Waste Water Engineering Multiple Choice Questions on “Oil and Grease Removal Methods”.
1. Oil and grease is the presence of inorganics in wastewater.
a) True
b) False
Answer: b
Clarification: Oil and Grease (O&G) are a common occurrence in wastewater. An EPA commissioned study recently concluded that O&G is an indicator of the presence of numerous other organics in a wastewater, the types that partition into oil.
2. The surfactants have ______ chains.
a) Linear
b) Hydrocarbon
c) Complex
d) Carbon
Answer: b
Clarification: Oil is chemically emulsified in water when emulsifiers such as surfactants or soaps are present. The surfactants have hydrocarbon chains. The simplest ones are sodium laurel sulphate or stearic acid which has a hydrophilic (water loving) and a lipophilic (oil loving) end.
3. Lipophilic end is water liking.
a) True
b) False
Answer: b
Clarification: The surfactants have hydrocarbon chains. The simplest ones are sodium laurel sulphate or stearic acid which has a hydrophilic (water loving) and a lipophilic (oil loving) end. The lipophilic end enters the oil droplet, while the hydrophilic end remains in the water.
4. The lipophilic end ______
a) Remains in water
b) Enters oil droplet
c) Dissolves
d) Stays on top of water surface
Answer: b
Clarification: The lipophilic end enters the oil droplet, while the hydrophilic end remains in the water. Since this creates a charge on the otherwise neutral oil droplet, the droplets will repel each other and disperse.
5. What is the size of the oil droplets?
a) Less than 50 microns
b) Less than 40 microns
c) Less than 30 microns
d) Less than 20 microns
Answer: d
Clarification: The droplets are less than 20 microns, while the colour of the water is white. The white colour is an indicator that the emulsion must be split to allow removal of the oil. The source of such oils is metal working fluids, coolants, lubricants, motor oil, hydraulic fluids, etc.
6. What is the colour of the emulsion?
a) White
b) Grey
c) Black
d) Yellow
Answer: a
Clarification: The white colour is an indicator that the emulsion must be split to allow removal of the oil. The source of such oils is metal working fluids, coolants, lubricants, motor oil, hydraulic fluids, etc.
7. What is the size of dissolved oil droplets?
a) Less than 10 microns
b) Less than 8 microns
c) Less than 7 microns
d) Less than 5 microns
Answer: d
Clarification: These are oils from the light end of the oil spectrum such as benzene, toluene and xylene. The molecules are less than five microns in size. They are removed very effectively by activated carbon.
8. _________ act as a coupling agent between oil oil/water phases.
a) Oil
b) Water
c) Emulsifier
d) Disinfectants
Answer: c
Clarification: Emulsifier act as a coupling agent between the oil/water phases. Because the emulsifier is polar on one end (i.e., it has a charge) and is non-polar at the other end, it prevents the oil from approaching and coalescing.
9. An emulsion is a _____ system.
a) Homogenous
b) Heterogenous
c) Natural
d) Oxidized
Answer: b
Clarification: An emulsion is a heterogeneous system that consists of at least one immiscible liquid intimately dispersed in another liquid in the form of droplets, whose diameter generally exceeds 0.1 microns.
10. Surfactants and finely divided solids _____the stability of the emulsion.
a) Increase
b) Decease
c) Neutralize
d) Nullify
Answer: a
Clarification: Surfactants and finely divided solids increase the stability of the emulsion. An emulsion is a heterogeneous system that consists of at least one immiscible liquid intimately dispersed in another liquid in the form of droplets, whose diameter generally exceeds 0.1 microns.
11. Organo clays are manufactured by modifying _____with quaternary amines.
a) Nitrogen
b) Bentonite
c) Potassium
d) Sodium
Answer: b
Clarification: Organo clays are manufactured by modifying bentonite with quaternary amines, a type of surfactant that contains a nitrogen ion. The nitrogen end of the quaternary amine (the hydrophilic end) is positively charged and ion exchanges onto the clay platelet for sodium or calcium.
12. What is the range of charge present on bentonite?
a) 20-30 meq/gram
b) 30-40 meq/gram
c) 40-60 meq/gram
d) 70-90 meq/gram
Answer: d
Clarification: The bentonite has a charge of 70-90 meq/gram. After it is treated with the quaternary amine, some 30-40 meq/gram remain, resulting in the organo clay also removing small amounts of the common heavy metals such as lead, copper, cadmium and nickel.
13. The design of oil/water separators is based on _________
a) Stoke’s law
b) Newton’s law
c) Boyle’s law
d) Charles’s law
Answer: a
Clarification: The design of oil/water separators is based on Stoke’s Law. The lighter oil droplets impact on the slant ribs of the media, coagulate and rise to the surface. The principle of air flotation is that oil droplets will adhere to air and gas bubbles and rise to the surface of the tank.
14. Compute the required pressure for the flotation thickener without recycle for the following information:
A/S = 0.008 mL/mg
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 302 kPa
b) 640 kPa
c) 380 kPa
d) 680 kPa
Answer: a
Clarification: The pressure is calculated by this formula A/S = 1.3 Sa (fP-1)/Sa. Then the P= p+101.35/101.35. p= 302 kPa.
15. Determine the surface area for the following information.
A/S = 0.008 mL/mg
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 34.7 m2
b) 67.7 m2
c) 38 m2
d) 72 m2
Answer: a
Clarification: Area = Flow/ surface loading rate. Area = 400 m3/d/8 L/m2/min x 1440 min/d. Therefore area = 34.7 m2.
16. Compute the required recycled rate for the flotation thickener with recycle for the following information:
A/S = 0.008 mL/mg
p = 275 atm
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 462 m3/d
b) 642 m3/d
c) 380 m3/d
d) 680 m3/d
Answer: a
Clarification: P= P+101.35/101.35. p= 3.73 atm.The pressure is calculated by this formula A/S = 1.3 Sa (fP-1)R/Sa. Q. R = 462 m3/d.
17. Determine the surface area with recycle for the following information.
A/S = 0.008 mL/mg
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 40.1 m2
b) 47.7 m2
c) 38 m2
d) 72 m2
Answer: a
Clarification: P= P+101.35/101.35. p= 3.73 atm.The pressure is calculated by this formula A/S = 1.3 Sa (fP-1)R/Sa. Q. R = 462 m3/d. Area =Recycle Flow/ surface loading rate. Area = 462 m3/d x 1000 L/m3/8 L/m2/min x 1440 min/d. Therefore area = 40.1 m2.
18. What is the velocity considered for a flotation thickener to remove oil?
a) 8 L/m2/min -160 L/m2/min
b) 180 L/m2/min
c) 200 L/m2/min-250 L/m2/min
d) 2 L/m2/min-8 L/m2/min
Answer: a
Clarification: Velocity to be considered for a floatation thickener to remove oil is 8 L/m2/min-160L/m2/min. With the velocity the surface area is found out. Mostly the velocity considered is 8 L/m2/min.