Re: [DML] Re: She starts (painfully), she runs (mostly), she doesn't go
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Re: [DML] Re: She starts (painfully), she runs (mostly), she doesn't go anywhere...



Tom
I just picked up an Eagle Premier 3.0 l PRV engine that I want to build up. Do you have a source for forged pistons over there?  I'm looking at $1,000 for a set here in the states. 

Ian Yanagisawa

On Jun 20, 2013, at 1:42 PM, "Tom" <dmctom@xxxxxxxxx> wrote:

> Elvis is right. A simple calculation of the voltage, current and resistance 
> can easily prove this.
> The resistance of the fuel pump stays the same and voltage will change. 
> Let's see what happens with the current.
> This is just an example as I don't know the actual resistance of the pump.
> 
> Let's assume that ideally the pump runs at 12.5V and its resistance is 1.5 
> ohm. A simple formula of I=U/R (current = voltage divided by resistance) 
> will help determine the current. Thank you Mr. Ohm ;)
> I = 12.5 / 1.5
> I = 8.33
> So, under ideal conditions our example pump will use 8.33 amps.
> To calculate power use we can multiply voltage by current. Thank you Mr. 
> Watt ;)
> P = 12.5 * 8.33
> P = 104.125 Watts
> 
> Now, if the voltage drops due to bad connections, weak battery, failed 
> alternator the result will change. So, lets assume that the voltage at our 
> example pump drops to 10.5 volts.
> I = 10.5 / 1.5
> I = 7
> The current drops to 7 amps.
> And the total power....
> P = 10.5 * 7
> P = 73.5 Watts
> 
> You can check my math to make sure I got it right.
> As you can see from these calculations if the resistance stays the same and 
> the voltage drops the current will drop with it. As a result the power use 
> will drop and the power OUTPUT will drop as well. This means lower fuel 
> pressure and lower fuel volume. As some point the engine will simply stall 
> because of lack of fuel.
> 
> Greetings from Poland!
> Tom Niemczewski
> Vin 6149 plus 2418, 3633, 5030, 16473, 17086
> Google earth: 52°25'17.66"N, 21° 1'58.40"E
> 
> --------------------------------------------------
> From: "Elvis" <elvisnocita@xxxxxx>
> Sent: Thursday, June 20, 2013 4:38 PM
> To: <dmcnews@xxxxxxxxxxxxxxx>
> Subject: [DML] Re: She starts (painfully), she runs (mostly), she doesn't go 
> anywhere...
> 
> >
> > Gentlemen,
> >
> > there is nothing inside the pumpt that controls the output power of it !
> > This is a stupid brushed motor that spins depending of the counterpressur 
> > that it sees.
> >
> > The unused pressure is transformed into fluid flow by the pressure 
> > regulator.
> >
> > And - a higher resistance (bad connector, thin wire, oxydized fuse, melted 
> > fuse holder) reduces the current.
> >
> >
> > NO need to believe me, I just make my money with fans - which is nothing 
> > else but an air-pump.
> > BTW - powerfull fans do have electronics inside to limit the max speed by 
> > PWM - therefore they can keep the power constant. When the voltage goes 
> > down - in certain limits - it draws more current.
> > Again - nothing like this in a fuel pump.
> > Yes, I even cut an old pump open to see how it works.
> >
> > Oh another example - the interior fan - it uses resistors to reduce the 
> > power, position 1 and 2 don't even require an extra relay...strang if - 
> > according to you - the current goes up ?!?!?!? :-P
> >
> > Have fun ;-)
> > Elvis & 6548
> >
> >
> >
> > --- In dmcnews@xxxxxxxxxxxxxxx, Ian <texas.twister@...> wrote:
> >>
> >> Ditto!
> >>
> >> Ian Yanagisawa
> >>
> >> On Jun 18, 2013, at 12:53 PM, JP Hindin <jplist2008@...> wrote:
> >>
> >> >
> >> >
> >> > On Tue, 18 Jun 2013, Elvis wrote:
> >> > > This is a very strange theory, that a lower voltages kill something 
> >> > > and draws more current.
> >> >
> >> > This one I /can/ answer - the work of the pump doesn't change, so it's
> >> > fixed - the laws of physics state that by lowering the voltage, the
> >> > amperage must go up (Volts x Amps = Watts). This damages the pump 
> >> > because
> >> > it's not designed to run at higher amperages, which would require it to 
> >> > be
> >> > more sturdy with heavier duty wiring and stators, thus it burns out.
> >> >
> >> > > Anyhow - is the piston in the fuel distributor stuck ?
> >> > >
> >> > > Push the air metering plate down and you 'll see.
> >> >
> >> > I'll check this also, thank you.
> >> >
> >> > I've never actually dug into the engine bay - I only redid the tank - 
> >> > so
> >> > I'm a little gunshy of tearing off covers and digging into it. I guess 
> >> > I'd
> >> > best get over it and get cracking.
> >> >
> >> > - JP
> >> >
> >> > > --- In dmcnews@xxxxxxxxxxxxxxx, Ian <texas.twister@> wrote:
> >> > > >
> >> > > >...
> >> > > > The low voltage ends up requiring more amperage to run the pump. 
> >> > > > This will eventually burn out the pump. There have been several 
> >> > > > threads on this issue. ...
> >> > >
> >> > > >
> 
> 
> 


[Non-text portions of this message have been removed]



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