a). The TLV of the mixture would be as follows:
[tex]0.643[/tex] ppm
b). The rate of evaporation would be as follows:
[tex]0.016 lbm/feet^2-min[/tex]
c). The ventilation rate would be as follows:
[tex]21097.096 ft^3/min[/tex]
a). Given that,
TLV values of the mixture(Benzene) [tex]= 0.5 ppm[/tex]
Toluene [tex]= 20 ppm[/tex]
so,
The equation becomes =
[tex](TLV - TWA)_{mix}[/tex] [tex]=[/tex] ∑[tex]ci[/tex]/∑[tex]ci/TLvi[/tex] ...(i)
Pressure of Benzene [tex]= 0.5[/tex] × [tex]94.44[/tex]
[tex]= 47.22 mm Hg[/tex]
Pressure of Toluene [tex]= 0.5[/tex] × [tex]27.9 mm Hg[/tex]
Total pressure [tex]= 760 mm Hg[/tex]
Now,
[tex]C_{benzene} = (47.22 * 10^6)/760 = 62131.579 ppm[/tex]
[tex]C_{toluene} = 18355.263 ppm[/tex]
∵ TLV = [tex]0.643 ppm[/tex]
b). The rate of evaporation can be determined via
[tex]Q_{m}/A[/tex]
For benzene, we get [tex]= 0.012 lbm/ft^2 - min[/tex]
for Toluene, we get = [tex]= 0.004 lbm/ft^2 - min[/tex]
Total rate of evaporation
[tex]= 0.012 + 0.012 lbm/ft^2 - min\\= 0.016 lbm/ft^2 - min[/tex]
Thus, the above answers are correct.
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