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+---
+title: "Bernoulli's theorem"
+firstLetter: "B"
+publishDate: 2021-04-02
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-04-02T15:05:08+02:00
+draft: false
+markup: pandoc
+---
+
+# Bernoulli's theorem
+
+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 $\va{v}$ and $\va{g}$ are constant in $t$,
+it becomes clear that a higher $\va{v}$ requires a lower $p$:
+
+$$\begin{aligned}
+    \frac{1}{2} \nabla \va{v}^2
+    = \va{g} - \frac{\nabla p}{\rho}
+\end{aligned}$$
+
+
+## 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.