# Intensity of Sunlight

As we found in the last quiz, sunlight typically reaches Earth’s surface at an intensity of about $$\SI[per-mode=symbol]{1000}{\watt\per\meter\squared}$$.

This isn't very high — to meet the electricity demand of countries we looked at like Costa Rica or Estonia would require tens of square kilometers of collector area.

What if we could increase this intensity to a much higher value, so that we could capture the same amount of total energy flow while using a smaller collector area? In this quiz, we will explore the benefits and limitations of increasing the intensity of sunlight.

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

Suppose you’re designing a solar car that needs $$\SI{1500}{\watt}$$ of electrical power to run (this is about 2 horsepower, so your car wouldn't impress many gearheads). You have access to a solar collector that can convert sunlight to electricity at $$25\%$$ efficiency.

For now, we won't worry about the specifics of how this solar collector converts sunlight to electricity. We'll learn about some real methods for converting sunlight to electricity in later chapters of this exploration.

If the incident sunlight has an intensity of $$\SI[per-mode=symbol]{1000}{\watt\per\meter\squared}$$, how much collector area do you need (in $$\si{\meter\squared}$$) in order to power the car?

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

Let's assume for a moment that we have a way of increasing the solar intensity to $$\SI[per-mode=symbol]{10000}{\watt\per\meter\squared}$$.

With this increased solar intensity, how much collector area (in $$\si{\meter\squared}$$) would be required to power the $$\SI{1500}{\watt}$$ car using the $$25\%$$ efficient collector?

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

Our solar car example shows that increasing solar intensity is helpful because it can reduce the collector area required for a certain power output. One way to increase radiation intensity is with a lens.

If we have a magnifying glass with a radius of $$\SI{10}{\centi\meter}$$ that focuses sunlight to a spot with a radius of $$\SI{1}{\centi\meter},$$ what is the intensity at that spot?

Assume that the intensity incident on the magnifying glass is $$\SI[per-mode=symbol]{1000}{\watt\per\meter\squared}$$ and that the magnifying glass doesn’t absorb or reflect any of the incident solar radiation, it transmits all of it to the spot.

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

We can characterize a method for concentrating sunlight with a concentration ratio. There are two concentration ratios that are commonly used: the area concentration ratio and the flux concentration ratio.

The area concentration ratio $$C_\textrm{area}$$ is the ratio of input area $$A_\textrm{in}$$ to output area $$A_\textrm{out}$$: $C_\textrm{area} = \frac{A_\textrm{in}}{A_\textrm{out}}$ The flux concentration ratio $$C_\textrm{flux}$$ is the ratio of output intensity $$I_\textrm{out}$$ to input intensity $$I_\textrm{in}$$: $C_\textrm{flux} = \frac{I_\textrm{out}}{I_\textrm{in}}$

In the previous magnifying glass example, both the area concentration ratio and the flux concentration ratio were $$100.$$

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

Another way to increase radiation intensity is with a curved mirror that reflects sunlight to a smaller spot. If we have a long curved mirror that is $$\SI{10}{\meter}$$ long and $$\SI{1}{\meter}$$ wide that reflects sunlight to a rectangular spot that is $$\SI{10}{\meter}$$ long and $$\SI{5}{\centi\meter}$$ wide, what is the area concentration ratio?

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

If the mirror reflects $$90\%$$ of the incident sunlight (and absorbs the other $$10\%$$), what is the flux concentration ratio?

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

Since we can use lenses or mirrors to concentrate sunlight, you might wonder: why don’t we always maximize the concentration ratio so we can have minimal solar collector area while still collecting the same amount of energy? The reason why is a bit complicated, and has to do with a concept called conservation of etendue.

Optics are reversible: if you shine ray a light through an optical system, it will trace a certain path and exit at a specific location and angle. If you shine another beam from the other direction at that location and angle, the reverse beam must follow the same path.

This means that there needs to be enough "room" for ray paths to be distinguishable in an optical system. If you reduce the area taken up by a bundle of beams, it is accommodated by increasing its angular spread, and vice versa.

Mathematically, this is expressed in the equation for conservation of etendue: $A_\textrm{out} \sin^2(\theta_\textrm{out}) = A_\textrm{in} \sin^2(\theta_\textrm{in})$ Where $$A_\textrm{in}$$ and $$A_\textrm{out}$$ and the input and output areas of the optical system, respectively, $$\theta_\textrm{in}$$ and $$\theta_\textrm{out}$$ are the input (or acceptance) angle and output angle, respectively. This equation applies for an optic that successfully transmits all light incident on the input area within the acceptance angle to the output area. You can make an optical system with a higher area concentration ratio than this equation allows, but then your system would not transmit all the incident light which you intended to concentrate.

Note, this form of the conservation of etendue equation is simplified, because it assumes the angular distribution is uniform at all inlet and outlet points. This would not be true for most real optics, but this form of the equation is easier to work with than the full form and is accurate enough for the exercises in this course.

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

Suppose we are designing a lens that has an output angle of $$60^\circ$$ and our target concentration ratio is $$10 000$$. What is the maximum input acceptance angle (in degrees) that our lens can have?

Recall that the equation for conservation of etendue is: $A_\textrm{out} \sin^2(\theta_\textrm{out}) = A_\textrm{in} \sin^2(\theta_\textrm{in})$

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

Conservation of etendue also leads to a maximum solar concentration ratio that is achievable on Earth, if you want to concentrate all of the sunlight incident on your concentrator. This is because the Sun appears in the terrestrial sky as a disc with a half angle of approximately $$0.27^{\circ}$$. Considering the apparent size of the Sun and conservation of etendue, what is the maximum concentration ratio that can be achieved on Earth?

Recall that the equation for conservation of etendue is: $A_\textrm{out} \sin^2(\theta_\textrm{out}) = A_\textrm{in} \sin^2(\theta_\textrm{in})$

## Intensity of Sunlight

### Introduction

# Intensity of Sunlight

We can increase the intensity of sunlight using a concentrating lens or mirror, which reduces how much solar collector area we need in order to generate a certain amount of power, but there are limits to how much we can concentrate sunlight. The next quiz will further explore what we need in order to have an operational concentrator, and will introduce additional practical limitations beyond the theoretical limit of solar concentration which we established in this quiz.

## Intensity of Sunlight

### Introduction

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