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Unit IV Scholarly Activity

Instructions

  1. Exposure Concentration in mg/m3      
  2. Manganese Fume
  • Analytical result – 5 µg
  • Sample volume – 30 L

Hence,

  • (5 µg)/(30 L) = 0.1667 µg/L

Convert to mg/m3

  • 1667µg/L = 0.1667 mg/m3
  1. Copper Fume
  • Laboratory result – 140 µg
  • Volume of sample – 960 L

Hence,

  • (140 µg)/(960L) = 0.1458 µg/L

Convert to mg/m3

  • 1458 µg/L = 0.1458 mg/m 3
  1. Lead Fume
  • Analytical result – 40 µg
  • Sample volume – 960 L

Hence,

  • (40 µg)/(960L) = 0.0417 µg/L

Convert to mg/m3

  • 0417 µg/L = 0.0417 mg/m 3
  1. Metal Working Fluids
  • Analytical result – 500 µg
  • Sample’s volume – 720 L

Hence,

  • (500 µg)/(720L) = 0.6944 µg/L

Convert to mg/m3

  • 6944 µg/L = 0.6944 mg/m 3
  1. Exposure in Parts per Million for Vapors
  2. 1,2,4 trimethylbenzene
  • Analytical result – 5 µg
  • Sample volume – 48 L
  • Compound’s molecular weight – 120.19

Hence,

  • (5 µg)/(48L) = 0.1042 µg/L
  • 1042 µg/L = 0.1042 mg/m 3
  • 1,2,4 trimethylbenzene ppm = [(0.1042 mg/m 3)(24.45 L)]/(120.19)
  • 1,2,4 trimethylbenzene = 0.021 ppm
  1. Toluene
  • Analytical results – 125 µg
  • Sample volume – 48 L
  • Compound’s molecular weight – 92.14

Hence,

  • (125 µg)/(48 L) = 2.6042 µg/L
  • 6042 µg/L = 2.6042 mg/m 3
  • Toluene ppm = [(2.6042 mg/m 3)(24.45 L)]/(92.14)
  • Toluene = 0.691 ppm
  1. Xylene
  • Analytical results – 20 µg
  • Sample volume – 48 L
  • Compound’s molecular weight – 106.16

Hence,

  • (20 µg)/(48L) = 0.0417 µg/L
  • 0417 µg/L = 0.0417 mg/m 3
  • Xylene ppm = [(0.0417 mg/m 3)(24.45 L)]/(106.16)
  • Xylene = 0.010 ppm
  1. Introduction of Errors in Results

Based on the present conditions, the errors might have resulted from failure to follow storage and shipping procedures. Additionally, the chemicals might also react with other chemicals in the atmosphere, resulting in the errors in the results.

Part II

dBA conversion to Percentage

  1. Shipping/Receiving
  • Use Table A-1
  • Time Weighted Average = (16.61 [log (10) (Dose/100)] + 90
  • 3 – 90 = (16.61 [log (10) (Dose/100)] + (90 -90)
  • -11.7 = (16.61 [log (10) (Dose/100)]
  • (-11.7)/ (16.61) = (16.61 [log (10) (D/100)]/ 16.61
  • -0.704 = log (10) (Dose/100)
  • 20 = Dose/100
    (0.2)(100) = (Dose/100)(100)
  • Thus, an 8-hour shift is equivalent to =20%.
  1. Hydraulic Press
  • D = (Length of shift)/T
  • T = (8)/2[(L-90)/5]
  • SL is measured TWA in dBA
  • An exposure of 93.0 dBA Solve [(SL – 90)/5]
  • (93.0 – 90)/5 = 0.6
  • Then solve 2(0.6)= 1.5157
  • 8/1.5157 = 5.3
  • = 12/5.3 = 2.3
  • A 12-hour shift Dose is equivalent to 2.3%
  1. Metal Working Line
  • D = (shift Length)/T
  • T = (8)/2[(L-900/5]
  • SL is the measured TWA in dB
  • An exposure of 84 dBA Solve [(L -90)/5]
  • (84 – 90)/5 = -1.2
  • 2(-1.2)= 0.4353
  • 8/0.4353 = 18.4
  • = 12/18.4
  • Thus, a 12-hour shift Dose is equivalent to = 0.6%
  1. Robotic Welding
  • D = (shift Length)/T
  • T = (8)/2[(L-900/5]
  • SL is the measured TWA in dBA
  • An exposure of 80.5 dBA[(L – 90)/5]
  • (80.5 – 90)/5 = -1.9
  • 2(-1.9)= 0.2679
  • 8/0.2679 = 29.9
  • = 12/29.9
  • Thus, a 12-hour shift Dose is equivalent to 0.4%
  1. Hand Welding
  • D = (shift Length)/T
  • T = (8)/2[(L-900/5]
  • SL is the measured TWA in dBA
  • Solve [(L – 90)/5] first for an exposure of 81.3 dBA
  • (81.3 – 90)/5 = -1.74
  • 2(-1.74)= 0.2994
  • 8/0.2994 = 26.7
  • = 12/26.7
  • For a 12 hour shift Dose = 0.4%
  1. Paint Booth
  • D = (shift Length)/T
  • T = (8)/2[(L-900/5]
  • SL is the measured TWA in dBA
  • An exposure of 79.5 dBA[(L – 90)/5]
  • (79.5 – 90)/5 = -2.1
  • 2(-2.1)= 0.2332
  • 8/0.2332 = 34.3
  • 12/34.3 = 0.3%
  • For a 12 hour shift Dose is 0.3%
  1. QA/QC Laboratory
  • Use Table A-1
  • TWA = (16.61 [log (10) (Dose/100)] + 90
  • 70 – 90 = (16.61 [log (10) (Dose/100)] + (90 -90)
  • -20 = (16.61 [log (10) (Dose/100)]
  • (-20)/ (16.61) = (16.61 [log (10) (D/100)]/ 16.61
  • -1.204 = log (10) (Dose/100)
  • 06 = Dose/100
  • (0.06)(100) = (Dose/100)(100)
  • For an 8 hour shift Dose is 6%
  1. Final Inspection
  • Use Table A-1
  • TWA = (16.61 [log (10) (Dose/100)] + 90
  • 5 – 90 = (16.61 [log (10) (Dose/100)] + (90 -90)
  • -16.5 = (16.61 [log (10) (Dose/100)]
  • (-16.5)/ (16.61) = (16.61 [log (10) (D/100)]/ 16.61
  • -0.99 = log (10) (Dose/100)
  • 10 = Dose/100
  • (0.10)(100) = (Dose/100)(100)
  • For an 8 hour shift Dose is 10%

 

 

References

Centers for Disease Control and Prevention. (1998). Occupational noise exposure. Revised Criteria 1998. Retrieved fromhttps://www.cdc.gov/niosh/docs/98-126/pdfs/98-126.pdf

Fuller, T.P. (2015). Essentials of industrial hygiene. Itasca, IL: National Safety Council

Themann, C. L., & Masterson, E. A. (2019). Occupational noise exposure: A review of its effects, epidemiology, and impact with recommendations for reducing its burden. The Journal of the Acoustical Society of America, 146(5), 3879. https://doi.org/10.1121/1.5134465

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