Water requirements miscellaneous Easy Questions and Answers | Page - 1

Water requirements miscellaneous

A water treatment plant of capacity, 1 m³/s has filter boxes of dimensions 6 m × 10 m. Loading rate to the filters is 120 m³/day/m². When two of the filters are out of service for back washing, the loading rate (in m³/day/m²) is

1. 144 m³/d/m²
1. 147 m³/d/m²
1. 146 m³/d/m²
1. 145 m³/d/m²

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144
Total water filtered/ day = 1 × 3600 × 24 = 86400 m³/ day
Surface area of filter bed = 86400 = 720 m²
120

Surface area of filter bed = 86400 = 720 m²
120

Area of one filter = 6 × 10 = 60 m²
∴ No. of filter = 720 = 12 filters
60

filters are out of service, no. of filters = 12
∴ Total surface area of 10 filters = 10 × 60 = 600 m²
∴ Loading rate = 86400 600 = 144 m³/d/m²

Correct Option: A

144
Total water filtered/ day = 1 × 3600 × 24 = 86400 m³/ day
Surface area of filter bed = 86400 = 720 m²
120

Surface area of filter bed = 86400 = 720 m²
120

Area of one filter = 6 × 10 = 60 m²
∴ No. of filter = 720 = 12 filters
60

filters are out of service, no. of filters = 12
∴ Total surface area of 10 filters = 10 × 60 = 600 m²
∴ Loading rate = 86400 600 = 144 m³/d/m²

In a wastewater treatment plant, primary sedimentation tank (PST) designed at an overflow rate of 32.5 m³/day/m² is 32.5 m long, 8.0 m wide and liquid depth of 2.25 m. If the length of the weir is 75 m, the weir loading rate (in m³/day/m) is

1. 111.67 m³/d/m
1. 110.67 m³/d/m
1. 109.67 m³/d/m
1. 112.67 m³/d/m

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112.67
SOR = 32.5 m³/d/m²
L = 32.5 m
B = 8 m, D = 2.25 m
SOR = Q
BL

∴ Q = SOR × B.L
= 32.5 × 8 × 32.5
= 8450 m³/d
q = Q
b

b = 75 m (weir length)
q = 8450
75

= 112.67 m³/d/m

Correct Option: D

112.67
SOR = 32.5 m³/d/m²
L = 32.5 m
B = 8 m, D = 2.25 m
SOR = Q
BL

∴ Q = SOR × B.L
= 32.5 × 8 × 32.5
= 8450 m³/d
q = Q
b

b = 75 m (weir length)
q = 8450
75

= 112.67 m³/d/m

A groundwater sample was found to contain 500 mg/L total dissolved solids (TDS). TDS (in%) present in the sample is ____

1. 1.75%
1. 0.05%
1. 1.05%
1. 1.25%

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0.05
TDS = 500 mg/L = .5g/L
1 litre = 1000g
∴ TDS = 0.5 × 100 = 005 %
1000

Correct Option: B

0.05
TDS = 500 mg/L = .5g/L
1 litre = 1000g
∴ TDS = 0.5 × 100 = 005 %
1000

Consider a primary sedimentation tank (PST) in a water treatment plant with Surface Overflow Rate (SOR) of 40 m³/m²/d. The diameter of the spherical particle which will have 90 percent theoretical removal efficiency in this tank is ________ mm. Assume that settling velocity of the particles in water is described by stokes’s Law. Given: Density of water = 1000 kg/m³; density of particle = 2650 kg/m³; g = 9.81 m/s²; Kinematic viscosity of water (v) = 1.10 × 10^–6m²/s

1. 20.58 μm
1. 21.58 μm
1. 23.58 μm
1. 22.58 μm

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22.58
% removal = V'_s × 100 = 90 %
V_s

∴ V'_s = 0.9 V_s

Correct Option: D

22.58
% removal = V'_s × 100 = 90 %
V_s

∴ V'_s = 0.9 V_s

An effluent at a flow rate of 2670 m³/d from a sewage treatment plant is to be disinfected. The laboratory data of disinfection studies with a chlorine dosage of 15 mg/l yield the model Nt = No e–0.145t where Nt = number of microorganisms surviving at time t (in min.) and No = number of micro-organisms present initially (at t = 0). The volume of disinfection unit (in m³) required to achieve a 98% kill of micro-organisms is _______

1. 55 m³
1. 52 m³
1. 53 m³
1. 50 m³

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49 to 51
N^t = 2% N_o = 0.02 No (remaining microorganisms)
N^t = N_o .e^–0.145t
0.02 N_o = N_o .e^–0.145t
∴ t = 26.979 minutes = 0.0187 days
Flow rate = 2670 m³/d
∴ Volume of disinfection unit = 2670 × 0.0187 = 50 m³

Correct Option: D

49 to 51
N^t = 2% N_o = 0.02 No (remaining microorganisms)
N^t = N_o .e^–0.145t
0.02 N_o = N_o .e^–0.145t
∴ t = 26.979 minutes = 0.0187 days
Flow rate = 2670 m³/d
∴ Volume of disinfection unit = 2670 × 0.0187 = 50 m³