Water requirements miscellaneous

Water requirements miscellaneous

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Water Requirements

Water requirements miscellaneous
  1. A surface water treatment plant operates round the clock with a flow rate of 35 m³/min. The water temperature is 15 °C and jar testing indicated an alum dosage of 25 mg/l with flocculation at a Gt value of 4 × 104 producing optimal results. The alum quantity required for 30 days (in kg) of operation of the plant is __________








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    37800
    Q = 35 m³/min
    = 35 × 10³l/min = 35 × 10³ × 60 × 24 l/day
    Alum dose = 25 mg/l = 24 × 106 kg/l
    ∴ For 30 days,
    Alum required = (35 × 10³ × 60 × 24) × (26 × 106) × 30 = 37800 kg

    Correct Option: A

    37800
    Q = 35 m³/min
    = 35 × 10³l/min = 35 × 10³ × 60 × 24 l/day
    Alum dose = 25 mg/l = 24 × 106 kg/l
    ∴ For 30 days,
    Alum required = (35 × 10³ × 60 × 24) × (26 × 106) × 30 = 37800 kg

  1. A suspension of sand like particles in water with particles of diameter 0.10 mm and below is flowing into a setting tank at 0.10 m³/s. Assume g = 9.81 m/s², specific gravity of particles = 2.65, and kinematic viscosity of water = 1.0105 × 10–2 cm²/s. The minimum surface area (in m²) required for this settling tank to remove particles of size 0.06 mm and above with 100% efficiency is _________








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    31 to 32

    Vs =
    9
    		(s - 1)d²
    18J

    =
    9.81
    		 × (2.65 - 1) × (6 × 10-2)²
    18 × 1.0105 × 10-2

    = 3.204 × 10–3m/s
    Vs =
    Q
    BL

    BL = surface area
    ∴ BL =
    0.1
    		 = 31.21 m²
    3.204 × 10-3

    Correct Option: C

    31 to 32

    Vs =
    9
    		(s - 1)d²
    18J

    =
    9.81
    		 × (2.65 - 1) × (6 × 10-2)²
    18 × 1.0105 × 10-2

    = 3.204 × 10–3m/s
    Vs =
    Q
    BL

    BL = surface area
    ∴ BL =
    0.1
    		 = 31.21 m²
    3.204 × 10-3

  1. A straight 100 m long raw water gravity main is to carry water from an intake structure to the jack well of a water treatment plant. The required flow through this water main is 0.21 m³/s. Allowable velocity through the main is 0.75 m/s. Assume f = 0.01, g = 9.81 m/s². The minimum gradient (in cm/100 m length) to be given to this gravity main so that the required amount of water flows without any difficulty is __________.








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    4.7 to 4.9
    Q = 0.21 m³/s
    V = 0.75 m/s

    π
    		d²
    Q
    4V

    ∴
    0.21
    		 = 0.28 m²
    0.75

    ∴ d = 0.597 m
    hf =
    flv²
    		 =
    0.01 × 100 × (0.75)²
    		 = 0.098m = 4.8 cm
    2gd2 × 9.81 × 0.597

    Minimum gradient =
    hf
    		 =
    4.8 cm
    l100 m

    ∴ Answer is 4.8

    Correct Option: A

    4.7 to 4.9
    Q = 0.21 m³/s
    V = 0.75 m/s

    π
    		d²
    Q
    4V

    ∴
    0.21
    		 = 0.28 m²
    0.75

    ∴ d = 0.597 m
    hf =
    flv²
    		 =
    0.01 × 100 × (0.75)²
    		 = 0.098m = 4.8 cm
    2gd2 × 9.81 × 0.597

    Minimum gradient =
    hf
    		 =
    4.8 cm
    l100 m

    ∴ Answer is 4.8

  1. The amount of CO2 generated (in kg) while completely oxidizing one kg of CH4 to the end products is _________.








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    2.7 to 2.8
    CH4 + 2O2 → CO2 + 2H2O
    Take molecular weights:
    CH4 : 12 + (4 × 1) = 16 g
    CO2 : 12 + (2 × 16) = 44 g
    This implies 16 g of CH4
    on oxidation produces 44 g of CO2.

    ∴ 1g of CH4 will give
    44
    		 × 1 = 2.75 kg CO2.
    16

    Correct Option: B

    2.7 to 2.8
    CH4 + 2O2 → CO2 + 2H2O
    Take molecular weights:
    CH4 : 12 + (4 × 1) = 16 g
    CO2 : 12 + (2 × 16) = 44 g
    This implies 16 g of CH4
    on oxidation produces 44 g of CO2.

    ∴ 1g of CH4 will give
    44
    		 × 1 = 2.75 kg CO2.
    16