re: wood stove "puffs"
28 dec 1999
>...i find that one of the most common problems causing backdraft, is when
>you have the house bottled up and you have been burning wood for a while,
>the air pressure inside the house drops from the air going up the chimney,
>and is much lower than the outside pressure... for me i open the front door
>for a minute or two to equalize the pressure...
and i responded...
>i wonder how long a "while" is, before the house pressure stabilizes,
>and how long a minute or two of door-opening makes a difference after
>the door is closed. i suspect it's a very short time in both cases.
>say a woodstove draws 20 cfm, and 10 house air volumes per hour rush in with
>a 50 pa depressurization (about 0.5 ach under natural conditions), and the
>house contains 32x32x16 ft^3 of air, and the rate of airflow into the holes
>and cracks is proportional to the square root of the pressure difference...
looks like about 10 milliseconds (1/100 of a second),
or maybe a little more, given the speed of sound...
10 v=32*32*16'house air volume (ft^3)
20 ach=10'house air volumes per hour at 50 pa
30 cfm=20'woodstove airflow (cfm)
40 dt=1'time step (ms)
50 for t = 0 to 8*dt step dt
60 chimflow=cfm/60*dt/1000'air up chimney (ft^3)
70 netflow=chimflow+netflow'initial air missing from house (ft^3)
80 dp=netflow/v*101325!'outdoor-to-indoor pressure difference (pa)
90 inflow=sqr(dp/50)*ach*v/3600*dt/1000'house air inflow (ft^3)
100 netflow=netflow-inflow'final air missing from house (ft^3)
110 print 200+t;"'";t;netflow;dp;inflow
120 next t
140 dpss=(cfm*60/(ach*v))^2*50'steady-state pressure difference (pa)
150 print"dp steady-state..";dpss
time out-inflow diff press inflow
ms ft^3 pa ft^3
0 4.110646e-05 2.061463e-03 2.922269e-04
1 6.471801e-05 2.315681e-03 3.097218e-04
2 7.871358e-05 2.461704e-03 3.193378e-04
3 8.714362e-05 2.548258e-03 3.249033e-04
4 9.226688e-05 2.600392e-03 3.282101e-04
5 9.539668e-05 2.632077e-03 3.302035e-04
6 9.731459e-05 2.651432e-03 3.314154e-04
7 9.849205e-05 2.663294e-03 3.321559e-04
8 9.921571e-05 2.670576e-03 3.326097e-04
dp steady-state.. 2.682209e-03 pa
the eventual steady-state pressure difference is equivalent to a
0.00001 inch water column, enough to push in a 3'x6' door with
0.01 pounds of pressure, the weight of about 10 drops of water.