Morris Dovey wrote:

*> I'd like to quantify the output for a passive solar heating panel I've *

*> built; but am not sure how to proceed.*

*> *

*> I can determine the exact volume of air in the collector (v, in cubic *

*> inches), the input temperature (i, in degrees Fahrenheit), the output *

*> temperature (o, in degrees Fahrenheit), and how long it takes for a unit *

*> volume of air (v) to pass through the collector (t, in seconds).*

*> *

*> Am I correct in assuming that if I introduce a puff of smoke at the *

*> input and measure the amount of time required for the smoke to appear at *

*> the output - that the measured interval is at least a good approximation *

*> to t?*

*> *

*> What formula(e) can I use to determine energy delivery in btu and watt *

*> units?*

*> *

*> [ For anyone interested, the numbers are close to:*

*> *

*> v = 4758.75 cu in*

*> i = 85.3ºF*

*> o = 181.0ºF*

*> t = 16 sec*

*> *

*> Is this result good, bad, or indifferent? ]*

*> *

*> Help would be much appreciated.*

*> *

If I am getting your numbers right, 4759 in^3 of air go through the

collector in 16 secs -- This gives a flow rate of:

Flow Rate = (Volume of air per unit time)

(4759 in^3)(1 ft^3/1728 in^3) /((16 sec)(1 min/60 sec))

= 10.32 ft^3/min (cfm)

Heat Output

= (Flow Rate)(air density)(Temperature rise)(specific heat of air)

= (10.32ft^3/min)(0.076 lb/ft^3)(181.0F - 85.3F)(0.24 BTU/lb-F)

= 18 BTU/min or 1080 BTU/hr or 317 watts

To get a rough idea how efficient the panel is, we would have to know:

1) what is the collection area (glazed area) of the panel

2) how the panel is oriented to the sun (e.g. pointed due south and

titled 90 degs).

2) How much sun is shinning on the panel -- if you don't have a way to

measure the sun intensity, then you could do the test on a clear day,

and we could look up the sun intensity knowing your location, and the

date/time. Doing the test around solar noon (when the sun is due

South), and with the panel pointed due south would be a good setup.

The efficiency is:

efic = (panels heat output)/(solar input)

where heat output is the 1080 BTU/hr from above.

The outlet temperature of 181F seems pretty high to me -- I think you

might get more efficiency (more heat out of the panel) if you figured

out a way to increase the air velocity. Perhaps larger vents, or less

restriction in the panel. The effect of this would be to lower your

output temperature, but increase the flow rate -- you should come out

ahead because your losses out the glazed side of the panel will be

less. The losses depend on the difference between your collectors

internal temperature, and the outside air temperature.

Another way to get the flow rate through the panel is to measure the

flow velocity at the exit vent. The little hand held devices Kestrel

(and others) sell to measure wind velocity work pretty well as long as

the velocity out of the vent is 90 ft/min or more. The one I have

cost about $0. But, timing the smoke puff through the panel might

actually be more accurate?

What are you using to generate smoke? I've been looking for a good

way to do this.

Gary

Gary wrote:

*> Another way to get the flow rate through the panel is to measure the *

*> flow velocity at the exit vent. The little hand held devices Kestrel *

*> (and others) sell to measure wind velocity work pretty well as long as *

*> the velocity out of the vent is 90 ft/min or more. The one I have cost *

*> about $0. But, timing the smoke puff through the panel might actually *

*> be more accurate?*

*> *

*> What are you using to generate smoke? I've been looking for a good way *

*> to do this.*

Sorry. My way isn't a healthy one and I certainly won't recommend

it to someone who's just helped me...

I blow a tiny puff of cigarette smoke through an 18" length of

1/4" ID tubing.

I would guess that the airflow though my collectors is uneven

(faster through some paths than others) and that the smoke method

may be accurate only for the shortest/fastest paths.

--

Morris Dovey

DeSoto, Iowa USA

Morris Dovey wrote:

*> I'd like to quantify the output for a passive solar heating*

*> panel I've built; but am not sure how to proceed.*

<snip>

Nick, Ecnerwal, and Gary...

Wow! A usenet first for me. I asked a newby question and got not

one but three informative and totally useful responses!

My thanks to all!

--

Morris Dovey

DeSoto, Iowa USA

> I'd like to quantify the output for a passive solar heating panel I've> built; but am not sure how to proceed.>> I can determine the exact volume of air in the collector (v, in cubic> inches), the input temperature (i, in degrees Fahrenheit), the output> temperature (o, in degrees Fahrenheit), and how long it takes for a unit> volume of air (v) to pass through the collector (t, in seconds).>> Am I correct in assuming that if I introduce a puff of smoke at the> input and measure the amount of time required for the smoke to appear at> the output - that the measured interval is at least a good approximation> to t?>> What formula(e) can I use to determine energy delivery in btu and watt> units?>> [ For anyone interested, the numbers are close to:>> v = 4758.75 cu in> i = 85.3ºF> o = 181.0ºF> t = 16 sec>> Is this result good, bad, or indifferent? ]>> Help would be much appreciated.>