Question

Hi,

I want the solution to the below-mentioned question.  

Ques: The accidental release of toluene has killed 3 out of 4 people in a laboratory. You are leading the investigation. You discover a toluene tank is stored at 40 pisg and 80F, and the room has ‘average’ ventilation at a rate of 5000 ft3/min and it at 1 atm. In the safety book you saved from college (Table 2-5) you find the k1 and k2 probit correlation parameters for toluene are -6.79 and 0.41, and the molecular weight of toluene is 92 lbm/lbmole with a specific gravity of 0.865. When interviewing the survivor of the accident you conclude the exposure was approximately 1 hour.

[1] Determine the toluene concentration required for the fatalities.

[2] If the lab ventilation maintained the toluene concentration at the fatal ppm found in [1], what is the evolution rate of toluene from the process?

[3] Determine the diameter of the sharp puncture in the toluene tank in inches?

Thanks,

Answer 1

1. Toluene concentration required for the fatalities:

By equation for probit analysis : Y = k1 + k2(ln Cⁿ t)

When Y = 2.67, It represent 1% fatalities

For Toluene, K1 = -6.79 and K2= 0.41 , n = 2.5, t is in minutes, C is ppm

Hence C = 1981 mg/m3

1. Evolution rate of toluene can be calculated by desired concentration of 1981 ppm

=1981*141.5842/10^6) mg ( 5000 ft3/min = 141.5842 m3/min)

=0.28 mg/min

=0.28*10^-6/867 m3/min

= 3.23*10^^-10

1. diameter of the sharp puncture in the toluene tank :

Exit Velocity from leakage = √(2* ( P2-P1)/ρ) Where, ρ = density, P are pressure

Hence V = 20 m/s

Hence Area = Q/V = 3.23*10^^-10 /20

Hence diameter = 0.00017857 inch

Similary you can calculate for 99% fatality. For 99% fatality , Y= 8.09