# Fluid mechanics

### 1 個解答

• 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.

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