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What is the theoretical potential for solar power generation?
If in theoretical, then multiply the land area by the fraction which is suitable for solar installations, multiply by the efficiency of the installation, and the peak solar insolence, to get the peak performance in W. Multiply by the number of daylight hours in a year to get the yearly power in Whr.
Example: Negev desert in Israel. Land area 13,000 sq Km. Perhaps 3050% is suitable for solar installations. Solar insolence is around 900 W/sq m. Let's use an optimistic solar installation which will convert 15% of the collected energy into useful power. One gets: (13x10^9 sq m) x (0.30 useful area) x (0.15 eff) x (900 W/sq m) = 527 GW. The Negev is about 55% of Israel's area. Israel accounts for about 0.3% of the land in the ME (small, isn't it  one wonders why the radicals want such a small place), so if you want to multiply the 527 GW by about 300, you get a rough estimate of the solar potential for the entire ME.
Comments:
1. No one will realistically cover the entire available ME land area with solar installations. So a better figure to use is the amount of energy available per some unit area and then figure how many of these areas are available for an installation. So let's calculate the amount per sq Km. Thus: (1 sq Km) x (10^6 sq m/sq Km) x (900 W/sq m) x (0.15 eff) = 135 MW/sq Km.
2. This figure is also not realistic since no one will cover a full sq Km with the collector panels. Spacing is needed for access such as cleaning (extremely important but little discussed), sun following, and/or maintenance. So let's say half: 70 Mw/sq Km. Similarly, this is peak with all parts of the installation working properly with clean panels. This never happens so let's say that a realistic figure is more like 50 MW/sq Km (an easier number to work with anyway). If year average sunlight is effectively 10 hrs per day (we will ignore the first and last hours of the day for all sorts of reasons such as lower insolence, interference from mountains, the few days it rains, etc), then the power produced is (50 MW/sq Km) x (10 hrs/day) x (365 days/yr) = 182.5 GWhrs/sq Kmyr.
Now multiply this number times the number of square Km you think solar installations will occupy to get the estimated power for any area.
3. I used the figure 900 W/sq m for the solar insolence because that is what we get in Israel (at least in the Tel Aviv area). If you know what the number is for the area in which you are interested, you should use that instead. Similarly, if you know the effective efficiency of the solar installation, you should use that.
Example: Negev desert in Israel. Land area 13,000 sq Km. Perhaps 3050% is suitable for solar installations. Solar insolence is around 900 W/sq m. Let's use an optimistic solar installation which will convert 15% of the collected energy into useful power. One gets: (13x10^9 sq m) x (0.30 useful area) x (0.15 eff) x (900 W/sq m) = 527 GW. The Negev is about 55% of Israel's area. Israel accounts for about 0.3% of the land in the ME (small, isn't it  one wonders why the radicals want such a small place), so if you want to multiply the 527 GW by about 300, you get a rough estimate of the solar potential for the entire ME.
Comments:
1. No one will realistically cover the entire available ME land area with solar installations. So a better figure to use is the amount of energy available per some unit area and then figure how many of these areas are available for an installation. So let's calculate the amount per sq Km. Thus: (1 sq Km) x (10^6 sq m/sq Km) x (900 W/sq m) x (0.15 eff) = 135 MW/sq Km.
2. This figure is also not realistic since no one will cover a full sq Km with the collector panels. Spacing is needed for access such as cleaning (extremely important but little discussed), sun following, and/or maintenance. So let's say half: 70 Mw/sq Km. Similarly, this is peak with all parts of the installation working properly with clean panels. This never happens so let's say that a realistic figure is more like 50 MW/sq Km (an easier number to work with anyway). If year average sunlight is effectively 10 hrs per day (we will ignore the first and last hours of the day for all sorts of reasons such as lower insolence, interference from mountains, the few days it rains, etc), then the power produced is (50 MW/sq Km) x (10 hrs/day) x (365 days/yr) = 182.5 GWhrs/sq Kmyr.
Now multiply this number times the number of square Km you think solar installations will occupy to get the estimated power for any area.
3. I used the figure 900 W/sq m for the solar insolence because that is what we get in Israel (at least in the Tel Aviv area). If you know what the number is for the area in which you are interested, you should use that instead. Similarly, if you know the effective efficiency of the solar installation, you should use that.