The recommended time weighted average air concentration for occupational exposure to water soluble hexavalent chromium (Cr VI) i
s 0.05 mg/m3. This concentration is based on the assumption that the individual is generally healthy and is exposed for 8 hours per day, 5 days per week, 50 weeks per year, over a working lifetime (that is from age 18 to 65 years). Assuming a body weight of 78 kg and inhalation rate of 15.2 m3/d over the working life of the individual, calculate the lifetime (75 years) CDI.
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Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
<u>Determine the friction angle at each depth</u>
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
Answer:
a) 1
b) 1813.96 MJ/kmol
c) 32.43 MJ/kg , 1980.39 MJ/Kmol
Explanation:
molar mass of ethanol (C2H5OH) = 46 g/mol
molar mass of octane (C8H18) = 114 g/mol
therefore the moles of ethanol and octane
ethanol = 0.85 / 46
octane = 0.15 / 114
a) determine the molar air-fuel ratio and air-fuel ratio by mass
attached below
mass of air / mass of fuel = 12.17 / 1 = 12.17
b ) Determine the lower heating value
LHV of ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol
LHV of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol
LHV ( MJ/kmol) for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol
c) Determine higher heating value ( HHV )
HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol
HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol
HHV in MJ/kg = 0.85 * 29.7 + 0.15 * 47.9 = 32.43 MJ/kg
HHV in MJ /kmol = 0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol
