The state of California is trying to increase the percentage of their electrical

Question

The state of California is trying to increase the percentage of their electrical power supplied by renewable technology. Noting that the Mojave Desert is very sunny, they consider placing a vast array of solar panels at this remote location. For structural simplicity, the panels will be oriented horizontally on the ground. Based on prior history, the expected peak electricity load for the state is approximately 48 GW. a. On the summer solstice, what is the peak insolation (in W/m^2) on the panels? Note that the center of the desert is at a latitude of 35 degree N. b. On the same date, what is the average insolation (in W/m^2) for the entire day? c. Assuming the panels have a 14% efficiency, how much surface area is required to power California at its maximum electricity load? Use the average insolation found in part (b), assuming that this input can be stored over time using batteries.

Explanation / Answer

Using Location Latitude and Longitude we can find solar insolation data using "NASA Surface meteorology and Solar Energy "

a)

Using this we can find Peak insolation is 5.7342 kWh/m2/day. on the summer solstice.

Peak radiation is Gp = 5.7342*1000/14.4 = 398 W/2

b)

Average insolation on the same day is 5.03 kWh/m2/day on the summer solstice.

Average Radiation data is Sunshine hours is 14.4 hours GA = 5.03*1000/14.4 = 349 W/2

c)

Given efficiency of solar panels is 14%

Area required to meet Peak demand P =48* 109 W

Energy required for one day E = P*24 = 48*109  * 24 = 1152 *109 Wh

E = Area of Panels * 0.14 * GA *14.4

Area of Panels = E/(0.14*GA *14.4)

= 1152 * 109 /(0.14 * 349 * 14.4 ) = 1,637,331,150 m2

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