Chapter 3 Toxicology at Workplace Assignment Solution 14-16

Question 14The 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 t 1i TWA   C (t )dt 8t We assume concentration Ci is averaged over the period of time Ti. Then TWA is then computed by equation below. TWA  C1T1  C2T2  ...  CnTn 8 We assume at 8 am the concentration is 110 ppm. Zaki 000 159.800 0.201   147.217 0.917 1.000 162.542 102.000 141.000 Exposed (min) Exposed (hr) (T) 10 55 62 73 40 12 48 17 46 70 48 59 0.000 43.000 143.000 162.167 0.310 0.000 110. the worker is not overexposed.142 1178.000 130.000 136.800 0.000 127.201 * Calculation obtained by interpolated values TWA  Sum(CT ) 1178.500 Sum CT 18.283 0.000 130.000 120.000 146.167 0.050 185.275 ppm 8hrs 8 Because the TLV of the substance is 150 ppm.983 CAverage (CAverage) X (T) 110.033 1.667 0.620 * 142.000 165.928 121.200 0.040 158.000 141.080 * 157.000 0.500 154.873 0.000 153.000 159.767 1.133 189.000 155.000 0.500 152.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. Zaki .000 153.333 110.200 139. 1-dichloroethane (TLV-TWA of 100 ppm). The workers are over exposed under these circumstances.Question 15 Air contain 4ppm carbon tetrachloride. We can use 2 equation First Equation n (TLV  TWA) mix  C i 1 n i Ci  (TLV  TWA) i 1  i 4  25 29   27. Compute the mixture TLV and determine if this value has been exceeded. Second Equation n Ci 4 25    1. Zaki . 25ppm 1.6 ppm 4 25 1.1-dichloroethane. Answer Mixture data: 4 ppm carbon tetrachloride (TLV-TWA of 5 ppm). the TLV-TWA has been exceeded.05  5 100 The total mixture concentration is 4 + 25 = 29 ppm.05  5 100 i 1 (TLV  TWA)i Because the quantity is greater than 1. 25 ppm 1. We use water as reference substance. Mass transfer coefficient of 0.83    0.949 ft min  30.83 cm/s 1/ 3 M  K  Ko  o  M  1 3 cm  18   1   0. Estimate the toluene concentration (ppm) near the vessel is the local ventilation rate is t000 ft3/min.482cms 1    0. Psat >> p. Determine time required to evaporate all the toluene in the vessel.48cm  1 min  Area  d 2   (3. Answer 3 Feet We must first calculate the evaporation rate. therefore the equation used is MKAP sat Qm  RgTL We must then determine K (Mass transfer coefficient (length/time) for an area A).13    1 ft  60s  K  0.482cms s  92.065 ft 2 A   4  4    Zaki . The lid of the vessel is left open (lid diameter = 3ft). The temperature is 85oF.14)(3 ft )2       7.Question 16 A vessel contains 42 gallons toluene. Assume for this case. The pressure is 1atm. 574  35. NJ: Prentice Hall.574 sat Ptoluene  e3.67 Rankine Kelvin = (Fahrenheit + 459.59 sat ln Ptoluene  3. B.67 [K] = [°R] × 5⁄9 Kelvin [°R] = [°F] + 459.67) X (5/9) = (85 + 459. 7 Ed(Upper Saddle River.52 C = -53.0137  3096.15) × 9⁄5 Celsius Fahrenheit [°F] = [°R] − 459.67 [°R] = [K] × 9⁄5 TL – Absolute temperature of Toluene TL  85  459.67  302.67 Retrieved from th David Himmelblau.65mm  Hg Zaki .67  544.67) × 5⁄9 [°R] = ([°C] + 273. C are constant A = 16.52  3.574  53.67) X (5/9) = 302. 2003) page 1057     sat ln Ptoluene  16.59 Kelvin P saturation of toluene   sat ln Ptoluene  A B C T T is temperature in kelvin A. Basic Principles and Calculations in Chemical Engineering.Rankine temperature conversion formulae from Rankine to Rankine [°C] = ([°R] − 491.0137 B = 3096. 179 min   69.065 ft 2 0.00131578947 atm sat Ptoluene  35.7302 ft 3 atm lb  mol R 544.65  0.179 min 0.0727 lbm min    The initial mass of Toluene  42 gal  7.65hour 60 min  Zaki .00131578947atm  0.949 ft min  7.234lb  303.83lbm  4.0469atm The Evaporation rate is MKAP sat Qm  Rg TL  92 lbm    lb  mol 0.83lbm 1gal Time to evaporate 303.67 R  0.0727 lbm min 1hour  4.1 mm hg = 0.0469atm  0.65mm  Hg  35. 144.9 ppm 0.949 ft    min  7.9 ppm to 3.1  K  0.5 ppm depending on the value of K.1 K=0.The concentration is estimate using equation below KAP sat C ppm  106 kQv P 0. the concentration is all greater than TLV.065 ft 2 0.45  628.5 The concentration will range from 628. Zaki .0469atm  6  10 k 1000 ft 3 min 1atm     314.5 ppm 0. Because the TLV for toluene is 20 ppm.5 K=0.45 k K is varies from 0.144.5 C ppm  314.5 0.1 C ppm  314.45  3.1 to 0.