AP Physics B : Fluid and Gas Mechanics

Study concepts, example questions & explanations for AP Physics B

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Example Questions

Example Question #1 : Understanding Bernoulli's Equation

An incompressible fluid flows without viscosity at \(\displaystyle 10 \frac{m}{s}\) through a pipe of radius \(\displaystyle 50 cm\). Without changing height, it flows into a pipe whose radius is \(\displaystyle 40 cm\). If the water pressure is \(\displaystyle 130 kPa\) in the \(\displaystyle 20 cm\) diameter pipe, what is its pressure in the \(\displaystyle 10 cm\) diameter pipe?

The mass density of the fluid is \(\displaystyle 1000 \frac{kg}{m^{3}}\).

Possible Answers:

\(\displaystyle 155.2 kPa\)

\(\displaystyle 57.9 kPa\)

\(\displaystyle 127.8 kPa\)

\(\displaystyle 201.3 kPa\)

\(\displaystyle 104 kPa\)

Correct answer:

\(\displaystyle 57.9 kPa\)

Explanation:

If the fluid is incompressible and flows without viscosity in a confined space, then an equal volume must move past any point in that space in the same time interval, even if the geometry of the confines changes.  An equal volume must pass Points \(\displaystyle A\) and \(\displaystyle B\) in the same time in the figure.

Pipe_jpeg

Since the diameter of the section on the right is smaller, the fluid must move faster in that region for an equal volume to pass by point \(\displaystyle B\) and point \(\displaystyle A\) in the same time.

In this problem, the flow speed in the larger section of pipe is \(\displaystyle 10 \frac{m}{s}\), so a \(\displaystyle 10 m\) cylinder of fluid passes point \(\displaystyle A\) in \(\displaystyle 1 s\). What length cylinder passes point \(\displaystyle B\) in that same second? It must be an equal volume, so we can use the equation for volume of a cylinder, where height is equal to the length of the pipe.

\(\displaystyle \pi r_{1}^{2}h_{1}=\pi r_{2}^{2}h_{2}\)

\(\displaystyle h_{1} = 10 m\)

\(\displaystyle h_{2} = (\frac{r_{1}}{r_{2}})^{2}h_{1} = (\frac{0.5 m}{0.4 m})^{2}(10 m) = 15.625 m\)

Hence, the flow speed in the smaller section of pipe is \(\displaystyle 15.625 \frac{m}{s}\), since this is the distance in the pipe that the fluid must travel per second.

Now use Bernoulli’s principle, which expresses conservation of energy in fluid flow as: 

Pressure + kinetic energy of flow + fluid potential energy = constant

Fluid potential energy depends on height (like gravitational potential energy), and so does not change from one part of the pipe to the other. So, for this case, Bernoulli's principle becomes:

\(\displaystyle p_{1} + \frac{1}{2}\rho v_{1}^{2} = p_{2} + \frac{1}{2} \rho v_{2}^{2}\)

\(\displaystyle p_{1}\) is the pressure in the larger pipe, \(\displaystyle \rho\) is the mass density of the fluid (constant for incompressible fluid), and \(\displaystyle v_{1}\) and \(\displaystyle v_{2}\) are, respectively, the flow speeds in the larger and smaller sections of the pipe. Putting in the known values and solving for \(\displaystyle p_{2}\) yields:

\(\displaystyle p_{2} = 130,000 Pa + \frac{1}{2}(1000 \frac{kg}{m^{3}})(100 \frac{m^{2}}{s^{2}} - 244.14 \frac{m^{2}}{s^{2}})\)

\(\displaystyle p_2= 130 kPa - 72.1 kPa\)

\(\displaystyle p_2= 57.9 kPa\)

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