Energy and Human Ambitions on a Finite Planet, 2021a

13 Solar <strong>Energy</strong> 200
1e8
1.0
0.9
0.8
0.7
6000 K
B¸ (W/m 2 /¹m)
0.6
0.5
0.4
0.3
0.2
0.1
UV visible IR
4500 K
3000 K
0.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0
wavelength, ¸ (¹m)
Figure 13.1: Planck spectra, orblackbody
spectra for three temperatures, indicating
where the ultraviolet, visible, <strong>and</strong> infrared
regions lie. The shapes of the three curves
(spectra) are described by Eq. 13.4, the star
locations are found by Eq. 13.5, <strong>and</strong> the total
power radiated, per square meter of surface
is the area under each curve, as captured in
Eq. 13.3. The dotted line relates to Example
13.2.2. Note the 1e8 factor on the vertical
axis, meaning that the axis goes from 0 to
1.0 × 10 8 W/m 2 /μm.
The shape of each spectrum is a plot of the function in Eq. 13.4 for three
different temperatures. If comparing output of Eq. 13.4 to Figure 13.1,
be aware that the units have been manipulated to favor microns over
meters. 12
Let’s come at this again with numbers to help us make sense of things.
Looking at the curve (spectrum) for 6,000 K, we will verify that each
equation makes some sense.
Example 13.2.1 First, Eq. 13.5 says that the wavelength where emission
peaks should be about 2.898×10 −3 /6000 ≈ 0.483×10 −6 m, or 0.483 μm.
12: Eq. 13.4 uses units of meters for λ, but
Figure 13.1 uses microns (μm, or10 −6 m)
for convenience. Also, Eq. 13.4 delivers an
answer in units of W/m 2 per meter of wavelength,
but for the plot we divided by 10 6 so
it would be W/m 2 per micron of wavelength.
By taking care of this detail, the area under
each curve in Figure 13.1 should match σT 4
as in Eq. 13.3.
Now look at the graph to see that the peak of the blue curve is indeed
just short of 0.5 μm, denoted by the red star at the top.
Example 13.2.2 Let’s now verify a point on the Planck spectrum,
picking 6,000 K <strong>and</strong> 1 μm to see if Eq. 13.4 l<strong>and</strong>s in the same spot as
indicated in Figure 13.1.
If we go through the laborious exercise of plugging in numbers to
Eq. 13.4 for T 6000 <strong>and</strong> λ 1 × 10 −6 (1 μm), we find 13 the overall 13: Numerically, the numerator is 3.74 ×
outcome is 3.73 × 10 13 W/m 2 per meter of wavelength. Once we adjust 10 −16 , the denominator is 10 −30 , <strong>and</strong> the
argument in the exponential is 2.4, so that
by 10 −6 for the units on the plot (see earlier margin note), we expect
the second fraction is 0.1.
0.373 × 10 8 W/m 2 per micron.
Indeed, the blue curve passes through this value at a wavelength of
1 μm, as indicated by the dotted line in Figure 13.1.
© 2021 T. W. Murphy, Jr.; Creative Commons Attribution-NonCommercial 4.0 International Lic.;
Freely available at: https://escholarship.org/uc/energy_ambitions.