We define the solar collector efficiency as the ratio of solar energy collected and divide by the solar energy available.
Is a term expressing the solar energy available.
It also stated that this energy in terms of 9kw-hr/m²/day).
If we need to calculate the solar energy collector efficiency so its output will be BTU/ft²/day. Insulator is varying according to the geographic location, time of the day/year, landforms, weather condition and orientation, direction. If the sunlight is clearly average available on the earth it mean no cloud scattered then the output a result that we obtain is 1m²/hr is 1kw. It is also better in the calculation of PV efficiency. If it is required to get the average solar heat then it is necessary that our assumption and expectation will be based on 300-BTU/ft²/hr are provided for this experiment. With the help of this assumption if we calculate he solar energy available to collector with a surface area of 1000 ft² perpendicular to the sun ray over period of 1 hr. If we put this process of calculation into practical, 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 contrary your answer is negative then you should read the information again. 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 the 100% accurate result Sunlight heat intensity is necessary to get the 100% output but it is not the only factor of energy development..
Is also an essential part. You will need a higher measuring thermometer to measure the high temperature. And you will also require taking estimate of the flow rate of the pump. To calculate the flow rate of the loop system a person used as 5/gallon pale and stopwatch. But for a closed loop system you will have to certainly depend on the manufacturer’s specification and their useful instruction. Once you get the knowledge about that then obtain the flow rate changing in the storage temperature over a period of time to get the output result you will need a display monitor. Heat collected equally exceeding in both of the system in the storage temperature (T2-T1) one gallon of water weigh 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/hr are available).
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 collector may be 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 heat intensity according to the change of weather or climate situation.
Cosign correction is essential part of the Pyranometer, because the Insulation is different as the angle of radiant energy. Regrettably to inform you about the professional Pyranometer is costly but no problem I have a 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.