One for the budding engineer or mathematician...

I'm trying to calculate the minimum oil pressure required in an engine. Pretty much all oil pressure is needed for is to overcome centripetal force in the main bearings, to get into the drillings to the big end bearings. But how to calculate how much?

I know that Force = mass x angular velocity squared x radius (F=mrw^2)

But the oil pressure acts over the whole radius, from 0 (or effectively zero, but it's pretty negligable as D approaches zero), so I guess I need to integrate that - e.g. to m*w^2*(0.5*D^2) - using mw^2 as a constant.
But the mass varies over the radii too, as there isn't much oil at the middle, but there is the whole diameters worth at the extremes... So does mass need integrating too, at the same time as radius, and if so how would I do that?
We can, at least, assume w is constant...

I also tried doing it a different way, and assuming 1cc of oil in a tube of cross sectional area of 1cu/cm, and then taking F=mrw^2 at discrete radii...

But I'm not thoroughly confused!

Anybody want to help out with some simple maths (that I'm too stupid to recall)? I've not been able to find this done before on the net (but I'm sure it has been done, as it's an important calculation for real life engine design). It's not on Wikipedia either that I'm aware of

Ta!

Why? just fill up the oil to required level...

I think this goes a bit beyond the "easy" math (whatever that is).

So far as I understand it you need to match the pressures comming from rotaion and displacement of the bearing "comming form the engine". You can either calculate them seperately or use the "direct" way by solving a reynolds differential equation.

also consider shear stress along the radius (acting against the centripetal force), vortices at the boundary layer (which will have a different viscosity than the rest of the oil), inertia of fresh oil passing through the inlets, and backpressure from oil passing through the outlets (both of which will depend on the geometry of the journal drillings).

Nope, more than happy to simplify it a lot and ignore viscosity effects, inertia, backpressure, boundary layer effects and all of that. I just want to know what oil pressure is needed to overcome the rotation of the crank in the main bearings as if it were a simple closed system etc etc.

vrooom - because the oil level isn't the issue. We have more than enough oil in the oil tank, but we want to make sure that the pressure never drops below the required 'estimated (but calculated)' figure, especially in 2.5g left hand corners whilst braking (or accelerating).

Any more ideas please? Thanks so far.

i can't help with the calculations, but what kinda engine is it? Dry sump or not? If wet sump, is it baffled? Surely someone has minimum specs for this engine somewhere?

Dry sumped Toyota 3S-GE in an F3 car with Tom's oil pumps (probably originally made by Pace).

Would the minimum specs, do you think, have the pressures at 3000, 4000, 5000, 6000 and 7000rpm? Most of the time minimum oil pressure is quoted at hot idle (when you NEED virtually nothing), and have a warning light using a simple '10psi switch' or similar. I want more info than that.

i will ask on the toymods forums, and see if anyone there can help. Might take a while for a reply as its only 630am here But i'll see what i can do.

are you sure its a rotational effect? after all the bearing is completely blocking the way for the oil to leave the area in question radially

Is there a problem using a 'big ass' pump ?
Is too much pressure a problem ? weight ?

have you try F1technical.com?
they got some really talented people there i believe you would get more help from there.

I'll give them a try, thanks for the link.

now, i don't know what year your 3s-ge is, but here is a Toyota Repair Handbook for 1993 Celica ST202 and SW20 MR2 3S-GE

Refer Page 393, it only specs idle and 5000rpm, but it's a start i spose.

http://www.celicagt.nl/Docs/st202_rm396e.pdf

Thanks a lot, it is indeed a start - our engine is, iirc, a 1994 one, so probably the same... I wonder why the 5000rpm has such a 'tolerance' on the pressure?

nfi, but it would be interesting to know what the oil press regulator valve is capped at... that would give you a high point..

i honestly have nfi on the tolerance for it, i'm not a mechanic, just a toyota fan

Also there was a change in generation of 3S motors between 93(gen2) and 94(Gen 3), so might pay to get specifically which your engine is.

http://www.beams-redtop.com/history.htm

Brief history on the 3S series.

It's a Generation 3, and the relief valve is set at 80psi. I know quite a lot about our engine - more than I'm letting on, because it's not actually relevant. I just want to know what the minimum oil pressure is at various RPMs to make sure that the oil scavenging is working properly and that the high g-forces are not causing problems.

Shot - yes, it is a rotational effect. The oil is pumped through the block to the main bearings, where it has to go through the shell bearing (via the hole(s) in them), into the journal to the approximate centre of the journal against centripetal force, and is then 'flung' to the big ends by the rotation.

Therefore the oil needs enough pressure to get from the outside of the bearing journal to the middle, and I'm fairly sure this pressure (albeit simplified) is calculatable.

Attached is a quick sketch of 'the problem'.

How about we pretend it has nothing to do with engines or racing etc. It's just a simple mathematics problem involving a varying mass and radius, and hence an integral equation of more complexity than my head can cope with.

Unfortuantely I can't help with the calculation as it's no way a simple calc. as many variables need to be taken into account to get anywhere like an authentic result. You certainly cannot ignore the viscosity of the oil as this is paramount to flow and pressure drop calculations, the viscosity will vary greatly with temperature so you will need the viscosity/temperature curves and the specific gravity for the oil you are using.

I'll be very interested to see what you come up with for this so please keep us posted, good luck

The fail criteria you want to calculate is the lack of oil in the center of the bearing?
Let's give it a try:
- See the locations of bearing and journal diameters D and d in attached file.
- Consider the journal filled with fluid of density v from the buttom to the top.
Attention: Be aware that you're actually dealing with a more dynamic problem. The journal actually empties when it's at the top due to centrifugal force (...or lack of centripetal, blah!) and has to be refilled when it reaches the bottom. If you'd like to calculate that you'd have to set up a CFD simulation because the resistance to the flow changes over time.
- Since the journal is filled fluid motion relative to the journal is neglectible.
- I calculate the amount of pressure only for the given position of the bearing. Any other position wouldn't allow fluid from inside the journal to be affected by oil pressure.

Force dF acting on a particle of fluid dV located somewhere between the centre of the bearing and the lower end of the journal, positive direction of dF is downwards:
dF = r*v*dV*w² + dV*v*g
r: Distance of dV from centre of the journal. r = 0..D
w: Rotational velocity.
g: Gravitational acceleration.

Since dF is (rather...) constant for each radius r the force acting on a disk of fluid of given distance from the centre of the journal is simply:
(Use conversion dV = dr*dA and integrate along dA)
dFa = (r*v*dr*w² + dr*v*g)*pi*d
Integrate dFa along dr from 0 to D to receive the sum of forces F acting on the cylinder of fluid:
F = pi*d*(v*w²*0.5*D² + v*g*D)
We only need the pressure and thus we receive
p = v*w²*0.5*D² + v*g*D
as a result to the pressure being necessary to keep the column of fluid from flowing back into the oil delivery system due to centrifugal forces.

Hint: That's pretty simple math and I'm sure you already wrote that down yourself. I'm just putting it up for discussion so people can comment on how valid the solution is. My personal opinion is that, even though I'm a fan of analytically solving engineering problems, this isn't one where such a solution holds much information.
Minimum oil pressure is, as far as I know, usually found by running a couple of engines on low oil pressures until they die - or hopefully survive.

Vain

No, I didn't get to that. I'm too stupid to get that far with real problems

Approx values
v = 810kg/m³
d = don't know, as I never measured it before building the engine! But it doesn't matter in the pressure calculation anyway
D = 0.0275 (I believe, from memory, the journal is 55mm dia)
w = let's say 5000rpm, which is 523 rads/sec

Which gives 2527 Pa, which is (for my old fashioned brain; I can't do SI units of pressure and understand what they mean in the real world) 0.36psi - which is wrong.


So, either the calculation is wrong (doesn't look like it), my units are wrong (don't think so), or you are correct that the solution is a uselessly small part of the whole thing, and that overcoming the rotation isn't much of a problem and isn't the main requirement of oil pressure.

Me likes interesting topics. Thanks.

Anyone else just get "blah blah blah" when you read these type of threads?

I consider myself reasonably intelligent, however, I feel as dumb as a chimp after reading that.

Quote from danowat :

Anyone else just get "blah blah blah" when you read these type of threads?

I consider myself reasonably intelligent, however, I feel as dumb as a chimp after reading that.

:iagree:

Yarp... all goobledegook..

I haven't done any maths since I started my placement. I'm screwed when I go back to finish.

maybe you should try calculating the the oil pressure required to float the crank in the bearing. when at rest, the crank will sit at the bottom of the bearing, and the space around it will be in the shape of a crescent (touching at the bottom, obviously). when operating, the crank should center itself, with a nearly ring shaped space all around.

if you know the surface area of the bearings (/2), the mass of the crank and the forces acting on it, you ought to be able to calculate how those forces are distributed to the bearing surface and thus infer the oil pressure required to resist them.

It's not a bad idea. I wonder if that is likely to be closer to reality than my first method, as the hydrodynamic oil wedge will be quite different to a static floating condition...

I'd love to run several engines, all at different rpm, under lateral G, gradually lowering the oil pressure until each fails. But alas I only have two engines (one in use, one a spare that we don't want to use if we don't have to), and a very limited budget. Hence finding out everything we can, from chassis behaviour and setup balance, to fuel behavious in the float chambers etc. Yes, it's all very simplified, but gradually you build up a picture of simplifiedness versus real life on-track experience (with some LFS-learnt theory too ), and come out at pretty close to the right answers.

At least, that's how it's worked over the last two years, and going from the back of the grid 2 years (and one day) ago to winning a championship 8 months ago, to instantly being on the pace of the experienced Dallara people 6 weeks ago (and hopefully again two weeks from now).

At Snetterton, during our test day, we experienced low oil pressure (I forget the actual figure, but worryingly low) in the two left hand bends - braking for the Esses and turning through Russells. This was traced to the car being underfilled, but no harm was done (phew). At the race meeting at the same circuit the pressures stayed much higher, but still a little low for comfort - only on left hand bends it seems, as though the dry sumping arrangement, which isn't easily modified, isn't good enough to maintain oil flow/pressure in left handers.

Mallory, the next race, is primarily right handers, but does have one fast left hander of any significance. Donington has a couple of left handers that might be a problem. Croft has two or three places where a problem might occur. Silverstone has one, and then back to the relatively safe Snetterton... I'd like to kill a potential problem before it does anything bad - but to know if we do actually have a problem I wanted to know what the minimum oil pressure is to ensure that the oil gets where it needs to be (and I was under the impression that overcoming 'centrifugal force' was the only requirement for oil pressure really.

My Dad has done a calculation, taking a small mass of oil at several different radii, and adding up the pressures to come to a value of 36psi at 7000rpm. This adding up of chunks is essentially what integration is, which is why I thought it might be possible to get a more accurate figure via mathematics... And I still hope to (as I don't fully trust his method yet).

Quote from danowat :

Anyone else just get "blah blah blah" when you read these type of threads?