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Getting code for egr circuit malfunction, but the replacement part only comes up being mechanical vaccuum type...........anyone familiar, or thoughts.

2007-01-11 21:26:05 · 4 answers · asked by helenlane_kia 2 in Cars & Transportation Maintenance & Repairs

4 answers

If you have this valve replaced by a professional mechanic, the labor would be the greatest expense. The part, its self, is relatively cheap.

It's a relatively simple repair. You or a friend / relative that you trust to work on your car could easily do this and save you some money.

2007-01-11 22:29:28 · answer #1 · answered by Peedlepup 7 · 0 0

It's that flying saucer looking thing attached to the intake. Check for vaccum leaks and dry-rotted lines. Pull the EGR and clean the carbon out of the intake. You may need to replace the EGR. But remove all of the carbon from the EGR port and fix all vaccum leaks first. If it runs rough, unplug the EGR and see if the engine smoothes out. If so, then your EGR has a problem. You can get by with just putting a screw in the vaccum line that goes to it. Some cars it won't work on, but I'm pretty sure it will work on the 2.8 V-6. Good luck.

2007-01-12 07:15:51 · answer #2 · answered by jeff s 5 · 0 0

This part is most likely on the rear of the engine block. It should not take more than an hour to replace it. Those vehicles did not have the sophisticated systems of today. It would be a mechanical valve, and most of the ones in use today are still mechanical units. It should be fairly easy to replace. Get the part, a book and do the replacement yourself.

2007-01-12 06:02:01 · answer #3 · answered by Anonymous · 1 0

Sooner or later all EGRs go bad. They get plugged with carbon soot, they corrode and leak, or someone blows compressed air into them and blows out the diaphram.
.
If the EGR fails, you get a bad N/o ratio which sets the oxy sensor to sending a fault code.
Either way, they are easy to replace. Just be careful, the bolts can be so frozen from corrosion they can break off.

2007-01-12 05:41:01 · answer #4 · answered by MechBob 4 · 1 0

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