Question

Suppose air, with a density of 1.29 kg/m3 is flowing into a Venturi meter. The narrow section of the pipe at point A has a diameter that is 0.402 times the diameter of the larger section of the pipe at point B. The U-shaped tube is filled with water and the difference in height between the two sections of pipe is h = 1.96 cm.

How fast is the air moving at point B?

Answer 1

"The equation states that
1/2 ρ v2 + P + ρgH = constant
As we are at a the same height, we can neglect the H part. "

" So, we have
1/2ρv2 + P = constant

Using the equation of continuity, we have
Velocity at A * Cross-section of A = Velocity at B * Cross section at B

So we have
Velocity at A = (1/0.402)² * Velocity at B (As the diameter is 0.402 times, the cross section is (0.402)² times) "

Let the velocity at B be x.

Now apply the derived Bernoulli equation.
(A) (B)
1/2ρ ((1/0.402)v)2 = 1/2 ρ (v)2 + P

2.6 ρ v2 = P

P is the pressure difference as shown by the Venturi meter. This is equal to
ρ * h * g = 1000 * 0.0196 * 9.8 = 192.08

So,
2.6 * 1.29 * v *v = 192.08
v*v = 57.2689
air velocity at B =v = 7.57 m/s "