Known torque required to operate a supercharger?

[C.A. Russo]

Hi everyone,

So this is my first post and I have high hopes for the members of this website and forum. I didn't find any result from my searches here or the internet. So here goes nothin'.

I'm trying to find out how much torque is required to turn any given supercharger. I don't care if it's about an old axial flow from 1960 or a brand new F-3A-123. I've asked the techs from manufacturers and I never get a satisfactory answer. I'm afraid they're cautious of calling me stupid or something. If I'm way off then, please, call me stupid.

Ideally, if anyone knows of a test that has been done with an electric motor to get a raw reading of output values and torque requirements. I say this because electric motors are all about torque. The only horsepower output rating is very low and relates to electricity instead of mechanical energy. Any thoughts? References? Comments? Please say whatever is on your mind about this topic.

[blownzoom440]

i can relate to the electric motor.there was a writeup where it was used on the gears of a ford 9 inch.

[Joo]

@Russo:
you can calculate the Power and torque taken from a charger . It's easy and accurate enough.
The Power you need to drive a charger is = boost * cfm of the blower @ max. rpm/ ( blower mechanical efficiency* thermal efficiency* mechanical efficency of the cog belt )

Set blower mechanical efficiency as 0.95
set thermal efficiency as 0.75
set mechanical efficency of the cog belt as 0.98

I prefer the SI-system, therefore:
Before calculating the power convert psi in N/m² and cfm in m³/s, than you get Power in Watt. Divide the power through 1000 and you geht Power in kW. Miltiplicate with 1.36 and you get Power in HP just for information.

Last step: Torque = Power / ( 2* PI* rpm/60 ).
PI= 3.1415927
Take power in Watt and you get Torque in Nm.

[C.A. Russo]

Thanks a lot, Joo. I hope I use your calculations wisely. Is there a name for these calculations that I may reference?

[C.A. Russo]

Do I use the efficiencies to divide the cfm or the product of boost*cfm?

[C.A. Russo]

Please help me. The values I start with are 3100 cfm (1.46 m³/s) , 40 psi (275790.2917271 N/m²), 11,217 rpm (before blower internal step-up).

Somehow, I keep coming up with a negative value for torque in the end. What am I doing wrong?

[Joo]

Please help me. The values I start with are 3100 cfm (1.46 m³/s) , 40 psi (275790.2917271 N/m²), 11,217 rpm (before blower internal step-up).

Somehow, I keep coming up with a negative value for torque in the end. What am I doing wrong?

I don't now negative is not possible with these equations. But it was good that you have added the values. With this extremely high boost these equations are not valid anymore. That's because of the polytropic exponent.

I will look in some books and if I find something I will post it here. You will need al lot of power to drive such a blower...

Kind regards

[james redman]

With these numbers and formulae I get

Power Nm/s 576660.8

Torque Nm 490.7691

Jim

[Joo]

With these numbers and formulae I get

Power Nm/s 576660.8

Torque Nm 490.7691

Jim

That's absolutely correct Jim. Though these formulas are not valid for those high boosts.

[C.A. Russo]

I didn't realize the final value was in Nm. Could you say whether the proper equations would result in greater or lesser values?

[Joo]

I didn't realize the final value was in Nm. Could you say whether the proper equations would result in greater or lesser values?

The more correct formula for a roots-type blower with an isentropic compression is:

P = p1* V. * κ/(κ-1) * { (p2/p1)^[(κ-1)/κ]-1} / η

note that (p2/p1)^[(κ-1)/κ] is one term and then take -1

P = Power
κ = Isentropic exponent (1.4 )
V. = Flow at inlet (calculated from the volume of the blower and the rpm
p1 = inlet pressure e.g. 14 psi
p2 = outlet pressure, your 40 psi
η = efficiency, with 0.7 you should be right.

with η you try to consider, that the compression in reality is not isentropic but polytropic.

As my native language is not english it's difficult for me to explain thermodynamics and further more it's over 20 years ago....

Hope this helps.
Joo