**Problem statement:**

The figure below shows a pipe that has two different cross-sectional areas, A_{1} = 25 cm^{2} and A_{2} = 4 cm^{2} respectively. The volumetric flow rate through the pipe is Q = 5 10^{-3} m^{3}/s. A Venturi meter filled with mercury is placed in the pipe as shown in the figure. Determine:

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

__Givens__: ρ = 10^{3} kg/m^{3}; ρ_{Hg} = 13.6 10^{3} kg/m^{3}; g = 10 m/s^{2}

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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 A_{1}, as shown in the figure above, we can write:

Isolating v_{1} and substituting the givens of the problem:

**Make sure you use SI units (m ^{2}) for both cross-sectional areas**.

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Using the continuity equation we can determine the speed of the fluid through A_{2}:

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 h_{1} – h_{2} is the height h of the mercury column, therefore: