My AFPR was in line with my stock FPR, so I decided to just get rid of the stock FPR and manufacture (because I can) a fitting to attach a -AN line directly from the oulet of the fuel rail to the AFPR inlet.

The fittings on the AFPR are all -04AN. My current fuel line setup is identical to the Stevetek (all -06AN from the pump to the rail)

My question is should the fittings on the AFPR also be -06 AN, or will keeping the -04 AN fittings be ok? I need to know what size fitting to manufacture for the end of my rail and what size line to use to connect to the AFPR inlet. (they will be the same of course) I am just going to leave the return line to the tank the way it is for now.

Thanx

I think the -4 should be fine.

If you're running a high-flowing pump, you'll want to be able to return the full amount of unused fuel to the tank at low/no boost. This is where the -6 would be more useful, perhaps.

However, if you're using a stock return line, that's going to be your rate-determining component. Unused fuel leaving the fuel rail will not be held up by -6, but it will eventually hit the stock return line, and get congested (if that were even to occur; I think it's unlikely). Therefore, the -4 probably wouldn't impose any more restrictions than the stock return line already does.

Just my thoughts...

BTW, I would like to commend Rob for being an excellent member of the board; for his insights and contributions, and his efforts to make quality posts.

- b.

Thanx Bernie. I try.

The fuel pump I am using is a Walbro 255. I have been using the AFPR for a while with the -04 fittings on it with no ill effects. (to my knowlege anyways) I just needed confirmation from some one that my idea wasnt actually a big mistake. What would the ill effects of fuel not returning to the tank be, just for future knowlege.

By eye, it looks as though the stock return line is actually smaller on the inside diameter than the stock fuel input line. (I'll have to confirm with a vernier caliper later) Does this smaller line actually help raise the fuel presure in the rail without having to adjust the AFPR so much?

By "fuel input line", I'm going to assume you mean the line from the pump to the rail (well, filter in between, too). I would call this the "feed" line, but that may not be the "official" name.

Anyway, the fuel input line is bigger than the return line because much of the fuel sent from the tank will be lost through the injectors, and only the surplus will need to pass through the return line. So, suppose the input line can flow 4L / minute, and the injectors shoot out 2L in a minute, you'll probably need a return line that can flow 2L / minute (the remainder). It's probably a bit more complicated, but that's the general idea.

If the AFPR/return line combination can't return the "correct" (adequate) amount of unused fuel, there are two counter-actions:

1) The fuel pressure will go up, and the injectors will be forced to allow more fuel to pass through them per "squirt".

2) The fuel pump motor will be reduced in efficiency, heat up, and burn out (over a long period of time, probably). The opposite effect can be observed with a vacuum cleaner: plug the "sucking end" completely, and you'll hear the motor's RPM decrease as it strains to fight the increased vacuum (in the car's case, it would be increased pressure, but both amount to a form of physical resistance). The motor will begin to heat up as well.*

Chances are, you'll get a combination of the two things... when all injectors are closed, the fuel pump will be strained; when open, an injector itself will be stressed from the pressure, and you may have problems... but probably not until very high pressures (90 PSI). The other potential problem is decreased fuel efficiency caused by injectors passing more fuel than necessary. But it's better than running too lean.

- b.

* Why will an electric motor burn out when contending with massive physical resistance?

A typical motor will have a few coils of wire that act as electromagnets, and they are activated in sequence by contacts mounted on the motor shaft itself. An active coil will magnetically coax the motor to turn part of a revolution, switching the contacts and activating the next coil, which will cause further turning, and so on. Now, the coils are just big loops of wound wire; not too much resistance, and high-current in large motors. The wire in the coils gets hot as it tries to dissipate the excess power (measured in Watts; this power lost as heat can be used to calculate the motor's efficiency). Anyway, in a three-coil motor, as long as the motor keeps turning, each coil is only passing current 1/3 of the time, but dissipating heat all the time. No problem. But when you force the motor to run really slowly (in other words, when the physical load on the motor is too large), each coil stays active longer during a cycle, heating up to "unsafe" levels. At some point, the excess heat will damage the coils (perhaps burning through the insulation on the wires -- usually enamel on bare copper -- causing a short, and reducing the coils ability to produce a strong magnetic field). Then the motor is "burnt out". Note: this is simply my understanding, and I welcome corrections.

[Edit] Fixed some wording and some mismatched parentheses. I'm German; I'm pedantic. Sue me.

I couldnt imagine a better explaination, Thanx for your time!

Cheers!
Rob