Water requirements miscellaneous


Water requirements miscellaneous

  1. 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









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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


  1. 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









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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



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









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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


  1. 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









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    22.58

    % removal =
    V's
    × 100 = 90 %
    Vs

    ∴ V's = 0.9 Vs

    Correct Option: D

    22.58

    % removal =
    V's
    × 100 = 90 %
    Vs

    ∴ V's = 0.9 Vs



  1. 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 _______









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