Outline. Solar Photovoltaic Applications. Midterm Problem One. Midterm Results. Midterm Problem Two. Midterm Problem One II

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1 Photovoltaic Solar April 6, 010 Solar Photovoltaic Application Larry Cartto Mchanical Enginring 483 Altrnativ Enrgy Enginring II April 6, 010 Outlin Midtrm xam rult Mt on Wdnday at PPM Confrnc room for photovoltaic tour Photovoltaic baic Application of photovoltaic ytm Cot and cll fficinci Concluion Midtrm Rult Numbr of tudnt: 30 Maximum poibl cor: 90 Man: 43.8 Mdian: 43 Standard Dviation: Midtrm Problm On Givn: Spctral miivity Find: Solar radiation aborbd αolar = εolar = 0.88[ f ( λ = 0.88 Emiivity μm Wavlngth (μm) T = 0,300μm K) 0] + 0.4[ 1 f ( λt = 0,300μm K) ] ( ) + 0.4( ) = For 800 W/m incoming olar radiation th aborbd radiation i (0.8710)(800 W/m) = W/m. b. If th urfac i opaqu, what happn to th radiation that i not aborbd? It i rflctd. 4 Midtrm Problm On II c. What i th miiv powr of th urfac if it tmpratur i 340 K? α340 K = ε 340 K = 0.88[ f ( λt = 1,190 μm K) 0] [ 1 f ( λt = 1,190 μm K) ] = 0.88( ) + 0.4( ) = x10 W E = εσt = ( 0.413) ( 340K ) 4 E = 18.8 W/m 4 m K d. What i th total amount of radiant nrgy laving th urfac? It i um of th miiv powr, 18.8 W/m, plu rflctd nrgy, 800 W/m W/m = 103. W/m. Total radiant nrgy laving th urfac i 86.1 W/m. 5 Q top Midtrm Problm Two Givn: problm on urfac i aborbr plat in olar collctor at 340 K. Gla plat ha T = 30 K and ε = h gap = 3.5 W/m K. Find: hat tranfr, pr unit arabtwn th aborbr plat and th lowr gla plat? 4 4 Acσ ( ) ( TP Tg ) Qtop 3.5W = hp g Ac TP Tg + = ( 340 K 30 K ) Ac m K ε P ε g x10 W 4 4 [( 340 K ) ( 30 K ) ] 4 m K 107.8W + = m ME 483 Altrnativ Enrgy Enginring II 1

Photovoltaic Solar April 6, 010 Midtrm Problm Two II I Q in part a l than, gratr, than or qual to Q top? It i qual Find: hat tranfr cofficint, U P-g, btwn th aborbr plat and th lowr gla plat?

It i gratr; it ha th am hat flow, Q top, with a mallr tmpratur diffrnc 7 Midtrm Problm Thr Givn: olar collctor rating ht.

2 Photovoltaic Solar April 6, 010 Midtrm Problm Two II I Q in part a l than, gratr, than or qual to Q top? It i qual Find: hat tranfr cofficint, U P-g, btwn th aborbr plat and th lowr gla plat? Qtop 107.8W Qtop A c m 5.39W U P g = = = = A T T T T 340 K 30 K m c ( ) ( ) ( ) K P g P g I U P-g l than, gratr, than or qual to U top? It i gratr; it ha th am hat flow, Q top, with a mallr tmpratur diffrnc 7 Midtrm Problm Thr Givn: olar collctor rating ht. Find: fficincy for T in = 40 o C, I = 800 W/m and ambint tmpratur = 0 o C P P 0 0 η = = = 6.1% I I Find: T out for watr uing tt flow rat from ht and prviou data 1 Qu ηia c o m W o Tout = Tin + = Tin + = 40 C + = C mc & p mc & 3 p 48mL 1000kg m o m 10 ml kg C 800W ( 6.1% ) (.411m ) 8 Midtrm Problm Thr II Find: fraction of 90 thrm/month from collctor for olar radiation = 4 kwh/m /day for 30-day month with τα/(τα)n = 0.94, T a = 17.8 o C, and F R /F R = F R U c = ngativ lop = W/m o C and F R (τα)n = intrcpt = 0.715; 90 thrm/mo = (90x10 5 Btu/thrm)(1.055 k/btu)(g/10 6 k) = G/mo. H total = (30 day) (4 kwh/m /day) ( G/kWh) = 0.43 G/m /month. F R U c = (3.975 W/m o C)(10-9 G/W )(3600 /hr)(4 hr/day)(30 day/mo) = G/m /mo, 9 Midtrm Problm Thr III G mo o o ( T T ) = (.411m ) ( 0.97) ( 100 C 17.8 ) = ' F Δt X = A F U R c R c rf a C FR D m mo G ' F τα H, Y = A F ( τα) R i total c R n = FR ( τα) n D 0.43G (.411m ) ( 0.715)( 0.97)( 0.94) m mo = G mo f = 1.09Y 0.065X 0.45Y X Y 3 = 1.09(0.0715) 0.065(0.086) 0.45( ) (0.086) (0.0715) 3 = 5.89% of th monthly hating load i 10 upplid by olar Wdnday Cla Mting Mt in PPM Confrnc room Entranc from Lot C6 Nar cornr of Haltd and Etiwanda 11 Slid with NIST logo hown hr takn from thi prntation 1 ME 483 Altrnativ Enrgy Enginring II

Photovoltaic Solar April 6, 010 Cll in Modul in Array Solar cll Solar modul Solar array 13 14 Baic Concpt Dirct convrion of light nrgy into lctrical currnt Bad on adorption of light photon with nrgy

6607x10-3 Light pd: c = 99,79,458 m/ frquncy( -1 ) = ν wavlngth (m) = λ E = hν alo bai for pctrum of blackbody radiation 15 Smiconductor Conductor hav lot of fr lctron to conduct hat and lctricity

Silicon ha 4 lctron in valnc band Adding phophoru with 5 lctron in it valanc band mak it air for th dopd matrial to rla fr lctron Thi i calld an n-typ matrial Adding aluminum with 3 lctron in it

3 Photovoltaic Solar April 6, 010 Cll in Modul in Array Solar cll Solar modul Solar array Baic Concpt Dirct convrion of light nrgy into lctrical currnt Bad on adorption of light photon with nrgy E = hν = hc/λ Planck contant: h = x10-3 Light pd: c = 99,79,458 m/ frquncy( -1 ) = ν wavlngth (m) = λ E = hν alo bai for pctrum of blackbody radiation 15 Smiconductor Conductor hav lot of fr lctron to conduct hat and lctricity Inulator hav almot no fr lctron Pur miconductor can produc om fr lctron laving lctrondficint atom known a hol Dopd miconductor add matrial with diffrnt numbr of valnc lctron 16 Smiconductor II Silicon ha 4 lctron in valnc band Adding phophoru with 5 lctron in it valanc band mak it air for th dopd matrial to rla fr lctron Thi i calld an n-typ matrial Adding aluminum with 3 lctron in it valanc band mak it air for th dopd matrial to produc hol Thi i calld a p-typ matrial 17 Smiconductor III oining n-typ and p-typ matrial rult in a pn junction Elctron and hol flow acro junction to rjoin Extrnal circuit i mor fficint mchanim for lctron to rjoin hol Solar cll i pn junction whr photon provid nrgy for lctron to bcom fr lctron 18 ME 483 Altrnativ Enrgy Enginring II 3

Photovoltaic Solar April 6, 010 Smiconductor to Solar Cll Phyic of olid viualiz valanc band whr lctron ar linkd to atom and conduction band for fr lctron Fr lctron hav highr nrgy Band gap i nrgy

4 Photovoltaic Solar April 6, 010 Smiconductor to Solar Cll Phyic of olid viualiz valanc band whr lctron ar linkd to atom and conduction band for fr lctron Fr lctron hav highr nrgy Band gap i nrgy diffrnc btwn valanc band and conduction band Photon from unlight mut hav nough nrgy to mov lctron from valanc to conduction band 19 Silicon Band Gap = 1.11 V E = 1.11 V = hc/λ giv λ = 1.1 μm L nrgtic photon (λ > 1.1 μm) will not rla an lctron Highr nrgy photon (λ < 1.1 μm) will rla lctron and produc hat Radiant nrgy not rlaing lctron will b rflctd or aborbd For λt (1.1 μm)(5800 K) = μm K olar fraction i 77.5% (not all ud) 0 Link to Dmo olarcll/javaolarcllfigur1.jpg 1 Solar Cll Currnt Dnity L = + o o S Hodg pp 31-3 for drivation Currnt dnity L = I L /A dlivrd to load Intrnal currnt dniti acro p-n junction: p-to-n, 0, and n-to-p, r, nt j = r o Cll output currnt dnity ; L = j 0 = lctron charg = x10-19 /V k = Boltzmann contant = 1.381x10-3 /K V = cll voltag Solar Cll Output L = 0 giv opn circuit voltag = + = 0 = L o o Voc 1 0 o V = 0 giv hort circuit currnt 00 L = + o o = + o o = L = + o o L o = 1+ o 3 Solar Cll Output II For givn opn circuit voltag, 0 and tmpratur, T, find 0 / 0 1 = 1 = 0 o oc + 1 U thi 0 / to plot L / vru V L o 1 = 1 1 = 1 oc ME 483 Altrnativ Enrgy Enginring II 4

Photovoltaic Solar April 6, 010 dp dv Solar Cll Powr P = I L

V 0 1 = A1 + AV = A 0Voc 0Voc + 1 + 1 + 1 V + 1+ 0Voc + 1 oc

Iload/Ihort circuit and P/Pmax Solar Cll Paramtr 1.1 1 0.9 0.

1 Currnt P/Pmax V at Pma I at Pmax for V Pmax = V at P max 5

5 Photovoltaic Solar April 6, 010 dp dv Solar Cll Powr P = I L V = L AV L 1 P = L AV = AV = AV 1 oc + 1 dp/dv = 0 giv P max V 0 1 = A1 + AV = A 0Voc 0Voc V Voc + 1 oc 0 0 oc P max dp max = P 1 = 0 dv Solv numrically Iload/Ihort circuit and P/Pmax Solar Cll Paramtr Currnt P/Pmax V at Pma I at Pmax for V Pmax = V at P max 5 V/V opn circuit ME 483 Altrnativ Enrgy Enginring II 5

6 Photovoltaic Solar April 6, ME 483 Altrnativ Enrgy Enginring II 6

Photovoltaic Solar April 6, 010 Photovoltaic Shipmnt http://www.ia.do.

rnwabl/pag/olarrport/olarpv,html 37 Modul Typ Crytallin ilicon from a wdg of

of miconductor matrial, uch a amorphou ilicon (a- Si), cadmium tllurid (CdT), or

onto th cll 38 Typ of Modul Shippd Cll and Modul Cot From manufacturr urvy

7 Photovoltaic Solar April 6, 010 Photovoltaic Shipmnt 37 Modul Typ Crytallin ilicon from a wdg of ingl-crytal or polycrytallin ilicon Cat, thin-film, or ribbon Thin-film from layr of miconductor matrial, uch a amorphou ilicon (a- Si), cadmium tllurid (CdT), or coppr indium gallium lnid (CIGS) Concntrator u ln that gathr and concntrat unlight onto th cll 38 Typ of Modul Shippd Cll and Modul Cot From manufacturr urvy ME 483 Altrnativ Enrgy Enginring II 7

8 Photovoltaic Solar April 6, 010 SF Vally Photovoltaic Cot From Supply 5% of avrag 860 kwh/mo 1.49 kw (pak) 148 ft ara Cot $10,416 ($,531 aftr incntiv) $7000/kW(pak) for ntir ytm Saving $/month, $10,995 for 5 yar with 4%/yar cot incra Rturn on invtmnt = 9.3% with incntiv, 3.4% without incntiv ME 483 Altrnativ Enrgy Enginring II 8

Goal: 10% and 10,000 h http://www.robaid.

9 Photovoltaic Solar April 6, 010 I Efficincy Important? Y and no Solar nrgy i fr What i cot pr quar mtr of olar photovoltaic cll? Solar cll cot mot xpniv Low fficincy, low cot could b bttr Auxiliary cot for tructur High fficincy will alway b bttr hr Organic Photovoltaic η < 6% Lowr cot than Silicon Liftim a fw 1000 hour Goal: 10% and 10,000 h Solar Concluion Promi of larg amount of pollutionfr nrgy with ignificant limitation Low nrgy flux rquir larg ara Cot for currnt olar matrial ar not jutifid by cot aving Solar pak match lctricity dmand and thrmal torag xtnd gnration tim Scinc of olar nrgy wll undrtood, but problm with prviou commrcial Still nd ignificant cot rduction to mak olar PV cot ffctiv x300_nit-organic-photovoltaic.jpg 51 5 ME 483 Altrnativ Enrgy Enginring II 9

Solid State Device Fundamentals

Solid State Device Fundamentals Solid Stat vic Fundamntal ES 345 Lctur Cour by landr M. Zaitv alandr.zaitv@ci.cuny.du Tl: 718 98 81 4101b Collg of Statn Iland / CUY p-n Junction p-n junction i a phyical contact of p- and n-typ miconductor.