Originally Posted by
Ron
I don't think it's crude...let's see if I can play mathemagician (with a rusty pencil ;-)
SumpLevel = FuelIn - FuelOut
FuelIn = CkValveFuelIn + Return
FuelOut = PumpOut
Return = PumpOut - Used
By substitution;
SumpLevel = (CkValveFuelIn + Return) - PumpOut; [~your equation]
SumpLevel = (CkValveFuelIn + (PumpOut - Used)) - PumpOut
SumpLevel = CkValveFuelIn - Used *
SumpFillRate > 0 IF TankLevel > SumpLevel AND CkValveFuelIn >= Used
IF SumpLevel > TankLevel Then SumpFillRate <= 0
Suppose you start with a full sump and a half tank. If you turn the key on, the fuel in the pump will simply circulate because the check valve will not let any in the sump (after purge and pressure up...) If you then start the engine, it will lower the sump level at the rate the fuel is being used, until the levels match (check valve about to open). From here on is where I differ with either ideal, so far.
I believe saying that the pump draws fuel out of the sump and the check valve will instantly let more in make sense except the fuel being returned (already) is not accounted for (no time to rush in before return)...or maybe better said - the return is Constant and will replace what fuel the pump removes minus the fuel used, so the check valves will only allow the net loss, matching the tank level, which is reflected in * above.
:hmm:Me thinks :tongue: