Condensation on A/C ducts.

[Carolina Breeze HVAC]

That pinhole, can indeed cause you to fail a blower door test.... yes....wasting energy...why do you think that ducts, installed correctly, are sealed from start to finish?

[nicksanspam]

No way.

Nick

[Stretch]

Nick, If you put in a hole so small it would not cause problems, it would not do anything either. A pinhole will do nothing. Stretch

[Carolina Breeze HVAC]

Way...but then, you would have to be dealing with some real crappy install anyway...

[nicksanspam]

Then why are we led to believe it will fail a blower door test? :-)

Nick

[nicksanspam]

Sure, with a 2' diameter pin :-)

Nick

[Carolina Breeze HVAC]

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?...

[nicksanspam]

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?

Nick

[nicksanspam]

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.

Nick