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Fotoboltaikoak 1SEA3D

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(3)
Played 20

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Fotobo - 3º erronka

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Spain

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Fotoboltaikoak 1SEA3D
 

Fotoboltaikoak 1SEA3DOnline version

Fotobo - 3º erronka

by Markel
1

Panel baten potentzia 250 W da. Eguzki-orduak 5 badira, kalkulatu ekoiztutako energia.

Respuesta escrita

2

P/V kontsulta egindakoan, P=180 W eta V=30 V badira, zein da I-a?

Escoge una o varias respuestas

3

Modulu bat Vmp = 32 V eta Imp = 6 A baditu, zein da Pmp?

Respuesta escrita

4

Ze eragin dauka irradiantzia handiagoak panel baten Pmax-ean:

Escoge una o varias respuestas

5

Energia-efizientziaren kalkulu sinplea: Pmax = η × G × A, non G globala irradiazioa den, A azalera, η efizientzia. Zein da faktore zuzena?

Escoge una o varias respuestas

6

Gainazko kontaktuan, panelak Vmp=28 V, Imp=7 A baditu, zein da Pmp?

Escoge una o varias respuestas

7

Panelaren azalera erabilgarri A=1,5 m² da eta irradiazioa G=1000 W/m², Pmax %= 0.2? Zein da potentzia?

Respuesta escrita

8

Bi panel serietan konektatuta, V = 2 × 30 V = 60 V, Imp = 6 A da. Pmp = ?

Respuesta escrita

9

Eguzki-ordu murritza: 4 orduko eguraldipean eta 2 panel ditugu. Panel bakoitza Pmax = 200 W. Zein inzango da 2 panelen ekoizpena?

Escoge una o varias respuestas

Explicación

P = Pmax × orduak → 250 W × 5 h = 1250 Wh = 1,25 kWh.

I = P/V = 180 W / 30 V = 6 A.

Pmp = Vmp × Imp = 32 V × 6 A = 192 W.

Irradiantziak handiagoa denean moduluek gehiago poztuta daude; Pmax handitzen da.

Formula zuzena: Pmax = η × G × A, beraz η = Pmax/(G×A).

Pmp = Vmp × Imp = 28 V × 7 A = 196 W.

Pmax ≈ η × G × A = 0,2 × 1000 × 1,5 = 300 W.

Pmp = V × I = 60 V × 6 A = 360 W.

Bi panelek, erabat, 200 Wp each → 400 Wp.

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