# Bernoulli's Principle (Fluids)

**Bernoulli's principle** states that the pressure of a fluid *decreases* when either the velocity of the fluid or the height of the fluid *increases*. Bernoulli's principles is integral to the design of airplane wings and ventilation systems. The actual equation itself resembles conservation of energy, however, in lieu of studying the motion of an individual particle, Bernoulli's principle generalizes for a collection of particles with a uniform density.

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## Derivation

Bernoulli's principle actually relates pressure to two separate phenomena. One is the idea that if a fluid moves faster, the individual particles will spread out more, decreasing the pressure on the surroundings. The other is the idea that as the amount of fluid above decreases, the particles will be less compacted, thus exhibiting less apparent pressure.

In both cases, the density of the fluid provides a "default" closeness of the particles in the fluid, so density is incorporated into the principle as well.

$P_A + \frac12 \rho v_A^2 + \rho gh_A = P_B + \frac12 \rho v_B^2 + \rho gh_B$

Fluid flows from one pipe into a second of smaller cross-sectional area. Which pipe feels a greater pressure from the fluid? (Assume they are on level ground.)

By continuity, $A_1 v_1 = A_2 v_2$. Since $A_1 > A_2$, $v_1 < v_2$. Bernoulli's principle for two sections of pipe at the same height is just $P_1 + \frac12 \rho v_1^2 = P_2 + \frac12 \rho v_2^2.$ Since equality must be maintained, $v_1 < v_2$ implies $P_1 > P_2$. $_\square$

Note:Perhaps counterintuitively, the pressure is greater in the wider section with the lower fluid velocity. As an analogy, consider a traffic jam caused by road construction which takes two lanes down to one. In this scenario, the "congestion" (or pressure) is worse in the wider two-lane section, where drivers are being forced to consolidate into just one lane, than in the skinnier one-lane section later.

A chimney near the surface of the earth is used to draw the smoke out of a fireplace and carry it up and out of the building. If it is determined that a $38 \text{ Pa}$ pressure difference is required to pull the smoke through a chimney, how tall in meters should the chimney be? (Assume the air around the fireplace and the top of the chimney is still.)

Since the air is still at the base and the top, $\frac12 \rho v^2 = \frac12 \rho (0)^2 = 0.$ So Bernoulli's equation simplifies to $P_A + \rho g h_A = P_B + \rho g h_B.$ Let the fireplace be location A and the top of the chimney be location B. The height of the chimney is then $h_B - h_A$. Solving the equation above for his relation yields $h_B - h_A = \frac{P_A-P_B}{\rho g}.$ The pressure difference provided is $38 \text{ Pa}$, density of air is $1.29 {\text{ kg}^3}/\text{m}$, and the acceleration due to gravity near the surface of the earth is $9.8 \text{ m}/\text{s}^2$. So, $h_B - h_A = \frac{38}{(1.29)(9.8)} = 3.0 \text{ (m)}.$

$1~\mbox{atm}$? Give your answer in **meters**.

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**Details and Assumptions:**

- $1~\mbox{atm}=101,325~\mbox{Pa}$.
- The acceleration of gravity is $-9.8~\mbox{m/s}^2$.
- The density of water is $1~\mbox{g/cm}^3$.

$0.1~\mbox{m}$ and a spigot of radius $0.01~\mbox{m}$. Initially the lemonade is at a height of $0.5~\mbox{m}$ in the vat. You then open the spigot. How long does it take for all the lemonade to flow out of the vat in **seconds**?

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**Details and Assumptions:**

- The vat is open to the air at the top.
- The acceleration due to gravity is $-9.8~\mbox{m/s}^2$.

## Formal Proof

In order to describe fluids in active flow, start with **the conservation of matter**: when fluid moves from one position to the next, it must do so in such a way that no fluid matter is destroyed.

For instance, if fluid is injected into the mouth of a tube at a rate of $J_\text{in}=1\text{ L}$ per second, $1\text{ L}$ of the fluid should come out the other side every second, i.e. $J_\text{in}=J_\text{out}$.

For a toy model, consider the device below that consists of one level section of tube $\Gamma_\text{A}$ with radius $r_A$, connected by a linker section to another level tube $\Gamma_\text{B}$ with radius $r_B$.

Find a relationship that connects the velocity and pressure of the fluid in either section of tube.

To start, take the system to be the fluid that's between the discs $\partial_\text{A}$ and $\partial_\text{B}$ at time zero, called $\Sigma$. In order for $\Sigma$ to flow to the right, there must be a net force to push it along. If the fluid pressure in $\Gamma_\text{A}$, $P_A$, is greater than the fluid pressure in $\Gamma_{B}$, $P_B$, then the fluid to the left of $\Sigma$, $\Sigma_L$, will push with greater force than the fluid to the right, $\Sigma_R$, and hence $\Sigma$ will flow. These two forces will be $P_A \pi r_A^2$ and $P_B\pi r_B^2$, respectively.

Appealing to the work-energy principle, $W = \Delta\text{K.E.}$ Here, the work is performed by the two pressures in moving $\Sigma$ along the tube. Consider the flow undertaken in some span of time $\Delta t$. In $\Gamma_A$, $\Sigma$ will move the distance $v_A\Delta t$, and in $\Gamma_B$, $\Sigma$ will move the distance $v_B\Delta t$.

Hence, the net work $W$ on $\Sigma$ is given by the work done on $\Sigma$ by $\Sigma_L$, $P_A \pi r_A^2 v_A \Delta t$, minus the work done by $\Sigma$ to $\Sigma_R$, $P_B \pi r_B^2 v_B \Delta t$:

$W = P_A \pi r_A^2 v_A \Delta t - P_B \pi r_B^2 v_B \Delta t.$

However, the conservation condition $J_\text{in} = J_\text{out}$ applies for the discs $\Gamma_A$ and $\Gamma_B$. Hence, it must be true that $\pi r_A^2 v_A \Delta t = \pi r_A^2 v_A \Delta t$.

In other words, the volume of fluid $\Delta V$ that flows into $\Gamma_A$ is equal to the volume of fluid that flows out from $\Gamma_B$. Hence,

$W = \Delta V \left(P_A-P_B\right) = \frac{\Delta m}{\rho} \left(P_A-P_B\right).$

Now, the work done on $\Sigma$ is equal to the change in kinetic energy of the fluid. The kinetic energy of $\Sigma$ should be the same as before with the exception that the quantity $\Delta m$ of liquid which initially moved with velocity $v_A$ in $\Gamma_A$ is now travelling with velocity $v_B$ in $\Gamma_B$, and therefore $\Delta \text{K.E.} = \frac12 \Delta m v_B^2 - \frac12 \Delta m v_A^2$.

So, $\frac{\Delta m}{\rho} \left(P_A-P_B\right) = \Delta m \left( \frac12 v_B^2 - \frac12 v_A^2 \right)$, or $P_A + \frac12 \rho v_A^2 = P_B + \frac12 \rho v_B^2$, which is Bernoulli's relation for fluid flow in an arbitrary tube of level height.

In this derivation, the tubes were kept at equal level for simplicity's sake. It is trivial to recalculate the relation for the case when the two tube sections are of differing heights in a gravitational field, as occurs for the plumbing system in an apartment building. In this case, the work-energy principle is given by $W = \Delta \text{K.E.} + \Delta \text{P.E.}$ and thus the full Bernoulli relation is

$P_A + \frac12 \rho v_A^2 + \rho gh_A = P_B + \frac12 \rho v_B^2 + \rho gh_B.$

Note that the calculation did not depend in any way on the particular setup that we used (the two tubes and linker section). The same calculation applies to a tube of arbitrary shape which carries out an arbitrary trajectory through a gravitational field. Thus, the relation can be used to connect any two cross sections of a fluid's flow.

**Cite as:**Bernoulli's Principle (Fluids).

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/bernoullis-principle-fluids/