The time to failure in operating hours of a critical solid-state power unit has

Question

The time to failure in operating hours of a critical solid-state power unit has the following hazard rate function Lamda(t) = 0.003( T/500) to the 0.05 power, for t is greater than or equal to 0.

(a) What is the reliability if the power unit must operate continuously for 50 hr?

(b) Determine the design life if reliability of 0.90 is desired.

(c) compute the MTTF

(d) Given that the unit has operated for 50 hr, what is the probability that it will survive a second 50 hr of operation?

(e) the power unit is also subject to chance failures at a rate 0.002 failure per operating hour because of the external environment. What is the power unit's reliability for the first 50 hr of operation considering both failure modes?

Explanation / Answer

(a) R(t) = 0 exp ( ) t t dt R(50) = 50 0.5 0 exp 0.003 500 t dt = 50 3 2 0 0.003 2 exp . 500 3 t

= 0.003 2 exp 50 50 500 3 = exp[–0.03162] = 0.9689

(b) R(tD) = 0.90 i.e., 0.5 0 exp 0.003 500 tD t dt = 0.90 i.e., 1 2 0 0.003 500 tD t dt = – 0.10536 i.e., 0.003 500 × 2 3 2 3 Dt = 0.10536 tD = 2 3 3 500 0.10536 2 0.003 = 111.54 hours.

(c) MTTF = 0 R t dt ( ) = 0.003 2 3 2 500 3 0 t e dt = 3 2 0 , at e dt where a = 0.003 2 3 500 = 1 3 2 3 0 2 . 3 x e x a dx, on putting x = at3/2 = 2 3 2 (2 3) 3a = 2 3 2 3 (5 3) 3a 2 = 2 3 0.9033 a , from the table of values of Gamma function. = 45.65 hours. (d

(d) P(T 100/T 50) = ( 100) ( 50) P T P T = (100) (50) R R = 100 50 exp ( )t dt = 0.002 0.002 32 32 exp 100 50 500 500 = exp[{– 0.08944} – {– 0.03162}] = 0.9438

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