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27 January, 15:52

1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. To flush this material out of the tank, water is pumped into the vessel at a rate of 40,000 liter/h. Liquid containing solids leaves at the same rate. Estimate the concentration of suspended solids in the tank at the end of 5 h.

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  1. 27 January, 17:03
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    Concentration = 10.33 kg/m³

    Explanation:

    We are given;

    Mass of solids; 10,000 kg

    Volume; V = 440,000 L = 440 m³

    Rate at which water is pumped out = 40,000 liter/h

    Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³

    Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;

    200m³/440m³ = 5/11

    Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;

    Thus,

    Amount left; = 10000 x (5/11) = 4545 kg

    The concentration would be calculated by:

    Concentration = amount left/initial volume

    Thus,

    Concentration = 4545/440 = 10.3 kg/m³
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