Suppose air, with a density of 1.29 kg/m3 is flowing into a Venturi meter. The n
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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 1.50 cm. How fast is the air moving at point B? m/s
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 1.50 cm. How fast is the air moving at point B? m/sExplanation / 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 * 0.015 * 9.8 = 147 So, 40 * 1.29 * v *v = 147 v*v = 2.8488 v = 1.68 m/s
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