1. The efficiency of an array of solar panels is related to the fraction of the

Question

1. The efficiency of an array of solar panels is related to the fraction of the number of electrons ejected from the array to the number of incident photons on the array, e. Take the intensity of the sun to be 1?1000 W/m2 and the wavelength of an average energy photon to be ? 500 nm. A photovoltaic array with area A 31 m2, contains 30 panels in parallel. Each panel contains 50 cells in series. Each cell has work function ? = 2.0 eV. Take the fraction of ejected electrons to incident photons to be 0.016. The array is connected to a large number of switched resistors, R-6.0 ?, in parallel. Note he 1240 eV-nm, 1 w = 6.2 x 1018 eV/s and 1 A = 6.2 x 1018 e/s. (a) How many photons strike the array each second? b) How many electrons are ejected from all the cells in the array each second? (c) The maximum current occurs when every electron ejected from the array flows through the circuit. What is the maximum current? (d) What is the voltage across, i. one cel one panel, ii the array? (e) Assume the voltage of the array is essentially unchanged until the current reaches is maximum value. How many switches can be closed before reaching the maximum current? f) What power is supplied by the array when it is operating at the maximum current? (g) Compute the ratio of the power supplied by the array when it is operating at the maximum current to the incident solar power on the array. This quantity is the efficiency of the array.

Explanation / Answer

Solution :-

Given data ;

intensity, I = 1000 W/m^2

lambda = 500 nm

A = 31 m^2

containing, n = 30 panels in parallel

each panel has m = 50 cells, in series

work function, phi = 2 eV

e = 0.016

R = 6 ohm

hc = 1240 eV nm

1W = 6.2*10^18 eV/s

1 A = 6.2*10^18 e/s

a. number of photons striking the array each second = I*A/energy of each photon

energy of each photon = hc/lambda

hence

number of photons striking the array each second N = 1000 * 6.2*10^18 eV * 500 nm/ s m^2 * 31 m^2 /1240 eV nm

N = 1000 * 6.2*10^18 * 500 / s * 31 /1240 = 775*10^20 photons per second

b. electrons ejected from all the cells = eN = 12.4*10^20 electrons per second

c. maximum current = 12.4*10^20 /6.2*10^18 = 200 A

d. i. voltage across one cell = V

for work function phi

hc/lambda = phi + qV

V = (hc/lambda - phi)/q = (1240/500 - 2)/q = 0.48 V

ii. votage across one panel = m*V = 24 V

iii. Voltage across array = 24 V

e. Vmax = 24 V

Imax = 200 A

hence

Vmax = Imax*Reff

Reff = Vmax/Imax = 0.12 ohm

hence

Reff = R/w = 6/w = 0.12

w = 50

hence we can close 50 switches once the current reaches maximum level

f. power supplied = Vmax*Imax = 4800 W

g. power ratio = 4800 W / 1000*31 = 0.15483870967741935

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