Repetition. Universität Karlsruhe (TH)

Transcription

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4 Repetition 4 Typical Ru complex sensitziers developed by Grätzel s group

5 Course schedule 5 Preliminary schedule 1. Introduction, The Sun 2. Semiconductor fundamentals 3. Solar cell working principles / pn-junction solar cell 4. Silicon solar cells 5. Copper-Indiumdiselenide solar cells 6. Cell optimization and highly efficient device concepts 7. Modules and system integration 8. Organic photovoltaics 9. Dye sensitized solar cells 10. Economics and profitability 11. Other renewable energies 12. Excursion

6 6 Weekend home near Montelimar (France) occupied only during summer. Energy consuming devices: Energy saving lamps, a small TV, a water pump and a energy-saving refrigerator. Daily energy demand: 680 kwh Daily energy production: 4kWh/kW p during summer 0.18-kW p -plant would be fully sufficient (supplemented by a rechargeable battery storage with a capacity of about 280 Ah 12 V DC, enough for 2.5 days.)

7 7 PV system near Cambridge (UK) Nominal output: 1 kw p Daily power: 3.6 kwh (July) Daily power: 0.7 kwh (December) Losses od 24% through cables and battery Availability 3.6 * 0.76 = 2.7 kwh/day (July) Availability 0.7 * 0,76 = 0.5 kwh/day (Dec) Summer: oversized Option: Smaller system plus generator (e.g. biomass) for winter season.

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12 12 Photovoltaics and Renewable Energies are growing markets. Rentability of a photovoltaic power plant? Cost efficiency is regulated by the market -> what is the cheapest way to generate e.g. electricity Unit for cost efficiency is /kwh η ( /kwh) = Investment + Maintainance costs Total generated kwh

13 13 Investments + Ground!

14 14 Total generated kwh Solar module efficiency Lifetime (20-30 a) Losses upon electrical conversion (-10% rel.) Solar irradiation / weather Orientation and Tilt Material Mono-Si Poly-Si a-si η (Lab) approx. 24 approx. 18 approx. 13 η (Fab) 14 to17 13 to15 5 to7

15 15 Assumption: No capital return! Total costs C Ges = A 0 + ΣA i ( ) Investment + annual operating costs Annual costs C a = C Ges / n ( /a) Lifetime n years Energie costs C E = C a / E a ( /kwh) Annual energie amount

16 16 Example: 1kW p photovoltaic generator connected to grid A 0 = 6500 (Year 2001 for PV systems < 5kW p ) A 10 = 1500 (Exchange of defective inverter module) Lifetime 25 years C a = (A 0 + A 10 )/25 = 320 /a Germany: E a = 800 kwh C E = 0,40 /kwh Sahara: E a = 2100 kwh C E = 0,16 /kwh

17 17 Real Life Economy : Investor wants Capital Return as usual in the market (7% in energy markets)! Capital after n years: C n = A 0 *q n with q = 1+p Including maintainance: C n = A 0 *q n + Σ A i *q n-i With A i constant: C n = A 0 *q n + A*(q n -1)/(q-1) (geometric series) Assuming A 0 =4500, A 10 =1000, p=6% (typical data from 2007) Germany: E a = 800 kwh C E = 0,49 /kwh Sahara: E a = 2100 kwh C E = 0,21 /kwh

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19 19 Coil power plant: Gas/steam turbine: C E = 0,05-0,08 /kwh C E = 0,03-0,04 /kwh Space heating (Oil/Gas): Water, domestic (Oil/Gas): C E = 0,06-0,07 /kwh (therm) C E = 0,10 /kwh (therm) Crude Oil Forward curve

20 20 Strong need to reduce costs, at least by factor 3! Thin film technologies (CIGS, a-si, CdTe, Polymers, ) Cheap processing 1. Printing 2. String ribbon (Evergreen Solar, -50%) Higher production volume Higher efficiency and lifetime location (storage, transport!)

21 21 Energy payback time Currently years in central Europe years in southern Europe (Source: EU project crystal clear) Systems, assumption: No maintainance (only 1 inverter) 1.63 years for mono-si 1.67 years for CIS thin film technology 1.78 years for multi-si 1.96 years for a-si Costs: 1/3 module, 1/2 inverter (exchange every 5 years)

22 22 Comissioning solar power in Germany

23 23 Rentability in light of the EEG Requested capital yield = 7,4% (capital return due to energy economy) EEG payoff = 49,21 Ct/kWh Annual maintaince: 1,5% of investment Germany: 800 kwh/kw p Investment limit: 3900 /kw p

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