What are solar collectors?
Collector peak efficiency is only achieved when ambient temperature and water temperatures are the same. During normal use, this is only likely to happen for a short period of time each day, and usually only when ambient temperatures are high (summer). Therefore during normal use, the solar collector can not always perform at such a high level of efficiency. This is true for all evacuated tube and flat plate collectors, not only solar collectors.
- When making comparisons with other products please take the above point into consideration. Do not simply use the peak efficiency values for energy output, as this will provide inflated figures. CLL values also play an important role in determining total energy output from a solar collector. CLL values and formulae's are described below:
- Monthly and annual values are calculated using 28 days and 336 days respectively to account for days of very low solar radiation.
- Energy output values are approximations. Actual energy output and overall system efficiency will depend upon installation location, climate, insulation, system configuration and many other factors. On rainy or heavily overcast days energy output will be greatly reduced.
- Energy is produced in the form of heat. In transporting and converting this energy, such as for air conditioning or central heating, some energy (heat) will be lost, as no system or insulation is 100% efficient.
CLL Formula:
The three performance variables for the solar collector as provided by the SPF testing laboratory in Switzerland (SPF report C632LPEN) are as follows (for metric calculations, based on absorber area):
Conversion Factor: h0 = 0.717
Loss Coefficient: a1 = 1.52 W/(m2K)
Loss Coefficient: a2 = 0.0085 W/(m2K2)
As well as the three performance variables shown above, isolation level (G) in Watts/m2, ambient temperatures (Ta) and average manifold temperature (Tm) must be know. These values give the value x, also sometimes presented as T*m, used in the formula below.
(Other slightly different forms of this formula are used, but provide the same result)
How to use the formula?
Based on the ambient temperature, average manifold temperature and isolation level; firstly
calculate the value for X.
At 2:40pm; ambient temperature of 25oC (77oF); average water temp [(T inlet +T exit)/2] of 50oC (122oF); isolation level of 800Watts/m2 (252Btu/ft2).
x = (50-25)/800 = 0.03125
Now enter all the values into the formula:
h(x) = 0.717 - (1.52*0.03125) - (0.0085*800*0.031252)
h(x) = 0.717 - 0.0475 - 0.0066 = 0.663
The solar conversion efficiency for that specific point in time and set of environmental conditions is 66.3%. That is: 66.3% of the energy provided by the sun is actually used to heat the water.
Based on the assumption that those three environmental factors (G, Tm and Ta) are stable for a period of one hour, then 800 x 0.663 = 530.4 Watts of energy per m2 of absorber area will be used to heat the water (168Btu/ft2). 530.4 Watts is equivalent to 456kcal, which could heat 100L of water by 4.56C (20 Gallons by 10.9F)