Download Lab 4. Solar energy

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 Theory
2.1 Heat transport
Heat transport is one form of energy flux (units W/m2). 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⋅10-34Js. 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/m2) of the radiation and the temperature
of the body is given by the Stefan-Boltzmann law:
S=εσT4
with T in K, ε being the emittance of the body (a number between 0 and 1) and
σ=5.67⋅10-8W/(m2K4).
The hotter a body, the shorter the wavelength for which the intensity is largest, as described by
Wien’s principle:
λmaxT = 2.90⋅10-3mK
Plotting the distribution for different temperatures shows on the one hand the fast increase with
T4 (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
q = η0b K b (θ )Gb + η0d Gd − a1 (Tf − Ta ) − a2 (Tf − Ta ) 2
with
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 = η0b Gb − a1 (Tf − Ta )
For the solar collector efficiency η, the ratio between thermal output and solar input, we receive:
q
a
η=
= η0b − 1 (Tf − Ta ) .
Gb
Gb
Exercise: Sketch the efficiency as a function of ∆T /Gb = (Tf − Ta ) /Gb .
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 mC p ∆T V f ρC p (Tout − Tin )
=
=
A
A
A
Vf: volume flow rate in [m3/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-1K-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 dm2 large plates, are 0.5V and 3A with full
irradiation (1000W/m2), 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
F=
Pmax
U 0 I0
A typical value for F is 0.8.
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/m2, solfångarens area är 1 m2. 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.
2.
3.
4.
5.
6.
7.
8.
9.
Studera hur solfångaren är uppbyggd och fungerar.
Studera mätuppställningen
Starta pumparna
Starta ”solen” (halogenlamporna)
Mät förändringen i temperatur i de båda reservoarerna och över solfångaren.
a. Mät temperaturerna var femte minut under 90 minuter.
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.
Beräkna effekten in
Beräkna effekten ut för varje mätintervall
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