For a photovoltaic efficiency of 20% and an average peak insolation of 350
W/m2, calculate the additional number of km² of solar collectors that would need to be installed in the United States to equal the per-capita photovoltaic capacity of Germany.

Question

Quizplus · Accepted Answer

The Answer is From the table the installed solar photovoltaic capacity in the Unitted States is

2.56 imes 10 ^ { 10 } \mathrm {~W}

. The present population of the U.S. is

3.22 imes 10 ^ { 8 }

so the per capita installed capacity is

\left( 2.56 imes 10 ^ { 10 } \mathrm {~W} ight) / \left( 3.22 imes 10 ^ { 8 } ight) = 79.6 ext { W per capita }

From the table the installed capacity in Germany is

3.97 imes 10 ^ { 10 } \mathrm {~W}

and the current population is

8.07 imes 10 ^ { 7 }

. So the per capita installed capacity in Germany is

\left( 3.97 imes 10 ^ { 10 } \mathrm {~W} ight) / \left( 8.07 imes 10 ^ { 7 } ight) = 491.9 \mathrm {~W} ext { per capita }

So to have the same per capita capacity as Germany, the U.S. needs an additional

491.9 \mathrm {~W}

- 79.6 W

= 412.3 \mathrm {~W}

per capita. This is a total additional capacity of

( 412.3 \mathrm {~W} ) imes \left( 3.22 imes 10 ^ { 8 } ight) = 1.33 imes 10 ^ { 11 } \mathrm {~W}

At

20 \%

efficiency and an average insolation of

350 \mathrm {~W} / \mathrm { m } ^ { 2 }

, the additional required collector area is

\left( 1.33 imes 10 ^ { 11 } \mathrm {~W} ight) / \left[ ( 0.2 ) imes \left( 350 \mathrm {~W} / \mathrm { m } ^ { 2 } ight) ight] = 1.895 imes 10 ^ { 9 } \mathrm {~m} ^ { 2 }

For a Photovoltaic Efficiency of 20% and an Average Peak