Energia Solar Fotovoltaica Problem Sheet 1 - Solutions (Ver. 2d: )

Transcription

1 1 harging a mobile hone with small V anel 11 The examle here uses 1800 mah caacity tery This caacity rating means that the tery can deliver 1800mA for 1h, or 900 ma for h, or 600mA for 3h, etc i / s it s s To know the energy which the tery can deliver, we are missing the voltage V A tyical lithium ion tery for mobile devices works at 37 V Knowing the caacity=1800mah and the Voltage=37V we can work out the energy contained in the tery However, this rating only tells us how much HARG the tery can deliver It does not tell us the energy How do we go about solving the roblem? e know that: Vi is ower, i is current and V is voltage e also know that: t here is energy and t is time Hoefully this equation is well known by you ith these two equations we can write: Vit Notice that we have current x time (it) in the equation above, which we can be defined as the caacity of the tery, ie the charge it can deliver If you look at the units used for current (oulomb er second) and time (in seconds), this is clearly evident V 37V it 1800mAh mVAh 6660mVAh Vit Vit 37V 1800mAh 6660mVAh Are you confused by the units? You are well aware that a good measure of electrical energy for ractical uroses is not Joules but something like or h Do you remember what these values actually mean? Remember that the units of ower are or J/s o why is Vi? e have looked at the units of i, so let us look at the units of V and the definition of voltage "The voltage between two oints is equal to the work done er unit of charge against a tic electric field to move the charge between two 1

2 oints and is measured in units of volts (a joule er coulomb)" ie V J / V i J s J s If we go back to our answer of the energy stored in the tery of =6660mVAh we now know we can write: That is: mVAh 666VAh VAh J h h J s h 6 66h Knowing the energy that can be stored in the tery we now want to work out how long our solar anel will take to charge the tery Our ule has an of 50cm and an efficiency of 15%, that is, it converts 15% of the incident ower onto in into electricity To know the actual incident ower on our ule (inc), we need to take into account its A: A 50cm m inc cm cm 5000m inc inc 1000 / cm 5 A 100m / cm Knowing the incident ower on the ule, and the ule's efficiency, we can now calculate how much ower the ule roduces (out) 5 15% out out out inc inc 075 To charge the tery, the ule has to deliver 666h Thus the iece of information missing to comlete our task is; "that is how long it takes to charge the tery?" can be found by rearranging the equation below and calculating: Tyically, a ule's characteristics are defined under ndard conditions Normally the ndard incident ower (or eak ower) is =1000/m

3 t t out 666h h t 888h 075 t 888h cycle cycle cycle m m insolation, insolation, cm m 95/ Next, the ule can convert 15% of this incident insolation to useful electrical energy: 15% 95 rod, rod, 14 insolation, Always take notice of how the units lay out In this case, the cancel to leave time in h 13 The question is best worded in terms of how often could you could charge your hone, and will be changed for the future e have an annual insolation of: insolation 1900/ m / That is, over the course of one, 1900 of energy is incident on 1m e must now undernd how much of the solar insulation we can convert into electrical energy, and for that we require the ule efficiency =15% and ule A=50cm e now know how much energy the ule can deliver over one, and we know the energy we require to charge the tery As such, we can now work out how many n times we could charge the tery with our V charger over the course of one rod, 666h / cycle n h 666h cycle 666h 10cycle n 10cycle 14 rod, Firstly the insolation on an of 50cm (the ule ) is: 3

4 1 In this question we have insolation defined using other units: 15//, that is the amount of energy a system will roduce over the course of the for every of ower inlled Thus a =00 system will over the course of a roduce: As a V system is tyicall guaranteed now for 30 s; 300/ Now the question tes that the cost of electricity goes u exonentially with an increment of % annually 15/ 15 00/ insolation rod rod 00 15/ insolation / / / To discuss the savings we have to think carefully about what how "self-consumtion" can occur You will notice that here we have deliberately chosen a small size system of 00 Let us assume that our baseload is over 00 whenever the V system is active (ie during daylight hours) As such we can then assume that all the electricity roduced by the ules (rod) is consumed and thus is a saving from the grid (d), ie we are not consuming electricity costing for each unit of energy 015 / rod rod 300/ 300/ A system with an exonential growth described by a % annual increase x, the value for the n can be found using the exression: Or in our case: T T c / c / y ( n) y n1 n 1 1 x ( n) 15c / ( n) 15c / 1 % n 10 1 n1 For your calculations, you will need to build a table showing the cost of electricity for each Year n Tariff c / Year n Tariff c / =15c /*10 n = onsidering that the self-consumtion for each remains connt, but the cost of electricity increases, to find the saving we must sum the savings for each individual 4

5 accumulated saving ( ) n T n 15c / n / And so the savings for the 30 s is: n T n 15c / 10 n30 n1 n30 n 15c / 10 n1 n1 300/ n1 300/ Because the value of the quantity of electricity self-consumed each does not vary, we can write: 300/ n 30 n1 15c / 10 n1 inst / The accumulated saving on can be calculated and the easiest way to visualise it is in a grah (n) And then erforming the calculation: 3 n n c / 10 n / 608 / One way of looking at this roblem is to visualise how long it takes for the savings to be savings are equal to the cost of inllation inst n1 n The investment for a 00 system, is if the inllation cost is /; inst where the blue line is the -on- accumulated saving, and the red line the cost of inllation From the grah we can see that the ayback time is just over 8 s 4 The oints to think about are: ower inlled, and hat is the exected temoral rofile for the ower consumtion? o Is it continuous? o an we consume everything we roduce all the time? Under what circumnces is the "selfconsumtion" law a viable business otion? 5

6 3 31 hat do we know: Fuel cost: fuel, 15 / Fuel energy density: fuel, 10 / ngine mechanical eff: 5 eng % ngine electrical eff: 70% alter A car engine has 5 % mechanical efficiency, that is how efficient it is converting the chemical energy of the fuel into mechanical energy ouled to a car engine is what is usual called an "alternator" which converts mechanical energy to electrical energy hat we want to know: elec, / elec 15 / 175 / elec, elec, elec elec, 086 / How does this comare with the cost of electricity you ay at home? 3 Let us rt by working out the useful energy electrical energy one can extract from our fuel source, ie elec, elec, eng 10 5% 70% elec elec, 175 Knowing how much electrical energy we can extract er unit of and knowing the cost of unit of the we can now work out the unit cost of the energy alt The question asks to comare the cost of the electricity roduced by the engine of the car to the cost of electricity roduced by a V system that has a of only 7 s The reason for this is that the idea is to comare the cost of electricity roduced by a solar anel mounted on the car As such, one can exect that the V system only last as long as a tyical car, and that the inllation cost be higher (5 /) To calculate the cost of a unit of energy roduced by a system we need to know, how much energy a system roduces over its, and cost of that inllation The inllation cost is given as: 6 / inst 5

7 e want to find the cost of unit of energy elec, in / V, Having the cost of inllation we now need to work out how much this will roduce over its e are given that 1 will roduce 15 on an annual basis, that is:, 15 / / Knowing how much is roduced in one, we can now work out how much is roduced over a t of 7 s:,,, t 15 7 e now know how much will roduce over a, and we know how much it costs to inll that 1 and as such we can work out the cost er unit of energy roduced 33 5 / 105/ hat we know: elec, V, elec, V,, inst 048 / D t elec avg 300 0,000km/ 7 50km/ h Our vehicle when moving requires elec 300 To undernd how much it consumes over a of the vehicle we need to know how far it moves D and for how long t on, so: D 140,000km D t on 0,000km/ 7 140,000km D 140,000km 50km/ h t 800h on avg Knowing how long t on the electrical ower elec is used, we can work out how much electrical energy elec is consumed: h elec elec elec t on 840 Knowing the electrical energy consumtion we can now work out how much this elec electricity costs: elec elec elec, hich, will deend on its source, V or fuel/engine ( elec, V,, elec, ) 7

8 elec, V, / elec, V, elec 403 elec, V, 4 41 hat we know: elec, / elec, elec 7 elec, It is then evident that the savings will be the difference between the costs: savings savings 319 elec, elec, V, BO, 15% 00 1 / That is, the ule efficiency inllation er unit, the cost of and the cost of inllation er unit of ower inlled To a first order aroximation, the BO scales with the of inllation, and so to decrease this associated cost, it is advantageous to minimise the of inllation This is only ossible (if we maintain the same ower) by increasing the efficiency of the ules so that er unit they roduce more As such we need to know how much ower we are inlling er unit and its associated cost, V The ower er unit, can be worked out by taking into consideration the ule efficiency and undernding that the rating is the ower a ule will roduce under ndard conditions, ie when 1000 m inc, / :, 15% 1000 inc, 8, 150

9 e now know how much ower is inlled er unit, and we know the cost of each, unit of inlled ower 1 /, and thus can work out the cost of the ules er unit :, 150, 1 /, / That is, our V ule technology cost er unit of ower has to be less than: 3 /,, 150 And so the inllation cost er unit is the sum of the BO and the ule inllation: 150, BO, 00 Now we can work back on the basis of how the arameters are defined:,,,,,,,, BO,,, BO, 350 And remembering that we know: 4 In this case we have a technology who's ule cost is significantly lower, at 0 m, / rather than 150 For this ule technology to be cometitive, its cost has to be less or equal than that we calculated beforehand However, because the ule cost is lower er unit, we can get away with inlling a greater of ules As such, we have to revert to calculating on a ower basis and not basis Knowing the ower inlled er unit and the cost of er unit work out the cost er unit of ower:, we can BO,,, 00 Next we can substitute in for,,, 0,,, : inc, BO, inc, e now have an exression with only one unknown (by definition 1000 m inc, / ), that is efficiency e can therefore now use the condition for minimum cost to define the minimum efficiency requirement: 9

10 , 3 / / ,, 3 / BO, inc, BO, inc, 3 / 5 510, rod, 10 1 / / 15/ / cons inst 15% 1000 cons h / inc, Moral of the story is: even if we someday invent a very chea way to manufacture ules, we still have to find ways of decreasing the cost of inllation of the suorting systems A inst, m 0km Think about how this comares to the of ortugal? It is a small fraction However, what would the investment cost be? How does this comare to GD? 10