# Bernoulli's equation - Venturi meter

Problem statement:

The figure below shows a pipe that has two different cross-sectional areas, A1 = 25 cm2 and A2 = 4 cm2 respectively. The volumetric flow rate through the pipe is Q = 5 10-3 m3/s. A Venturi meter filled with mercury is placed in the pipe as shown in the figure. Determine:

1. the speed of water in both cross sectional areas of the pipe.
2. the pressure difference between them.
3. the difference in height h of mercury on the two sides of the U-tube.

Givens: ρ = 103 kg/m3; ρHg = 13.6 103 kg/m3; g = 10 m/s2

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Solution:

Let’s assume a steady flow through the pipe. In this conditions we can use both the continuity equation and Bernoulli’s equation to solve the problem.

The volumetric flow rate is defined as the volume of fluid flowing through the pipe per unit time. This flow rate is related to both the cross-sectional area of the pipe and the speed of the fluid, thus with the continuity equation.

If the fluid travels a distance Δx in a time interval Δt through the cross-sectional area A1, as shown in the figure above, we can write:

Isolating v1 and substituting the givens of the problem:

Make sure you use SI units (m2) for both cross-sectional areas.

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Using the continuity equation we can determine the speed of the fluid through A2:

To determine the pressure difference between both cross-sectional areas we use Bernoulli’s equation:

And substituting the two speeds calculated before as well as the density of water:

To determine the difference in height h of mercury on the two sides of the U-tube we are going to make use of Pascal’s principle, taking h = 0 at mercury’s surface (see figure below).

From Pascal’s law it follows that at h = 0, pressure is the same on both sides of the U-tube, so:

But h1 – h2 is the height h of the mercury column, therefore:

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