7 replies to this topic

[nobbycameron]

Hello here is one which will get you thinking.

i am superchargeing my 1380 over the summer.

i am trying to determine the size of the pulley for my 1380 and also my dads 998 which he is also supercharging.

i have a eaton m45 perfomance map from the web which has flow rate in m^3/hour plotted against pressure.

I need to calculate the flow rate the engine requires at different engine speeds, this bit is easy, just swept volume mulitplied by the number of inlet strokes... This gives theoretical volume induced but i need to know actual volume induced. There are obviously losses during the iduction stroke.

On a n/a engine this can be calculated by volumetric efficieny which is about 85%. However i need to work out volumetric efficiency for a forced induction engine. I have read on the internet that volumetric efficiency can go over 100% for a forced induction engine. I can't see how this is possible as in my view the volume of the cylinder is fixed so the induced volume cannot go above 100%. This gets me thinking it is mass flow or volume the air fuel would take up if it were at atmospheric pressure.

i hope i havn't lost you so what i have come up with is the expression below, can anyone confirm it is correct.

volumetric eff. = V1(volume of fuel/air mixture if at atmospheric pressur) / swept volume.

this leads to another question, the values on the performance map for flow rate (m^3/hour) - are these flow rate at say 7 psi or if the fuel air mixture is at atmospheric pressure.

i hope this all makes sense, anyone who can shed some light on this would be very usefull

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[nobbycameron]

ello, just to add to the above post, the green lines on the graph are supercharger speed, the contour lines are efficiency and the pressure relates to atmosphere so 1 = atmospheric and 2 = 2x atmospheric.

well this is my interpretation of it....

[Wil_h]

You need to work the other way around.

So calculate the what flow and psi is required to make the bhp that you want. For the eaton I'd work the maths so that the psi comes out at no higher than 10psi.

Onece you have the flow and pressure ratio at maximum rpm you can then plot it on the comp map to get the required rpm; and then work out the pulley sizes required..

You still need to use 0.85 VE for your calulations.

Just quickly I get the folowing, all in lb/min, you'll have to do the conversion yourself for the Eaton map:

1380cc, 140bhp, 10psi, 15.5 lb/min

998cc, 100bhp, 10psi, 11.0 lb/min

[nobbycameron]

thanks for your reply wil

i have come up with the following procedure which i think works, can anyone confirm

1. choose a desired pressure at a desired rpm, say 9.8psi which equals 1.7 on the pressure ratio at 6000rpm

2. With this chosen pressure ratio and rpm calculate the amount of air fuel mixture induced on one stroke = swept volume x pressure ratio and multiply this by the number of inlet strokes at 6000rpm. this should give required flow from supercharger at 6000 rpm

3. from the performance map match the calculated flow with the flows on the map to get super charger speed

4. then pulley size is just a ratio of the 6000rpm:supercharger speed

what does everyone think about this, am i making too big a assumption with regard to the amount of air/fuel filling the cylinder on each induction stroke???

cheers

[Wil_h]

Not too sure about your maths.

the is a great expination of how to calculate flow and psi here http://www.turbobyga...ech_center.html

[nobbycameron]

hello again,

i think i am gradually getting to the bottom of this,

my next question is what sort of compression ratio does everyone use? if i am going to go for 10psi then i guess the comp ratio needs to be lower to stop pre-ignition?

What sort of comp ratio do you use on turbo engines, i suppose it would be the same problem for turbo engines. how do you avoid pre-ignition if you are running your turbo at say 15psi?

My engine is 1380 with quite high compression ratio so i'm going to need to do quite a bit to lower the comp ratio.

your views on this would be good

cheers

[nobbycameron]

thanks for your reply wil

i have come up with the following procedure which i think works, can anyone confirm

1. choose a desired pressure at a desired rpm, say 9.8psi which equals 1.7 on the pressure ratio at 6000rpm

2. With this chosen pressure ratio and rpm calculate the amount of air fuel mixture induced on one stroke = swept volume x pressure ratio and multiply this by the number of inlet strokes at 6000rpm. this should give required flow from supercharger at 6000 rpm

3. from the performance map match the calculated flow with the flows on the map to get super charger speed

4. then pulley size is just a ratio of the 6000rpm:supercharger speed

what does everyone think about this, am i making too big a assumption with regard to the amount of air/fuel filling the cylinder on each induction stroke???

cheers

hello, i have had another thought, with reference to point 2 in the quote above this assume that on each induction stroke the cylinder is getting filled with fuel air at say 10 psi. in reality i dont think this will occur as there will be losses in the induction process and not enough time for the whole cylinder to fill.

by using the fact that most normally aspirated engines have 85% volumetric efficiency, can you assume that only 85% on the cylinder volume gets filled with the 10psi fuel/air mix???

is this a reasonable assumption??

regards

rob

[mini-majic]

you get over 100% VE on a forced induction engine because you are filling the cylinder under pressure. So the whole cylinder will get filled and then a little extra due to the pressure that its being filled at. this is why you can get over 100% VE from a forced induction engine.

i find this site good for getting figures to plot onto comp maps:

http://www.motorgeek...ndex.php?page=6

use a VE of 85%.