A horizontal pipe of diameter 0.81 m has a smooth constriction to a section of diameter 0.486 m . The density of oil flowing in the pipe is

Question

A horizontal pipe of diameter 0.81 m has a smooth constriction to a section of diameter 0.486 m . The density of oil flowing in the pipe is 821 kg/m3 . If the pressure in the pipe is 7970 N/m2 and in the constricted section is 5977.5 N/m 2 , what is the rate at which oil is flowing

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Khang Minh 2 weeks 2021-07-22T17:28:56+00:00 1 Answers 3 views 0

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    2021-07-22T17:29:59+00:00

    Answer:

    2.06 m³/s

    Explanation:

    diameter of pipe, d = 0.81 m

    diameter of constriction, d’ = 0.486 m

    radius, r = 0.405 m

    r’ = 0.243 m

    density of oil, ρ = 821 kg/m³

    Pressure in the pipe, P = 7970 N/m²

    Pressure at the constriction, P’ = 5977.5 N/m²

    Let v and v’ is the velocity of fluid in the pipe and at the constriction.

    By use of the equation of continuity

    A x v = A’ x v’

    r² x v = r’² x v’

    0.405 x 0.405 x v = 0.243 x 0.243 x v’

    v = 0.36 v’ …. (1)

    Use of Bernoulli’s theorem

    P+\frac{1}{2} \rho v^{2}=P' +\frac{1}{2}\rho'v'^{2}

    7970 + 0.5 x 821 x 0.36 x 0.36 x v’² = 5977.5 + 0.5 x 821 x v’²    from (i)

    1992.5 = 357.3 v’²

    v’ = 5.58 m/s

    v = 0.36 x 5.58

    v = 2 m/s

    Rate of flow = A x v = 3.14 x 0.405 x 0.405 x 2 x 2 = 2.06 m³/s

    Thus the rate of flow of volume is 2.06 m³/s.

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