Lab 4. Solar energy: Solar Collectors and Photovoltaic Cells

FYSA15 Miljöfysik Lab 4. Solar energy: Solar Collectors and Photovoltaic Cells 1 Introduction This lab shall give an overview over the two main techniques of using the energy coming from the sun: solar collectors and solar cells (photovoltaic cells or PV cells). 2.1 Heat transport 2 Theory Heat transport is one form of energy flux (units W/m 2 ). It can occur via three different phenomena: conduction, convection and radiation. 2.1.1 Conduction In a solid substance the transport occurs through conduction: the atoms and/or molecules move randomly and fast, bounce against each other and thereby cause their neighbours to move and so on. The thermal energy spreads as motional energy through the material not due to particle transport but due to particle interaction. Spontaneous net energy flow always occurs from the higher to the lower temperature; that is the second principle of classical thermodynamics. 2.1.2 Convection Unlike the case of a solid, particles of a fluid can move throughout the material. This means that another transfer of thermal energy, known as convection can take place besides conduction. Convection is an energy flow due to a bulk displacement of a material. 2.1.3 Radiation basic terms To understand what influences the efficiency of a collector or a solar cell, it is important to become familiar with the basics of radiation and its properties. In our case we will be concerned with different parts of the electromagnetic spectrum, in order of decreasing energy: UV, visible light and infrared (IR) radiation. Electromagnetic radiation can be described either as a wave or as a beam of particles, so-called photons. Both models are correct. Which one is used depends on the phenomenon that shall be described: the propagation of electromagnetic radiation (including interference or diffraction) can be conveniently described with the wave model, whereas the particle model helps to describe the interaction with matter such as the photoelectirc effect. The two models are connected: A wave has a wavelength λ and a frequency f. They are connected with the velocity of propagation c according to c = λ f. The energy of a photon is given by E = h f= hc/λ, where h is 1

the Planck constant, h = 6.626 10-34 Js. A higher energy therefore corresponds to a smaller h wavelength and vice versa. The photon carries a momentum p = which is a typical particle λ property. This relation holds for all particles and relates the typical wave property λ with the typical particle property p. The relation was proposed by de Broglie and has no classical analogue, since the Planck constant is a typical feature of quantum physics. Question: Which of the parts of the spectrum that are mentioned above (UV visible, IR) has the largest wavelengths? Which the shortest? All bodies have a temperature higher than absolute zero and therefore continuously emit energy in form of electromagnetic radiation distributed more or less over all wavelengths according to Planck s radiation principle. The relation between the intensity S (power per area, W/m 2 ) of the radiation and the temperature of the body is given by the Stefan-Boltzmann law: S=εσT 4 with T in K, ε being the emittance of the body (a number between 0 and 1) and σ=5.67 10-8 W/(m 2 K 4 ). The hotter a body, the shorter the wavelength for which the intensity is largest, as described by Wien s principle: λmaxt = 2.90 10-3 mk Plotting the distribution for different temperatures shows on the one hand the fast increase with T 4 (area under the curve), on the other hand the shifting of the maximum to smaller wavelengths with higher temperature. Figure 1 Apart from radiating each material has a certain capacity to reflect, transmit or absorb radiation in different fractions, labelled ρ, τ and α. However, conservation of energy requests ρ + τ + α = 1. Different materials or coatings can have different properties concerning reflection and absorption of different wavelengths. Question: The plates used in a solar collector are black. Can you think of a reason, why? 2

2.2 Solar collectors Solar collectors, producing hot water, are maybe the most natural way of making use of solar energy. The basic principle is the following: black plates are heated by the solar radiation and are in contact with a pipe in which a fluid is pumped. The fluid is heated by the plates and in most cases the hot fluid is heat exchanged in a storage tank. Alternatively the fluid can be let into the district heating system. The fluid in the solar collector is most often a water glycol mixture where the glycol prevents freezing in winter. Figure 2 Whenever sun radiation hits a dark plain, most of the energy is converted into thermal energy. However, there are always losses of energy to the surroundings by convection, radiation and conduction. To prevent the irradiated plates (absorber plates) to be cooled by the surrounding air, they are placed in a box with heat isolation at the back and glass in front. Since glass is not completely transparent one part of the incoming light is reflected and another small fraction is absorbed. These losses are called optical. Another optical loss is the reflection at the absorber plates. A different kind of losses are the so-called thermal losses. They are due to the fact that the absorber plates radiate and thus lose energy, the more the hotter they are. #picture, M. Areskoug, Miljöfysik, p123, fig 8.3# 3

Considering all incoming and outgoing energy fluxes, the thermal output of the collector can be written as with q = η 0b K b (θ)g b + η 0d G d a 1 (T f T a ) a 2 (T f T a ) 2 η0b: optical efficiency for beam solar irradiance η0d: optical efficiency for diffuse solar irradiance a1 : constant part of the overall heat loss coefficient a2 : temperature dependent part of the overall heat loss coefficient Gb: beam irradiance on the collector Gd: diffuse irradiance on the collector Tf: mean fluid temperature Ta: ambient temperature θ: angle of incidence Kb(θ): angle of incidence modifier Kb(θ)=1-b0(1/cosθ - 1);θ <60, this is needed because the reflection of the glass depends on the angle of incidence b0: characteristic parameter of Kb(θ) a=a1 + a2(tf - Ta): overall heat loss coefficient Since the solar simulator we are going to use has almost no diffuse radiation Gb can be put to zero. Moreover, we assume the heat loss factor to be independent of the temperature (a 2= 0). We will measure with normal incident, so that K(θ) becomes 1. With these presumptions the equation for the thermal output can be simplified to q = η 0b G b a 1 (T f T a ) For the solar collector efficiency η, the ratio between thermal output and solar input, we receive: η = q G b = η 0b a 1 G b (T f T a ). Exercise: Sketch the efficiency as a function of T /G b = (T f T a ) /G b. How can a1 and η0b be determined from the plot? We can see that the collector operates the worse the hotter it becomes (increasing Tf). The explanation is that the heat losses grow with growing working temperature. One solution to keep the losses small is to try and keep the working temperature low. But this has other consequences that must be taken into consideration when choosing the parameters of a system. Let us, for example, look at a small house where the domestic hot water shall be heated with a collector. One can keep the temperature low and heat a lot of water with a high efficiency to maybe 25 C. But what if people want to take a bath? Then they probably would prefer some 36-40 C. The water would need to be heated up somehow else, maybe with electricity. So it 4

could be more convenient to heat a couple of litres with the collector to this higher temperature, for the price of greater losses. Maybe it is even better to heat a smaller amount (with even greater losses) to an even higher temperature of let us say 60 C. This can then be used for heating and bathing (when mixed with colder water). We can see here how complex the subject becomes in detail. To be able to find the optimum solution for a certain system computer simulation programs are of immense use. The thermal output of the collector gives that fraction of the incoming energy that is actually of use, i. e. which is converted into thermal energy and transferred to the water, heating it. Therefore, we can calculate the thermal output (and thereby the efficiency), by calculating the energy that is transferred to the water per second, the so-called heat flow rate Q. The energy that is needed to warm up a certain mass m of water by a certain difference in temperature T is given by E=mCp T where Cp is the specific heat of water (energy needed to heat 1kg by 1K). The mass, in turn, is given by m=vρ, with ρ being the density of water and V the volume. The rate which heat is transferred with is therefore given by the volume flow rate Vf. The thermal output power determines how much energy (heat) is gained per second with a collector area A: q = Q A = mc p T A = V f ρc p (T out T in ) A Vf: volume flow rate in [m 3 /s] (a very large unit more appropriate for a river) ρ: density of the heating fluid (water, about 990kgm -3 ) Cp: specific heat of water (4180Jkg -1 K -1 ). Tin: temperature of ingoing water Tout: temperature of outgoing water 2.3 Solar cells Another way of making use of the sun s energy is to gain electric energy. This can be done with so-called solar cells or photovoltaic cells, PV. The exact explanation how they work requires quite a bit of understanding of semiconductor physics which is beyond the scope of this experiment. The important thing is, that the incoming photons can give their energy to electrons in the semiconductor material and set them free so that they can move easily. Inside the cell, there is a small electric field (the explanation where this comes from would lead too far, it is semiconductor physics again). In this field, the freed electrons gain electric potential energy, corresponding to a voltage (voltage is energy per charge). This makes it possible for a current to flow in an attached circuit. 5

The processes may be pictured with help of the following simple model: Water is filled into buckets from a reservoir and lifted up step by step to a higher lying point. With each step it is lifted, it gains more (gravitational) potential energy. From that higher point it can be let streaming down again and the stored potential energy can be converted into useful energy, for example by driving a turbine. Typical values for solar cells, that are mostly about 1 dm 2 large plates, are 0.5V and 3A with full irradiation (1000W/m 2 ), which corresponds to about 15% efficiency. Often, several single cells are coupled together to a so-called panel or module. 2.3.1 Current-voltage-characteristic The U-I characteristic of a common voltage source is a straight line and has two main parameters: the open-circuit voltage U0 and the inner resistance Ri. U=U0-RiI U0 gives the voltage when no outer circuit is connected (no current flows) and Ri accounts for the fact, that some of the voltage loss occurs already in the source, decreasing the actual voltage at the poles, when a current is flowing. For a solar cell, the characteristic looks different. It can be approximated with two straight lines: one can be regarded as the characteristic of a normal voltage source, starting at the U-axis at U0 (typically 0.55V for a single cell) and having a small negative slope, corresponding to Ri. If the resistance in the outer circuit is decreased, the current I increases according to Ohm s law and therefore the voltage at the poles U decreases. The current can, however, only grow until a certain maximum value I0. This is due to the fact that the electrons are lifted with the energy of the incoming light. A certain intensity corresponds to a certain number of photons per second, which means a certain number of freed electrons per second that gives a certain current. This current I0 is called short-circuit current, since it flows when the outer resistance is zero. We cannot have a higher value for the current, even if the value I0 is reached already for non-zero resistance. If we reduce the resistance further, the current will not increase, but the pole voltage will decrease since the cell is drained of electrons. Figure 3 6

Normally, one looks at the I-U characteristic. It can be seen that the maximum current is proportional to the incoming intensity. Figure 4 2.3.2 Power The power the cell delivers is given in each point by P=I U, corresponding to the area of a rectangle in the graph. This area varies rather strongly with the resistance one connects to the cell. The optimum point lies somewhere near the kink of the curve. The maximum power delivered by the cell also depends on the intensity of the light, because the maximum current does. The optimum voltage, however, is rather independent of the intensity (as can be seen from the I-U-plot). Therefore, fixing the voltage to a value near the kink helps to get a good value for the gained power at any incoming intensity. This can be done with the help of an accumulator: When the cell can give a greater current than the load resistance needs the surplus current charges the accumulator. On the other hand, if the current from the cell becomes too low, the accumulator can provide an additional current. The pair of I-U values that the cell actually works at is called working point. It is determined by two requirements: on the one hand it has of course to lie on the cell s characteristic. On the other hand the connected (constant) resistance R forces it to lie on a straight line with the equation U=RI, according to Ohm s law. The working point is given by the intersection of the two curves. The crucial thing is now to choose the resistance that way, that makes the working point lie as close as possible at the optimum point. Figure 5 7

Question: Can you imagine why this correlation can make it difficult to succeed with apparently easy experiments with solar cells? Hint: Think of what happens when the intensity changes 2.3.3 Fill factor As indicated above, an ideal solar cell should have a completely rectangular characteristic with the maximum power P=U0I0. The better the solar cell, the nearer its characteristic comes that shape. A measure for the quality is therefore the so-called fill factor F, which is given by A typical value for F is 0.8. F = P max U 0 I 0 Förberedelseuppgifter 1. En solfångare värmer 20 liter vatten under 5 timmar, från 15 C till 40 C. a. Hur stor är ökningen i termisk energi (inre energi) i vattnet (svara i joule)? b. Hur stor är uppvärmningseffekten (nettoeffekten) (svara i watt)? c. Effekten in mot solfångaren är 200 W/m 2, solfångarens area är 1 m 2. Hur stor blir verkningsgraden? 2. Två voltmetrar är kopplade till en solcell som bestrålas med ljus från en halogenlampa. Strömmen är 0,5 A och spänningen är 5 V, denna ström och spänning ger den största uteffekten, Pmax. a. Hur stor är uteffekten (svara i watt)? b. U0 är 7,5 V och I0 är 0,6 A, vad blir fyllnadsfaktorn? 8

3 Experiment Solfångare Denna del går ut på att undersöka hur en solfångare fungerar och hur stor andel av det inkommande ljuset som kan omvandlas till nyttig värme. Uppställningen består av en liten riktig solfångare med halogenlampor som solersättning. 1. Studera hur solfångaren är uppbyggd och fungerar. 2. Studera mätuppställningen 3. Starta pumparna 4. Starta solen (halogenlamporna) 5. Mät förändringen i temperatur i de båda reservoarerna och över solfångaren. a. Mät temperaturerna var femte minut under 90 minuter. 6. Mät instrålningsintensiteten med pyranometern på olika delar av solfångaren. OBS! Håll pyranometern med ett fast grepp i glaset över den bestrålade ytan. Tag ett medelvärde. 7. Beräkna effekten in 8. Beräkna effekten ut för varje mätintervall 9. Beräkna verkningsgraden som funktion av tiden. Gör följande diagram: Temperaturen som funktion av tiden Verkningsgraden som funktion av temperaturskillnaden mellan solfångaren och rumstemperaturen Diskutera: Hur utvecklas verkningsgraden för solfångaren med tiden, och varför? Solceller Solceller omvandlar solljus till elektrisk energi. Experimenten går ut på att undersöka vilken verkningsgrad solcellen har samt undersöka hur den uppför sig för olika belastning och instrålning. 1. Studera uppställningen, se Fig. 6 2. Starta solen (halogenlamporna) 3. Mät instrålningsintensiteten, I, över solcellerna med pyranometern. Tag ett medelvärde. Beräkna instrålningseffekten, Pin=I*A, där A är solcellernas area. 4. Variera resistansen, R, och mät spänningen och strömmen. Ta extra många mätvärden där värdena varierar kraftigt. 5. Bestäm Pmax 6. Beräkna fyllnadsfaktorn för solcellen 7. Beräkna verkningsgraden vid maxeffekten, Pmax, d.v.s. Pmax/Pin 9

8. Upprepa för ett annat värde på strålningsintensiteten. Enklast är att ändra avståndet mellan halogenlamporna och solcellen Gör följande diagram: Strömmen som funktion av spänningen Uteffekten som funktion av resistansen Figur 6. Mätuppställning för solcell Diskutera: Tolka diagrammen och kommentera resultaten Vad kan man säga om effektens beroende av resistansen? Hur skiljer sig diagrammen åt för de olika strålningsintensiteterna? Slutdiskussion Jämför de olika metoderna att använda solljuset. Är det någon skillnad i verkningsgrad? Kan man förstå detta på något sätt? Vid energiomvandlingar finns nästan alltid förluster. Vilka är de i de olika fallen och hur skulle de kunna minimeras? Vilka osäkerheter förekommer i mätningarna? 10