Chapter 3 Toxicology at Workplace Assignment Solution 14-16

Question 14

The TLV-TWA for a substance is 150ppm. A worker begin a work shift at 8 am and complete the shift at 5 pm. 1 hr lunch break is included between 12noon  –  1 pm, where it can be assumed that no exposure to the chemical is occur. The following data were taken in the work area at the times indicated. Has the worker exceed the TLV specification? Concentration PPM 110 110 130 143 162 147 142 153 157 159 165 153 130

Time 8:00 AM 8:10 AM 9:05 AM 10:07 AM 11:20 AM 12:00 AM 12:12 PM 1:00 PM 1:17 PM 2:03 PM 3:13 PM 4:01 PM 5:00 PM Answer

TLV-TWA for a substance is 150ppm. TLV-TWA is defined by basis of 8 hours of work, independent of total time exposed. The data given involves 9 hours from 8 AM to 5 PM. The procedure is to integrate the concentration vs time and then divide by 8 hours

TWA 

1

t i

C (t )dt   8 t 

We assume concentration Ci is averaged over the period of time ti me Ti. Then TWA is then computed by equation below.

TWA 

C 1T 1  C 2T 2  ...  C nT n 8

We assume at 8 am the concentration is 110 ppm.

Zaki

Time 8:00 AM 8:10 AM 9:05 AM 10:07 AM 11:20 AM 12:00 AM 12:12 PM 1:00 PM 1:17 PM 2:03 PM 3:13 PM 4:01 PM 5:00 PM

Concentration PPM 110.000 110.000 130.000 143.000 162.000 146.620 * 142.000 153.080 * 157.000 159.000 165.000 153.000 130.000

Exposed (min)

Exposed (hr) (T)

10 55 62 73 40 12 48 17 46 70 48 59

0.167 0.917 1.033 1.217 0.667 0.200 0.800 0.283 0.767 1.167 0.800 0.983

C Average

(C Average) X (T)

110.000 120.000 136.500 152.500 154.310 0.000 0.000 155.040 158.000 162.000 159.000 141.500 Sum CT 

18.333 110.000 141.050 185.542 102.873 0.000 0.000 43.928 121.133 189.000 127.200 139.142 1178.201

* Calculation obtained by interpolated values

TWA 

Sum(CT ) 8hrs



1178.201 8

 147.275 ppm

Because the TLV of the substance is 150 ppm, the worker is not overexposed.

Zaki

Question 15

 Air contain 4ppm carbon tetrachloride, 25ppm 1,1-dichloroethane. Compute the mixture TLV and determine if this value has been exceeded. Answer

Mixture data: 4 ppm carbon tetrachloride (TLV-TWA of 5 ppm). 25 ppm 1,1-dichloroethane (TLV-TWA of 100 ppm).

We can use 2 equation First Equation n

(TLV   TWA) mix 

 C  i



i 1 n

C i

 (TLV   TWA) i 1

i

4  25 29   27.6 ppm 4 25 1.05  5 100

The total mixture concentration is 4 + 25 = 29 ppm. The workers are over exposed under these circumstances. Second Equation n

C i

 (TLV   TWA) i 1

 i

4 5



25 100

 1.05

Because the quantity is greater than 1, the TLV-TWA has been exceeded.

Zaki

Question 16

 A vessel contains 42 gallons toluene. The lid of the vessel is left open (lid diameter = 3ft). Determine time required to evaporate all the toluene in the vessel. The temperature is 85oF. Estimate the toluene concentration (ppm) near the vessel is the local ventilation rate is t000 ft3/min. The pressure is 1atm. Answer 3 Feet

We must first calculate the evaporation rate.  Assume for this case, Psat >> p, therefore the equation used is

Qm 

 MKAP  sa t   R g T  L

We must then determine K (Mass transfer coefficient (length/time) for an area A). We use water as reference substance, Mass transfer coefficient of 0.83 cm/s 1/ 3

  M     K    K o  o    M   

1

cm   18   3     0.83    0.482cms1  s   92.13   

  1 ft     60 s    K   0.482cms     0.949  ft  min  30.48cm  1 min  1

 Area

   d 2    (3.14)(3 ft )2        7.065 ft 2  A   4   4      

Zaki

Rankine temperature conversion formulae from Rankine

to Rankine

[°C] = ([°R] − 491.67) × 5  ⁄  9 [°R] = ([°C] + 273.15) × 9  ⁄  5

Celsius

Fahrenheit [°F] = [°R] − 459.67

[K] = [°R] × 5  ⁄  9

Kelvin

[°R] = [°F] + 459.67 [°R] = [K] × 9  ⁄  5

TL – Absolute temperature of Toluene

T  L  85  459.67  544.67 Rankine Kelvin = (Fahrenheit + 459.67) X (5/9) = (85 + 459.67) X (5/9) = 302.59 Kelvin

P saturation of toluene



   A 

 sa t  toluene

ln  P 

 B C   T 

T is temperature in kelvin  A, B, C are constant  A = 16.0137 B = 3096,52 C = -53.67 Retrieved from th

David Himmelblau, Basic Principles and Calculations in Chemical Engineering, 7   Ed(Upper Saddle River, NJ: Prentice Hall, 2003) page 1057



  16.0137 

 sa t  toluene

ln  P 

3096.52

 53.67  302.59

 3.574

 sat   3.574 ln  P toluene  sat   P toluene  e3.574  35.65mm  Hg  

Zaki

1 mm hg = 0.00131578947 atm  sat   35.65mm  Hg   35.65  0.00131578947atm  P toluene 

 0.0469atm The Evaporation rate is

Qm 

 MKAP  sat 



 R g T  L

92 lbm

lb  mol 0.949  ft  min  7.065 ft 2 0.0469atm 

0.7302  ft  atm 3

lb  mol  R 544.67  R  



 0.0727 lbm min The initial mass of Toluene

 42 gal 

7.234lb 1 gal 

 303.83lbm

Time to evaporate



303.83lbm 0.0727 lbm min

 4,179 min 

 4,179 min

1hour  60 min

 69.65hour 

Zaki

The concentration is estimate using equation below

C  pp m 

 KAP  sa t  kQv P 

106

0.949  ft  min 7.065 ft 2 0.0469atm  6  10 3 k 1000  ft  min 1atm  

314.45 k 

K is varies from 0.1 to 0.5

0.1  K   0.5 K=0.1

C  ppm 

314.45 0.1

 3,144.5 ppm

K=0.5

C  ppm 

314.45 0.5

 628.9 ppm

The concentration will range from 628.9 ppm to 3,144.5 ppm depending on the value of K. Because the TLV for toluene is 20 ppm, the concentration is all greater than TLV.

Zaki