# On Photovoltaic Generators and Scattering Cross Sections

Subtitle: Dimensional Analysis again.

Our photovoltaic generator has about 5 kW rated ‘peak’ power – 18 panels with 265W each.

South-east oriented part of our generator – 10 panels. The remaining 8 are oriented south-west.

Peak output power is obtained under so-called standard testing condition – 1 kWp (kilo Watt peak) is equivalent to:

• a panel temperature of 25°C (as efficiency depends on temperature)
• an incident angle of sunlight relative to zenith of about 48°C – equivalent to an air mass of 1,5. This determines the spectrum of the electromagnetic radiation.
• an irradiance of solar energy of 1kW per square meter.

Simulated spectra for different air masses (Wikimedia, User Solar Gate). For AM 1 the path of sunlight is shortest and thus absorption is lowest.

The last condition can be rephrased as: We get 1 kW output per kW/minput. 1 kWp is thus defined as:

1 kWp = 1 kW / (1 kW/m2)

Canceling kW, you end up with 1 kWp being equivalent to an area of 1 m2.

Why is this a useful unit?

Solar radiation generates electron-hole pairs in solar cells, operated as photodiodes in reverse bias. Only if the incoming photon has exactly the right energy, solar energy is used efficiently. If the photon is not energetic enough – too ‘red’ – it is lost and converted to heat. If the photon is too blue  – too ‘ultraviolet’ – it generates electrical charges, but the greater part of its energy is wasted as the probability of two photons hitting at the same time is rare. Thus commercial solar panels have an efficiency of less than 20% today. (This does not yet say anything about economics as the total incoming energy is ‘free’.)

The less efficient solar panels are, the more of them you need to obtain a certain target output power. A perfect generator would deliver 1 kW output with a size of 1 m2 at standard test conditions. The kWp rating is equivalent to the area of an ideal generator that would generate the same output power, and it helps with evaluating if your rooftop area is large enough.

Our 4,77 kW generator uses 18 panels, about 1,61 m2 each – so 29 m2 in total. Panels’ efficiency  is then about 4,77 / 29 = 16,4% – a number you can also find in the datasheet.

There is no rated power comparable to that for solar thermal collectors, so I wonder why the unit has been defined in this way. Speculating wildly: Physicists working on solar cells usually have a background in solid state physics, and the design of the kWp rating is equivalent to a familiar concept: Scattering cross section.

An atom can be modeled as a little oscillator, driven by the incident electromagnetic energy. It re-radiates absorbed energy in all directions. Although this can be fully understood only in quantum mechanical terms, simple classical models are successful in explaining some macroscopic parameters, like the index of refraction. The scattering strength of an atom is expressed as:

[ Power scattered ] / [ Incident power of the beam / m2 ]

… the same sort of ratio as discussed above! Power cancels out and the result is an area, imagined as a ‘cross-section’. The atom acts as if it were an opaque disk of a certain area that ‘cuts out’ a respective part of the incident beam and re-radiates it.

The same concept is used for describing interactions between all kinds of particles (not only photons) – the scattering cross section determines the probability that an interaction will occur:

Particles’ scattering strengths are represented by red disks (area = cross section). The probability of a scattering event going to happen is equal to the ratio of the sum of all red disk areas and the total (blue+red) area. (Wikimedia, User FerdiBf)

# Rowboats, Laser Pulses, and Heat Energy (Boring Title: Dimensional Analysis)

Dimensional analysis means to understand the essentials of a phenomenon in physics and to calculate characteristic numbers – without solving the underlying, often complex, differential equation. The theory of fluid dynamics is full of interesting dimensionless numbers –  Reynolds Number is perhaps most famous.

In the previous post on temperature waves I solved the Heat Equation for a very simple case, in order to answer the question How far does solar energy get into ground in a year? Reason: I have been working on simulations of our heat pump system since a few years. This also involves heat transport between the water/ice tank and ground. If you set out to simulate a complex phenomenon you have to make lots of assumptions about materials’ parameters, and you have to simplify the system and equations you use for modelling the real world. You need a way of cross-checking if your results sound plausible in terms of orders of magnitude. So my goal has been to find yet another method to confirm assumptions I have made about the thermal properties of ground elsewhere.

Before I am going to revisit heat transport, I’ll try to explain what dimensional analysis is – using the best example I’ve ever seen. I borrow it from theoretical physicist – and awesome lecturer – David Tong:

How does the speed of a rowing boat depend in the number of rowers?

References: Tong’s lecture called Dynamics and Relativity (Chapter 3), This is the original paper from 1971 Tong quotes: Rowing: A similarity analysis.

The boat experiences a force of friction in water. As for a car impeded by the friction of the surrounding air, the force of friction depends on velocity.

Force is the change of momentum, momentum is proportional to mass times velocity. Every small ‘parcel’ of water carries a momentum proportional to speed – so force should at least be proportional to one factor of v. But these parcel move at a speed v, so the faster they move the more momentum is exchanged with the boat; so there has to be a second factor of v, and force is proportional to the square of the speed of the boat.

The larger the cross-section of the submerged part of the boat, A, the higher is the number of collisions between parcels of water and the boat, so putting it together:

$F \sim v^{2}A$

Rowers need to put in power to compensate for friction. Power is energy per time, and Energy is force times distance. Since distance over time is velocity, thus power is also force times velocity.

So there is one more factor of v to be included in power:

$P \sim v^{3}A$

For the same reason wind power harvested by wind turbines is proportional to the third power of wind speed.

A boat does not sink because downward gravity and upward buoyancy just compensate each other; buoyancy is the weight of the volume of water displaced. The heavier the load, the more water needs to be displaced. The submerged volume of the boat V is proportional to the weight of the rowers, and thus to their number N if the mass of the boat itself is negligible:

$V \sim N$

The volume of something scales with the third power of its linear dimensions – think of a cube or a sphere; so the surface area scales with the square of the length, and the cross-section A scales with V – and thus with N:

$A \sim N^{\frac{2}{3}}$

Each rower contributes the same share to the total rowing power, so:

$P \sim N$

Inserting for A in the first expression for P:

$P \sim v^{3} N^{\frac{2}{3}}$

Eliminating P as it has been shown to be proportional to N:

$N \sim v^{3} N^{\frac{2}{3}}$
$v^{3} \sim N^{\frac{1}{3}}$
$v \sim N^{\frac{1}{9}}$

… which is in good agreement with measurements according to Tong.

Heat Transport and Characteristic Lengths

In the last post I’ve calculated characteristic lengths, describing how heat is slowly dissipated in ground: 1) The wavelength of the damped oscillation and 2) the run-out length of the enveloping exponential function.

Both are proportional to the square root of a simple number:

$l \sim \sqrt{D \tau}$

… the factor of proportionality being ‘small’ on a logarithmic scale, like π or 2 or their inverse. τ is the period, and D was a number expressing how well the material carries away heat energy.

There is another ‘simple’ scenario that also results in a length scale described by
$\sqrt{D \tau}$ times a small number: If you deposit a confined ‘lump of heat’, a ‘point heat’ it will peter out and the average width of the lump after some time τ is about this length as well.

Using very short laser pulse to heat solid material is very close to depositing ‘point heat’. Decades ago I worked with pulsed excimer lasers, used for ablation (‘shooting off) material from ceramic targets.This type of lasers is used in eye surgery today:

Heat is deposited in nanosecond pulses, and the run-out length of the heat peak in the material is about $\sqrt{D \tau}$ with tau being equal to the very short laser’s pulse length of several nanoseconds. As the pulse duration is short, the penetration depth is short as well, and tissue is ‘cut’ precisely without heating much of the underlying material.

So this type of $\sqrt{D \tau}$ length is not just a result of a calculation for a specific scenario, but it rather seems to encompass important characteristics of heat conduction as such.

The unit of D is area over time, m2/s. If you accept the heat equation as a starting point, analysing the dimensions involved by counting x and t you see that D has to contain two powers of x and one of t. Half of applied physics and engineering is about getting units right.

But I pretend I don’t even know the heat equation and ‘visualize’ heat transport in this way: ‘Something’ – like heat energy – is concentrated in space and closely peters out. The spreading out is faster, the more concentrated it is. A thin needle-like peak quickly becomes a rounded hill, and then is flattened gradually. Concentration in space means curvature. The smaller the space occupied by the lump of heat is, the smaller its radius, the higher its curvature as curvature is the inverse of the radius of a tangential circular path.

I want to relate curvature to the change with time. Change in time has to be measured in units including the inverse of time, curvature is the inverse of space. Equating those, you have to come with something including the square of spatial dimension and one temporal dimension – something like D [m2/s].

How to get a characteristic length from this? D has to be multiplied by a characteristic time, and then we need to take a the square root. So we need to put in some characteristic time, that’s a property of the specific system investigated and not of the equation – like the yearly period or the laser pulse. And the resulting length is exactly that $l \sim \sqrt{D \tau}$ that shows up in any of of the solutions for specific scenarios.

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The characteristic width of the spreading lump of heat is visible in the so-called Green’s functions. These functions described a system’s response to a ‘source’ which resemble a needle-like peak in time. In this case it is a Gaussian functions with a ‘width’ $\sim \sqrt{D \tau}$. See e.g. equation (14) on PDF-page 14 of these lecture notes.