*C.

Torque(turbo)/Torque(NA) = Power(turbo)/Power(NA) = density(turbo)/density(NA)
= [P/T(turbo)] / [P/T(NA)]
= [2/(80+273)] / [1/(20+273)]
= 1.66
ie the turbocharged engine has 66% more power (and torque) at the same revs.

This is only a rule of thumb. many other parameters will change when an engine is turbocharged - exhaust back pressure, detonation tendency etc, and these will affect the final power result.

[PlatinumZealot]

Hi all, anyone able to give me a basic rule of thumb power and torque increase for various boost pressures on a turbo
e.g. 20lbs- 10% power and 7% torque increase <<<Just an example (not meant to be correct)
Boost pressures between 8 and 86lbs is the goal (8,14,18,26,32,40) - whatever you want to do be awesome if you could do every one between 8 and 86lb or give me a derivable formula (rule of thumb) for torque and horsepower based off of boost pressure.

Cheers guys and Thanks in advance

I bar of boost doubles your power.
So if your engine was making 150hp before boosting.. 1 bar will send it to 300hp roughly.

[Vortex347]

Hi all thanks for your replies,

I bar of boost doubles your power.So if your engine was making 150hp before boosting.. 1 bar will send it to 300hp roughly.

So then at 14.5lb of boost, your engine hp is doubled? That sounds a bit high...but I don't know seems logical when you think about it.

Example. NA engine with intake air at 1 atm and 20*C. Same engine with turbo - 2 atm and 80*C.Torque(turbo)/Torque(NA) = Power(turbo)/Power(NA) = density(turbo)/density(NA)= [P/T(turbo)] / [P/T(NA)]= [2/(80+273)] / [1/(20+273)]= 1.66ie the turbocharged engine has 66% more power (and torque) at the same revs.

So at 2 atm or 29lb of boost, the engine has 66% more power this seems pretty reliable in relation to what I know. What about for different boost pressures?

Cheers for your replies guys and thanks in advance

[gruntguru]

So at 2 atm or 29lb of boost, the engine has 66% more power this seems pretty reliable in relation to what I know. What about for different boost pressures?

Whoa there. That is for (the completely arbitrary) CAT of 80*C. If you intercool back to 20*C you get 100%.

2 atm is absolute pressure ie 1 atm (14.5 psi) boost.

[Vortex347]

Hi guys thanks for your replies

So as a general increase what would be a good relational constant to match boost with power and torque
e.g. for every 1lb of boost you gain 4% hp and torque?
I'd assume this is a fairly linear increase.

Cheers guys and thanks in advance

[PlatinumZealot]

Hi all thanks for your replies,

I bar of boost doubles your power.So if your engine was making 150hp before boosting.. 1 bar will send it to 300hp roughly.

So then at 14.5lb of boost, your engine hp is doubled? That sounds a bit high...but I don't know seems logical when you think about it.

This is a real world rule of thumb. For newer engines it is even higher.

[gruntguru]

Hi guys thanks for your replies

So as a general increase what would be a good relational constant to match boost with power and torque
e.g. for every 1lb of boost you gain 4% hp and torque?
I'd assume this is a fairly linear increase.

More like 6.6% and that means 6.6% of the original power level ie if you have a 50 psi motor making 1000hp it wont make 1066 at 51 psi. The calculation would be 1000 x (51 + 14.5)/(50 + 14.5) = 1015 hp.

Yes it is fairly linear but always tapers off at some higher boost level depending on the engine, fuel, turbo etc.

[flynfrog]

Hi all thanks for your replies,

I bar of boost doubles your power.So if your engine was making 150hp before boosting.. 1 bar will send it to 300hp roughly.

So then at 14.5lb of boost, your engine hp is doubled? That sounds a bit high...but I don't know seems logical when you think about it.

This is a real world rule of thumb. For newer engines it is even higher.

that's not possible.... If you don't double the mass airflow you can't double the HP

[Vortex347]

Hi all thanks for your replies

More like 6.6% and that means 6.6% of the original power level ie if you have a 50 psi motor making 1000hp it wont make 1066 at 51 psi. The calculation would be 1000 x (51 + 14.5)/(50 + 14.5) = 1015 hp.Yes it is fairly linear but always tapers off at some higher boost level depending on the engine, fuel, turbo etc.

So then that would mean the engine has a base hp of 227hp?
So if we take a Subaru WRX for example which has a power of 302hp at 14.5lb of boost I think (not too sure if this is there stock boost pressure or not). How would you go about working out it's base engine power without turbo?
Or what about if you didn't know boost pressure but you knew it had 302hp turbo-charged and 140hp naturally aspirated, I take it you could find out boost pressure?

Also what about turbo-diesels as opposed to turbo petrols? I take it they have a different power increase?

Cheers mate and thanks in advance

[flynfrog]

So then that would mean the engine has a base hp of 227hp?

So if we take a Subaru WRX for example which has a power of 302hp at 14.5lb of boost I think (not too sure if this is there stock boost pressure or not). How would you go about working out it's base engine power without turbo?

divide by 2

Or what about if you didn't know boost pressure but you knew it had 302hp turbo-charged and 140hp naturally aspirated, I take it you could find out boost pressure?

In theory but this is just a rough approximation there are many other factors like intercooler efficiency pumping losses ect.

Also what about turbo-diesels as opposed to turbo petrols? I take it they have a different power increase?

The air doesn't care what fuel is being burnt.

Cheers mate and thanks in advance

[gruntguru]

Hi all thanks for your replies

More like 6.6% and that means 6.6% of the original power level ie if you have a 50 psi motor making 1000hp it wont make 1066 at 51 psi. The calculation would be 1000 x (51 + 14.5)/(50 + 14.5) = 1015 hp.Yes it is fairly linear but always tapers off at some higher boost level depending on the engine, fuel, turbo etc.

So then that would mean the engine has a base hp of 227hp?

New power = (Old power) x (new MAP)/(old MAP)
= 1000 x (0+14.5)/(50+14.5)
= 225 hp

So if we take a Subaru WRX for example which has a power of 302hp at 14.5lb of boost I think (not too sure if this is there stock boost pressure or not). How would you go about working out it's base engine power without turbo?

New power = (Old power) x (new MAP)/(old MAP)
= 302 x (0+14.5)/(14.5+14.5)
=151 hp

Or what about if you didn't know boost pressure but you knew it had 302hp turbo-charged and 140hp naturally aspirated, I take it you could find out boost pressure?

New MAP = (Old MAP) x (New power)/(Old power)
= (0+14.5) x 302/140
= 31.3 PSIA (16.8 PSI boost)

Also what about turbo-diesels as opposed to turbo petrols? I take it they have a different power increase?

Providing the AFR is the same before and after, the power increase can be calculated the same way. The problem is that diesels operate with a wide range of AFR (always with excess air). Turning up the boost only increases the mass air flow. That will not increase the power at all unless the fuel injection is recalibrated to add more fuel.

[Vortex347]

Vortex347 wrote:
Hi all thanks for your replies
gruntguru wrote:
More like 6.6% and that means 6.6% of the original power level ie if you have a 50 psi motor making 1000hp it wont make 1066 at 51 psi. The calculation would be 1000 x (51 + 14.5)/(50 + 14.5) = 1015 hp.Yes it is fairly linear but always tapers off at some higher boost level depending on the engine, fuel, turbo etc.

So then that would mean the engine has a base hp of 227hp?
New power = (Old power) x (new MAP)/(old MAP)
= 1000 x (0+14.5)/(50+14.5)
= 225 hp

I was pretty close

And torque increases by the same percentage (6.6%) as power too right?

New power = (Old power) x (new MAP)/(old MAP)

Definitely simple enough.

Cheers for that