= 1050°F = 565°C = 839 K
T=554°F = 295°C = 563 K
This gives an ideal Carnot efficiency of 100 ×(1-563K/839K) = 33 %
Thus the total solar insolation must be 1 GW/0.33 = 3 GW. The installation in Figure 9.4 is in California and from
Figure 8.2 the average insolation can be estimated to be 225 W/m. The area needed to produce the necessary power
is 1.3 x 10 m. For a circular array this corresponds to an area about 4 km in diameter. This is too large of an area.
Why?