1997 Savana Van has my mechanic STUMPED!

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

Reply to
billpocz
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I have a 1997 GMC Safari (V6 engine but basically the same). When I bought it a year ago I new it would soon need a fuel pump (pressure was a little low and it was noisy).

One thing that was recommended when I replaced it was to check and run a new ground wire directly from the pump wiring to a good ground. If the tech hasn't done it already, it would be a good idea to check the pump current draw. If it's high, it may be the ground and could cause the pump to fail.

Good luck Larry

Reply to
Larry W

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

Reply to
<HLS

Pull the fuel pump. Take it to a rebuilder, or take it apart, and find out why it failed.

It may have failed because it was originally defective. Even an original GM one can be bad out of the box sometimes.

It may have failed because it is clogged up with gunk that is stuck at the bottom of the tank and getting past the filter. I would definitely check the filter anyway, too.

But taking it apart will let you know what happened.

--scott

Reply to
Scott Dorsey

The connector for these pumps is the same as for an oxygen sensor. They are way overloaded and tend to slightly melt. That's what I'd be looking at first.

Reply to
Steve Austin

Back in the late 1960s I drove a parts truck for an auto parts store.There were quite a few new defective parts. cuhulin

Reply to
cuhulin

A high resistance circuit will draw more amps than normal.

Reply to
Noozer

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.

Reply to
M.M.

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

Reply to
Erik

Larry, thanks for the tip, I will defintely pass it on!!

Bill

Reply to
billpocz

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

Bill

Reply to
billpocz

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 snipped-for-privacy@q66g2000hsg.googlegroups.com, billpocz at snipped-for-privacy@lcpcc.org wrote on 6/4/07 4:14 PM:

Reply to
AWN

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

SteveL

Reply to
pakeha

exactly wrong

Reply to
AZ Nomad

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.

Reply to
AZ Nomad

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...

Reply to
M.M.

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.
Reply to
Noozer

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

Reply to
Scott Dorsey

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.

Reply to
clifto

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