I'm posting this up for anyone who wants to work out how much boost can be made by various pulley sizes on a Supercharger.

Most of this is from a PM between me and Lee and I don't want to have to rewrite it all.

We predicted that there will be an approximate 3psi loss on final boost figure due to inefficiencies.
In the message I doubt my calculations however it looks as if these figures where in fact accurate as the predicted boost figures are what I'm currently achieving.

Also in3 is meant to be cubic inches but there's no subscript/superscript tags on the forum.

[quote="AeroX"][quote="Lerch"]you any good at maths? lol[/quote]
Challenge Accepted!

[b]Car red line:[/b] 6,500rpm
[b]Eaton M45 red line:[/b] 14,000rpm
[b]Derived max pulley ratio:[/b] ~2.15:1

[b]M45 air flow:[/b] 45in³/rev = ~737cc/rev
[b]in³ to cc:[/b] 1in³ = ~16.387cc

[b]Engine Drive Pulley Diameter:[/b] 130mm
[b]M45 Pulley Diameter:[/b] ~80mm (cir: 250mm, dia=cir/pi, 250mm/~3.142=79.57mm)
[b]Pulley Ratio:[/b] 130mm:80mm = 1.625:1

1.5L engine has ~1,498cc across 4 cylinders, of which 2 are open per rev so actual capacity per rev:
~1,598cc / 2 = ~749cc

M45 blows ~737cc of air per rev so multiply by the pulley ratio for actual:
~737cc * 1.625 = ~1,198cc

So ~1,198cc of air is compressed into ~749cc of space per rev.
The only bit I'm not sure on is how to calculate the psi from these figures.

------------------------------------------------------------------------

So I expect this next bit to be completely wrong...

Atmospheric pressure is 1bar (14.5psi), N/A engines run at atmospheric pressure so:

14.5psi / ~749cc = 0.01935
~1,198cc * 0.01935 = 23.2psi - 14.5psi = 8.7psi boost ???

It would be best to check the last bit with some sort of online calculator as I don't know if this is the correct way to calculate psi (i doubt it is lol)

This calculation doesn't take into account other places the air can go so the predicted 8.7psi boost will actually be less due to space in the intake manifold and induction piping etc... So if the calculation is correct then the current pulley would be quite well matched to the engines in our cars, which makes me doubt the calculations because nothing is ever that easy lol[/quote]


EDIT: Corrected atmospheric pressure in psi within calculations.

NOTE: Charger efficiency needs to be taken into account, at the 1.625 Pulley ratio the Eaton M45 will operate with 60% efficiency up to around 10,800 SC RPM which is around 6650 RPM, so at this ratio the charger should be constantly running at 60% efficiency. When the final psi is then multiplied by this we get 8.7psi * 0.6 = 5.2psi (which is fairly close to what I've been able to log).

in³...edited

Ah cheers for that Marc

I subscribed at first thinking that your pressure calculations were wrong as I thought you were trying to solve for the 2.15 max pulley ratio and I was going to point that out. Then I realized your actual ratios and it looks like your calculations for pressure are correct but your understanding of efficiency is not.
There are a number of different types of efficiency. You are referring to volumetric efficiency as depicted here.
[img]http://www.capa.com.au/pics/eaton_Mp62Graph.gif[/img]

Now to get real perfect calculations you would need to know the VE curve of the motor and the supercharger. If the motor is at 80% ve and the SC is at 90 then the boost will be higher than calculated. If the motor is at 90% VE and the SC at 80% then the boost will be lower than calculated.
Since it's hard getting VE curves for motors and since motors usually operate around 80% to 90% VE and so does the SC you can just drop VE from the equation to do rough calculations.
If you are seeing that drastic of a reduction in boost you are either seeing the difference in SC VE to engine VE or you are loosing boost somewhere. One thing to remember is that any pressure drop before the SC (Restrictions from throttlebody, piping, air filter etc) will be compounded by the pressure ratio so if you are running a 2:1 pressure ratio you should be making a bar (14.5 PSI) of boost however if you have a 2 PSI pressure drop before the SC then you are starting with 12.5 PSIA on the inlet of the SC times your 2:1 PR = 25 PSIA -14.5 = 10.5 PSIG.

What you are referring to when you say 60% efficiency I have to assume you are talking about adiabatic efficiency.
This map can be seen here.
[img]http://www.superchargerforums.com/attachments/awww-eaton-com_ecm_groups_public__pub__eaton__per_documents_content_ct_128483-gif.551/[/img]

http://www.stealth316.com/2-adiabat1.htm
That can be calculated here
http://www.stealth316.com/2-turbotemp.htm

So leaving temp at 100 degrees and putting boost at one bar where the M45 map peaks you end up with about 58% adiabatic efficiency so your supercharger outlet temps will be about 309f.

Now for drastic comparison we could look at a nice perfectly sized turbo that would have around 75% adiabatic efficiency. So now you enter that value leaving the others the same and this particular turbo would have outlet temps of 262f.


As for double checking your math, or just making it easier to do calculations you can just use an ideal gas law calculator.
http://www.calctool.org/CALC/chem/c_thermo/ideal_gas

The only problem with this is that it only addresses one thing at a time. You can calculate what the pressure will be if you put 1200cc of air in a 749cc container but it assumes the temp stays the same but as you see above the temp will not stay the same. The 309 degree air will want to expand so in the fixed system the pressure will increase further.
You can estimate closer by taking the temps seen in the outlet temp calculator and changing the temp from ambient to that output to see how that further effects pressure.
If you have an intercooler then you would use the outlet temp of the intercooler.

http://www.calctool.org/CALC/chem/c_thermo/ideal_gas