Question

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)

Answer 1

The air flow necessary to remain at the lower explosive level is 4515. 04cfm

How to solve for the rate of air flow

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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