Share
Explore BrainMass

Calculations & Formulas MCQs

Can anyone help with these formulas / calculations. If so, please so your work and tell me how you arrived with your answer.

61. If the face opening for a laboratory hood installation is to be 2 feet high and 5 feet wide, what volumetric flow rate is required to generate a face velocity of 100 ft/min?
a. 2000 ft3/min
b. 500 ft3/min
c. 1000 ft3/min
d. 10000 ft3/min

63. The volume of air flowing through a 2 foot by 2 foot duct with a velocity of 150 fpm is:
a. 300 fpm
b. 300 cfm
c. 600 cfm
d. 150 cfm
e. 600 fpm

64. Given that the molecular weight of Toluene is 92.1, what is the equivalent value for a measure value of 150 ppm?
a. 565 mg/m3
b. 40 mg/m3
c. 150 mg/m3
d. 75 mg/m3
e. 300 mg/m3

65. Calculate the eight hour TWA for the following asbestos exposure results collected on one individual during a shift:
- 2 hours at 0.05 f/cc
- 2 hours at 0.10 f/cc
- 30 minutes at 0.30 f/cc
- 30 minutes lunch outside the work area - no sampling
- 3 hours at 0.02 f/cc

a. 0.51 f/cc
b. 0.068 f/cc
c. 0.064 f/cc
d. 0.38 f/cc
e. None of the above - it can not be determined from the available information

Attachments

Solution Preview

Please see the attached file.

61. If the face opening for a laboratory hood installation is to be 2 feet high and 5 feet wide, what volumetric flow rate is required to generate a face velocity of 100 ft/min?
a. 2000 ft3/min
b. 500 ft3/min
c. 1000 ft3/min
d. 10000 ft3/min

We just need to calculate the velocity given area of cross section. The area of cross section (A) in this case is 2*5=10 ft2. The velocity (V) required is 100ft/min. Now the flow rate Q =A*V so Q= 10 ft2*100100ft/min =1000 ft3/min

63. The volume of air flowing through a 2 foot by 2 foot duct with a velocity of 150 fpm is:
a. 300 fpm
b. 300 cfm
c. 600 cfm
d. 150 cfm
e. 600 fpm

Remember that fpm is the unit of ...

Solution Summary

500+ words explain 4 multiple choice questions that test the ability to use formulas and calculations.

$2.19