A sample of municipal sewage is diluted to 1% by volume prior to running a BOD5 analysis. After 5 days the oxygen consumption is
1 answer:
Answer:
BOD5 = 200 mg/L
Explanation:
given data
diluted = 1% = 0.01
time = 5 day
oxygen consumption = 2.00 mg · L−1
solution
we get here BOD5 that is BOD after 5 day
and here total volume is 100% = 1
so dilution factor is = 100
so BOD5 is
BOD5 = oxygen consumption × dilution factor
BOD5 = 2 × 100
BOD5 = 200 mg/L
You might be interested in
Answer:
the MTTF of the transceiver is 50.17
Explanation:
Given the data in the question;
failure modes = 0.1 failure per hour
system reliability = 0.85
mission time = 5 hours
Now, we know that the reliability equation for this situation is;
R(t) = [ 1 - ( 1 - )³]
so we substitute
R(5) = [ 1 - ( 1 - )³]
= 0.85
[ 1 - ( 1 - )³]
= 0.85
[ 1 - ( 0.393469 )³] = 0.85
[ 1 - 0.06091 ] = 0.85
0.9391 = 0.85
= 0.85 / 0.9391
= 0.90512
MTTF = 5 / -ln( 0.90512 )
MTTF = 50.17
Therefore, the MTTF of the transceiver is 50.17