From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 14 Oct 2022 23:25:28 +0200 Subject: Initial commit after migration from Hugo --- source/know/concept/bernoullis-theorem/index.md | 85 +++++++++++++++++++++++++ 1 file changed, 85 insertions(+) create mode 100644 source/know/concept/bernoullis-theorem/index.md (limited to 'source/know/concept/bernoullis-theorem') diff --git a/source/know/concept/bernoullis-theorem/index.md b/source/know/concept/bernoullis-theorem/index.md new file mode 100644 index 0000000..c7f7483 --- /dev/null +++ b/source/know/concept/bernoullis-theorem/index.md @@ -0,0 +1,85 @@ +--- +title: "Bernoulli's theorem" +date: 2021-04-02 +categories: +- Physics +- Fluid mechanics +- Fluid dynamics +layout: "concept" +--- + +For inviscid fluids, **Bernuilli's theorem** states +that an increase in flow velocity $\va{v}$ is paired +with a decrease in pressure $p$ and/or potential energy. +For a qualitative argument, look no further than +one of the [Euler equations](/know/concept/euler-equations/), +with a [material derivative](/know/concept/material-derivative/): + +$$\begin{aligned} + \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v} + = \va{g} - \frac{\nabla p}{\rho} +\end{aligned}$$ + +Assuming that $\va{v}$ is constant in $t$, +it becomes clear that a higher $\va{v}$ requires a lower $p$. + + +## Simple form + +For an incompressible fluid +with a time-independent velocity field $\va{v}$ (i.e. **steady flow**), +Bernoulli's theorem formally states that the +**Bernoulli head** $H$ is constant along a streamline: + +$$\begin{aligned} + \boxed{ + H + = \frac{1}{2} \va{v}^2 + \Phi + \frac{p}{\rho} + } +\end{aligned}$$ + +Where $\Phi$ is the gravitational potential, such that $\va{g} = - \nabla \Phi$. +To prove this theorem, we take the material derivative of $H$: + +$$\begin{aligned} + \frac{\mathrm{D} H}{\mathrm{D} t} + &= \va{v} \cdot \frac{\mathrm{D} \va{v}}{\mathrm{D} t} + + \frac{\mathrm{D} \Phi}{\mathrm{D} t} + + \frac{1}{\rho} \frac{\mathrm{D} p}{\mathrm{D} t} +\end{aligned}$$ + +In the first term we insert the Euler equation, +and in the other two we expand the derivatives: + +$$\begin{aligned} + \frac{\mathrm{D} H}{\mathrm{D} t} + &= \va{v} \cdot \Big( \va{g} - \frac{\nabla p}{\rho} \Big) + + \Big( \pdv{\Phi}{t} + (\va{v} \cdot \nabla) \Phi \Big) + + \frac{1}{\rho} \Big( \pdv{p}{t} + (\va{v} \cdot \nabla) p \Big) + \\ + &= \pdv{\Phi}{t} + \frac{1}{\rho} \pdv{p}{t} + + \va{v} \cdot \big( \va{g} + \nabla \Phi \big) + \va{v} \cdot \Big( \frac{\nabla p}{\rho} - \frac{\nabla p}{\rho} \Big) +\end{aligned}$$ + +Using the fact that $\va{g} = - \nabla \Phi$, +we are left with the following equation: + +$$\begin{aligned} + \frac{\mathrm{D} H}{\mathrm{D} t} + &= \pdv{\Phi}{t} + \frac{1}{\rho} \pdv{p}{t} +\end{aligned}$$ + +Assuming that the flow is steady, both derivatives vanish, +leading us to the conclusion that $H$ is conserved along the streamline. + +In fact, there exists **Bernoulli's stronger theorem**, +which states that $H$ is constant *everywhere* in regions with +zero [vorticity](/know/concept/vorticity/) $\va{\omega} = 0$. +For a proof, see the derivation of $\va{\omega}$'s equation of motion. + + +## References +1. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. -- cgit v1.2.3