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# Water Treatment (Part - 1) Civil Engineering (CE) Notes | EduRev

## Topic wise GATE Past Year Papers for Civil Engineering

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## Civil Engineering (CE) : Water Treatment (Part - 1) Civil Engineering (CE) Notes | EduRev

The document Water Treatment (Part - 1) Civil Engineering (CE) Notes | EduRev is a part of the Civil Engineering (CE) Course Topic wise GATE Past Year Papers for Civil Engineering.
All you need of Civil Engineering (CE) at this link: Civil Engineering (CE)

Q. 1 Chlorine is used as the disinfectant in a municipal water treatment plant. It achieves 50 percent of disinfection efficiency measured in terms of killing the indicator microorganisms (E-Coli) in 3 minutes. The minimum time required to achieve 99 percent disinfection efficiency would be   [2019 : 2 Marks, Set-II]
(a) 19.93 minutes
(b) 11.93 minutes
(c) 9.93 minutes
(d) 2 1.93 minutes

Ans: (a)

During disinfection variations of micro-organism is given by
Nt = N0e-kt
Nt = No. of micro-organism at time t
N0 = No. of micro-organism at time 0
So, disinfection efficiency at any time 't',

Q. 2 A water treatment plant treats 6000m3 of water per day. As a part of the treatment process, discrete particles are required to be settled in a clarifier. A column test indicates that an overflow rate of 1.5 m per hour would produce the desired removal of particles through settling in the clarifier having a depth of 3.0 m. The volume of the required clarifier, (in m3, round off to 1 decimal place) would be________.    [2019 : 2 Marks, Set-II]
Ans:
500m3

Q. 3  A wastewater is to be disinfected with 35 mg/L of chlorine to obtain 99% kill of micro-organisms. The number of micro-organisms remaining alive (Nt) at time f, is modelled by Nt = N0e-kt, where N0 is number of micro-organisms at t = 0, and k is the rate of kill. The wastewater flow rate is 36 m3/h and k = 0.23 min-1. If the depth and width of the chlorination tank are 1.5 m and 1.0 m respectively, the length of the tank (in m, round off to 2 decimal places) is_______ .   [2019 : 2 Marks, Set-I]
Ans: 8.01m

Q. 4 A 0.80 m deep bed of sand filter (length 4 m and width 3 m) is made of uniform particles (diameter = 0 40 mm, specific gravity = 2.65, shape factor = 0.85) with bed porosity of 0.4. The bed has to be backwashed at a flow rate of 3.60 m3/min. During backwashing, if the terminal settling velocity of sand particles is 0.05 m/s, the expanded bed depth (in m, round off to 2 decimal places) is_______ .    [2019 : 2 Marks, Set-I]
Ans:  1.21m

Backwash velocity,

Q. 5 Sedimentation basin in a water treatment plant is designed for a flow rate of 0.2 m3/s. The basin is rectangular with a length of 32 m. width of 8 m and depth of 4 m. Assume that the settling velocity of these particles is governed by the Stokes' law. Given: density of the particles = 2.5 g/cm3, density of water = 1 g/cm3, dynamic viscosity of water = 0.01 g/(cm.s) gravitational acceleration = 980 cm/s2. If the incoming water contains particles of diameter 25 |am (spherical and uniform), the removal efficiency of these particles is  [2019 : 2 Marks, Set-I]
(a) 100%
(b) 51%
(c) 78%
(d) 65%
Ans:

Flow rate, Q0= 0.2 m3/sec
Plan area,
(PA) = LB = 32 x 8 = 256 m2
(OFR) over flow rate

Now, settling velocity of particle of size 25 Î¼m be us

Q. 6 A completely mixed dilute suspension of sand particles having diameters 0.25, 0.35, 0.40. 0.45 and 0.50 mm are filled in a transparent glass column of diameter 10 cm and height 2.50 m. The suspension is allowed to settle without any disturbance. It is observed that all particles of diameter 0.35 mm settle to the bottom of the column in 30 s. For the same period of 30 s. The percentage removal (round off to integer value) of particles of diameters 0.45 and 0.50 mm from the suspension is_______ .    [2019 : 1 Mark, Set-I]
Ans: 100%

Since sand particle of size 0.35 mm settles to the bottom of the column in 30 sec particles having size greater than 0.35 mm i.e. 0.45 and 0.50 mm will also settle in suspension at the bottom of column by 100% in 30 sec, infact these bigger sized particle will settle by 100% in less than 30 sec. So answer is 100%.

Q. 7 At a small water treatment plant which has 4 Filters, the rates of filtration and backwashing are 200 m3/d/m2 and 1000 m3/d/m2, respectively. Backwashing is done for 15 min per day. The maturation, which occurs initially as the filter is put back into service after cleaning, takes 30 min. It is proposed to recover the water being wasted during backwashing and maturation. The percentage increase in the filtered water produced (up to two decimal places) would b e _________ .    [2018 : 2 Marks, Set-II]
Ans: 7.525%

Let total area of filters be 1 m2
Water used for backwashing

Water used for maturation

Total water wasted for backwashing and maturation
= 10.4166 + 4.166 = 14.58 m3
Water to be treated by filtered

Q. 8 A flocculation lank contains 1800 m3 of water, which is mixed using paddles at an average velocity gradient Gof 100/s. The water temperature and the corresponding dynamic viscosity are 30Â°C and 0.798 x 10-3 Ns/m2, respectively. The theoretical power required to achieve the stated value of G (in kW, up to two decimal places) is____.    [2018 : 2 Marks, Set-II]
Ans: 14.36 kW

Power required P = Î¼VG2

= 0.798 x 10-3 Ns/m2 x 1800 m3 x (1005)2
= 14364 Nm/s or Watt
= 14.364 kW
= 14.36 kW

Q. 9  A rapid sand filter comprising a number of filter beds is required to produce 99 MLD of potable water. Consider water loss during backwashing as 5%, rate of filtration as 6.0 m/h and length to width ratio of filter bed as 1.35. The width of each filter bed is to be kept equal to 5.2 m. One additional filter bed is to be provided to take care of break-down repair and maintenance. The total number of filter beds required will be  [2018 : 2 Marks, Set-I]
(a) 19
(b) 20
(c) 21

(d) 22
Ans: (c)

Total water to be filtered = 99 x 1.05 MLD = 103.95 MLD
(Addition of 5% to be used for backwashing)

âˆ´ Surface area of each filter = 36.504 m2
Total surface area required

Total no. of working units required

1 unit is to added as standby, thus total no. of units required = 21

Q. 10 A municipal corporation is required to treat 1000 m3/day of water. It is found that an overflow rate of 20 m/day will produce a satisfactory removal of the discrete suspended particles at a depth of 3 m. The diameter (in meters, rounded to the nearest integer) of a circular settling tank designed for the removal of these particles would be _____ .    [2017 : 2 Marks, Set-II]
Ans: 8m

Assuming diameter of tank to be 'd'

Q. 11 The spherical grit particles, having a radius of 0.01 mm and specific gravity of 3.0, need to be separated in a settling chamber. It is given that

• g = 9.81 m/s2
• the density of the liquid in the settling chamber = 1000 kg/m3
• the kinematic viscosity of the liquid in the settling chamber =10-6m2/s

Assuming laminar conditions, the settling velocity (in mm/s, up to one decimal place) is ________ .       [2017 : 2 Marks, Set-I]
Ans: 0.436 mm/sec

Q. 12 A water supply board is responsible for treating 1500 m3/day of water. A settling column analysis indicates that an overflow rate of 20 m/day will produce satisfactory removal for a depth of 3.1 m. It is decided to have two circular settling tanks in parallel. The required diameter (expressed in m) of the settling tanks is____ .    [2016: 1 Mark, Set-II]
Ans: 6.91 m

Discharge to be treated by one tank,

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