Define "pinhole" :-)
Every leak adds to the drop, and yes, you can have one too many..
I mean, if its ok to have one, lets add another, it wont hurt anything.
While we are at it, hell, those 20 didnt add anything, but its sure getting
cooler up here....lets add another.
You also realize some of that Flex out there is so cheap that you can indeed
cause the inner duct to come apart when you pin hole it right?...
OK. A hole made by a pin. I just miked 5 pins out of a sewing drawer,
0.0256, 0.0252, 0.0288, 0.0252, and 0.0286" diameter.
You seem to know little about airflow or blower door tests...
An "airtight" 0.2 ACH house leaks about 4 ACH at 50 Pa, about
1280 cfm for a 2400ft^2x8' house. The measurement accuracy is
on the order of 100 cfm. So...
1. How much pressure is needed to make 100 cfm flow through a pinhole?
2. How many pinholes are needed for 100 cfm of airflow at 50 Pa?
3. How large must one "pinhole" be for 100 cfm of airflow at 50 Pa?
4. How large must it be to remove 2% moisture by weight (which halves
the R-value) from 40 ft^3 of fiberglas insulation in a month, with
45 F AC air at 100% RH flowing from 1" H20 duct pressure, if the
insulation is in an 85 F attic at 50% RH?
5. How much energy would it "waste," if combined with an extra R19 layer
of insulation, compared to exposing a 20' R8 duct to hot attic air?
If 10 cfm = 118Pi(0.025/2)^2sqrt(dP), dP = 30K psi :-)
If dP = 0.00725 psi makes 1 pinhole leak Q=0.00491 cfm, 10/Q = 20,362.
If 100 = 118Pi(D/2)^2sqrt(0.00725), D = 3.6", a large pin :-)
ie 40x0.5x0.02 = 0.4 pounds of water...
and wi = 0.0064, warming to 85 F at 100% RH (wd = 0.0267), so
60C0.075(wd-wi) = 0.09154C lb/h leaves the fiberglass, making
C = 0.4/(0.09154x30dx24h) = 0.006 cfm...
with 0.006 = 118Pi(D/2)^2sqrt(0.0361), and D = 0.0186", a small pin :-)
Using your numbers, a 6" x 20' duct with 31 ft^2 of surface would lose
24h(85-45)31ft^2/(R0.2+R8+R1.67) = 3015 Btu/day. The pinhole would lose
about 24h(85-45)0.006 = 6 Btu/day. The extra insulation would reduce
the duct loss to about 24h(85-45)31/(R0.2+R8+R19) = 1094 Btu/day, for
a net savings of 3015-(1094-6) = 1927 Btu/day, with the pinhole.
If there is not too much water dripping inside I have seen large cheap aluminum
broiler pans used to catch the water, which eventually evaporates. Outside, see
if you can make a catch pan to divert and drain water away. BTW is it possible
the A/C installer can be convinced he should fix the problems?
"BTW is it possible
the A/C installer can be convinced he should fix the problems?"
I don't know where the problem lies. That's why I posted.
I was referring to where the ducts cross. Big tubes in the ceiling.
Where the ducts cross each other.
The big silver tubes in the attic. One particular spot in question, the
duct tubing was criss crossed and where they made contact, it seemed to
create more condensate there.
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