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6 October, 00:52

Kerosene is mixed with 10 ft3 of ethyl alcohol so that the volume of the mixture in the tank becomes 14 ft3. Determine the specific weight and the specific gravity of the mixture.

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  1. 6 October, 03:45
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    specific weight = 49.7 lb/ft^3

    specific gravity = 0.797

    Explanation:

    From Appendix A,

    pk = 1.58 slug/ft^3

    pea = 1.53 slug/ft^3

    volume of kerosene is

    Vk = 14 ft^3 - 10 ft^3 = 4 ft^3

    The total weight of the mixture is therefore

    W = pk g Vk + pea g Vea

    = (1.58 slug/ft3) (32.2 ft/s^2) (4 ft^3) + (1.53 slug/ft^3) (32.2 ft/s^2) (10 ft^3)

    = 696.16 lb

    The specific weight and specific gravity of the mixture are

    γm = W / V

    γm = 696.16 lb / 14 ft^3 = 49.73 lb / ft^3 = 49.7 lb/ft^3

    Sm = γm / γv

    Sm = 49.73 lb/ft^3 / 62.4 lb/ft^3 = 0.797
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