re: solar heating for barn
13 feb 2004
>...dallas semi ds 1920 "1-wire" sensors... take the form of a stainless
>steel cylinder that is about 0.6 inch dia by 0.25 inches high.
with v = 0.000041 ft^3 of volume and thermal capacitance c = 64v = 0.0026
btu/f of thermal capacitance (like water, by volume) and s = 0.0072 ft^2 of
surface and g = 2s = 0.014 btu/h-f of thermal conductance to slow-moving
air and rc = c/g = 0.18 hours...?
>i can glue(?) a carbon resistor rated at 0.25 watt to the top of the sensor,
>and connect it to a 5v regulated supply -- so for 0.1 watt (as a starting
> p = 0.1 watt = i e = (e/r) (e)
> r = (e) (e) / p = (5) (5)/ (0.1) = 250 ohm
...0.1w is 0.34 btu/h, which might raise the sensor temp 0.34/0.014 = 24 f
in still air.
>under steady solar conditions, i can read the temperature sensor with and
>without the voltage applied to the resistor, and the temperature difference
>should be proportional to the flow velocity.
>i'm not sure if the temperature sensors have good enough sensitivity
>for this, but it seems easy to try... my estimate for the flow velocity
>from the 4 inch by 20 inch collector vents is only about 1 ft/sec...
that's 0.68 mph, which might increase the sensor-to-air conductance to
(2+0.68/2)/2g = 0.016, raising the sensor temp 0.34/0.016 = 21 vs 24 f.
not a big difference. can we make it bigger and lower the time constant?
use 1/2 w and stick a 2" copper foil square on the sensor?
>this is based on:
>vent size is 4 inches by 20 inches = 0.56 ft^2
>the 14 ft^2 of collector area per vent generates (14ft^2)(200 btu/hr -ft^2)
>= 2800 btu/hr, or 0.8 btu/sec
looks like some of that will be lost to the outdoors through the glazing.
>assume that the collector generates a temperature rise around 100f (e.g.
>something like 40f up to 140f).
that would make the average collector temp 90 f. if it's 26 f outdoors,
you might lose (90-26)14ft^2/r1 = 896 btu/h through the glazing...
>so, what is the weight flow of air to reqd to absorb 0.8 btu/sec with a 100f
air's specific heat is about 0.24 btu/f-lb, about 1/4 of water...
>and, what velocity through a 0.56 ft^2 vent does this weight flow result in?
>q = 0.8btu/sec = (wflow) (cp) (dt)
>wflow = (0.8 btu/sec) /( (0.2 btu/lb-f) (100f) = 0.04 lb/sec
>volume flow = vflow = (wflow)/ (density) = (0.04 lb/sec) / (0.076lb/ft^3) =
>velocity = vflow/a = (0.52 ft^3/sec) / (0.56 ft^2) = 0.93 ft/sec
>surprisingly slow?? does this seem about right?
hmmm. if 40 f barn air flows into the collector and t f air flows out,
the average temp of the air in the collector is (t+40)/2, so it loses
((t+40)/2-26)14ft^2/r1 = 7t - 84 btu/h through the glazing, and the rest
of the heat goes into the barn, ie 250x0.9x14 = 3150 btu/h = 7t-84+qb,
or 3234 = 7t+qb. meanwhilst, cfm = 16.6x0.56sqrt(7(t-40)) = 24.6sqrt(t-40),
and a 1 cfm airstream with a 1 f temp diff carries about 1 btu/h of heat, so
qb = 24.6(t-40)^1.5, ie t = 40+(132-0.285t)^0.666. plugging in t = 100 on
the right and raising to the evil power makes t = 62 on the left. plugging
in 62 on the right makes t = 63.5 on the right, then 63.4, then 63.4. so,
cfm = 24.6sqrt(63.4-40) = 119, which makes v = cfm/ft^2 = 213 feet per min,
or less, since the air needs to turn two corners, or more, if the screen
keeps 40 f air near the glazing, with less heat loss to the outdoors.
this could be a very efficient collector.