promotion image of download ymail app
Promoted
匿名
匿名 發問於 科學及數學其他 - 科學 · 6 年前

Fluid mechanics

1 個解答

評分
  • 天同
    Lv 7
    6 年前
    最愛解答

    Apply Bernoulli's equation to the first floor and the basement,

    (Pb) + pg(Hb) + (1/2).(p)(vb)^2 = (P1) + pg(H1) + (1/2)(p)(v1)^2

    where Pb and P1 are the pressures at the basement faucet and 1st floor faucet respectively

    p is the density of water, and g is the acceleration due to gravity

    Hb and H1 are the heights of the faucets, referenced to ground level, at the basement and 1st floor respectively.

    Because at the faucet opening, the water is under free flow. Hence,

    Pb = P1 = 0

    Thus, pg(Hb) + (1/2).(p)(vb)^2 = pg(H1) + (1/2)(p)(v1)^2

    i.e. (vb)^2 = (v1)^2 + 2g[(H1 - Hb)]

    (vb)^2 = 6.5^2 + 2 x 9.81 x [1.5 - (-2.5)] (m/s)^2

    i.e. vb = square-root[6.5^2 + 2 x 9.81 x 4] m/s = 11 m/s

    [Note: Hb = -2.5 m because the faucet at the basement is 2.5 m below ground level]

    Consider the 1st floor and 2nd floor, apply Bernoulli's equation again,

    (P1) + pg(H1) + (1/2)(p)(v1)^2 = (P2) + pg(H2) + (1/2).(p)(v2)^2

    Because P1 = P2 = 0,

    thus, pg(H1) + (1/2)(p)(v1)^2 = pg(H2) + (1/2).(p)(v2)^2

    (v2)^2 = (v1)^2 + 2g[(H1) - (H2)]

    i.e. (v2)^2 = 6.5^2 + 2 x 9.81 x [1.5 - (1.5+2.5+1.5)] (m/s)^2

    (v2)^2 = 42.25 - 49.05 (m/s)^2 = -6.8 (m/s)^2

    There is no real value for v2. Therefore, there is no water coming out from the faucet on the 2nd floor.

    • Commenter avatar登入以回覆解答
還有問題嗎?立即提問即可得到解答。