13 Solar Energy

Now we come to the main attraction. As we saw in Chapter 10, all of the strictly renewable energy options are ultimately derived from the sun. The two resources of the last two chaptershydroelectricity and windrepresent tiny crumbs of the overall solar input to the planet.

Practically speaking, it is difficult to concoct ways to harness more than a few terawatts of power from hydroelectric or wind-based resources. Similar limitations apply to biologically-derived energy, geothermal, tidal, ocean currents, wave energy, etc. This may be worrisome considering that human society currently demands 18 TW. Meanwhile, the sun delivers 83,000 TW to the earth’s surface (Fig. 10.1; p. 167, Table 10.2; p. 168). That’s almost 5,000 times more than the demand. By the numbers, then, the sun seems to offer all we might ever need. In fact, the quantitative imbalance is so extreme as to make one wonder why we would ever mess around doing anything else.

Yet at present, the U.S. gets only about 0.9% of its energy from solar power, or 2.3% of its electricity. Similarly for the world, 1.2% of global energy is solar (2.1% of the electricity). It would seem to be vastly underutilized.

This chapter explains the nature of solar energy, its potential for use, practical considerations, and looks at the state of installations. While most of the focus is on photovoltaic (PV) panels that directly generate electricity from light, solar thermal power generation is also covered.

13.1 The Energy of Light

Sec. 5.10 (p.79) introduced the basics of the energy of light. This section acts as a refresher and lays the foundation for the rest of the chapter.

Photovoltaic cells, showing grid contacts and crystal domains. Photo Credit: Tom Murphy

Light - a form of electromagnetic radiationis composed of photons— individual particles of energy each having a characteristic wavelengthwhat we might call color. Photons are such tiny quanta of energy that familiar environments are awash in unfathomably large numbers of them. A typical light bulb, for instance, emits quintillions¹ of photons every second.

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Definition 13.1.1
The energy of a single photon, in various forms, is

E=hν = hc/λ ≈ 2 × ${10}^{-19}$J/λ(in μm) ≈ 1.24 eV/λ(in μm)
(13.1)

where h = 6.626 × ${10}^{-34}$ J.s (Planck’s constant), and v is the frequency of the light in Hertz (Hz, or inverse seconds).

The second form (hc/λ) is useful, as we more commonly characterize the “color” of light by its wavelength, λ. The speed of light, c≈ 3 x ${10}^8$ m/s, connects frequency to wavelength via λν = c. (13.2)

The third form in Definition 13.1.1 makes it easy to compute photon energy in Joules given the wavelength in microns.² Visible light has a wavelength around 0.4–0.7 μm (violet-to-red), so a typical photon energy, at 0.5 μm, is about 4 x ${10}^{-19}$ J. It’s a tiny number!

Example 13.1.1

About how many photons strike a 0.4 $m^2$ patch of sidewalk per second if the overhead sun is delivering 1,000 $W/m^2$?

For the visible light characteristic of sunlight, we can use a convenient wavelength of 0.5 μm, amounting to 4 × ${10}^{-19}$ J of energy per photon. The patch of sidewalk we describe receives light energy at a rate of 400 W or 400 J/s.³ How many 4 x ${10}^{-19}$ J photons does it take to amount to 400 J? Divide⁴ to get ${10}^{2}1$.

The final form in Definition 13.1.1 relates to the fact that photons frequently interact with electrons as we will see in Section 13.3, making it convenient to convert to another energy unit called the electron-volt, or eV (introduced in Sec. 5.9; p. 78). One electron volt is the energy it takes to move an electron through an electric potential of one Volt. The conversion is 1 eV = 1.602 × ${10}^{-19}$ J. For instance, the 0.5 μm (blue-green) photon we used in the previous example would have an energy around 2.5 eV.

Why should we care about unthinkably small quantities of light? Three reasons come to mind:

1. Eq. 13.1 elucidates that bluer photons⁵ have higher energy than red photons, which is important to know;

2. Individual photons interact with matter at the microscopic scale and are relevant to understanding solar photovoltaics and photosynthesis;
3. It’s how nature really works.

13.2 The Planck Spectrum

We should first understand where photons originate, which will help us understand how solar panels work and their limitations. Until recent technological advances, photons tended to come from thermal sources. It’s true for the white-hot sun,⁶ and true for flame and incandescent light bulb filaments⁷. Likewise, hot coals, electrical heating elements, and lava are all seen to glow.

Physics tells us how such hot sources radiate, as covered by the next three equations. The first (with units) is:

$P=A\sigma T^4$ (W). (13.3)

We already saw this equation in the context of Earth’s energy balance in Sections 1.3 and 9.2. It is called the Stefan-Boltzmann law, describing the total power (in W, or J/s) emitted from a surface whose area is A (in square meters) and temperature, T in Kelvin.⁸

The constant, $\sigma \approx 5.67\times {10}^{-8}W/m^2/K^4$ is called the Stefan-Boltzmann constant, and is easy to remember as 5-6-7-8.⁹

$$B_{\lambda }={\displaystyle \frac{2\pi h{с}^2}{{\lambda }^5(ehc/\lambda k_{B}T-1)}}\text{(}W/m^2\text{)}/m$$

(13.4)

Eq. 13.4 might look formidable, but only λ and T are variable. It describes the Planck spectrum, otherwise known as the blackbody¹⁰ spectrum.

For some temperature, T, this function specifies how much power is emitted at each wavelength, λ. Three fundamental physical constants from key areas of physics make an appearance: c≈ 3 x ${10}^8$ m/s is the familiar speed of light from relativity; h≈ 6.626 x ${10}^{-34}$ J. s is Planck’s constant from quantum mechanics, and kB≈ 1.38 x ${10}^{-33}$ J. K is the Boltzmann constant of thermodynamics.¹¹

${\lambda }_{m}ax\approx 2.898\times {10}^{-3}/T(inK)$ (m).

(13.5)

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Eq. 13.5 is called the Wien law and is a numerical solution identifying the peak of the blackbody spectrum as a function of temperature. Higher temperatures mean higher kinetic energies at a microscopic scale, so that higher-energy (shorter-wavelength) photons can be produced. This is why as objects get hotter, they move from red to white, and eventually to a blue tint.

All this may seem overwhelming, but take a breath, then just look at Figure 13.1. Everything so far in this section is captured by Figure 13.1.

Figure 13.1: Planck spectra, or blackbody spectra for three temperatures, indicating where the ultraviolet, visible, and 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, and 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 $1.0x{10}^{8}W/m^2/\mu m$.

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 $2.898\times {10}^{-3}/6000\approx 0.483\times {10}^{-6}m$, or 0.483 μm.

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 and 1 μm to see if Eq. 13.4 lands 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 and A = 1 × ${10}^{-6}$ (1 μm), we find¹³ the overall outcome is $3.73\times {10}^{1}3W/m^2$ per meter of wavelength. Once we adjust by ${10}^{-6}$ for the units on the plot (see earlier margin note), we expect $0.373 x 10^8 W/m^2$ 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.

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Example 13.2.3

Finally, to assess Eq. 13.3, we can crudely estimate the area under the blue curve by drawing a rectangle that we think has about the same total area. We put the top of the rectangle at the top of the blue curve and ask how wide it would need to be to approximately match the area under the blue curve.

If we make it 0.5 μm wide, it seems too thin: the area is smaller than what’s under the blue curve. 1.0 μμm wide seems like too much area. So we pick something in the middle like 0.75 μm (Figure 13.2).

[Figure 13.2: Attempts to crudely match the area under the 6000 K Planck spectrum using rectangles of three different widths. The widest (1 μm; dashed line) is too much area; the narrowest (0.5 μm; dotted line) is too small. The middle (0.75 μm; pink area) seems closest, by eye.]

This choice makes the area about $1.0 x 10^8 W/m^2$/μm (value at the top of the rectangle) times 0.75 μm, coming to $7.5 x $10^{7}$ W/m^2$. Since this is power per area, we make a minor rearrangement of Eq. 13.3 to $P/A = σT^4$ and evaluate for T = 6000 K to find $7.35 x 10^7 W/m^2$. Hey, not bad! So it all hangs together.

Now that we’ve batted the equations around a little bit, like a cat might do if given a new mouse toy, let’s absorb some key takeaways (Follow along using Figure 13.1.). First, as the source gets hotter, the area under the curve¹⁴ increases drastically-as the fourth-power of T, according to Eq. 13.3. This is seen in Figure 13.1 in that going from 3,000 K to 6,000 K¹⁵ leads to a tremendous increase in area under the curve: 16 times, in fact.(4)

Second, as an object gets hotter, it emits at shorter wavelengths, going from red-hot to yellow-hot to white-hot. The sun, at 5,800 K, peaks at 0.5 μm, in the blue-green region. We don’t see it as blue-green because it emits plenty of red light as well, making a blend. Notice how well the 6,000 K spectrum in Figure 13.1 covers the visible colors. A cooler star at 3,000 K has a red tint to it, since the 3,000 K spectrum (Figure 13.3) shows a marked red-heavy tilt: blue is not as well represented as red is. Conversely, a hot star at 10,000 K will have a blue tint to it, since it peaks around 0.3μm and has considerably more flux at the blue end of the spectrum than at the red end.

Figure 13.3: A 10,000 K star (or any blackbody) has a spectrum that tilts blue in the visible spectrum, while a cooler star (object) at 3,000 K has a red slant. Spectra are normalized to the same peak for easier comparison.

Finally, it is worth absorbing the overall lesson that photons from a glowing source emerge as a distribution, spanning a wide range of wavelengths (colors). This will turn out to be important in understanding why solar panels have the efficiencies that they do.

13.3 Photovoltaics

We are now prepared to dig into how photovoltaic (PV) panels actually work, and what governs panel efficiency.¹⁶ The word “photovoltaic” can be loosely read as: volts from photons, or electricity from light.

Various materials are used as the principal component in PV panels, but the vast majority are made of high-purity silicon, so we will speak in these terms alone. The basic physics will be the same for other materials as well. Getting too far into the weeds in describing the semiconductor physics is outside the scope of this course, but it is worth painting a general picture enough to appreciate how much we can expect to get out of a PV panel.

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Figure 13.4: PV cell structure and function. A junction between n-doped and p-doped semiconductors sets up an electric field across the junction. If an electron promoted to the conduction band by an incoming photon wanders into the junction, it is swept across (red arrow) and successfully contributes to current. Electrons do not contribute if created above the junction-as is more probable for blue photons that are not likely to penetrate as far. Electrons do not contribute to the external (useful) current if they recombine (fill a hole) before finding the junction (red “poof”).

Figure 13.4 illustrates the basic scheme. The underlying idea is that adjoining two slabs of silicon into which small amounts of two kinds of impurities have been deliberately introduced¹⁷ that either contribute extra electrons (n-type, for negative charge donors) or create vacancies for electrons (p-type, for effective positive charge donors).

Putting pdoped and n-doped materials together creates a junction¹⁸ exhibiting a contact potential. The result is that “donated” electrons right at the junction in the n-doped material decide to relocate across the junction to vacancies in the p-doped material, creating a wall of negative charge on the p-side of the junction and leaving behind “holes” (missing electrons) effectively creating positive charges¹⁹ on the n-side of the junction.

In the region right around the junction²⁰ an electric field is set up between the separated charges. Any electron wandering into this depletion region will be swept across the junction, across the contact potential, and will contribute to a current that is then driven around the external circuit to return to its p-side home.²¹ Figure 13.4 shows additional salient features that will be pointed out as the story develops below.

13.3.1 Theoretical Efficiency of Photovoltaics

We will now follow the fate of one photon as it encounters the photovoltaic material. Doing so will expose the physical process of photovoltaics and simultaneously track losses to elucidate efficiency expectations.

The basic scheme is that a photon knocks an electron away from an atom in the PV cell, and this electron has some chance of being swept across the junction upon which it contributes to a useful current.²² The goal is to get an electron across the line. It is not unlike some sports where crossing a line is the goal, but many factors are lined up defending against successful attainment of this goal. Efficiency is related to the chance that a photon will produce a “win.”

A photon leaves the hot solar surface aimed right at a PV panel on Earth. The photon can be any “color,” distributed according to the Planck spectrum²³ in Figure 13.1. The most probable wavelength for a 5,800 K blackbodyaccording to Eq. 13.5is ~0.5 μm, but it could reasonably be anywhere from 0.2-3 μm. The atmosphere will knock out (absorb or scatter) most of the ultraviolet light before it reaches the panel, and some of the infrared light is absorbed in the atmosphere as well. But almost 75% of the energy²⁴ makes it to the panel. What happens next depends on the wavelength.

First, we must understand something about the silicon material. The atoms in a typical silicon PV cell are arranged in an orderly lattice, grown as a single crystal. Expensive panels have mono-crystalline silicon, meaning that each 15 cm square cell comprising the panel is a thin slice of one giant crystal. Less expensive poly-crystalline (or multi-crystalline) panels have cells that are a patchwork²⁵ of randomly-oriented crystals at the millimeter to centimeter scale. But microscopically, both types are orderly crystals.

Silicon has four electrons in its valence shell (outermost shell), so that a “happy” silicon atom is home to a four-outer-electron family. These electrons are said to exist in the valence band.²⁶ But provided a sufficient energy kick, an electron can leave home and enter the conduction band,²⁷ where it can freely move through the crystal and can potentially contribute to an electric current, if it finds the junction. The threshold energy level to promote an electron from the valence to the conduction band is called the band gap,²⁸ which for silicon is 1.1 eV (1.8 × $10^{-19}$ J).(5)

Infrared photons at a wavelength of > 1.1 μm have an energy of E < 1.1 eV,²⁹ according to Eq. 13.1. The energy falls below the band gap of silicon, and as such is not capable of promoting an electron within the silicon from the valence band to the conduction band. These longer-wavelength photons sail right through the silicon crystal as if it were transparent glass. Since these photons are not absorbed, the part of the incident energy in the infrared beyond 1.1 μm is lost. For the solar spectrum, this amounts to 23%, and is portrayed in Figure 13.5.

Figure 13.5: Energy budget in silicon PV cell. The areas of the four regions represent the fraction of the total incident energy going to each domain. All light at wavelengths longer than 1.1 μm (infrared; 23%) passes through the silicon without being absorbed. The photons that are absorbed give excess kinetic energy to electrons, losing 33% of the incident energy as heat. This effect is progressively more pronounced the shorter the wavelength. Of the remaining 44%, about a quarter disappear as electrons “recombine” with vacancies (holes) in the silicon before getting a chance to contribute to a useful current by crossing the junction, leaving 32% as the maximum theoretical efficiency.

For the 77% of sunlight whose photons are energetic enough to bump an electron into the conduction band,³⁰ it’s game-on, right? Well, not so fast literally. Photons whose energy is higher than 1.1 eV have more energy than is needed to lift the electron into the conduction band. It only takes 1.1 eV to promote the electron, so a blue-green photon at 0.5 μm (~ 2.5 eV) has an excess of about 1.4 eV. The lucky electron is not just lifted out, but is given a huge boost in the process, rocketing out of the atom. It’s going too fast! It knocks into atoms in the crystal and generally shakes things up a bit before settling down.(5) We call this heat, or thermal energy:³¹ its excess kinetic energy is transferred to vibrations (randomized kinetic energy of atoms) in the crystal lattice. The blue curve in Figure 13.5 reflects this loss: we get to keep all the energy at 1.1 μm (1.1 eV), thus the blue curve joins the overall black curve here.

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But as the wavelength gets shorter and the energy gets higher, a greater fraction is lost to heat. Overall, 33% of the incident photon energy is lost to heat as the boosted electrons rattle the crystal before being tamed.

Now we’re down to 44% of the original incident energy in the form of conduction-promoted electrons that have shaken off their excess kinetic energy. But then here’s the rub: electrons are dumb. They don’t know which way to go to find the junction, so aimlessly bounce around the lattice, in a motion called a random walk.³²

Some get lucky and wander into the junction, where they are swept across³³ and contribute to external current.

Others fall into an electron vacancy (a hole) in a process called recombination: game over.³⁴ Roughly speaking, about three-quarters³⁵ of the electrons get lucky by wandering into the junction before being swallowed by a hole. So of the 44% available, we keep 32% (called the Shockley-Queisser limit ³⁶).

Another significant loss arises because some photons are absorbed in the very top layer above the junction, so that the resulting electrons do not have the opportunity to be swept across the junction to contribute to useful energy. The shorter the wavelength, the shallower the photon is likely to penetrate into the cell.³⁷

Meanwhile, photons around 1 μm are likely to penetrate deep-well past the junction-making it less likely that the liberated electrons will find the junction before settling into a new home (lattice site) via recombination. Figure 13.4 reflects this color-dependence, and also depicts one electron from the blue photon being generated above the junction, which will not have an opportunity to do useful work by crossing the junction.

In total, the basic physics of a PV cell is such that 20% efficiency is a reasonable expectation for practical implementations.³⁸ Indeed, commercial silicon-based PV panels tend to be 15-20% efficient, not far from the theoretical maximum.

This may seem like a low number, but don’t be disappointed! Biology has only managed to achieve 6% efficiency in the best-case photosynthesis (algae). PV technology beats that by a factor of three! And as we’ll see in Section 13.6, the only thing higher efficiency really does besides driving up the price is it makes the footprint (area occupied) smaller for the same power delivery. But it’s already small enough to comfortably fit on most roofs, so efficiency is not a chief limitation at this point.

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Figure 13.6: PV panels (modules) are constructed of typically 18, 36, 54, or 72 cells in series, two of which are depicted here. Cells are usually ~15 cm squares layered exactly as depicted in Figure 13.4. The bottom sides are covered by a continuous metal contact and the tops host a fine grid of metal contacts that minimize blockage of incident sunlight. Each cell presents ~0.5 V, arranged in series to add up to tens of volts per panel. To accomplish this, the top grid of one cell is connected to the bottom contact on the next, all down the line.

PV panels are usually constructed of many individual PV cells wired in series, as depicted in Figure 13.6. Each cell delivers maximum power when it’s at a voltage around 0.45 V, and cells are usually arranged in strings of 18, adding to about 8 V. Panels commonly have 2, 3, or 4 such strings of 18-thus 36, 54, or 72 cells total-becoming 16 V, 24 V, or 32 V devices at peak power. Figure 13.7 shows typical performance curves for a PV cell (or whole panel) operating in various light levels. Recalling from Eq. 5.2 (p. 77) that electrical power is current times voltage, the power put out by a PV panel can be found as the area of the rectangle formed from the origin to the operating point somewhere on the curve. The point that maximizes area (power) is shown in Figure 13.7 as the “maximum power point.” A battery being charged might hold the panel to a lower voltage, whose corresponding rectangle has a smaller area, thus operating at less than the panel’s maximum power.

Figure 13.7: Current-voltage (I–V) curve for a PV cell. The cell in full sunlight will operate somewhere on the thick blue curve, and on a lower red curve under weaker illumination. The maximum power point (mppt) is about 0.45 V for silicon, while the nominal design might be for 0.35 V so that a 36-cell panel is sized to charge a 12 V battery. Rectangle area is proportional to power delivered, since P = IV.

One serious downside of panels is that because cells are wired in series, partial shading of a single cell limits the current the whole panel delivers to that of the weakest link in the chain. In other words, 17 of 18 cells in a chain might be in full sun, but if the shadow from a roof vent, chimney, or tree shades one cell and limits its current to 10% of its full value, the whole chain³⁹ is knocked to 10%. Bypass diodes can isolate problem sections, but usually in chunks of 18-24 cells, so that the panel can still be seriously impaired by partial shading. Connecting panels in series also creates vulnerability to partial shadowing.

This problem is sometimes mitigated via micro-inverters: each panel has an inverter so that higher-voltage outputs are combined in parallel.

Box 13.1: Why Not Parallel? Given the downsides of series connection of cells, why not connect cells in parallel-the only other option for connecting many cells together?

Series combination adds voltages, keeping the same common current. Parallel combination shares a common (low) voltage but adds currents. The same power (P = IV) obtains either way. But two problems arise from a parallel combination of cells. First, the ~0.5 V voltage is too small to be useful for most devices. Second, the power lost in connecting wires scales as the square of current, so designing a system with a large current is asking for trouble.⁴⁰ That said, PV installations often combine panels in both series and parallel-like 10 panels in series in parallel to another 10 in series. By this time, the voltage is plenty high to offset the losses.

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13.4 Insolation

Let’s start our journey from the physics principles we covered in Section 13.2. The sun’s surface is a sweltering 5,770 K, meaning that it emits $σT^4≈ 6.3 x 10^7 W/m^2$ over its surface. The sun’s radius is about $10^9$ times that of the earth’s,⁴¹ which itself is 6,378 km at the equator. Multiplying the radiation intensity by the area gives total power output: $4πRσт^4≈ 3.82 × 10^26$ W. That’s one bright bulb!

Sunlight spreads out uniformly into a sphere expanding from the sun. By the time it reaches Earth, the sphere has a radius equal to the Earth-Sun distance, which is $r = 1.496 x 10^11$ m.⁴² Spreading $3.82 × 10^26$ W over a sphere of area $4πr^2$ computes to 1,360 $W/m^2$. That’s what we call the solar constant ⁴³, and it’s a number worth committing to memory.(5)

Earth intercepts sunlight over the projected area presented to the sun: a disk of area R. Bright features like clouds and snow reflect the light back to space without being absorbed, and even darker surfaces reflect some of the light. In all, 29.3% of the incoming light is reflected, leaving 960 $W/m^2$ absorbed by the R projected area of the planet. But now averaging the 960 $W/m^2$ input over the $4πR^2$ surface area of Earth cuts the number down by a factor of four,⁴⁴ to 240 $W/m^2$.

High latitude sites suffer more from low sun angles, and obviously cloudier locations will receive less sun at the surface. Taking weather into account, a decent number for the average amount of power from sunlight reaching the ground is about 200 $W/m^2$.(5) This is called insolation⁴⁵-the “sol” part of the word stemming from solar.

Solar Flux Context $W/m^2$
Arriving at Earth 1,360
Full, overhead sun (no clouds) ~1,000
Absorbed by $πR^2$ 960
Absorbed by $4πR^2$ 240
Typical insolation, includes weather ~200
Typical delivered by 15% efficient PV 30

Table 13.1: Summary of solar power densities. Full overhead sun can be larger than the global absorbed number because the global number includes reflection from clouds, while overhead direct sun corresponds to a local cloud-free condition.

Table 13.1 summarizes these various power densities, the last line being typical insolation multiplied by 0.15 to represent the yield from a 15% efficient photovoltaic panel lying flat in a location receiving an insolation of 200 $W/m^2$.

Figure 13.8 shows global insolation, variations arising from a combination of latitude and weather.

Figure 13.8: Insolation onto locally horizontal surfaces for the world (for flat plates facing directly upward), in units of $W/m^2$ and kWh/$m^2$/day. The area of the blue square in the middle of the Atlantic ocean is enough to satisfy current global energy demand, using 15% efficient solar collection (but distributed, of course). Source: The World Bank.

Figure 13.9: Horizontal insolation for the U.S. for a flat plate facing directly upward. Native units for the graphic are in kWh/$m^2$/day, the break-points between colors running from 4.0 to 5.75 kWh/$m^2$/day in steps of 0.25. These values can be converted to $W/m^2$ by multiplying by 1,000 W/kW and dividing by 24 h/day. Annotations are added once in each color band (in black or yellow) to indicate the equivalent measure in $W/m^2$ ⁴⁶. Alaska is not even close to scale. From NREL.

Figure 13.9 shows the variation of insolation across the U.S. The latitude effect is evident, but also weather/clouds make a mark, giving the southwest desert the highest solar potential. Even so, the variation from best to worst locations⁴⁷ is not even a factor of two.

Figures 13.8 and 13.9 are in the context of a flat surface.⁴⁸ For solar panels, it makes sense to tilt them to an angle equaling the site latitude and oriented toward the south.⁴⁹ The noon-time sun is always high in the sky near the equator, so panels there should lie flat.⁵⁰

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But at high northern latitudes, the sun is lower toward the southern horizon, so the panels should tilt up to best face the sun. Tilting at an angle equal to the latitude is the best compromise, as Figure 13.10 illustrates.

Figure 13.10: The left globe shows the sun’s view of 21 panels of the same size sitting flat on the ground at their various sites. In the middle globe, the panels are all tilted up toward the equator. Notice the improvement in how much panel area is visible to the sun by doing this especially at higher latitudes. At right is the side view, from which it is easier to appreciate why the best tilt angle is equal to the site latitude.

Photovoltaic Solar Resource of the United States

Tilting panels toward the equator at an angle equal to site latitude optimizes annual yield, and the results are shown in Figure 13.11. Note that the numbers in Figure 13.11 are not strictly insolations, since that’s defined as what reaches flat ground. In this case, the area (square meters) is that of the panel, not of the land.

The fact that the numbers in Figure 13.11 are higher than in Figure 13.9 is not to say that the land offers more solar energy if the panels are tilted: just that an individual panel can get more light. But in this case, panels need to be spaced out to avoid shadowing,⁵¹ as Figure 13.12 illustrates.

Figure 13.12: On a fixed piece of land receiving a fixed amount of sunshine at a slant angle, the amount of energy received is independent of whether the panels are flat or tilted. Just tilting the flat panels up (middle) results in self-shading. It makes the most sense to tilt and separate panels (right), one benefit being that fewer panels are needed to collect the same incident energy.

Figure 13.11: Solar potential for flat panels tilted to latitude, oriented south-relevant to PV panel installations. The graphic is presented in units of kWh/$m^2$/day, the breakpoints between colors running from 3.0 to 6.5 kWh/$m^2$/day in steps of 0.5. Annotations are added once in each color band (in black or yellow) to indicate the equivalent measure in $W/m^2$ ⁴⁶. From NREL.

Some applications need to track the sun, like those that concentrate solar power, and only work when the sun is not blocked by clouds.⁵² This brings us to Figure 13.13, showing the potential per square meter of collector (mirror or lens) used for the concentration (the topic of Section 13.8.2). The same pattern holds, in that the desert southwest dominates. But a look at the numbers indicates that the cloudier regions are not much better than just a flat panel facing upward (as is the case for Figure 13.9). In the southwest, where skies are often cloud-free, the boost can be about 30% over the flat, upward-facing panel. Concentration schemes make less sense away from such regions.

Figure 13.13: Solar potential for tracking panels, facing directly toward the sun’s position and requiring a cloud-free view of the sun (concentrating collectors). The graphic is presented in units of kWh/$m^2$/day, the break-points between colors running from 4.0 to 7.5 kWh/$m^2$/day in steps of 0.5. Annotations are added once in each color band (in black or yellow) to indicate the equivalent measure in $W/m^2$ ⁴⁶. From NREL.

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Stepping back, let’s appreciate a few big-picture facets from these maps. First, numbers tend to be in the general neighborhood of 150-300 $W/m^2$.(4) Burn this range in-it’s a useful context. Second, the variation from the most solar-intense places in the contiguous U.S. to the weakest areas 51 is not more than a factor of two on an annual basis.(5) This is astounding. The Mojave desert in California and the rain-forest Olympic Peninsula in Washington would seem to be practically day vs. night with respect to solar illumination. But not so much: only a factor of two.⁵³ Part of what this means is that if storage over annual timescales could be realized, solar power would become practical almost everywhere.⁵⁴

Box 13.2: Hours of Full-Sun Equivalent

A useful take-away comes from the native units used in the three maps presented here: kWh/$m^2$/day, as opposed to our preferred $W/m^2$. Although they look different at a glance, kWh is a unit of energy, so kWh/day is a power, just like W. Since a kilowatt is 1,000 W and a day is 24 h, 1 kWh/day is 1,000 Wh/24 h = 41.67 W.⁵⁵ So we can multiply 6 kWh/$m^2$/day by 41.67 to get 250 $W/m^2$.

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But the main purpose of this box is to point out the following. Full overhead sunshine bathes the ground in about 1,000 $W/m^2$.⁵⁶ So if you could contrive to keep the sun directly overhead for 5 hours, you’d get 5 kWh of solar energy for each square meter on the ground. Therefore, if your site is listed as getting 5 kWh/$m^2$/day, it’s the equivalent amount you’d get from 5 hours of direct overhead sun.⁵⁷

What actually happens is that the day is longer than 5 hours, but for much of the day the sun is at a lower angle so that the panel is not directly illuminated, and weather can also interfere. This leads to a concept of full-sun-equivalent-hours. A site getting an annual average of 5.4 kWh/$m^2$/day might be said to get 5.4 hours of full-sun-equivalent each day. It’s a pretty useful metric.

Box 13.2 leads to a crucial bit of understanding on characterizing a PV system. Panels are rated on what they would deliver when illuminated by 1,000 $W/m^2$ at a temperature of 25°C.⁵⁸ So the measure in $kWh/m^2/day$, or full-sun-equivalent hours tells you effectively what fraction of a day the panel will operate at its rated capacity.

Example 13.4.1

A 250 W panel at a location getting $4.8 kWh/m^2/day$, or 4.8 full-sun-equivalent hours, is basically operating at 250 W for 4.8 hours out of every 24, or 20% of the time. So the panel delivers an average power of 50 W, not 250 W.⁵⁹

The 250 W rating is referred to as “peak” Watts, sometimes denoted 250 Wp. Panels are sold this way, and now cost about $0.50/Wp.

A 30-year study by the National Renewable Energy Lab ⁶⁰ initiated in 1960 characterized solar potential across the U.S. and produced detailed statistics on what each location might expect to collect each month for panels in different orientations.

Table 13.2 is a subset of the complete data for St. Louis, Missouri.⁶¹ All cases in Table 13.2 correspond to a panel facing south, at various tilts (including flat, at 0° and vertical at 90°; other tilts are relative to the site latitude of ≈ 39°). From this, we see that tilting the panel at the site latitude delivers an annual average of 4.8 kWh/$m^2$/day, matching the graphic expectation from Figure 13.11. Also shown is the monthly breakdown and how different tilts translate to performance. We will visit this table again in Section 13.6 to help us establish an appropriate size for a residential installation.

Angle	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec	Year
0°	2.2	2.9	3.9	5.0	5.9	6.4	6.4	5.7	4.6	3.5	2.3	1.8	4.2
0 - 15°	3.2	3.8	4.6	5.4	5.9	6.3	6.3	6.0	5.3	4.5	3.2	2.7	4.8
0	3.6	4.2	4.7	5.3	5.6	5.8	5.9	5.7	5.3	4.8	3.5	3.1	4.8
0 + 15°	3.8	4.3	4.6	4.9	4.9	5.0	5.1	5.2	5.1	4.8	3.7	3.3	4.6
90°	3.5	3.7	3.4	3.1	2.6	2.4	2.6	3.0	3.5	3.8	3.2	3.0	3.2

Table 13.2: Solar exposure (kWh/$m^2$/day) for a south-facing panel in St. Louis, MO, at various panel tilts (0 is latitude, which happens to be 39° for St. Louis). 0° means a panel lying flat, pointing straight up (like on a flat roof), and 90° means vertical, like on a (south-facing) wall (see Figure 13.14).

Figure 13.14: Panel tilts for Table 13.2, for 0 = 39°.

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13.5 The Incredible Solar Potential

The potential of sunlight can be assessed by pieces we have already seen. Multiplying the solar constant of 1,360 $W/m^2$ by the projected area of Earth (R) and by 0.707 to account for the 29.3% reflection loss, we compute that Earth absorbs solar energy at a rate of #1.23 x 10^17$ W, or 123,000 TW (1 TW is $10^12$ W). Compared to the 18 TW societal scale, that’s huge! Notice that the non-reflected entries back in Table 10.2 (p. 168) add⁶² to this same value.

Placing solar panels on just 10% of the land (itself 29% of Earth’s surface ⁶³ area) capturing the incoming energy at 15% efficiency would produce solar power at ~500 TW: about 25 times today’s 18 TW usage rate. This, in turn, means that our current energy demand could be met by covering only 0.4% of land⁶⁴ with photovoltaic panels: see Figure 13.8 for a visual representation of how much this is. Solar is the only currently-available resource that can come anywhere close to satisfying our current energy appetite.⁶⁵ And it exceeds our demand by such a huge margin! We therefore have reason to be excited about solar energy: the raw numbers are great news, indeed.

So it seems like a done deal. Solar. Let’s get started! Wait, why aren’t we there yet?

Naturally, solar has some downsides.

First, the cost.
Panel cost has dropped to something like $0.50 or less per peak-Watt. To get 10 TW of delivered (average) power at typical locations getting 20% capacity factor⁶⁶ would require about 50 TWp (Example 13.4.1 defines Wp), costing 25 trillion dollars.

The cost of other necessary components and installation double the cost for utility-scale projects ⁶⁷, bringing the cost to $50 trillion.⁶⁸

The global annual economy is not quite twice this. To outfit the world with the requisite number of panels would take about 60% of the economy for a year, or 6% over 10 years, or 1.5% continuously.⁶⁹

For comparison, the world goes through 30 billion barrels of oil each year at a cost of $50/bbl, meaning $1.5 trillion per year. Installing enough panels to fully satisfy demand would cost three-decades-worth of the entire global petroleum budget. So it’s not going to happen fast.
To put things on a personal scale, Americans use power at a rate of 10,000 W. To cover this, we would need about 50 kWp of panel per person, costing $50k per person.⁷⁰ When can we expect your payment?

Another daunting realization is that even though only 0.4% of the land is needed to match today’s demand, this is comparable to the amount of area currently covered by roads and buildings.(5) A road trip across the country conveys a sense for how vast (and boring) all that pavement can be. And pavement is a fancy form of dirt. It is true that PV panels are also an ultra-pure, ultra-fancy form of dirt. But it’s a different level of high-tech. It becomes hard to fathom that much PV around the world.(4)

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A major impediment to solar power is its intermittency⁷¹. Figure 13.15 shows 31 days of solar capture, along with typical state-wide electricity demand. The two do not look very similar: not well matched. Demand is far more constant than the solar input, which is obviously zero at night. Even the peaks do not line up well, since demand is highest in the evening, well after solar input has faded away.

Figure 13.15: Solar input (red) and electricity demand (blue) look nothing alike. Solar data from the author’s home begins 31 March 2020, while demand is for California. Tick marks denote the start of each date, at midnight. April 22-27 are essentially perfect cloudless days, while the earlier part of the month had rainy periods. Note that even a very rainy day (April 10) provides some solar power (15% as much as a full-sun day). Intermittent clouds cause the “hair” seen on some days. The capacity factor for the month is 19%, while the perfect six days near the end perform at 27% capacity. From this, we infer that weather caused the yield to be 70% what it would have been had every day been cloudless.

Storage is required in order to mitigate the intermittency, allowing the choppy solar input to satisfy the demand curve of Figure 13.15. We don’t have good solutions for seasonal storage,⁷² so a complete reliance on solar energy would necessitate over-building the system to handle months of low-sun conditions through winter (see the annual variation in Table 13.2), making it cost all that much more.

Finally, all energy is not equivalent and substitutable. Solar PV cannot power passenger airplanes or power our cars down the road real-time (only via storage).⁷³ For all its potential, the hangups are serious enough that more than 60 years after the first demonstration of a photovoltaic cell, less than 1% of our energy derives from this ultra-abundant source.

Box 13.3: Why no Solar Planes?

Consider that full overhead sun delivers 1,000 $W/m^2$. The top surface area of a typical commercial airplane (Boeing 737) is about 450 $m^2$. If outfitted with the most expensive space-worthy multi-junction PV cells getting 50% efficiency, the plane would capture about 500 $W/m^2$ and a total of 225 kW. Sounds like a lot! The problem is that a Boeing 737 spends about 7 MW while cruising (and more during the climb). We’re shy by a factor of about 25, even in optimal⁷⁴ conditions! Any solar-powered airplane⁷⁵ would be very light and very slow by air travel standards. See also Box 17.1 (p. 290) on the difficulty of battery-powered planes.

Cars have a similar problem: a top area around 10 $m^2$ equipped with the most expensive PV money can buy would get 5 kW, or less than 7 horsepower. That’s about 20 times less powerful than typical cars, so again: light and slow.

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13.6 Residential Solar Considerations

Despite these drawbacks, it can still make a lot of sense⁷⁶ to invest in solar photovoltaics for the home. We’ll explore sizing and cost in this section.

13.6.1 Configurations

A typical household uses much or most of its energy when the sun is not shining: lighting, cooking, evening entertainment, charging an electric vehicle, etc. To get around this, the system would need to have local storage,⁷⁷ or be tied to the regional electrical grid so that excess production can be exported in the daytime and electricity produced by the utility used at night or when household demand exceeds solar production. The overwhelming majority of solar installations in the U.S. are grid tied, and very few mess with batteries, which can double the cost of a system and need replacement before the system has paid for itself in electricity bill savings.

Box 13.4: Disappointing Dependence

A disappointing surprise to many who “go solar” is that their house has no power when the electrical service to the house is disrupted— even in the light of day. A grid tied system needs the grid to operate. Safety concerns prohibit PV systems from continuing to energize a disabled grid.

Only “off-grid” battery systems continue to work in these scenarios, but then the disappointment shifts to the price tag, maintenance, and replacement of worn-out batteries after a few thousand cycles.

While a description of the components and practical workings are beyond the scope of this book, students might be interested in an article the author wrote after first getting started tinkering with PV systems ⁷⁸.

13.6.2 Sizing and Cost

How large does an installation need to be? If the goal is to cover annual or monthly electricity use in a grid-tied system, the only two pieces of information needed are the typical electricity consumption in the relevant period and the average solar input at that location for the period of interest.

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 per year) is from electricity. That’s 5 qBtu, or 5.3 × $10^{18}$ J in one year (3.156 x $10^{7}$ s), or 167 GW. Distributed among 130 million households in the U.S.,⁷⁹ average household electricity consumption is 1,285 W. Applied over 24 hours, this makes for just over 30 kilowatt-hour (kWh) per day for an average household.⁸⁰

The next piece is solar potential at the location of interest. We’ll use the excerpted data from ⁶⁰ for St. Louis, Missouri found in Table 13.2.

Example 13.6.1

Let’s design a grid-tied PV system for an average U.S. household in an average⁸¹ U.S. city (St. Louis). We’ll orient the panels facing south and tilted to the site latitude (39°) and 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.⁸²

But 18% panels will require about 35 $m^2$ of panel,⁸³ which would be a 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 $W/m^2$ and 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.⁸⁴

We get the same answer either way, which is a good check.⁸⁵

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, and less efficient as a result;
• The panels will get a little dirty;
• The equipment that converts panel output to AC electricity is not 100% efficient.

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So it’s a good idea to bump the number up by 20% or so, and order a 7.5 kW PV system for the case under study. A typical full cost (panels, electrical converters, installation) lately runs just shy of $3 per peak Watt (Figure 13.16), which in this case brings the price tag to roughly $20k. If electricity costs $0.15 per kWh—approximately the national average each 30 kWh day costs $4.5, accumulating to $20k after 12 years. Federal and state incentives can make the payback time shorter.

What would these numbers become if trying to meet monthly instead of annual demands? December is usually the worst month for PV in the northern hemisphere, when the sun is lowest in the south, and the days are shortest. Table 13.2 backs this up, showing 3.1 kWh/$m^2$/day for the chosen panel orientation in December. This is about two-thirds the annual average, so we would need to increase the size of the system (and thus cost and payback time) by a factor of 1.5 to produce enough in December.⁸⁶

If sizing for an off-grid system, we need to factor in some inefficiency for battery charge/discharge and design for poorer months, so should increase by another factor of at least 1.5. The cost of batteries can be rather large, too. A good rule of thumb is to have at least three days of storage in the event of no solar input for several days during a stormy period. For our 30 kWh per day target, we would want about 100 kWh of storage. As an easy way to get a cost estimate for storage, the Tesla powerwall 2 is 13.5 kWh⁸⁷ and costs about $7k apiece. If we follow along, the cost of the off-grid PV system for 30 kWh/day at an installation cost of $3/W will be 7500 W x 1.5 x 1.5 x $3≈ $50k for panels/installation plus $56k for batteries.⁸⁸ Then the batteries may be in need of replacement every 10-15 years.⁸⁹

If this seems rather alarming, don’t worry-there’s a trick to making it much more affordable/practical: don’t demand 30 kWh per day! Even though we picked 30 kWh/day due to the fact that it is the average American electricity demand, it is a worthwhile challenge³⁶ to seek ways to use far less energy than the average. Trying to make solar fit our present expectations may be the wrong approach. Also, expecting to get away from fossil fuels and have it be cheaper may be unrealistic.

Figure 13.16: Evolving price of PV installation per peak Watt. Yellow is for the panels; the two blues for electronics; peach is labor; and hashed is for the utility hookup, inspection, taxes, profits [89, 91]. From NREL.

13.7 Photovoltaic Installations

The Energy Information Administration’s Electric Power Monthly (EPM) ⁹⁰ provides detailed statistics on power generation in the U.S. Photovoltaic data is available in the EPM’s tables 1.17.B and 6.2.B. In the usual way, we first look at installed capacity, based on the actual average delivered power. Figure 13.17 shows the situation in the U.S. California is rocking it!

Figure 13.17: Photovoltaic power production by state, in GW, in 2018.

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The average solar power in California was 4.3 GW in 2018, far ahead of the next biggest: North Carolina at 0.82 GW. For California, this is 13% of its electricity. But electricity production is 38% of all energy in the U.S., so we might say that California gets about 5% of all its energy from solar.(4) This is far ahead of other states.⁴⁶ The U.S. as a whole gets about 0.9% of its energy from solar.

Next, we divide by area to get power density from photovoltaic installations. A site having an insolation of 200 $W/m^2$ and 15% efficient panels has access to 30 $W/m^2$ of production capability (Table 13.1).

Figure 13.18: Photovoltaic power production areal density by state, in milliwatts per square meter.

Figure 13.18 shows how much we’re actually getting. New Jersey has its moment in the sun, here. A few sites (NJ, MA) are pushing⁶⁰ 15 m$W/m^2$, which is a factor of 2,000 lower than the full potential. What this says is that only 1/2,000 of the land (0.05%) is covered by solar panels. This sort-of makes sense, right?

Figure 13.19: Per capita photovoltaic power production by state, in Watts per person.

On a per-population basis (Figure 13.19), Nevada shines brightest, at 180 W per person.⁶⁷ The southwestern U.S. is doing well overall, as is North Carolina on this measure.

Figure 13.20: Photovoltaic capacity factors by state. While we see lots of darker green, it’s because everybody has similarly low numbers, due to unavoidable nighttime and low sun angles. Somebody tell Wyoming, North Dakota, and Alabama to get with the program!

Finally, we look at capacity factor: how much was generated compared to installed capacity (Figure 13.20). We expect something like 20%, corresponding to 4.8 full-sun-equivalent hours per day. The best states top out at about 0.27, equating to about 6.5 full-sun-equivalent hours per day.

0

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States at higher latitude and/or having more clouds will do more poorly on this measure. Alaska clocks in just over 0.1, mapping to about 2.5 hours per day, on average.

Country, installed(GWp), average(GW), % of all energy. global share (%)
China 175 ~18 1.2 27
U.S. 62 10.6 0.9 16
Japan 56 6.5 3.5 10
Germany 46 5.0 3.3 7.5
India 27 4.1 1.5 6
World 510 67 1.5 100

Table 13.3: Top five global producers of PV power in 2018, accounting for two-thirds of the world’s total production [92, 93]. The installed PV corresponds to peak watts (Wp), or production in full overhead sun.

Globally, two-thirds of the photovoltaic capacity is represented by five countries, shown in Table 13.3. Note that delivered power is significantly lower than installed capacity because of the low capacity factor for solar.

13.7.1 Pros and Cons of Photovoltaics

Before advancing to solar thermal generation, let’s summarize the major advantages and disadvantages of solar photovoltaics. First, the good stuff:

• PV taps into a super-abundant resource-the only renewable that has such a margin; [[218]]
• PV technology has no moving parts or steam; panels are robust and last a long time;
• PV is one of the few resources that can fit on a rooftop and provide self-contained electricity generation;
• PV efficiency is rather good: close to theoretical expectations and much better than biology has managed at getting energy from sunlight;
• PV technology works well, and despite expense has been deployed on rooftops across the world;
• Life-cycle CO2 emissions are 15 times smaller than that of traditional fossil fuel electricity ⁷¹;
• PV is often a good solution when utility electricity is far away.

And now the less attractive aspects:

• PV is intermittent, and not well-matched to energy demand; it would be hard to “balance” the electrical grid if too much of the input came from such an intermittent source, and storage is difficult;
• PV is still expensive⁷⁸ relative to prevailing energy resources - especially important in terms of up-front cost;
• Electricity alone is not well-suited to many of our current energy demands, like transportation and industrial heat/processing; Stand-alone operation requires batteries, at least doubling the cost and adding maintenance/replacement demands;
• Even partial shading can be disproportionally disruptive;
• PV manufacturing involves environmentally unfriendly chemicals;
• PV deployment can harm habitats if installed in undeveloped areas.

13.8 Solar Thermal

Photovoltaics (Section 13.3) convert sunlight directly into electricity, but this is not the only way to harness energy from the sun. Solar energy can also be used for heat. We’ll first have a brief look at home heating, then turn to electricity generation from solar heat.

13.8.1 Passive Solar Heat

Full sun delivers something like 1,000 $W/m^2$ at the earth’s surface. Now imagine a window in a house intercepting 1.5 $m^2$ of sunlight, in effect admitting 1,500 W into the home-like a space heater, and it’s free! Depending on window construction, some of the infrared energy may be blocked, so maybe not all 1,000 $W/m^2$ will make it inside, but a sizable portion will. Clever design has south-facing windows for receiving low-angle winter sun, but an overhang to keep out the high summer sun (Figure 13.21).

Figure 13.21: A well-designed house has thick walls, thick insulation, and doublepaned windows. Even better, it can have south-facing windows that admit sunlight in the winter but not in the summer (the overhang shields the window). A large, dark thermal mass-stone or brick works wellcan absorb energy and continue to release heat into the evening.

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A dark and massive absorber⁹¹ inside the house capturing the heat can continue to provide warmth through the evening hours. The Passive House designs mentioned in the context of Box 6.1 (p. 87) attempt to maximize solar capture so that little active heating is required.

13.8.2 Solar Thermal Electricity

While 1,000 $W/m^2$ is nice, the power is too diffuse to get anything very hot and create a large enough ΔT to allow the operation of an efficient heat engine (Sec. 6.4; p. 88). More complex arrangements can concentrate solar power-think of a magnifying glass-to heat up a liquid in pipes. Figure 13.22 shows an example of a parabolic reflector that can track the sun to concentrate light onto the energy-absorbing central pipe. This shape can be extruded along a long cylinder-a “trough”-following the pipe.

Figure 13.22: Solar trough cross sections showing the focusing of sunlight onto a central pipe. The troughs can be oriented to follow the sun.

Figure 13.23: A common solar thermal power scheme uses parabolic “trough” reflectors to focus sunlight onto a central pipe, which carries oil that can be heated to very high temperatures for making steam to run a traditional electrical power plant very much like that of Fig. 6.2 (p. 90). Optional thermal storage can save heat for later use.

Figure 13.23 shows a schematic representation of a typical solar thermal (ST) collector, and a picture of one appears in Figure 13.24. A curved reflector tilts to track the sun, concentrating light onto a long pipe in front of the reflector carrying a fluid (usually oil) that can be heated to a high temperature by the absorbed sunlight. The hot oil pipes can then be run through water to boil it and make steam, thereafter driving a traditional steam power plant. Such ST arrangements are sometimes called concentrated solar power (CSP). Another common variant-called a “power tower”-is shown in Figure 13.25, in which an array of steerable flat mirrors on the ground direct sunlight to the top of a central tower to make steam.

Figure 13.24: Parabolic trough-based ST plant, in which part of the power generation facility is seen in the background. Reflectors must be spaced out to prevent self-shadowing. From U.S. DOE.

Figure 13.25: One of three “power towers” of the Ivanpah facility in California. By Craig Dietrich.

As for efficiency, solar thermal is at face value similar to PV: 15-20% is fairly typical. Broken down, roughly 50-75% of the available energy successfully transfers to the fluid, and then the heat engine delivers about 25-30% efficiency. But these numbers only apply if we count just the area of the reflective collector. Because they have to track the sun, and self-shadowing is to be avoided, only a small amount of the land area is occupied by the reflectors. Characterization of real facilities indicates that only 3% of the solar energy hitting the patch of land corresponding to the power plant is exported in the form of electrical energy.

But efficiency is not everything. 3% of a gigantic resource like solar energy input can still be tremendously large.(4) It translates to over 6 $W/m^2$ for a standard insolation of 200 $W/m^2$, which is about thirty-times better than wind, per land area.(5) While a field of PV panels outperform an ST installation by a factor of 5-6, the technologically simpler solar thermal designs can be more cost effective than PV. Reflectors and oil pipes are low-tech cheap devices, compared to photovoltaic material. The production cost for solar thermal is estimated to be about $0.06/kWh, which is lower than the typical retail cost of electricity, but still a factor of two higher than fossil fuel electricity production costs.

One disadvantage of solar thermal is that concentration only works when the sun itself is visible in the sky: no obscuring clouds.(4) One way to think of it is: if you can’t see your shadow, solar concentration will not work. Meanwhile, PV panels will still produce a meaningful amount of daytime electricity from the bright sky and clouds even if the sun itself is not “out.”

Balancing this disadvantage is the fact that solar thermal has some built-in storage capacity, in that the heated oil can be “banked” for some hours ⁹² and continue to produce electricity even during the passage of a cloud or for a few hours into the evening. In this sense, it can better match the peak of electrical demand (early evening: Figure 13.15) than can PV, which goes to zero once the sun sets.

As seen in Figure 13.13, the desert southwest is the best place in the U.S. for solar thermal electricity generation. It makes sense that deserts would be good spots, since effective concentration requires no interference from clouds. Incidentally, transmitting electricity over intermediate distances (across regions) is fairly efficient: typically better than 90% for distances shorter than ~1,000 km.

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In terms of implementation, solar thermal is a small player. In 2018, only four states produced solar thermal power, 68% from California and 22% from Arizona. Table 13.4 provides some context, comparing ST to PV in each of the four states that have any solar thermal. For the entire U.S., less than 0.1% of electricity derives from solar thermal, and PV is about 25 times bigger on the whole. Globally, ST averages about 1.1 GW⁹³ (2016), about half in Spain and a third in the U.S.⁹⁴

State; ST MW avg.; ST % elec.; PV MW avg.; PV % elec.; ST/PV % California 281 1.25 4,285 19.0 6.6
Arizona 89 0.08 765 5.1 11.6
Nevada 35 0.09 552 12.1 6.3
Florida 6 0.002 326 1.2 1.8
U.S. total 410 0.086 10,565 2.2 3.9 [[221]]

Table 13.4: Solar Thermal (ST) generation in the U.S. in 2018, compared to photovoltaic (PV); MW is megawatts.

13.8.3 Pros and Cons of Solar Thermal

Summarizing the pros and cons for solar thermal (ST), starting with the good aspects:

• ST taps into a super-abundant resource the only renewable that has such a margin;
• ST technology is low-tech and inexpensive, using well-developed power plant technologies;
• ST has built-in short-term storage capacity for covering evening power demands;
• Life-cycle CO2 emissions are 20 times smaller than that of traditional fossil fuel electricity ⁷¹.

And the less great stuff:

• ST requires direct sunlight; intolerant of clouds;
• ST is only possible at utility-scale, requiring a power plant;
• ST has a lower land-area efficiency than PV panels;
• Some disruption will be imposed on the local environment/habitat.

13.9 Upshot for Solar

Hands down, solar is the only renewable resource capable of matching our current societal energy demand. Not only can it reach 18 TW, it can exceed the mark by orders of magnitude. Finding space for panels is not a limitation. The efficiency of PV panels is perfectly respectable based on physics expectations, and beats the best that biology has done by a factor of 3-4. The efficiency is high enough that roof space tends to be more than sufficient to satisfy the demands of individual houses.

Pros and cons are listed separately for PV and ST in Section 13.7.1 and Section 13.8.3, respectively.

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Holding solar back is its intermittency⁹⁵ and high up-front cost. Intermittency can be solved by battery storage, but this can double the cost and require maintenance and periodic battery replacement. Additionallyas for many of our renewable optionsall of our society’s demands ⁹⁶ are not well met by electricity generation.

Sizing up a PV installation is fairly straightforward. Having first determined how many kWh per day are to be produced, on average, divide this by the kWh/$m^2$/day value for the site,⁹⁷ which is essentially the number of hours of full-sun ⁹⁸ equivalent, and tends to be in the 4-6 hour ballpark.

This says how many kilowatts the array should produce in full sun (peak Watts). For instance, if only 10 kWh/day are needed,⁹⁹ and the region in question gets 5 kWh/$m^2$/day, the system needs to operate at a peak power of 2 kWp, costing about $6k to purchase and install (grid tied). Inflating by 20% offsets unaccounted losses¹⁰⁰ to better match real conditions.

Solar thermal energy is another way to run a traditional steam-based power plant, using relatively low-tech mirrors and pipes to concentrate solar energy into a heat-carrying fluid that can later make steam. Effective efficiencies are relatively low,¹⁰¹ but on the bright side, the low-tech nature makes it fairly cheap, and the technique can accommodate some degree of thermal storage for use some hours into the evening. Anything¹⁰² starting from solar input has the potential to be a major player, given the ~100,000 TW scale of solar energy incident upon the earth.

13.10 Problems

1. If we had two monochromatic (single-wavelength) light sources-a green one at λ = 0.5 μm and a near-infrared one at A 1.0 μmeach emitting photons at an energy rate of 1 W,103 how does the number of photons emerging per second from each source compare? Is it the same number for each because both are 1 W sources, or is it a different number-and by what factor, if so?
2. Overhead sunlight arrives on the surface of the earth at an intensity of about 1,000 $W/m^2$. How many photons per second strike a solar panel whose area is 1.6 square meters, if the typical wavelength is λ = 0.5 μm?
3. Using the setup in Problem 2, how many photons enter your pupil every second if you look directly at the sun? When doing so, your pupil restricts to a diameter of about 2 mm.
4. The dimmest stars we can see with our eyes are thirteen orders-ofmagnitude104 dimmer than the intensity of the sun. Building off of Problem 3, how many photons enter your eye per second at this edge of detectability? 103: Hint: recall that 1 W is 1 J/s. 104: $10^{-13}$ times

223 5. Warm humans at ~300 K and glowing-hot light bulb filaments at ~2,400 K both radiate according to Eq. 13.3. How much more power per unit area ($W/m^2$) does an incandescent filament emit compared to human skin, roughly? 6. The outcome of Problem 5 indicates that a hot light bulb filament emits thousands of times more power per unit area than human skin. Yet both a human and a light bulb may emit a similar amount of light105—both around 100 W. Explain how both things can be true? 7. At what wavelength does the Wien Law say the Planck spectrum will peak for a temperature of 4,500 K? Express your answer in microns and compare to Figure 13.1 for confirmation. 8. Human bodies also glow by the same physics as the sun or a light bulb filament, only it is too far out in the infrared for the human eye to see. For familiar objects (and human skin) all in the neighborhood of 300 K, what is the approximate wavelength of peak blackbody radiation, in microns? 9. We might describe the efficiency of a light bulb as the fraction of its total light output that falls within the visible range. If using a thermal source106 why would you expect it to be impossible to reach 100% efficiency at any source temperature, based on what Figure 13.1 shows? 10. Using a technique similar to that in the text, approximate the height and width of a rectangle that has comparable area to the spectrum for T = 4500 K in Figure 13.1. Compute the area of your rectangle and compare to the expectation from oT4. 11. What are two reasons that blue photons are disadvantaged in terms of having their energy contribute to useful current in silicon photovoltaics? 12. Which photons are most responsible for heating up a silicon photovoltaic panel in full sun: blue photons or infrared photons (beyond 1.1 μm)? 13. If a blue photon having 3.3 electron-volt of energy liberates an electron in silicon, whose band gap is 1.1 eV,, what fraction of the photon’s energy is “kept” by the electron once it settles down from the excess?107 14. If a 2.5 electron-volt photon liberates an electron from silicon with a 1.1 eV band gap, how much kinetic energy does the emerging electron have? Express in both eV and Joules, and then determine the velocity of the electron if the electron mass is 9 × $10^{-31}$ kg. 15. Briefly summarize the sequence of events108 that results in a 105: Humans emit in the infrared, so we don’t see it with our eyes. Note that the spectrum is spread over so many microns and we’re being approximate about the temperature, so relax your answer to an easy, round number. 106: like the sun or an incandescent light bulb; one that obeys Eq. 13.4 107: The rest going into excess kinetic energy which just heats up the PV cell. 108: …consistent with the relevant physics, of course

224 successful contribution to PV current, starting with a green photon leaving the sun and ending with an electron crossing the junction. 16. Many people have an instinctive reaction to discount the « 20% PV panel efficiency as disappointingly low-perhaps thinking they should hold out for higher. Present a multi-point argument about why the efficiency is actually pretty good, and why in practice it is plenty good enough to be practical. 17. If aiming for a particular power output109 from a PV array, describe explicitly/quantitatively how PV panel efficiency interacts with the physical size (area) of the array. For instance, what happens if the efficiency doubles or is cut in half, while keeping the same target output? 18. Make the connection between Figure 13.4 and Figure 13.6 by drawing a zoom-in of the bottom left corner of one of the cells in Figure 13.6. 19. Figure 13.7 shows operational curves of a PV cell for different levels of illumination. If the illumination is low and the panel continues to operate at maximum power,[^110] which changes the most compared to full-sun operation: the voltage or the current? Why might lower light (fewer photons) directly connect to a lower current based on the physics of PV operation? 20. Replicate the calculation on page 206 (showing work) that starts with the surface of the sun being 5,770 K and finds that we receive 1,360 $W/m^2$ at Earth. 21. According to Figure 13.8, which continent appears to have the most solar potential? How would you rate China? Do the largest populations and/or largest energy consumers in the world tend to be well-aligned to the best solar resources? 22. Examine Figure 13.9 to determine the insolation at the “four corners” location where Arizona, New Mexico, Utah, and Colorado touch. Express this in both kWh/$m^2$/day and in $W/m^2$, showing how to convert from one to the other. 23. What is a typical value for hours of full-sun-equivalent111 of solar exposure in the U.S. based on the map and native units in Figure 13.9? Explain how you arrive at this. 24. A 30 year study by the National Renewable Energy Lab112 indicates that in San Diego, a typical year delivers an annual average of 5.0 kWh/$m^2$/day of insolation for a flat panel facing straight up. Convert this to $W/m^2$. 25. The same study mentioned in Problem 24 finds that worst year in San Diego delivered an annual average of 4.7 kWh/$m^2$/day 109: Make up your own number if it helps. 110: . largest rectangle that fits in curve Just using this location as an unambiguous spot on the map. 111: … full sun meaning 1,000 $W/m^2$ 112: … called the Redbook study: ⁶⁰

and the best gave 5.2 kWh/$m^2$/day for a flat horizontal panel. What, then, is the range of annual average insolation values in units of $W/m^2$ for San Diego, and what percentage variation is this, roughly (in round numbers)? 26. The study from Problem 24 finds that a flat panel facing south and tilted at various angles113 relative to the horizontal produce the following annual average yields in units of kWh/$m^2$/day: Angle Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Year

225 113: where we use to represent the site latitude 32.7° for San Diego 0° 3.1 3.9 4.9 0-15° 4.1 4.8 5.6 6.1 6.3 6.4 6.9 6.5 5.4 4.4 3.4 2.9 5.0 6.4 6.3 Ꮎ 4.7 5.3 5.8 6.3 +15° 5.1 5.5 5.7 5.9 5.9 5.2 90° 4.5 4.3 3.9 3.2 2.4 6.3 6.8 6.7 6.0 5.3 4.5 3.9 5.6 5.8 6.4 6.5 6.1 5.7 5.1 4.6 5.7 5.1 5.6 5.9 5.8 5.8 5.4 5.0 5.5 2.1 2.2 2.9 3.6 4.4 4.6 4.5 3.5 What tilt delivers the best yield for the year, and how much better is this (in percent) than a flat plate facing straight up? What tilt appears to result in minimal seasonal variation? 27. What if you adjusted the tilt of panels throughout the season to maximize yield? Reproduce the table above, but only writing in the highest number for each month.114. What average does this track produce for the year, and how much improvement (in percent) does this represent compared to the best fixed-tilt performance? 28. If typical insolation is 200 $W/m^2$, how much land area would be needed for a 15% efficient flat PV array supplying an average of 10 kW of power-which is the U.S. individual share? If arranged in a square, how large is the side-length of this array? Compare its size or area to something familiar. 29. If typical insolation is 200 $W/m^2$, how much land area would be needed for a 15% efficient flat PV array supplying an average of 10,000 W for every person in the U.S. (population 330 million). If arranged in a square, how large is the side-length of this array? Draw it on top of a state of your choice, to scale. 30. Based on what is presented in the text,[^115] why is solar power still such a minor player if it is so hugely abundant and the technology has been around for a long time? What are some of the challenges? 31. You look at a PV panel and mentally estimate it to measure about 0.8 m by 1.5 m. Knowing that PV panels tend to be 15-20% efficient, you guess that it is 18% efficient. How much power would you expect it to deliver in full sun (1,000 $W/m^2$ incident)? 32. According to the table in Problem 26, San Diego can expect an annual average solar yield of 5.7 kWh/$m^2$/day when the panel is tilted to the site latitude and facing south.[^116] If a household seeks to produce a modest 8 kWh per day using 16% efficient panels,

114: The pattern will “graph out” the tilt adjustments over time. This exercise provides insight into PV at a personal-scale that would cover an American’s total share of energy demand across all sectors. This exercise provides insight into the total PV area needed to cover America’s total energy demand across all sectors. 115: also fine to bring in prior/outside knowledge 116: … absent any shadows, of course

226 how large will the array need to be? Express as an area in square meters, and in side length for a square of the same area. 33. Using the parameters from Problem 32, interpreting the solar yield as daily hours of peak-sun-equivalent (at solar exposure of 1 k$W/m^2$) what should the array size be in terms of peak Watts in order to deliver 8 kWh per day? How much would the system cost to install at $3.00 per Wp? 34. One way to look at solar payback time time is to note that an installed system will cost something like $3,000 for each kWp (peak capacity), and that you’ll produce x kWh from that 1 kW, array if your region gets x hours of full-sun-equivalent on average. Since each kWh of electricity costs something like $0.15, it becomes straightforward to compute the value per day as $0.15x, and determine how long to match the $3k investment. The result is independent of the actual array size, depending only on the cost per Wp, the solar yield at your location, and the cost of electricity. What would the payback time be, in years, if the cost is $3/Wp p the yield is 6 hours per day of full-sun-equivalent, and electricity in your region costs $0.15/kWh? 117 35. From Table 13.3, compute the capacity factors for the countries listed, and for the whole world, based on average vs. installed power. What is a characteristic range of numbers, and why 118 is it so low? Which country does the best, and which does the worst? What clues does Figure 13.8 offer as an explanation? 119 36. How much power would a large window measuring 2 m wide and 1.5 m tall admit if the sun were shining straight119 into the window from a cloudless sky? How many kilowatt-hours does this translate to over a four hour period, and how much is it worth monetarily compared to electricity at $0.15/kWh? 37. Solar photovoltaics are practical for individual homes, but solar thermal is only to be found in large utility-scale installations. What is the practical reason why we should not expect solar thermal installations on peoples’ rooftops for electricity generation? 38. Solar thermal has a fairly low efficiency in terms of land area of about 3%, compared to 15-20% for PV. Many would shake their heads and say that’s too low to be of any use. What is the counter-argument that it may be fine? It is a good practice to round the final size up a bit to make sure additional inefficiencies do not prevent reaching the goal. 117: … consistent with the first sentence 118: Hint: a large part is insolation vs. overhead sun. 119: ..e.g., what is the most one might expect for direct sun?