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...



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. ...
>> > >
>> > > >
 



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