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the mass of a fluid system is .311 slug, its density is 30lb.ft^3 and g is 31.90fps^2. a) find the specific volume, b) the specific weight, and c) total volume

  • thermodynamics - ,

    c) Your "density" is a weight density. The mass density is
    (weight density)/g = 0.9405 slug/ft^3

    Total volume is (mass)/(mass density)
    = 0.311 slug/0.9405 slug/ft^3
    = 0.3307 ft^3

    Note: the correct value of g at the surface of the Earth is 32.2 ft/s^2, not 31.9. I will use your incorrect value.

    b) specific weight is the ratio of density to that of water. Water's density (in weight/volume units) is 62.4 lb/ft^3
    So the specific weight is 30/62.4 = 0.481 (dimensionless)

    (a) For the specific volume, I will assume they want volume (ft^3) per lb.
    relative to that of water. That is the reciprocal of specific weight, or 2.079 (dimensionless)

  • thermodynamics - ,

    The mass of a fluid system is 0.311 slug, its density is 30 lb/ft^3 and g is 31.90 fps^2. Find
    a. the specific volume
    v=1/p 1/30=.03333ft^3/s^2

    b. the specific volume=pg/k
    3olb/ft^3 times 31.90 ft/s^2 all over 32.174 lb.ft/lb.s^2 we get 29.7445 lb/ft^3

    c. the total volume
    .311 times 32.174 all over 30
    we get 0.3335 ft^3

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