A process involves the removal of oil and other liquid contaminants from metal parts using a heat-treat oven, which has a volume of 15,000 ft3. The oven is free of solvent vapors. The ventilation rate of the oven is 2,100 cfm, and the safety factor (K) is 3. The solvent used in the process evaporates at a rate of 0.6 cfm (cubic feet per minute). The operator would like to know how long it would take the concentration to reach 425 ppm.

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DeniceSandidge DeniceSandidge · Answer 1 · 2020-03-06T07:50:00+01:00

Answer:

time = 4.89 min

Explanation:

given data

volume = 15,000 ft³

ventilation rate of oven = 2,100 cubic feet per minute

safety factor (K) = 3

evaporates at a rate = 0.6 cubic feet per minute

solution

we get here first solvent additional rate in oven that is

solvent additional rate = [tex]\frac{0.6}{1500}[/tex]

solvent additional rate = 4 × [tex]10^{-5}[/tex] min

solvent additional rate == 4 × [tex]10^{-5}[/tex] × [tex]10^{6}[/tex]

solvent additional rate = 40 ppm/min

and

solvent removal rate due to ventilation will be

removal rate = [tex]\frac{2100}{1500}[/tex] × concentration in ppm

removal rate = 0.14 C ppm/min

and

net additional rate is

net additional rate (c) = m - r

so net additional rate (c) is = 40 - 0.14C

so here

[tex]\frac{dc}{dt}[/tex] = 40 - 0.14C

so take integrate from 0 to t

[tex]\int\limits^t_o {dt}[/tex] = [tex]\int\limits^{425/3}_0 \frac{dc}{40-0.14C}[/tex] ....................1

here factor of safety is 3 so time taken is [tex]\frac{425}{3}[/tex]

solve it we get

time = [tex][\frac{-50}{7} \ log(40-\frac{7c}{50} ]^{425/3} _0[/tex]

time = 4.89 min