I'm not an engineer and my hydraulics aren't bulletproof but some stuff to think about:
The oil pump has a bore and stroke of about .5" x 1" which would be displacement of .25 inch. At 4000 rpm with no resistance that would pump about 1000 cu in a minute or about 4 gallons, which is ridiculous. However we know that in the stock setup this volume has to pass through a 3/16" orifice at the top of the return pipe, which is .007 sq in, and let's say the rocker box feed pipe is the same, so this side of the system is pushing oil through .014 sq inch (smaller at the rocker bearing passage) at maybe 2.5 psi at cruising speed (above atmospheric pressure). Now I've opened the standpipe orifice by 33%, so at whatever pressure the pump makes at any speed about 33% more oil should come out of the sump and into the oil tank. Apparently, even if the pressure to the rocker feed has fallen a bit the volume seems sufficient to keep the top end happy. The stock oil pump at idle is supposed to push 10 to 12 inches of head pressure to the stock orifices -- I seem to be getting at least 50% more than that with an enlarged orifice and Morgo pump.
Anyone happen to know how many quarts of oil circulate through the oil tank each minute? We know that the other side of the pump pushes oil through the crankshaft at 60 to 80 psi but I don't know what that means in terms of volume per minute.
Hi PA,
If the oil pump was capable of moving 4 gallons or even 1.7 gallons of oil with the engine at 4000 rpm, it would be fantastic.
Unfortunately the pump is just not designed that way.
The calculations made so far have been made on the assumption that the pump is symmetrical in operation and that what is drawn into the pump from the sump on the downstroke, is then pumped back out of the return line on the upstroke. To do this, the pump would require two ball valves, one for inlet the other for outlet. Unfortunately the Triumph/morgo pumps have only one ball valve on the outlet. The other valve is a piston port, very high up the bore of the pump, this results in a pump that is asymmetrical in operation.
From the top of the stroke, the piston port is covered and the decending piston creates negative pressure pulling the outlet ball valve hard on to it’s seat, with some help from the ball return spring. The port is uncovered by the piston and as the piston coninues to decend oil and air is drawn into the bore.
The effective downward stroke will be the from the port to the top of the piston at it’s lowest point. Not the entire length of the stroke. For arguments sake let’s say 0.175 cu.in.
This is great it is actually quite a lot, enough for the pump to lift oil from the bottom of the sump, even if it is aerated.
Now the piston starts to ascend, and the pump pushes most of the oil it has just lifted, straight back out of the open port and into the scavenge pipe again. Once the piston closes the port on the upstroke, the remaining oil is forced past the ball valve and into the return line. If you measure the distance from port closure to top of stroke, it is only (very rough guess) 1/16 “ (0.625") and 1/8” (0.125”)
Flow now would be onle 0.4-0.8 gallons a minute at 4000rpm.
But, as Stuart has previously stated, you can only take out what you put in, so in reality the oil that passes out of the return is the same that is put in by the smaller diameter pressure piston. So we have been making the calculations on the wrong side of the pump.
The difference in volume on a standard pump is 30% between pump and scavenge sides.
With this in mind, there will be only 0.3 to 0.6 gallons passing back to the tank per minute at 4000rpm.
Of course I do not know the effective pressure stroke exactly, and I am working from memory/guesswork only, so the flow calculations are only indicative of a lowered volume flow rate.
I do hope you have cured your wet sumping issue and are out there having fun on the TR6.
Regards
Peg.
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