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Unformatted text preview: According to ElWakil (p. 541), the terrestrial radiation received by the continental U.S. is about
2.2×1016 kW·hr/yr. If a typical solar cell has a conversion efficiency of 12%, determine how long
it would be before the entire surface of the U.S. is covered with solar panels if all of the electricity
was produced this way. Utilize the electricity use datum from 2006 (see Homework #1) as the
baseline, and assume that the electrical consumption rate grows at a constant rate of 2.5% per year.
Solution
From Homework #1, the total electricity use in the U.S. in 2006 was
4,052,967,846 thousand kWe·hr = 4.053×1012 kWe·hr
First, calculate the solar electricity potential based on the incident energy and the solar cell
efficiency (ε)
E possible = ε Eincident = (0.12)(2.2 × 1016 kW·hr/yr)
= 2.64 × 1015 kW·hr/yr
We can use the financial type equations for the future value due to the growth of an initial
principal amount of money (in this case energy) in order to find the number of years until the
electric energy use equals the possible production (at today’s solar cell efficiency)
⎛
F = P ⎜1 +
⎝
15 2.64 × 10 i⎞
⎟
n⎠ nT 12 kW·hr/yr = (4.053 × 10 ⎛
0.025/yr ⎞
kW·hr/yr) ⎜1 +
⎜ 1 period/yr ⎟
⎟
⎝
⎠ 651.4 = (1.025) T
Now take the log10 of both sides
log10 (651.4) = log10 (1.025) T [ 2.81383 = T log10 (1.025)
T= 2.81383
= 262.4 years
log10 (1.025) (1 / yr ) T A solar thermal plant with energy storage capability is to be constructed at about a 40˚ latitude to
provide the electric energy to a system whose 24hr average load is 100 MWe during the winter.
Assume that the solar thermal power plant will have a 25% energy conversion efficiency; that the
nighttime load is 40% of the total load; and that 10% of the energy that must be stored is lost. (a)
Determine the total land area (in km2) required for the reflectors. (b) For comparison also
determine the area (in km2) required for a fixed horizontal system. In both cases, it is believed that
a parabolic reflector system can be set up so that the reflectors cover 45% of the land area on
which they are placed. The reflected energy collected at 40˚ in winter is 6 kW·hr/m²·day for a
tracking system, and only 1.85 kW·hr/m²·day for a horizontal fixed system.
Solution
Daytime load = (100 MWe) (0.6) (24 hrs/day) = 1440 MW·hr/day
Nighttime load = (100 MWe) (0.40) (24 hrs/day) = 960 MW·hr/day
Total energy needed to fulfill the nighttime load, including stored energy loss
= (960 MW·hr/day) / (0.9) = 1067 MW·hr/day
Total daily electric energy needs = Daytime load + Nighttime load
= 1440 + 1067 = 2507 MW·hr/day
Required thermal load = (2507 MW·hr/day) / (0.25) = 10,028 MW·hr/day
(a) For a tracking system using reflectors:
Energy collected = 6 kW·hr/m²·day
10,028 × 10 3 kW·hr/day
Area =
= (3.7 × 10 6 m 2 )(1 km / 1000 m) 2
(6 kW·hr/day)(0.45)
= 3.7 km 2 (b) For a fixed horizontal system:
Energy collected = 1.85 kW·hr/m²·day
Area = 10,028 × 10 3 kW·hr/day
= (12 × 10 6 m 2 )(1 km / 1000 m) 2
(1.85 kW·hr/day)(0.45)
= 12 km 2 ...
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This homework help was uploaded on 02/11/2012 for the course EEE 463 taught by Professor Holbert during the Spring '08 term at ASU.
 Spring '08
 HOLBERT

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