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does anyone know what the best fuel pump upgrade for a 1993 r32 gtst would be, running GTR injecotrs. Placed a Walbro gss342 in it a few weeks back and I would fail under load, guys at autobarn said most people run them in their 32s. Just wondering which one is best that I don't have to rewire and run constant 14v to it.

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Stock pump?

Seriously, it's a wire, a fuse and a relay, and there are multiple threads with pictures showing how to rewire a pump for 14v. It's not rocket science.

The only other pump that might work is the Nismo/Tomei pumps, but I wouldn't rely on the stock wiring for that kind of flow personally.

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Yeah stock pump feels like it's coming to its end. I've researched lots about that rewire but apparently you can have problems with fuel over heating and evaporating

Nonsense! Before I got my surge tank my in tank directly wired Walbro GSS 341 ran fine - never heard of anybody's fuel disappearing through evaporation!

The Walbro worked without rewiring but I put in a direct wire and relay because the pump will live longer that way and because I was getting a bigger turbo. There are a couple of tutorials in the DIY section.

The 341 would have been $400 over the counter. :P

Go to Jaycar, buy a relay, some wire and a fuse and run it to the battery. It need to be done, the Walbro's and most other aftermarket pumps require 12V minimum or the brushes will burn out. (The less the voltage, the more current is required to spin the pump.)

Your fuel temp will be fine with a 255L Walbro, it's when you go to a surge tank that it really becomes a problem.

I'm sorry but voltage and current go hand in hand. They are not 2 different entities.

V = I R

you failed at physics :)

why would resistance change?

you need to change the equation to:

P = VI, now say "if" the pump is run at 3bar base pressure + 1.5bar of boost (because yolo).. it would require ~11amp at 13.5V to run the Walbro 255.

This means, it will consume 148.5 watts

so put that back into the equation, now it becomes 148.5 = V x I, now as you reduce V, you need to increase I

Let's say we run the pump at 12V, that becoming 148.5 = 12 x I, current now has increased to 12.375AMP

A good real world example, high voltage power lines are in the 10s of thousands.. the reason behind this is to carry lower current, thus less resistance

I never said resistance will change. (Only partially true though because heat plays a big role on resistance)

Ohms law states that voltage and current are proportional.

I see what you are saying.

With your power line theory we can go 10kv over the line which lets say is 100 ohm now let's check current flow.

V = IR

10k = I x 100

I = 100 amp

Let's now reduce the voltage at the same resistance.

5kv = I x 100

I = 50 amp

Current is voltage flow. It will flow directly proportional to that of voltage if resistance does not change.

In order to calculate power with no other factors changing which you have done, you have to go back to the first step and see if any factors have changed ie resistance would most probably have to change in that factor.

I get a sense that some of this might be misinterpreted as "having a go" but if in person this would turn into a great discussion.

Correct in most cases, unless the resistance changes.

Running a pump at 8v will burn the brushes out, the motor is designed to run at 12+v, not 8v with load. I have removed many for that reason.

Yeah stock pump feels like it's coming to its end. I've researched lots about that rewire but apparently you can have problems with fuel over heating and evaporating

I use the Walbro GSS 341 wired directly to the battery. I can also confirm there are no fuel heating issues.

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