Supercharger

Marsho · Post by **Marsho** » Sat May 26, 2007 8:07 am

Hi, I am new to this forum and need some advice regarding a supercharger. The question is would it be feasable to fit the Eaton M90 to a 4.6 rover v8, ie would it be to powerful, or not suitable for the job, great web site and I look forward to any replies/advice many thanks, David.

The Original Tom · Post by **The Original Tom** » Wed May 30, 2007 12:00 am

"The Eaton Model 90 supercharger was engineered for 3.0L to 5.7L passenger cars and light truck engines."

The M90 shifts appx 1500cc of air per rotation. It has a maximum rpm of 13,650.

Assuming your engine redlines at 5500 rpm, to get max boost throughout the rev-range, you want the charger near max rpm:

As your engine is 4.6, it needs 4.6 litres of air every 2 rotations.
13,500 (for safety)/5,500 = 2.4545
So for every 1 engine rotation, you can make the 'charger run at a max of 2.4545 rotations. (1:2.4545 pulley ratio).

So at 5,500 engine rpm, the engine NEEDS:
(5,500 x 4.6) / 2 = 12,650 litres of air.

At 5,500 engine rpm, at a ratio of 1:2.5454, the charger will spin at 13,500 rpm, and produce:
13,500 x 1.5l (air mover per rotation) = 20,250 l of air

20,250 / 12,650 = 1.6x more air than required.

Air required enters inlet manifold at a little under 1 bar (atmospheric pressure, but lower, because of vacuum created)

So lets call the 'required pressure' 1 bar.

1.6 - 1 = 0.6 bar extra than required.

1 bar = 14.504 psi.

0.6 x 14.504 = 8.7 psi.

So, the short answer to your question is: yes it will be fine on your engine, as long as you don't want to run more than about 8.5 psi of boost.

You may wish to consider the M112, which shifts 1840cc of air per rotation, and although I don't know the max rpm for it, assuming it's the same as the M90, and all the losses etc are constant:

(8.7 / 1500) x 1840 = 10.7 psi of boost max.

If anyone knows different, please inform

Eliot · Post by **Eliot** » Wed May 30, 2007 5:40 am

Given that jaguar fitted the 112 to the 4.2 v8, i would select that model.

Marsho · Post by **Marsho** » Wed May 30, 2007 7:33 am

Many thanks for the reply, I understand the principles, but when it comes to the equation side of it all it leaves me cold but hopefully now I can carry on and get on with it, regards David.

JP. · Post by **JP.** » Wed May 30, 2007 5:05 pm

Sorry but these calculations are not correct. The correct formula to calculate boost is:

Boost (Psi) = (25.58 x Blower C.I.D x overdriver ratio) : Engine C.I.D - 14.7

A M90 is to small in displacement for feeding a 4.6.
for example: I am running a M110 on a 3500 with an overdriveratio of 1.6 (=60%) and I am seeing 6Psi boost.
60% overdrive means that our blower is running 9600rpm @ 6000rpm engine speed.

This formula is suplied bij Uk Blowers and is pretty accurate.

Marsho · Post by **Marsho** » Fri Jun 01, 2007 5:21 pm

Hi, thanks for the info, but if you say its to small, why is it quoted as being suitable for 3.o to 5.7 litre engines (now I am confused) regards David.

The Original Tom · Post by **The Original Tom** » Fri Jun 01, 2007 6:37 pm

JP. Could you talk me through that equation?
If I do it for Marsho's question, I get:
(25.58 x 90 x 2.4) = 5525.28. Then I'm confused by the colon? Unless it's meant to be / , in which case I get:
5525.28 / (280 - 14.7) = 20.8 psi of boost. Not right.

Any clues as running through my maths again, there's nothing that defies physics there

JP. · Post by **JP.** » Fri Jun 01, 2007 7:09 pm

Using your overdrive ratio of 140% (2.4)

25.58 x 90 x 2.4 = 5525.28
5525.28 : 280 = 19.733

19.733 - 14.7 = 5 Psi

And at 6000 rpm engine speed the blower will run 14400 rpm.

Marsho · Post by **Marsho** » Fri Jun 01, 2007 8:12 pm

Now I must lie down in a darkend room.