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Posted by Gary on February 13, 2004, 2:32 am
 
Hi Nick,

Just to be sure I understand your suggestion:

The temperature sensors that I have are Dallas Semi DS 1920 "1-wire"
sensors.
They take the form of a stainless steel cylinder that is about 0.6 inch dia
by 0.25 inches high.
The contacts are made to the top and bottom of the cylinder.
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
point):

        P = 0.1 Watt = I E = (E/R) (E)
        R = (E) (E) / P = (5) (5)/ (0.1) = 250 ohm

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.  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
Assume that the collector generates a temperature rise around 100F (e.g.
something like 40F up to 140F).

So, what is the weight flow of air to reqd to absorb 0.8 BTU/sec with a 100F
temperature rise?
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) =
0.53 ft^3/sec

Velocity = Vflow/A = (0.52 ft^3/sec) / (0.56 ft^2) = 0.93 ft/sec

Surprisingly slow??  Does this seem about right?
Seems like this will make any kind of velocity measurment difficult.

Gary









velocity


Posted by nicksanspam on February 13, 2004, 4:11 pm
 


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...?


...0.1W is 0.34 Btu/h, which might raise the sensor temp 0.34/0.014 = 24 F
in still air.


Somewhat non-linearly.


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?


Looks like some of that will be lost to the outdoors through the glazing.


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...
 

Air's specific heat is about 0.24 Btu/F-lb, about 1/4 of water...


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.

Nick


Posted by nicksanspam on February 8, 2004, 6:55 pm
 

Maybe more, with some thermal bridging?


NREL data indicate December is the worst-case month for solar
house heating in Helena, when 390 Btu/ft^2 of sun falls on the
ground and 810 falls on a south wall on an average 21.2 F day
with an average daily high and low of 11.2 and 31.3, for an
average daytime temp of about (21.2+31.3)/2 = 26.3.


NREL's data assumes a 0.2 ground reflectance. Their Solar
Radiation Data Manual for Buildings suggests adding Iadj
= 0.5(Rd-0.2)Ih(1-cos(B)), where Rd is a larger ground
reflectance (about 0.6 for snow) and B is the (90 degree)
collector tilt and Ih (390) falls on the ground, so the
collector might see 810+0.5(0.6-0.2)390(1-cos(90)) = 888
Btu/ft^2/day, or 0.9x888x103 = 82.3K Btu/day through (clear
vs tinted?) polycarbonate glazing with 90% solar transmission.

If the collector works for 6 hours on an average December day,
you might have 82.2K = 6h(T-26.3)103ft^2/R1 + 24h(T-21.2)160,
approximately, which makes the average barn temp T = 40.4 F.


Steve Baer might suggest H/15 = 7x12/15 = 5.6"...


Sounds good.


Sounds good.
 

Some kind of plugabble holes or closable doors at the top and bottom.
You want them to close tight in wintertime, and keep wasps out...


It's better to use data based on longitude too, ie historical data for
the actual location, for average vs clear days, including clouds and fog.


There will be some thermal resistance between the mass and the air.
This might not matter much, if the mass has a large area.


It might be closer to 20 than 40 F warmer.


Larger seems better.


A fan or blower might double it, and allow better air temp control, and let
you make the collector taller.


I'd say so.


Yes, with barn air next to the glazing and heated air on the north side
of the glazing.


Dry cleaner bags hinged at the top, hung over holes in the wall. They
work pretty well. They can get stuck or folded or torn. It's good to
inspect them every week or so. The holes can be covered with chicken
wire to keep the bags from sucking through in the wrong direction.
 

Same as winter? Polycarbonate plastic can take very high temps, 150 F air
on the inside, or more...


You might think about making your "sunspace" deeper, eg 4' or 8' out from
the barn, to decrease the airflow resistance and to better separate barn and
solar-warmed air and to slow down the airflow near the glazing and to allow
more summer venting and shading with an overhang, and to make more usable
floorspace, which might include a couple of chairs or a clothesline.

Nick


Posted by Duane C. Johnson on February 8, 2004, 11:50 pm
 Hi Nick;

nicksanspam@ece.villanova.edu wrote:
 

OK, I looked up the average december map for
"South Facing Vertical Flat Plate":
http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/atlas/serve.cgi
My guess shows the value to be about
(2.5kWh/m^2)/day

To convert this
(2.5kWh/m^2)/day * (3412BTU/kWh * 0.093m^2/ft^2)
(2.5kWh/m^2)/day * 317BTUm^2/kWhft^2 = 792BTU/ft^2/day

Close enough.

However, I think the NREL data assumes an ideal solar collector.
One that is 100% efficient at capturing the insolation.

Real collectors, whether they be PV (maybe 15%) or thermal,
always have some losses.

One needs to know the operating efficiency at the required
input, output, and ambient temperatures to asses what
amount of the insolation will be left.  


I agree, NREL gives us a method to estimate the extra input
from reflective sources.

However, I believe they are assuming ideal collectors.

If I'm wrong please show me where.

NREL has no way of knowing what type, configuration, or
quality collector one is using.
 

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Posted by nicksanspam on February 9, 2004, 12:39 am
 

I got these numbers from their Blue vs Red book on the same site.


Yes. They list the amount of sun that falls on the surface.


In a sense. I didn't explicitly calculate collector efficiency
in the rest of my posting, but I did take account of the ambient
temp, the R1 glazing with 90% solar transmission, and so on, in
calculating the indoor air temp. Care to go back and calculate
the collector efficiency, ie the net daily heat delivered to
the barn divided by the daily solar energy that falls on the outside
of the collector?

Nick


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