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mgbinspect

Exhausted...

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Here' one for the brain trust:

A 44,000 btu furnace with a draft inducing blower (right flue)

A 88,000 btu furnace with a draft inducing blower (center flue)

And a 75,000 btu gas water heater - gravity (left flue)

ALL share this flue. It seems to me that it should have stepped up another size after the third flue joined in. Undersized? Any thought?

Click to Enlarge
tn_201139143335_DSCN8959.jpg

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And, all three units are crammed into a 6 x 9 unvented room in the finished basement ... hmm....

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OK,

You know that with my second grade math skills I can't do the math for you but the flue must equal the largest connector plus 50% of all others. Get out a pencil.

ONE TEAM - ONE FIGHT!!!

Mike

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OK,

You know that with my second grade math skills I can't do the math for you but the flue must equal the largest connector plus 50% of all others. Get out a pencil.

ONE TEAM - ONE FIGHT!!!

Mike

That'll work.

Thanks.

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OK,

You know that with my second grade math skills I can't do the math for you but the flue must equal the largest connector plus 50% of all others. Get out a pencil.

ONE TEAM - ONE FIGHT!!!

Mike

Do you mean that the cross sectional area of the main flue must equal the area of the largest flue plus 50% of the combined areas of the remaining flues? Where does this rule come from?

Marc

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I believe we're talking volume - a little geometry ala pi. Richtig?

If memory serves, it's V = 2 pi R

Nah. That can't be right. That's got to be circumfrence.

V = piR squared. I think that's it, but I'll look it up.

Yes pi R squared is correct.

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Area of a circle = (Pi) times (Radius squared).

Pi = approximately 3.14

Sorry Mike O. I couldn't help it. [;)]

Marc

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