Solar Collector Efficiency

What is wrong with this animation?

**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.

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 ft^{2} 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%

** 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.