A. Calculate the concentration of a toxic (Cout) that is added to a lake with an input concentration of 100 μg/l. The flow rate
is constant at 10 m3/s. Assume that the system is at the steady state and that the toxic has a degradation rate (reactions term) of 50kg/day.
What is the removed fraction of toxic (in %)?
What is the removed fraction of toxic (in %)?
1 answer:
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Answer:
ordinary bulb total cost is $39.54
fluorescent bulb total cost is $13.05
amount save = 39.54 - 13.05 = $26.49
resistance = 626.1 ohm
Explanation:
in the 1st part
bulb on time = 3 year = 4380 hours
life of bulb = 750 h
so number of bulb required =
number of bulb required = 6
cost of 6 bulb is = 6 × 0.75 = $4.5
so
cost of operation is = 100 × 4380 ×
cost of operation = $35.04
so total cost will be = $4.5 + $35.04 = $39.54
and
when compare with florescent bulb
time = 3 year = 4380 h
life of bulb = 10000 h
so number of bulb required =
number of bulb required = 0.43 = 1
cost of 6 bulb is = 1 × 5 = $5
so
cost of operation is = 23 × 4380 ×
cost of operation = $8.05
so total cost will be = $5 + $8.05 = $13.05
in part 2nd
total amount save while compare bulb is
amount save = 39.54 - 13.05 = $26.49
and in part 3rd
resistance of bulb is
resistance =
resistance =
resistance = 626.1 ohm
Answer: 37.4K
Explanation:
See attachment
