# Flow change due to a 90 bend

1. Oct 10, 2009

### bsprowl

I'm not a student. I'm a 64 year old guy trying to determine the flow change (volume) in an V-8 engine oil gallery.

The feeding gallery comes directly from the oil pump and is 7/16 inch. It ends at a 90 degree intersection with a 3/8 inch gallery. I understand the loss of volume due to the size change and I understand the loss of volume due to the sharp 90 degree intersection. But how much loss is what I'd like to know.

Assuming a pressure of 50 pounds how much volume loss (flow rate?) happens between the feeding (7/16 inch) gallery and the output gallery? How much will the pressure drop at six inches from the intersection? If the pressure was increased to 100 pounds would we get close the the volume of oil available in the feeding gallery at 50 pounds?

Please remember that the pump has a pressure relief that limits the output pressure, so the oil pressure in the feeding gallery is limited by the pump's pressure relief spring. Also an engine's oiling system is just a collection of controlled leaks so this abrupt change in direction probably is the major flow restriction. This 3/8 inch gallery splits into several galleries that feed the cam and main bearings, the hydraulic lifters and the rocker arms.

If it matters we can use an oil temperature of 150 degrees and standard 90 weight motor oil.

Thanks,

Bob

2. Oct 10, 2009

### Q_Goest

Hi Bob. Please verify I understand you correctly: The 7/16" oil galley ends and makes a 90 degree turn into a single 3/8" galley, it doesn't 'T' into a 3/8" that allows flow in both directions, right?

Anyway, it doesn't really matter. There's no loss of volume (or mass flow) going through the 90. There's only a pressure drop. Nature has this rule called "conservation of mass" which means that you can't destroy or create mass. All the oil that goes through the 7/16" galley has to go through the 3/8" galley as well. No mass loss. Oil is essentially incompressible (density remains constant) so there's no volume loss either. As the oil makes the turn, it has to increase in velocity so that the same amount of oil per second that just exited the 7/16" galley can now go through the 3/8" galley.

There's a drop in pressure associated with the various restrictions in your engine, but no loss in volume or mass.

To determine the loss of pressure, one has to have the flow rate though. So I can't tell you what the pressure loss is without knowing the flow rate. Even this doesn't matter much since the pressure drop across the various 'leak points' in the engine isn't likely to be affected unless one is getting more oil than another due to changes in upstream pressure between the two leak points.

An oil pump in a motor is a positive displacement pump. It puts out a fairly constant volume of oil per second. There's a slight drop in volumetric flow with an increase in discharge pressure because there is some leakage past the gear that's rotating, but without having a curve (graph) for this, let's just assume it's a constant displacement device. So regardless of the pressure drop in the elbow, there won't be a significant reduction in oil flow unless the relief valve opens. But if the oil pressure is lower than the relief valve set pressure, it won't open and the elbow has no significant affect on the flow of oil. The only difference you would see by reducing pressure drop through this corner is a reduction in the discharge pressure of your oil pump.

3. Oct 10, 2009

### bsprowl

OK I’ve think I get it. The most restrictive point controls the pressure and therefore the volume. The pump’s release valve creates an upper limit, but the most restrictive point may also.

I was focusing on the restriction and how that limited the volume but that volume must be constant before and after the restriction. Improving that tight 90 turn (actually not possible due the engine design) would increase the volume, but the pressure before and after the improvement would be the same.

Thanks. This has been bugging me for years.