# Fluid mechanics

### 1 個解答

• 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