The air flow necessary to remain at the lower explosive level is 4515. 04cfm
<h3>How to solve for the rate of air flow</h3>
First we have to find the rate of emission. This is solved as
2pints/1.5 x 1min
= 2/1.5x60
We have the following details
SG = 0.71
LEL = 1.9%
B = 10% = 0.1 a constant
The molecular weight is given as 74.12
Then we would have Q as
403*100*0.2222 / 74.12 * 0.71 * 0.1
= Q = 4515. 04
Hence we can conclude that the air flow necessary to remain at the lower explosive level is 4515. 04cfm
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Answer:
1.10 %
Explanation:
The enrichment of the product can be calculated using the following equation:
(1)
<em>where W: is waste or tails = 26000, P: is the product = 32000, xp: is the wt. fraction of ²³⁵U in product, xf: is the wt. fraction of ²³⁵U in feed = 0.72% and xw: is the wt. fraction of ²³⁵U in waste or tails = 0.25%. </em>
By solving equation (1) for xp, we can find the enrichment of the product:
Therefore, the enrichment of the product is 1.10 %.
I hope it helps you!