Consider the following configuration of solar photovoltaic arrays consisting of

Question

Consider the following configuration of solar photovoltaic arrays consisting of crystalline silicon solar cells 12 3H4 There are two subsystems connected in parallel, each one containing two cells. In order for the system to function, at least one of the two parallel subsystems must work. Within each subsystem, the two cells are connected in series, so a subsystem will work only if all cells in the subsystem work. Consider a particular lifetime value to, and suppose we want to determine the probability that the system lifetime exceeds to. Let A, denote the event that the lifetime of cell i exceeds to ( 1, 2,4). We assume that the A's are independent events (whether any particular cell lasts more than to hours has no bearing on whether or not any other cell does) and that P(A) = 0.6 for every i since the cells are identical using P(A) 0.6, the probability that system lifetime exceeds to is easily seen to be 0.5904. To what value would 0.6 have to be changed in order to increase the system lifetime reliability from 0.5904 to 0.63? Hint: Let P(A) = p, express system reliability in terms of p, and then let x = p2.] (Round your answer to four decimal places.)

Explanation / Answer

Let the probability that the system lifetime exceeds t0 = p. Then,

P[lifetime of both sub-systems (1) and (2) exceed t0] = p2 [because (1) and (2) are in series and also independent].

Similarly,

P[lifetime of both sub-systems (3) and (4) exceed t0] = p2 [because (3) and (4) are in series and also independent].

Now, sub-systems (1) and (2) together and sub-systems (3) and (4) together are in parallel and hence, the system lifetime would exceed t0 if either or both have lifetime exceeding t0.

Thus,

P(system lifetime exceeds t0) = p2 + p2 – (p2.p2)

[because P(A or B) = P(A) + P(B) - P(A & B) = P(A) + P(B) - P(A).P( B), when A and B are independent]

= 2p2 + p4

We want this to be 0.63. Thus, we have:

2x – x2 = 0.63 or

x2 – 2x + 0.63 = 0, where x = p2

Then, x = {2 ± (4 – 2.52)}/2

              = 1 ± 0.37

              = 1 ± 0.6083

              = 0.3917 [1.6083 is not admissible since probability cannot exceed 1]   

=> p2 = 0.3917 or

p = 0.6259 ANSWER

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