I was bored in one of my classes and decided to find out how much horsepower I can get out of a certain about of boost from a turbo or supercharger.

The formula I came up with is this...

(wanted horsepower - starting horsepower) - 14.7 / 14.7 = psi for wanted hp

Now this took an hour of just trying random stuff and I would like to know if it is even close to right. (also, don't start saying there are already formulas like this and whatever. When you're sitting it class, you can't exactly get on the internet)

Here is my logic behind it all.

wanted hp - starting hp: Came up with this because it gives you how much the horsepower has to increase by.
- 14.7: Got this from another trial and error thing. Basically how I got it is the standard atmospheric pressure is 14.7, so above that is concidered boost
divide by 14.7: This came from thinking, well...in order to get needed psi, you need to get rid of the already stable atmospheric pressure (14.7)

I don't know how accurate it is or anything, I am looking to see if it is even remotely correct.

An example:

(500hp - 240hp) - 14.7 / 14.7 = psi
(245.3) / 14.7 = psi
psi needed to get 500 hp from 240 hp = 16.68 psi

or

(350 - 240) - 14.7 / 14.7 = psi
(95.3) / 14.7 = 6.48


Seem right?! Open to thoughts, comments, critisism (as long as it's nice/helpful).

 

Well I will tell you what I personally use to get close. Power is a function of airflow, and so is boost. It is a fairly accepted standard that it takes roughly 1.5 cfm to make 1 hp. (I generally introduce a fudge factor of .1 cfm)

First figure out how much airflow you need.

500 * 1.5 = 750 cfm

Now figure out how much air your engine inhales naturally aspirated. For this we'll assume your engine is going to rev no higher than 5500 rpm

350 * 5500 / 3456 (constant) = 557 cfm

557 * .85 (assumed volumetric effiency of your mostly stock engine) = 473 cfm

473 / 14.7 = 32.17 cfm per psi

750 / 32.17 = 23.31 psi

23.31 - 14.7 = 8.61 psi of "boost" at sea level into the intake

Now what this formula doesn't take into consideration is the change in air density due to the heat of compression and higher adibiatic temperatures of hot intake air. I use a spreadsheet for that but figure on running 2 psi more to make up for those losses. This is why an intercooler is so important (any charge cooling that will increase air density back towards the ideal is helpful)

It is also benificial to reduce or eliminate any pressure drop on the inlet of the compressor. (between the air filter and compressor inlet) This means cool air and a large filter will improve power output. A magnehelic is a great tool to use for optimization.

Hope it was helpful.

 

350/1728=.202
.202 x 6000 / 2 = 607.5
607.5 x .85 = 516.49

516.49 CFM at idle with 85% efficiency.

1.5 x 500 = 750 cfm
516.5/14.7=35.13
750/35.13=21.34
21.34 - 14.7=6.65

That is with my engine inputs, 350 size, .202 cubic feet, 516.5 cfm at 85% efficiency. So using your equations too, I need 6.5 psi of boost to reach 500 horsepower. But doesn't that only calculate what psi I need to reach the cfm level to reach the 500 horsepower? I would think you would need to find what the 85% is of the 750 too, and calculate that into it also.

750 x 1.15 = 862.5 (to compensate for the 15% loss)
862.5/35.13=24.55
24.55-14.7=9.85

Though also, I would think you need to calculate the N/A horsepower cfm into the cfm needed for that horsepower. So it'd be...
862.5+516.5=1379 (I also did 750 x 1.85 and got 1387.5, that a coincidence or something to use?)But now I know that isn't right...SOOOOOOOO.......

Here is another one going through my head

((wanted hp - starting hp) x 1.5 / 14.7) - 14.7 = boost needed

Cause I also thought it depends on what your starting horsepower is. So the wanted hp minus the starting lets say is

500-240 x 1.5 = 390
390/14.7=26.53 (psi including atmospheric pressure)
26.53-14.7=11.83 psi needed for 500hp

Now does that one sound right? Not saying you are wrong but I don't think it only takes my car 6.5 psi to get to 500hp. The roughly 12 psi sounds more like it. What are your thoughts?

 

I stand behind my original equation as it pretty much gets the job done. The 85% VE number is factored only once. The boost is increasing air density of the factored breathing characteristics. The turbo in your case is inhaling the required airflow. Once compressed the charge is denser by the compression. This denser charge passes through the same orifices the n/a charge does it is just more tightly packed. This is why a boosted engine does not require an automatic increase in throttle body or port size on thr inlet side if it is a mild combination. Case in point I have a customer who made a tick over 1000 hp with a single 700 dp holley carb. While 8-9 psi doesn't sound like much it increased your engine size by over 60% or to 560 cubic inches. How hard do you think it is to make 500hp with a 560 inch engine? Like falling off a log. If you factor VE% its still a 470+ inch engine.

 

It looks like may using using rear wheel HP in your calculation, since your "starting HP" is 240 HP in your calculations. That's not the way to do it. You need to work with flywheel HP, since that's what determines the air mass flow requirements.

Drivetrain loss is irrelavant, but if you want to work in terms of rear wheel HP, your formulas need to convert that number to flywheel HP. Since you've got an A4, drivetrain losses may increase as a function of flywheel HP, because of the converter.

If you did that, and assume 16% losses for your 240rwHP engine, and 18% losses for your 500rwHP engine, you'd get:

240 / 0.86 = 286 flywheel HP
500 / 0.82 = 610 flywheel HP

If you look at the ratio of the HP, you've got to increase the mass air flow by:

610 / 286 = 2.133

The only way to get 2.13 X the mass in the same volume is to increase the absolute pressure by the same ratio:

2.133 x 14.7psi = 29.9psi

meaning you are looking at 15.2psi boost, and that's without accounting for the compressors adiabatic efficiency and the heat of compression as pointed out in the earlier post. In actuallity, you'd need to run even more than 15psi to make up for the lost compression HP.

Your own formulas fall completely apart, since there is no continuity of units. You can't just subtract "psi" from HP and expect to get a meaningful number.

I'm not suggesting that my analysis is anything more than a "ballpark", highly simplified approach, but its at least rational in terms of maintaining the continuity of the units used in the calculations.

One other thing to think about.... it all relates to volumetric efficiency. Improve the air flow (= improve the HP) of the NA engine, and you improve the flow through the boosted engine. Start with an NA engine that can make 400 flywheel HP, and you need a lot less boost. It isn't likely that anyone would (or could) just strap a 15# boost system on a stock engine... it wouldn't last more than a few minutes.

 

Keep in mind when figuring the air mass flow required, that mass will deteriorate as the pressure increases. IOW, retaining the equality of all other density related factors needs to be addresed.

 

You might want to clarify what you mean by "mass will deteriorate as the pressure increases". As pressure goes up, air density goes up. If you are achieving that increased pressure through a higher pressure ratio in the compressor, air temperature will increase and some density will be lost. Is that what you are getting at?

 

Yes. Was probably obvious, but chose to mention it anyway. As more pressure is generated, a need to update the air cooling may be necesary, if marginal to begin with. I should have made this clearer.

 

how about your basehorsepower divided by 14.7. and that will give you how much horsepower your motor will get with every pound of boost.