Suppose air, with a density of 1.29 kg/m 3 is flowing into a Venturi meter. The

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 1/3 of 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 1.74 cm. How fast is the air moving at point B?

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 1/3 of 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 1.74 cm. How fast is the air moving at point B?

Explanation / Answer

Use the Bernoulli equation and the equation of constant volume flow.

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

So, we have
1/2?v2 + 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 = 9 * Velocity at B (As the diameter is 1/3 times, the cross section is 1/9 times)

Let the velocity at B be x.

Now apply the derived Bernoulli equation.
(A) (B)
1/2? (9v)2 = 1/2 ? (v)2 + P
1/2 ? (80) v2 = P
40 ? v2 = P

P is the pressure difference as shown by the Venturi meter. This is equal to
? * h * g = 1000 * 0.0174 * 9.8 = 170.52

So,
40 * 1.29 * v *v = 170.52
v*v = 3.304
v = 1.80 m/s

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