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匿名 發問於 科學及數學其他 - 科學 · 6 年前

Fluid mechanics

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  • 天同
    Lv 7
    6 年前
    最愛解答

    Let P1 and P2 be the pressure at the lower and upper opening ends of the manometer tube respectively.

    Hence, P1 = P2 + (1.5 - 0.25)(po)g + 0.25(p')g

    where po and p' are the densities of oil (S.G. = 0.86) and manometer liquid (S.G. = 2.5) respectively.

    i.e. P1 - P2 = (1.25 x 860 x 9.81 + 0.25 x 2500 x 9.81) Pa = 16 677 Pa

    Apply Bernoulli's Equation,

    P1 + (1/2)(po)(V1)^2 = P2 + (1/2)(po)(V2)^2 + 1.5(po)g

    where V1 and V2 are the velocities of oil near the lower and upper opening ends of the manometer tube respectively.

    thus, P1 - P2 = (po/2)[(V2)^2 - (V1)^2] + 1.5(po)g

    i.e. 16677 = (po/2)[(V2)^2 - (V1)^2] + 1.5(po)g ----------- (1)

    Now, apply the Continuity Equation,

    0.03(V1) = 0.012(V2)

    V1 = (0.012/0.03)(V2) = 0.4(V2)

    Equation (1) becomes,

    16677 = (po/2)[(V2)^2 - 0.16(V2)^2] + 1.5(po)g

    i.e 16677 = (860/2).[(1 - 0.16)(V2)^2] + 1.5 x 860 x 9.81

    16677 = 361.2(V2)^2 + 12655

    Hence, V2 = 3.337 m /s

    The flow rate of oil = 0.012 x 3.337 m ^3/s = 0.04 m ^3/s

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