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AP Physics B : Understanding Bernoulli's Equation

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Example Question #1 : Liquids

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

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

Possible Answers:

57.9 kPa

201.3 kPa

155.2 kPa

104 kPa

127.8 kPa

Correct answer:

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 and 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 and point in the same time.

In this problem, the flow speed in the larger section of pipe is $10 \frac{m}{s}$ , so a 10 m cylinder of fluid passes point in 1 s . What length cylinder passes point 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.

$\pi r_{1}^{2}h_{1}=\pi r_{2}^{2}h_{2}$

$h_{1} = 10 m$

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

Hence, the flow speed in the smaller section of pipe is $15.625 \frac{m}{s}$ , 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:

$p_{1} + \frac{1}{2}\rho v_{1}^{2} = p_{2} + \frac{1}{2} \rho v_{2}^{2}$

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

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

p_2= 130 kPa - 72.1 kPa

p_2= 57.9 kPa

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AP Physics B : Understanding Bernoulli's Equation

Study concepts, example questions & explanations for AP Physics B

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Example Question #1 : Liquids

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