Dec 6 2023.
I measured the airflow rates from the lower level vent when the exhaust fan was on full power, dome slit was open, door to dome was closed, and other vents were closed as well. This is the vent that roughly faces North, with fan installed.
2-d anemometer is currently installed in a temporary location sitting on the dome floor in front of this vent at about midlevel.
This happened at 8:30 pm local Chile time on Dec 6 2023. Measurements were taken with handheld propellor-type anemometer: Auputtriver model AP-846A.
The vent frame is basically 2.0 meters wide and 1.5 meters high. There are horizontal louvers placed about 11.5 cm apart. The measurements presented below were taken on a 30 cm x 30 cm grid with the anemometer held ~30cm away from the vent.
All measurements are in m/s. Normal component was measured, horizontal flow direction away from vent is positive.
| 0.8 | 0.1 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 | 0.0 |
| -0.5 | 0.0 | 0.7 | 0.5 | 0.5 | 1.3 | 0.9 | 1.9 |
| -0.2 | 0.0 | 0.0 | 0.5 | -0.7 | 0.1 | 0.2 | 2.5 |
| 0.0 | 0.0 | 0.4 | 4.2 | 0.0 | 4.1 | 2.7 | 1.7 |
| 2.8 | 3.3 | 3.0 | 3.1 | 1.8 | 3.9 | 2.0 | 1.2 |
| 1.5 | 2.5 | 5.0 | 0.3 | 5.5 | 3.9 | 0.2 | 4.0 |
There are 6x8 = 42 cells each of size 30cm x 30 cm. That's a total area of 900*42 cm^2 =3.78m^2. Each cell conributes 0.3m x 0.3m = A'= 0.09 m^2 of area and its measured velocity so each cell's flow rate is v*0.09 m^3/s.
Sum total of flux through this is sum(v_i*0.09) = 0.09*sum(v_i) = 0.09* 64.7 m/s so total air volume being moved is 5.8 m^3/s.
Let's estimate dome volume. Hemisphere is 9.3m diameter. Cylindrical height is about same as diameter.
So hemisphere dome volume is (4/3)*pi*(r^3)/2 = 210 m^3. where r=9.3/2.
cylindrical piece is pi*((9.2/2)^2)*9.3= 631 m^3. But about a third of that is taken up with pier so it's more like 400m^3.
At this level of accuracy we can estimate entire Aux Tel air volume as 600 m^3. At 5.8 m^3/s we get an exchange about every 100 seconds.
https://noirlab.edu/public/programs/vera-c-rubin-observatory/rubin-auxtel/
tract about 6 kW
Imagine the air leaves 1 degree warmer than it entered. How much heat does that correspond to? Heat capacity of air is around 1 kJ/m^3K. So it takes 1 kJ to raise temperature of 1 m^3 by 1K.
We move about 6 m^3 per second of air. So we would extract 6 kJ per sec per degree of air temp rise. It might be interesting to measure incoming and outgoing air temperature, but we need 0.1K accuracy or better.
During this time period the anemometer vent airspeed indicated 0.4 (presumably m/s?). During normal operations it's 4 times lower but the other vents are open then as well.