1997 Savana Van has my mechanic STUMPED!

On Jun 4, 11:41 pm, Erik > wrote:
> In article >,
>
> "M.M." > wrote:
> > Noozer wrote:
>
> > > A high resistance circuit will draw more amps than normal.
>
> > How so?
>
> > By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a
> > more or less constant voltage as you'd have in a car, as R increases, I
> > decreases.
>
> M.M.'s right...
>
> When these pumps 'fail', are you guys checking to be sure power is
> actually being delivered to the pump, and that a good reliable ground is
> provided?
>
> As another poster mentioned, are the connector/s in good shape.
>
> I have a feeling this one is going to be something simple along those
> lines...
>
> Erik

Thanks Erik! Yeah, I feel the same, that it will probably be a simple
thing, its just finding it!!

Bill

Does the fuel pressure check out as normal? Do you have any issues with
your oil pressure gauge reading? I know that if your oil pressure gauge is
faulty, you will have these symptoms. If you run 12V to the pump directly,
does it engage? If you **hear** the pump but there's not enough pressure, I
would suspect a damaged pulsator or check valve. If the amp draw is much
higher than normal (I think most GM pumps run around 9A +/- 1A), I would
suspect a blockage somewhere (either in the tank or leading into the fuel
filter or rail). Please tell us what the pressure of the fuel system
shows... As well, have you scanned the system? Do your short term/long
term fuel readings seem high or low? If sounds like an electrical problem or
a physical blockage to me (but I am not a mechanic - just someone who has
been in a similar boat with a 4.3 W series 95 vortec). Are you certain you
changed the correct EFI relay?? Does your vehicle have a red or pink wire
tap laying in the driver's side shock tower cavity? You should be able to
run a fused 12V lead to this test connection to accurately test the pump
itself (unless your vehicle doesn't have the test connection).
Good luck,
Andrew.



in article , billpocz
at wrote on 6/4/07 4:14 PM:

> recently purchased a 1997 Savana van (5r.8 L Vortec) that seemed in
> very good condition. The previous owner told me that the only problem
> they had was that about a year ago, the battery kept draining and the
> fuel pump 'went'. They had both changed and didn't have any more
> problems.
>
> Last week, I arrived home and shut the van off, later went to restart
> and it would crank but not start. My friend (who is a certified Ford
> mechanic) came over and determined that the fuel pump was not
> working.
> I had the van towed to his shop where he changed the fuel pump and
> filter (he removed an AutoZone pump and replaced it with a GM one.
> All
> seemed OK for two days, and the pump stopped again!
>
>
> He took the vehicle back, stated that it may be the pump relay and
> changed both. He also checked the wiring harness for any problems but
> said it looked great.
>
>
> The van seemed to work fine, drove it for about a week. It started to
> 'miss' at times (usually when letting off the throttle), and wanted
> to
> stall. It did stall a few times but would restart after a few
> minutes.
> Now, the fuel pump is gone again, and it is dead in the water.
>
>
> My mechanic friend is at wits end and is ready to throw in the towel.
> Is it time for a real GM mechanic?
>
>
> Does anyone have any directions to point my mechanic (since it is
> back
> at his garage)?
>
>
> Thanks!
>
>
> Bill
>

On Mon, 04 Jun 2007 19:09:22 -0700, "M.M." > wrote:

>Noozer wrote:
>>
>> A high resistance circuit will draw more amps than normal.
>>
>>
>
>
>How so?
>
>By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a
>more or less constant voltage as you'd have in a car, as R increases, I
>decreases.

So what happens when the V drops, R stays the same, I does what??

SteveL

On Tue, 05 Jun 2007 01:25:57 GMT, Noozer > wrote:


>> If the ground were bad, wouldn't you expect the current to be low, on the
>> average,
>> rather than high?

>A high resistance circuit will draw more amps than normal.


exactly wrong

On Wed, 06 Jun 2007 02:56:40 GMT, > wrote:


>On Mon, 04 Jun 2007 19:09:22 -0700, "M.M." > wrote:

>>Noozer wrote:
>>>
>>> A high resistance circuit will draw more amps than normal.
>>>
>>>
>>
>>
>>How so?
>>
>>By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a
>>more or less constant voltage as you'd have in a car, as R increases, I
>>decreases.

>So what happens when the V drops, R stays the same, I does what??

What happens when a flashlight battery goes dead (V drops to zero).
Same principle.
What does the current do? You should be able to answer this on
your own.

wrote:
>> By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a
>> more or less constant voltage as you'd have in a car, as R increases, I
>> decreases.
>
> So what happens when the V drops, R stays the same, I does what??
>

Well, let's see...suppose it was 12v and 12 ohms, just to make it easy.
That would make I be 12/12 = 1 amp.

Now, suppose the voltage drops to 6v, I would be 6/12 = .5 amp.

Looks to me that I would decrease then, too.

At least, that's the way I remember it from Circuits 101, way back when
they taught us about vacuum tubes, but I don't think the laws of physics
have changed much since then...

"M.M." > wrote in message
...
> wrote:
>>> By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a
>>> more or less constant voltage as you'd have in a car, as R increases, I
>>> decreases.
>>
>> So what happens when the V drops, R stays the same, I does what??
>>
>
> Well, let's see...suppose it was 12v and 12 ohms, just to make it easy.
> That would make I be 12/12 = 1 amp.
>
> Now, suppose the voltage drops to 6v, I would be 6/12 = .5 amp.
>
> Looks to me that I would decrease then, too.
>
> At least, that's the way I remember it from Circuits 101, way back when
> they taught us about vacuum tubes, but I don't think the laws of physics
> have changed much since then...

Yaya... I get it. I was wrong.

So why do you need to use a larger gauge wire in a longer extension cord?
Obviously there is more resistance in 500' of 14gauge wire than there is in
50'. According to what you're saying, you should be able to carry more
current over a longer distance due to the resistance.

Noozer > wrote:
>
>
>So why do you need to use a larger gauge wire in a longer extension cord?
>Obviously there is more resistance in 500' of 14gauge wire than there is in
>50'. According to what you're saying, you should be able to carry more
>current over a longer distance due to the resistance.

This is a voltage drop issue. Your load is in _series_ with the resistance
of the wire.

The total resistance of the system is (Rwire + Rload) so the total
current through the system can be found with

Vbattery = I(Rwire+Rload)

Now the idea here is that that additional resistance is _reducing_ the
total amount of current that can flow through the circuit.

Another way of looking at this is by saying that the voltage the load
sees is the voltage the battery produces MINUS the voltage drop due to
resistance.
--scott
--
"C'est un Nagra. C'est suisse, et tres, tres precis."

Noozer wrote:
> "M.M." > wrote in message
> ...
>> wrote:
>>>> By Ohm's law, current (amps) = voltage / resistance (I = E / R). With a
>>>> more or less constant voltage as you'd have in a car, as R increases, I
>>>> decreases.
>>> So what happens when the V drops, R stays the same, I does what??
>>>
>> Well, let's see...suppose it was 12v and 12 ohms, just to make it easy.
>> That would make I be 12/12 = 1 amp.
>>
>> Now, suppose the voltage drops to 6v, I would be 6/12 = .5 amp.
>>
>> Looks to me that I would decrease then, too.
>>
>> At least, that's the way I remember it from Circuits 101, way back when
>> they taught us about vacuum tubes, but I don't think the laws of physics
>> have changed much since then...
>
> Yaya... I get it. I was wrong.
>
> So why do you need to use a larger gauge wire in a longer extension cord?
> Obviously there is more resistance in 500' of 14gauge wire than there is in
> 50'. According to what you're saying, you should be able to carry more
> current over a longer distance due to the resistance.

He addressed what happens when resistance stays the same and something
else changes. You've somehow parlayed that into what happens when
resistance changes and the other parameters are unspecified. If you'll
specify what happens to either voltage or current, we'll tell you
which way the other parameter goes.