## Solar Savings

### What are the potential savings with SunScan solar geysers?

In a 12 month period you should save between 35 and 55% of your water heating bill.

Payback Period – Approximately 3 years

### Solar Efficiency

Example: 200L Retro Fit System

It achieved the following results on the SABS test:

Q-factor 27.9 MJ or 7.745 Kwh

Using NASA’s 22 year radiation tables, the average annual radiation for Cape Town (taking cloudy rainy days into account) is equal to 6.42 KWhr/sqm/day.

Our SunScan system will deliver 11.15 Kwhr of energy on the average day in the Western Cape.

11.15 KWhr per day equates to 4070 KWhr per annum

Present electricity prices including Vat for high-use Eskom Clients is R2 per KWhr.

4070 KWhr at R 2 per KWhr equates to a saving of R8 140 per year.

20 year cumulative savings = R505,232.00 ### Technical Analysis of Solar Water Heating SABS tests each system by “plugging” it into a very sophisticated performance monitoring station.

Each system is tested with varying amounts of radiation (measured in MJ/sqm) and differing input temperatures of water. The water is then pumped from the geyser via the monitoring station which measures the temperature as well as the volume of water. From this, a measure of the amount of energy that has been stored in the geyser can be obtained. This energy value is given in MJ and must not be confused with the amount of sunlight that the system was exposed to which was measured in MJ/sqm.

Each system is exposed to varying amounts of sunlight (irradiation) and different days (e.g. 10, 16, 20, 25 MJ/sqm) and then the results are plotted. The cold water supply temperature (Tc) to the geyser is also varied and graphs are plotted for the various irradiation amounts as well as different cold water supply temperatures and air temperatures(Ta). A line is plotted for each (Ta-Tc). For instance the red line above shows performance of a system for a temperature diff of (Ta-Tc = 10 deg) between the air temperature and the cold water supply temperature.

### Q-Factor

From the above graph, a specific set of conditions can be chosen to be able to compare different supplier’s efficiencies.

The values that ESKOM decided on are 16MJ/sqm irradiation and (Ta-Tc=10 deg), represented by the red line on the above graph. This provides us with the Q-factor for this system and is a measure of the energy delivered by a solar system on an average day of the year.

From the above graph the Q-factor for this system is measured at 27.9 MJ.

Converting MJ to KWh

SABS uses MJ to measure energy. We pay for our electricity in KWh. To convert from MJ to KWh proceed as follows:

1 KWh = 3.6 MJ Therefore a Q-factor of 27.9 MJ equates to (27.9 MJ / 3.6) = 7.745 KWh

### Annual energy production

NASA provides us with 22 year average Solar radiation values. This takes into account cloudy and rainy weather. So it is an actual value of the energy that a system is exposed to. The table below is for Cape Town. Cape Town receives and annual average of 6.42 KWh/sqm/day. This is equal to 6.42 X 3.6 = 23.1 MJ which is higher than SABS standard day of 16 MJ.

To calculate how much energy this system will accumulate, scale up the SABS value from 16 MJ to 23.1 MJ

23.1/16 = 1.44

Multiply the Q-factor by 1.44:

27.8 MJ X 1.44 = 40.13 MJ

40.13 MJ / 3.6 = 11.15 KWh

So using NASA’s 22 year average daily radiation values it can be seen that on an average day, this system will deliver 11.15 KWh of energy.

### Cost savings per year

11.15 KWh X 365 = 4070 KWh per year

4070 Kwh per year X R 2 KWh = R8 140 per year