1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
|
---
title: "Bernoulli's theorem"
sort_title: "Bernoulli's theorem"
date: 2021-04-02
categories:
- Physics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---
For inviscid fluids, **Bernouilli'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.
|