Rate of Air Flow Calculation to Prevent Explosion

How to calculate the air flow necessary to prevent an explosion?

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=0.71)

Calculation for the Rate of Air Flow

The air flow necessary to remain at the lower explosive level is 4515.04 cfm.

In order to determine the rate of air flow required to prevent an explosion, we first need to calculate the rate of emission. In this case, two pints of ethyl ether evaporate over 1.5 hours.

Given details:

• Specific Gravity (SG) = 0.71
• Lower Explosive Level (LEL) of ethyl ether = 1.9%
• Desired concentration = 10% or 0.1 (constant)
• Molecular weight of ethyl ether = 74.12

Firstly, we calculate the rate of emission:

(2 pints / 1.5 hours) x 1 hour x 60 minutes = (2 / 1.5) x 60 = 80 minutes = 403 minutes

Next, we calculate the air flow necessary to maintain the concentration below the explosive level:

Q = (403 x 100 x 0.2222) / (74.12 x 0.71 x 0.1)

Q = 4515.04 cfm

Therefore, the air flow necessary to prevent an explosion and maintain ethyl ether concentration at 10% or less of the LEL is 4515.04 cubic feet per minute (cfm).