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Lelu [443]
2 years ago
10

two pints of ethyl ether evaporate over a period of 1,5 hours. what is the air flow necessary to remain at 10% or less of ethyl

ether's lower explosive level? (LEL=1.9%, SG=.71)
Engineering
1 answer:
lesya692 [45]2 years ago
7 0

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

Read more on the rate of air flow on brainly.com/question/13289839

#SPJ1

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                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

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