This set of Waste Water Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Flocculant Settling”.
1. A process of contact and adhesion whereby the particles of a dispersion form larger-size clusters is called ________
Explanation: According to the IUPAC definition, flocculation is “a process of contact and adhesion whereby the particles of dispersion form larger-size clusters”. Agglomeration and coagulation lead to flocculation.
2. What is the other name for deflocculation?
Explanation: Deflocculation is the exact opposite of flocculation, also sometimes known as peptisation. Usually in higher pH ranges in addition to low ionic strength of solutions and domination of monovalent metal cations the colloidal particles can be dispersed.
3. How many types of solids are present in water?
Explanation: Settling and non-settling are two major types of solids found in water. The combination of these two types makes up the total suspended solids found in water. Larger visible particles are called settling particles and are generally bigger than 1 micrometer.
4. What is the movement of solids in water called?
b) Brownian motion
c) Horizontal motion
d) Viscous motion
Explanation: The random movement of these solids (colloids) in water is called Brownian motion. It is the Brownian motion of the solids that keep them suspended in water for so long.
5. The clustered particles formed in the presence of a flocculant are called ______
Explanation: The clustered particles formed in the presence of a flocculant are called flocs. Any material microscopically suspended in another is called colloid. A large molecule chain that is composed of many small parts.
6. What is the chemical formula of aluminium sulphate?
Explanation: Aluminium sulphate is a white, solid (powder) with a chemical formula of Al2(SO4)3. Flocculant is a choice because it disassociates in water to form high amount of charged compounds.
7. 1 micrometer = _______
a) 10-2 m
b) 10-3 m
c) 10-6 m
d) 10-9 m
Explanation: 1 micrometer is equal to 10-6 meter.
1 nanometer is equal to 10-9 meter.
8. What is the test used to select the type of coagulant required/
a) Bar test
b) Jar test
c) Stock test
d) Coagulant test
Explanation: The purpose of this test is to select types of coagulant (alum) and also to estimate the optimal dose needed in removing the charged particles that occurred in raw water.
9. What is the ratio of Alum/Phosphate while adding alum to a waste water in order to remove the phosphates present?
a) 1.4 : 2.5
b) 1 : 2
Explanation: The ratio of Alum /Phosphate 1.4:2.5. In this ratio, alum is added in order to remove the phosphate from the waste water. The exact application is done by onsite testing.
10. At what pH is the solubility of AlPO4 minimum?
Explanation: At the pH 6.3, the solubility of aluminium phosphate is minimum. This is brought about by adding Aluminium sulphate. This can also be followed by biological removal methods where the organic phosphates are already removed.
11. At what pH is the solubility of FePO4 minimum?
Explanation: At the pH 5.3, the solubility of ferrous phosphate is minimum. This is brought about by adding Ferrous salts. This can also be followed by biological removal methods where the organic phosphates are already removed.
12. Addition of which of these result in lowering the BOD level?
c) Aluminum salts
d) Ferrous salts
Explanation: Addition of polymers results in lowering the BOD value of the waste water. This also results in better and improved settling. This should be added finally.
13. What is the ideal mixing time required for polymers?
a) 60 secs
b) 120 secs
c) 10-30 secs
d) 30-60 secs
Explanation: The ideal mixing time for the polymers is around 10-30 secs. Polymers shouldn’t be subjected to insufficient or excessive mixing. If that is the case then the process efficiency will definitely decrease, resulting in poor settling and thickening characteristics.
14. What is the detention time assumed while designing a flocculator?
a) 10 mins
b) 20 mins
c) 45 mins
d) 60 mins
Explanation: The detention time for a flocculator is assumed as 20 mins. Flocculation is usually followed by clarification. This process is very vital in removing the TSS.
15. Calculate the volume of flocculator for the following data:
Flow rate: 2200m3/hr
a) 733 m3
b) 855 m3
c) 933 m3
d) 785 m3
Explanation: The volume of the flocculator is calculated as 733 m3. This is calculated by assuming the detention time as 20 mins. Thus volume = 2200 x (20 /60) = 733m3.
16. What is the ratio of Polymer/Phosphate while adding polymer to a waste water in order to remove the phosphates present?
b) 1 : 2
Explanation: The ratio of Polymer /Phosphate 1:3. In this ratio, polymer is added in order to remove the phosphate from the waste water. The exact application is done by onsite testing.
17. Calculate the volume of flocculator for the following data:
Flow rate: 500m3/hr
a) 355 m3
b) 155 m3
c) 167 m3
d) 285 m3
Explanation: The volume of the flocculator is calculated as 167 m3. This is calculated by assuming the detention time as 20 mins. Thus volume = 500 x (20 /60)= 167m3.
18. What is the detention time assumed while designing a flash mixer?
a) 2 mins
b) 10 mins
c) >10 mins
Explanation: The detention time considered would be 1 min for a flash mixer. A flash mixer is one in which chemical mixing is done to bring about coagulation. This is where coagulation occurs. It is prior to a flocculator.
19. Determine the size of a flash mixer for the following data. Round off to the nearest digit.
Flow rate: 2200 m3/hr
a) 37 m3
b) 73 m3
c) 367 m3
d) 183 m3
Explanation: The size of the flash mixer would be 37 m3. Volume = 2200 x 60/3600. Thus a 37 m3 volume size flash mixer is required.
Sanfoundry Global Education & Learning Series – Waste Water Engineering.
To practice all areas of Waste Water Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.