Solar Collector Efficiency

 

Collector efficiency is expressed as the ratio of solar energy collected divided by the solar energy available.  

Insolation is a term used to express the solar energy available. 
Photovoltaic systems express this energy in terms of (kW hr/(m²·day). 
Solar collector efficiency is calculated in terms of BTU/ft²·day. Insolation varies according to geographic location, time of day, time of year, landscape, weather and orientation. On a clear sunny day in an average area on our planet the amount of direct solar energy available to 1 m² per hour is 1KW. This is a good thing to know if your calculating PV efficiency but we’ll be calculating collector efficiency so well be more interested in the amount of heat available. For solar heat energy available our rule of thumb we’ll be based on the assumption that 300 BTU/ft²/hr are available on a sunny day..

300 BTU/ft²/hr is a bit high for most seacoast areas, but if you live on the south side of a mountain in Montana 300 BTU/ft²/hr is believable. 

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Under this assumption calculate the the solar energy available to collectors with a surface area of 1000 ft² perpendicular to the sun's rays over a period of 1 hr. ???

ANS:  If your answer to this question is 300,000 BTU's of heat energy available to a surface 1000 ft²per hour you understand the concept.    
      

So if our 1000 ft² array of collectors has an efficiency of 100% we would be able to collect 300,000 BTU's of heat per hour BUT of course a 100% efficiency rating is impossible. The intensity of sunlight striking collectors is often the most important factor involved in heat gain but it is not the only factor. Collector efficiency is also important. To measure collector efficiency you'll need at least one thermometer capable of handling high temperatures. You'll also need to estimate the flow rate of your pump.
I use a 5 gallon pale and a stop watch to calculate flow rate in an open loop system. For a closed loop system you’ll have to rely on the manufactures specifications. Once you know the flow rate all you need to do is monitor the change in storage temperature over a period of time. Heat collected equals the rise in storage temp (T2-T1) (weight of the water). One gallon of water weighs 8.3 lbs.

OK!  Ready for another problem"

Let’s say you have a collector surface area of 100 ft2 and a 200 gallon storage tank and you’re able to raise the temperature of this tank from 80 F to 90 F in one hour. What is the efficiency of your collector?  (Assume 300 BTUs/ft²/hr are available)

 ANS: Heat available per hour =  300 x 100  =  30,000 BTU
        Heat collected =  200 gal x 8.3 x   10  =  16,500 BTU
        Efficiency        =  30,000/16,500             =   55% 

This rough estimate of collector efficiency may be all you need BUT efficiency will vary with ambient temperature, differential temperature between collector and storage and flow rate. Since our flow rate remains constant our main concern should be the ambient temperature and the differential temperature (T1-T2). Just knowing the collector input temperature and the collector output temperature will give us T1 and T2. Now all we'll need is a pyranometer to verify sunlight intensity on the collector. With this rough calculation all we could do is guess at the solar flux intensity but with the aid of a pyranometer we would know the sunlight intensity and be free to measure collector efficiency under a variety of weather conditions.