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10 July, 07:23

A tank with a volume of 8 m3 containing 4 m3 of 20% (by volume) NaOH solution is to be purged by adding pure water at a rate of 3 m3/h. The solution leaves the tank at a rate of 2 m3/h. When the tank is full, the solution leave the tank at the same inlet flow of 3 m3/h. Assume perfect mixing. Specific gravity of pure NaOH is 1.22. Determine the time necessary to purge 95% of the NaOH by mass from the tank.

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  1. 10 July, 07:51
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    The time necessary to purge 95% of the NaOH is 0.38 h

    Explanation:

    Given:

    vfpure water (i) = 3 m³/h

    vNaOH = 4 m³

    xNaOH = 0.2

    vfpure water (f) = 2 m³/h

    pwater = 1000 kg/m³

    pNaOH = 1220 kg/m³

    The mass flow rate of the water is = 3 * 1000 = 3000 kg/h

    The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg

    When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg

    The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³

    The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h
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