Co (5. 00 g) and co2 (5. 00 g) were placed in a 750. 0 ml container at 50. 0 °c. The partial pressure of co in the container was
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Answer:
Failure rate = 20%
MTBF = 880 hours
Explanation:
given data
batteries = 10
tested = 200 hours
one failed = 20 hours
another fail at = 140 hours
solution
we know that Mean Time between Failures is express as = (Total up time) ÷ (number of breakdowns) ....................1
so here Total up time will be
Total up time = 200 × 10
Total up time = 2000
and here
Number of breakdown = 1 at 20 hour and another at 140 hour = 2
so it will be = (Total up time) ÷ (number of breakdowns) .......2
= = 1000
so here gap between occurrences is
gap between occurrences= 140 - 20
gap between occurrences = 120 hour
and
MTBF will be
MTBF = 1000 - 120
MTBF = 880 hours
and
Failure rate (FR) will be
Failure rate (FR) = 1 ÷ MTBF ................3
Failure rate (FR) = R÷T ......................4
as here R is the number of failures and T is total time
so Failure rate (FR) = 20%