Question

A Venturi meter is used to measure the speed of water flow in a pipe of cross-sectional area 0.01 m^2. A constriction (of cross-sectional area 0.002 m^2) is put in the pipe. Two vertical tubes, open to the atmosphere, rise from two points, one of which is in the constriction. Compare h1 and h2. Which one should be higher? Let’s say that h1 = 1.5 m and h2 = 0.6 m, find P1, P2, and the flow speed v1 and v2 in the pipe.

Answer 1

The problem applies Bernoulli's principle and continuity equation to calculate pressures P1 and P2, and flow speeds v1 and v2 of water in a Venturi meter with given cross-sectional areas and manometer fluid heights.

Step-by-step explanation:

The question involves applying the Bernoulli's principle and the continuity equation for incompressible fluid flow in a Venturi meter, which is a device used for measuring the flow speed of a fluid in a pipe. The cross-sectional areas of two sections of the pipe and the height difference in the manometer fluid enable us to calculate the pressures and velocities at those sections.

According to the continuity equation for an incompressible fluid, the product of cross-sectional area (A) and fluid velocity (v) is constant, so A1*v1 = A2*v2. Given that A1 is larger than A2, it implies that v2 must be greater than v1, meaning the fluid speeds up in the constricted part.

Applying Bernoulli's principle, we know that the total mechanical energy per unit volume is constant, so P1 + 1/2 ρv1^2 + ρgh1 = P2 + 1/2 ρv2^2 + ρgh2, where ρ is the density of water, P is pressure, v is velocity, g is the acceleration due to gravity, and h is the height of the fluid in the manometer. Rearranging the formula with the given h1 and h2 values, and knowing water is essentially incompressible, allows us to solve for P1 and P2. With P1 and P2, we can then find v1 and v2 using the continuity equation.