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14 July, 23:01

A liquid mixture containing 40.0 wt% (by mass) benzene (B) and 60.0 wt% toluene (T) is fed to a distillation column. A product stream leaving the top of the column (the overhead product) contains 90.0 wt% B, and 95% of B fed to the column is recovered in the overhead. The volumetric flow rate of the feed stream is 2000 L/h and the specific gravity of the feed mixture is 0.872. Determine the flowrates of all components in all streams using our 9-step process.

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  1. 15 July, 00:30
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    Flow rate of benzene = 706. 22kg/h

    Flow rate of toluene = 1003 kgT/h

    Explanation:

    m1 = (2000 ltr/h) * (0.872 kg/ltr) = 1744 kg/h

    Since 95% is recovered in the overhead, it means that the bottom stream contained 5% of benzene. Thus;

    M (b3) = (0.05) (0.4m1) = 0.05 x 0.4 x 1744 = 34.88 kgB/h

    Now in overhead product;

    Basis : 100 kmol overhead

    90 kmol B, 10 kmol T

    Molar masd of benzene is approximately 78g/mol

    Benzene = 90 kmolB * 78 kgB/kmolB = 7020 kgB

    Molar mass of toluene = 92g/mol. hence,

    Toluene = 5 * 92 = 461 kg T

    Y (b2) = 7020 / (7020 + 461) = 0.9384

    Benzene balance:

    0.4 m1 = m2 (yB2) + mB3

    0.4 * 1744 = m2 (0.9384) + 34.88

    ⇒ m2 = 706. 22kg/h

    Toluene balance:

    0.6 m1 = (1-yB2) m2 + mT3

    0.6 * 1744 = { (1-0.9384) * 706. 22} + mT3

    So, mT3 = 1003 kgT/h
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