Question

Water is flowing through a circular tube with a cross-sectional area 0.4m^2 at a rate of 8m/s.

Step 1: the cross-section of the tube gradually decreases to 0.2m^2,

Step 2: then the water shoots out upright from the tape (in the up vertical direction), and gains 4m in height.

What is the new velocity of the water after it gains 4m in height? [save to 1 decimal places]

Assume g=10m/s²; for step 1 and 2, the water is under the same atmosphere pressure;

Tips: Step 1 where the conservation of mass is applicable;

Step 2 where the conservation of energy is applicable.

Answer 1

v₁=8m/s

A₁=0.4m², A₂=0.2m²

Use continuity equation,

A₁v₁= A₂v₂

v₂=(A₁/A₂)v₁=(0.4/0.2)*8=16m/s

Now use Bernoulli’s principle

P_i+1/2 *ρv_i² +ρgh_i = P_f+1/2 *ρv_f² +ρgh_f

P_i= P_f

1/2 *ρv_i² +ρgh_i = P_f+1/2 *ρv_f² +ρgh_f

½ρ(v_i² – v_f²) = ρg(h_f - h_i)

(v_i² – v_f²) = 2g(h_f - h_i)

v_f²= v_i² –2g(h_f - h_i)

v_f=sqrt[v_i² –2g(h_f - h_i)]

Plug values, v_i= 16m/s, h_f - h_i=4m

v_f=sqrt[16² –2*10*10]

v_f=7.5 m/s