Previously I described the different practical alternate power sources that I could use to power my Deep Deep Dark Green computer system but I have not touched upon some items that need to be discussed. One item is how to get the maximum efficiency from the alternate power source. Just taking a solar array, slapping it on top of a house, connecting the array to a voltage regulator then connecting the voltage regulator to the battery seems like the thing to do but there are some issues that need to be taken into account to maximize the efficiency of a power delivery system. Since we are dealing with limited power resources as we only have sunlight for some part of the 24-hour "daily" cycle, and the total amount of sunlight varies from Summer to Winter we need to optimize our power system as much as practical.

Since I am using solar cell technology for the prime power source for the deep deep dark green computer system I need to determine approximately how much electrical power I will potentially have available for the project. In doing some research I ran across the chart you see to the right. I would suggest you right-click on the image and open either a new tab or window to view the image as it is rather large in size - about 40-km per pixel displayed but gives you a pretty good idea of how much energy is potentially available in your area of the United States. My locale is on the east coast of Florida - yes, I can watch space shuttle launches from the front yard (grin). My area receives about 4.0 - 4.5 KWh per day per meter of surface area. Not too bad - of course this is for a clear day so I expect it to be less. My hope is the solar array I have will be large enough to supply the power I need for the computer system to be used for about two to three hours a day. The total per day use of the system may be higher or lower but that is about the average per week I use the computer - and of course it is possible, with the system I am using to have the computer go into a power-saving mode when not actually being used.

Given all of the power derived from sunlight is supplied by the solar array we need to make sure we are using the most optimum methods to collect the sunlight for the solar array. We don't want to be wasteful here as the solar array is the power "input" for the rest of the system. We need to insure we derive as much usable power as we can since everything from the solar array to the power storage system induces some loss of the collected power. Just the nature of the real world. I figure the solar array I have is around 10% efficient - in other words it will convert around 10% of the solar energy to electrical power as long as it is receiving direct sunlight.

The "total" square meter area of the solar array I have is around 0.28 sq. meters in size ( 1-foot X 3-foot). Given the potential solar energy in my area of the world is around 4-kWh per day per square meter of surface area then the best possible conversion I can expect is around (0.28 X 4000-watts) 1120-watts per day. Sounds wonderful - but - there is a problem using such a simplistic calculation. The calculation assumes you would collect ALL the solar energy falling on the 0.28-sq. meter area on a given day AND you had perfectly clear days all the time. Not going to happen in my case or anyone else for that matter!

There is also the problem of incidence angle - the angle of the solar array surface relative to the Sun's position in the sky directly affects the total energy conversion process. To extract the maximum amount of energy from sunlight you want the sunlight to shine directly on the solar array. Any deviation from having the sunlight strike the solar array directly reduces the collected power output from the solar array. An example would be the solar array directly faces the Sun at Noon but it is 9-AM in the morning. The Sun's apparent angle in the sky at Noon is directly overhead (for this example) but is approximately only 45-degrees above the horizon at 9-AM (assuming the horizon is 0-deg and Noon is 90-deg). If the solar array is oriented to receive the maximum amount of sunlight at Noon then the angle of incidence with respect to the Sun will be 45-Degrees at 9-AM. The power output from a solar array referenced to the angle of incidence for the sunlight striking it follows the Sin of the angle of incidence. What this translates to is a reduction of 15 - 25% of the total energy potentially collected by the solar array. Of course to capture the maximum amount of energy possible there needs to be a method to "track" the Sun across the sky which increases complexity and power requirements (something has to drive the solar array to track the Sun).

You can calculate the loss due to the angle of incidence difference between directly on the solar array verses a different angle by just looking at the angle value of the difference in the table on the right and multiplying the total power output value for the solar array by the Sin value for that angle. It is pretty close and accurate for most practical applications. At least my solar array follows it pretty closely!

Given the above information we can get closer to what is actually expected in terms of total power potentially available from the solar array. I am using a fixed position solar array that is mounted on my home's roof (which is south facing) and the pitch of the roof is such that I do not need to adjust for the latitude location (28.2-deg. North) as the pitch is pretty close already ( about 24-deg.). I am making the allowance of the angle difference on the conservative side and reducing the possible energy collected by 5% to compensate. Given the angle difference is only about 4-deg. and based on the Sin Chart I would expect a reduction of around 1% of the value I am going with a very conservative estimate here!

The following are the percentage values of loss for the different variables previously described:

Since I am using solar cell technology for the prime power source for the deep deep dark green computer system I need to determine approximately how much electrical power I will potentially have available for the project. In doing some research I ran across the chart you see to the right. I would suggest you right-click on the image and open either a new tab or window to view the image as it is rather large in size - about 40-km per pixel displayed but gives you a pretty good idea of how much energy is potentially available in your area of the United States. My locale is on the east coast of Florida - yes, I can watch space shuttle launches from the front yard (grin). My area receives about 4.0 - 4.5 KWh per day per meter of surface area. Not too bad - of course this is for a clear day so I expect it to be less. My hope is the solar array I have will be large enough to supply the power I need for the computer system to be used for about two to three hours a day. The total per day use of the system may be higher or lower but that is about the average per week I use the computer - and of course it is possible, with the system I am using to have the computer go into a power-saving mode when not actually being used.

Given all of the power derived from sunlight is supplied by the solar array we need to make sure we are using the most optimum methods to collect the sunlight for the solar array. We don't want to be wasteful here as the solar array is the power "input" for the rest of the system. We need to insure we derive as much usable power as we can since everything from the solar array to the power storage system induces some loss of the collected power. Just the nature of the real world. I figure the solar array I have is around 10% efficient - in other words it will convert around 10% of the solar energy to electrical power as long as it is receiving direct sunlight.

The "total" square meter area of the solar array I have is around 0.28 sq. meters in size ( 1-foot X 3-foot). Given the potential solar energy in my area of the world is around 4-kWh per day per square meter of surface area then the best possible conversion I can expect is around (0.28 X 4000-watts) 1120-watts per day. Sounds wonderful - but - there is a problem using such a simplistic calculation. The calculation assumes you would collect ALL the solar energy falling on the 0.28-sq. meter area on a given day AND you had perfectly clear days all the time. Not going to happen in my case or anyone else for that matter!

There is also the problem of incidence angle - the angle of the solar array surface relative to the Sun's position in the sky directly affects the total energy conversion process. To extract the maximum amount of energy from sunlight you want the sunlight to shine directly on the solar array. Any deviation from having the sunlight strike the solar array directly reduces the collected power output from the solar array. An example would be the solar array directly faces the Sun at Noon but it is 9-AM in the morning. The Sun's apparent angle in the sky at Noon is directly overhead (for this example) but is approximately only 45-degrees above the horizon at 9-AM (assuming the horizon is 0-deg and Noon is 90-deg). If the solar array is oriented to receive the maximum amount of sunlight at Noon then the angle of incidence with respect to the Sun will be 45-Degrees at 9-AM. The power output from a solar array referenced to the angle of incidence for the sunlight striking it follows the Sin of the angle of incidence. What this translates to is a reduction of 15 - 25% of the total energy potentially collected by the solar array. Of course to capture the maximum amount of energy possible there needs to be a method to "track" the Sun across the sky which increases complexity and power requirements (something has to drive the solar array to track the Sun).

You can calculate the loss due to the angle of incidence difference between directly on the solar array verses a different angle by just looking at the angle value of the difference in the table on the right and multiplying the total power output value for the solar array by the Sin value for that angle. It is pretty close and accurate for most practical applications. At least my solar array follows it pretty closely!

Given the above information we can get closer to what is actually expected in terms of total power potentially available from the solar array. I am using a fixed position solar array that is mounted on my home's roof (which is south facing) and the pitch of the roof is such that I do not need to adjust for the latitude location (28.2-deg. North) as the pitch is pretty close already ( about 24-deg.). I am making the allowance of the angle difference on the conservative side and reducing the possible energy collected by 5% to compensate. Given the angle difference is only about 4-deg. and based on the Sin Chart I would expect a reduction of around 1% of the value I am going with a very conservative estimate here!

The following are the percentage values of loss for the different variables previously described:

- 5% for difference in solar panel angle vs latitude location.
- 3% for the cover glass on the solar panel (remember the cover glass?).
- 25% for the fact it is a fixed position solar panel (conservative here again).

This gives me a net loss of around 33% so I am going to go more conservative with a 35% loss to work with. Now, the total potential energy that falls in a square meter in my area works out to be between 4 - 4.5 KWh total for a day. Given the solar array is approximately 0.28 square meters is size that will reduce the total potential energy available to 1.12 - 1.26 KWh total for a day. We can not get that much power as the solar array would have to be 100% efficient in converting all the sunlight striking it to electrical power so we need to multiply the total potential energy by 10% (0.1) which yields around 112 - 126 watt-hours total per day Now you see why there is a big move to increase the efficiency of solar cells! 85 - 90% of the total potential power is lost due to the conversion efficiency factor of the current solar cell technologies.

Now we know the approximate total available power our solar array is capable of if everything is optimal for collecting the available sunlight. We do not have an optimal configuration as the solar array is in a fixed position (does not track the Sun) and is not at the optimal angle for the latitude location (24 vs 28.4 deg). Given these two inadequacies we have to multiply our previous calculation results by 0.65 (35 % loss) to account for those incurred losses. The calculated value of the total potential available power per day works out to 73 - 82 watts-hour for the day. That is a far cry from the total potential power available over a 0.28 square meter but it still does not take into account everything! This sure adds a whole new meaning to "losing daylight"!

Solar array specifications can be a little misleading (say it isn't so!) where the manufacturer specifies the "total power" the panel is capable of collecting. They use the maximum voltage the panel is capable of producing then will increase the amount of current drawn from the panel to reduce the voltage produced to about 80% of the total "open circuit" voltage - the current times the voltage is the "power" rating for the solar panel. This is well and good if you have the power storage system operating at the solar array 80% open voltage point but most battery storage systems do not. Case in point - the solar array I am using is capable of generating an open-circuit voltage of 20-VDC is direct sunlight. This number is empirically derived (took the panel out in the sunlight at 1-PM on a perfectly clear day and measured the open-circuit voltage). While it is nice to get that much voltage I don't have any battery storage systems that operate at the 80% value for that voltage ( 16-VDC ) but instead use around 13.8-VDC for the charging voltage. Since the solar array I am using is rated at 20-Watts at 16-VDC I have to "de-rate" the solar array to take into account the lower voltage of operation - which works out to around 14 % less power available, or around 17.2-Watts for the solar panel. What all of this means is the solar panel will not supply it's rated 20-watts of power in direct sunlight due to the lower voltage of operation required for the battery charging operations so I lose about another 14 % of the total potential power available per day (this one always trips people up!). So now my 73 - 82 Watt/hours per day potential energy drops down to 14 % less or about 63 - 71 Watts/Hour per day. We are starting to get a little "slim" on power here (grin). As you can see there are a good number of variables and trade-offs that can affect the total efficiency of a solar system. In my case I will be running at the lower end of the efficiency spectrum due to simplification of the system. Given the power limitations of the computer system to less than 20-watts total power draw the above calculations indicate I should be able to run the system for about three hours total on a really good day. Of course - the lower the power drain by the computer system the longer the operational time. I may be adding an additional solar panel to my system in order to get longer operational time. At $90 for a panel it is not too expensive to do this and may still be within budget.

Bottom line....

Some may think this is just not worth the effort that is required to build and install a solar array power system. And, for the most part, they would be correct if you did not take into account the possibility of no power being available! Case in point - I live in an area that is susceptible to power outages due to such things as Lightning induced or Hurricane induced power outages. During the 2004 Hurricane season I saw five different storms affect my area with a resultant power outage of three days for each storm. I personally know people who were without power for weeks on end due to the storms - downed power lines that were not restored for weeks are not uncommon from Hurricanes. My biggest issue with no power was the lack of communications during those power out days - even the phone lines and cell phones were not working for a couple of days after the storms. The phone lines were down for various reasons - no power at the telephone sub-station due to problems with generators or batteries being discharged. The cell towers had the same issues. Hopefully they have taken steps to correct those deficiencies but being telephone company owned I do have my doubts - does not add to their bottom line and they would just claim those are rare events anyway. Having a self-powered system would have been an advantage as I have a satellite Internet system so I would have had communications if I had a self-powered system. I would not be at the mercy of the local telephone companies for communications...

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