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2003 v6 accord engine bog under load

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  #11  
Old 11-06-2020, 06:26 PM
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Fuel trim is at 1 then goes to 1.1 under load. I timing is 10* initial and gets to 40* at 3,000 RPM. I’ll look at removing the O2’s tomorrow.
 

Last edited by Las27563481; 11-06-2020 at 09:49 PM.
  #12  
Old 11-14-2020, 07:26 PM
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Well I removed the downstream O2’s and the car ran so much better. There is a Y pipe joining the two exhaust pipes thn aanother shared cat and what appears to be a resonator, then iY’s out to two mufflers. I think the center, aft cat is clogged. The bolts on the flanges were rusty and I didn’t want to break anything else. I’ve broken enough. I’ve never had a clogged resonator nor clogged mufflers, they usually rust out and leak as a failure mode.

ideas?????
 
  #13  
Old 01-19-2021, 06:04 PM
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Well, what I call the aft cat, behind the front two was filled with material from at least one of the two forward cats. The foreword ones looked decent. I think one of the foreword ones fell apart and dumped into the aft cat, the one that failed was replaced, but what fell out of it clogged the aft cat. I cleaned out the aft cat with a pressure washer and blew a hole through it. The car now runs, maybe as good as new.
 
  #14  
Old 01-21-2021, 08:48 AM
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Hope you dont have to have it inspected anytime soon, don’t think it will pass.
 
  #15  
Old 01-21-2021, 09:30 AM
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Yeah I know. I’ll have to clear the code and see if it throws another. No O2s on this cat so it may not. I did get a code for O2 high current, but the battery was very low, not even a click when you turn the ignition. Low voltage will cause high currents in some cases I suspect. New battery this weekend to test theory.
 
  #16  
Old 01-21-2021, 09:56 PM
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Thanks for coming back with the findings and your fix for now.

I thought it was the other way around
High current will cause low voltage.
A battery's voltage will go down when drawing large amounts of current as a starter motor would.

But, I could be wrong??
 
  #17  
Old 01-22-2021, 09:33 AM
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Power is V * I to get the same power when V drops, I has to rise to meet the expected power. Again P=V*I. The O2 is going to use an expected amount of power like a variable resistor and when the voltage drops the current draw will be more. I am no expert, but understand power and energy. Degreed EE with 25 years in the power field. Worked on cars since the early 80’s.
 

Last edited by Las27563481; 01-22-2021 at 09:39 AM.
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