Energy and Human Ambitions on a Finite Planet, 2021a

13 Solar <strong>Energy</strong> 214
relevant period <strong>and</strong> the average solar input at that location for the period
of interest.
75: . . . makes sense for a population of 330
The first can be surmised from electricity bills, usually giving a monthly
total usage in kWh. We can get an approximate average scale from
Fig. 7.2 (p. 105), which indicates that 42% of residential energy (11.9 qBtu
in the U.S., 75 average household electricity consumption is 1,285 W.
per year) is from electricity. That’s 5 qBtu, or 5.3 × 10 18 J in one year
(3.156 × 10 7 s), or 167 GW. Distributed among 130 million households
million: translates to 2.5 people per household,
on average
Applied over 24 hours, this makes for just over 30 kilowatt-hour (kWh)
per day for an average household. 76
76: This is another case where students
might suggest replacing this whole paragraph
with the result. The point is to build
connections, context, <strong>and</strong> tools to apply previous
knowledge.
The next piece is solar potential at the location of interest. We’ll use the
excerpted data from [88] for St. Louis, Missouri found in Table 13.2. [88]: National Renewable <strong>Energy</strong> Lab (1994),
Example 13.6.1 Let’s design a grid-tied PV system for an average
U.S. household in an average 77 U.S. city (St. Louis). We’ll orient the
panels facing south <strong>and</strong> tilted to the site latitude (39 ◦ ) <strong>and</strong> purchase
PV panels at 18% efficiency (pretty typical).
Table 13.2 indicates that for this configuration we can expect an annual
average of 4.8 kWh/m 2 /day of input. If we’re shooting for 30 kWh per
day, we would need 6.25 m 2 of panel operating at 100% efficiency. 78
Solar Radiation Data Manual for Flat-Plate <strong>and</strong>
Concentrating Collectors
77: . . . solar-wise
78: . . . Divide 30 kWh/day by
4.8 kWh/m 2 /day
But 18% panels will require about 35 m 2 of panel, 79 which would be a 79: . . . Divide 6.25 m 2 by 0.18
square array about 6 meters on a side (about 20 feet) or a rectangle 5
by 7 meters, etc. The total area (400 square feet) is much smaller than
a typical house footprint, so that’s good news.
But panels are not marketed by the square meter. They are sold in terms
of peak Watts: what the panel would deliver in 1,000 W/m 2 sunlight
(see Example 13.4.1). How do we convert? Two ways are instructive.
Example 13.6.2 In one method, we multiply the 35 square-meter area
from Example 13.6.1 by 1,000 W/m 2 <strong>and</strong> then by the PV efficiency
(18% in this example) to get how much would be delivered: 6.3 kW.
Alternatively, we could adopt the interpretation of 4.8 kWh/m 2 /day
as the equivalent full-sun hours operating at peak output (Box 13.2).
To get our target 30 kWh in 4.8 hours of full-sun-equivalent, we would
need to produce 6.25 kW for those 4.8 hours. 80
We get the same answer either way, which is a good check. 81
80: 6.25 kW times 4.8 hours is 30 kWh.
81: The math is actually just the same, but
we rearranged the order <strong>and</strong> interpretation.
We should assume that the panels will not achieve their rated potential
due to the facts that:
◮ The 25 ◦ C specification is almost never realized for a PV panel in
the sun: PV panels in the sun get hot, <strong>and</strong> less efficient as a result;
◮ The panels will get a little dirty;
© 2021 T. W. Murphy, Jr.; Creative Commons Attribution-NonCommercial 4.0 International Lic.;
Freely available at: https://escholarship.org/uc/energy_ambitions.