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Carolina Breeze HVAC wrote:

Stretch wrote:

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.

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


Define "pinhole" :-)


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.

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.


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?


If 10 cfm = 118Pi(0.025/2)^2sqrt(dP), dP = 30K psi :-)

2. How many pinholes are needed for 100 cfm of airflow at 50 Pa?


If dP = 0.00725 psi makes 1 pinhole leak Q=0.00491 cfm, 10/Q = 20,362.

3. How large must one "pinhole" be for 100 cfm of airflow at 50 Pa?


If 100 = 118Pi(D/2)^2sqrt(0.00725), D = 3.6", a large pin :-)

4. How large must it be to remove 2% moisture by weight (which halves
the R-value) from 40 ft^3 of fiberglas insulation


ie 40x0.5x0.02 = 0.4 pounds of water...

in a month, with 45 F AC air at 100% RH...


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...

flowing from 1" H20 duct pressure...


with 0.006 = 118Pi(D/2)^2sqrt(0.0361), and D = 0.0186", a small pin :-)

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?


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