### Definition

We
define the solar collector efficiency as the ratio of solar energy collected
and divide by the solar energy available.

### Insulation’s

Is a
term expressing the solar energy available?

### PV System

It
also stated that this energy in terms of 9 kw-hr/m²/day).

### Collector efficiency

If
we need to calculate the solar energy collector efficiency, so its output will
be BTU/ft²/day. The insulator is varying according to the geographic location,
time of the day/year, land-forms, weather condition and orientation, direction.
If the sunlight is averagely available on the earth it means no cloud scattered
then the output a result that we obtain is 1m²/hrs is 1kw.

It
is also better in the calculation of PV efficiency. If it is required to get
the average solar heat then our assumption and expectation must be based on
300-BTU/ft²/hrs are provided for this experiment. With the help of this
assumption if we calculate the solar energy available to collector with a
surface area of 1000 ft² perpendicular to the sunray throughout 1 hr. If we put
this process of calculation into practice, can we get the required figure?

If
your results or output is 300,000 BTU of heat energy provided for a surface
1000 ft² then it is clear that you can understand the concept. If on the
If our output is 100% then it is quite clear that we are obtaining 300,000 BTU
of heat/hr. but it is impossible to get a 100% accurate result Sunlight heat
intensity is necessary to get the 100% output but it is not the only factor of
energy development.

It
is also an essential part. You will need a higher measuring thermometer to
measure the high temperature. Besides, you will require taking an estimate of
the flow rate of the pump as well. To calculate the flow rate of the loop
system a person used as 5/gallon pale and stopwatch. However, for a closed-loop
system, you will have to certainly depend on the manufacturer’s specifications
and their useful instruction.

Once
you get the knowledge about that then obtain the flow rate changing in the
storage temperature over a period to get the output result you will need a
display monitor. Heat collected equally exceeding in both of the systems in the
storage temperature (T2-T1) one gallon of water weight is 8.3lbs.

### For the Solution of another Problem

For example,
if you have a collector its surface area is 100-ft² and a 200-gallon
accumulator tank and you require that you increase the temperature of this tank
from 80°F to 90°F in one hour. The question is arising in the mind that what is
the efficiency of your collector. (Suppose, 300 BTU/ft²/hrs are available).

### Solution

Heat
available per hour=300×000=30,000 BTU. Heat collected=200 gall×8.3×10=16,500
BTU. Efficiency=30,000/16,500=55%.

The
estimated efficiency of the collector maybe you used, but the efficiency will
be different in ambient temperature, differential temperature between collector
and storage, and its flow rare. Since our flow rate remains constant, our
principal concern should be the temperature of ambient and differential
(T2-T1). By this process, we get to know about the collector input/output
temperature orderly provided us T1 and T2.

For
the verification of sunlight intensity on the collector, you will need a device
Pyranometer. With this useful device, we are capable to calculate the sun’s
heat intensity according to the change of weather or climate situation. Cosign
correction is an essential part of the Pyranometer because the Insulation is
different from the angle of radiant energy. Regrettably to inform you about the
professional Pyranometer is costly but no problem.

I
have developed such a device like that “Sky-Eye” its cosign correction work
such as Pyranometer does, but Sky-Eye is costly like Pyranometer it is cheap
and you can afford to buy and it’s also easy to place it correctly on the
collector.