summaryrefslogtreecommitdiff
path: root/source/know/concept/euler-equations/index.md
blob: 415e2f1b15e5c18a617404577e057695e47f0bf5 (plain)
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
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
---
title: "Euler equations"
sort_title: "Euler equations"
date: 2021-03-31
categories:
- Physics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---

The **Euler equations** are a system of partial differential equations
that govern the movement of **ideal fluids**,
i.e. fluids without [viscosity](/know/concept/viscosity/).



## Incompressible fluids

In a fluid moving according to the velocity field $$\va{v}(\va{r}, t)$$,
the acceleration felt by a particle is given by
the **material acceleration field** $$\va{w}(\va{r}, t)$$,
which is the [material derivative](/know/concept/material-derivative/) of $$\va{v}$$:

$$\begin{aligned}
    \va{w}
    \equiv \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
    = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
\end{aligned}$$

This infinitesimal particle obeys Newton's second law,
which can be written as follows:

$$\begin{aligned}
    \va{w} m
    = \va{w} \rho \dd{V}
    = \va{f^*} \dd{V}
\end{aligned}$$

Where $$m$$ and $$\dd{V}$$ are the particle's mass and volume,
and $$\rho$$ is the fluid density, which we assume
to be constant in space and time in this case.
Now, the **effective force density** $$\va{f^*}$$ represents the net force-per-particle.
By dividing the law by $$\dd{V}$$, we find:

$$\begin{aligned}
    \rho \va{w}
    = \va{f^*}
\end{aligned}$$

Next, we want to find another expression for $$\va{f^*}$$.
We know that the overall force $$\va{F}$$ on an arbitrary volume $$V$$ of the fluid
is the sum of the gravity body force $$\va{F}_g$$,
and the pressure contact force $$\va{F}_p$$ on the enclosing surface $$\partial V$$.
Using the divergence theorem, we then find:

$$\begin{aligned}
    \va{F}
    = \va{F}_g + \va{F}_p
    = \int_V \rho \va{g} \dd{V} - \oint_{\partial V} p \dd{\va{S}}
    = \int_V (\rho \va{g} - \nabla p) \dd{V}
    = \int_V \va{f^*} \dd{V}
\end{aligned}$$

Where $$p(\va{r}, t)$$ is the pressure field,
and $$\va{g}(\va{r}, t)$$ is the gravitational acceleration field.
Combining this with Newton's law, we find the following equation for the force density:

$$\begin{aligned}
    \va{f^*}
    = \rho \va{w}
    = \rho \va{g} - \nabla p
\end{aligned}$$

Dividing this by $$\rho$$,
we get the first of the system of Euler equations:

$$\begin{aligned}
    \va{w}
    = \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
    = \va{g} - \frac{\nabla p}{\rho}
\end{aligned}$$

The last ingredient is incompressibility:
the same volume must simultaneously
be flowing in and out of an arbitrary enclosure $$\partial V$$.
Then, by the divergence theorem:

$$\begin{aligned}
    0
    = \oint_{\partial V} \va{v} \cdot \dd{\va{S}}
    = \int_V \nabla \cdot \va{v} \dd{V}
\end{aligned}$$

Since $$V$$ is arbitrary,
the integrand must vanish by itself,
leading to the **continuity relation**:

$$\begin{aligned}
    \nabla \cdot \va{v} = 0
\end{aligned}$$

Combining this with the equation for $$\va{w}$$,
we get a system of two coupled differential equations:
these are the Euler equations for an incompressible fluid
with spatially uniform density $$\rho$$:

$$\begin{aligned}
    \boxed{
        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
        = \va{g} - \frac{\nabla p}{\rho}
        \qquad \quad
        \nabla \cdot \va{v}
        = 0
    }
\end{aligned}$$



## Compressible fluids

If the fluid is compressible,
the condition $$\nabla \cdot \va{v} = 0$$ no longer holds,
so to update the equations we demand that mass is conserved:
the mass evolution of a volume $$V$$
is equal to the mass flow through its boundary $$\partial V$$.
Applying the divergence theorem again:

$$\begin{aligned}
    0
    = \dv{}{t}\int_V \rho \dd{V} + \oint_{\partial V} \rho \va{v} \cdot \dd{\va{S}}
    = \int_V \dv{\rho}{t} + \nabla \cdot (\rho \va{v}) \dd{V}
\end{aligned}$$

Since $$V$$ is arbitrary, the integrand must be zero.
The new **continuity equation** is therefore:

$$\begin{aligned}
    0
    = \dv{\rho}{t} + \nabla \cdot (\rho \va{v})
    = \dv{\rho}{t} + \va{v} \cdot \nabla \rho + \rho \nabla \cdot \va{v}
    = \frac{\mathrm{D} \rho}{\mathrm{D} t} + \rho \nabla \cdot \va{v}
\end{aligned}$$

When the fluid gets compressed in a certain location, thermodynamics
states that the pressure, temperature and/or entropy must increase there.
For simplicity, let us assume an *isothermal* and *isentropic* fluid,
such that only $$p$$ is affected by compression, and the
[fundamental thermodynamic relation](/know/concept/fundamental-relation-of-thermodynamics/)
reduces to $$\dd{E} = - p \dd{V}$$.

Then the pressure is given by a thermodynamic equation of state $$p(\rho, T)$$,
which depends on the system being studied
(e.g. the ideal gas law $$p = \rho R T$$).
However, the quantity in control of the dynamics
is not $$p$$, but the internal energy $$E$$.
Dividing the fundamental thermodynamic relation by $$m \: \mathrm{D}t$$,
where $$m$$ is the mass of $$\dd{V}$$:

$$\begin{aligned}
    \frac{\mathrm{D} e}{\mathrm{D} t}
    = - p \frac{\mathrm{D} v}{\mathrm{D} t}
\end{aligned}$$

With $$e$$ and $$v$$ the specific (i.e. per unit mass)
internal energy and volume.
Using that $$\rho = 1 / v$$,
and substituting the above continuity relation:

$$\begin{aligned}
    \frac{\mathrm{D} e}{\mathrm{D} t}
    = - p \frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{1}{\rho} \Big)
    = \frac{p}{\rho^2} \frac{\mathrm{D} \rho}{\mathrm{D} t}
    = - \frac{p}{\rho} \nabla \cdot \va{v}
\end{aligned}$$

It makes sense to see a factor $$-\nabla \cdot \va{v}$$ here:
an incoming flow increases $$e$$.
This gives us the time-evolution of $$e$$ due to compression,
but its initial value is another equation of state $$e(\rho, T)$$.

Putting it all together,
Euler's system of equations now takes the following form:

$$\begin{aligned}
    \boxed{
        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
        = \va{g} - \frac{\nabla p}{\rho}
        \qquad \quad
        \frac{\mathrm{D} \rho}{\mathrm{D} t}
        = - \rho \nabla \cdot \va{v}
        \qquad \quad
        \frac{\mathrm{D} e}{\mathrm{D} t}
        = - \frac{p}{\rho} \nabla \cdot \va{v}
    }
\end{aligned}$$

What happens if the fluid is actually incompressible,
so $$\nabla \cdot \va{v} = 0$$ holds again? Clearly:

$$\begin{aligned}
    \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
    = \va{g} - \frac{\nabla p}{\rho}
    \qquad \quad
    \frac{\mathrm{D} \rho}{\mathrm{D} t}
    = 0
    \qquad \quad
    \frac{\mathrm{D} e}{\mathrm{D} t}
    = 0
\end{aligned}$$

So $$e$$ is constant, which is in fact equivalent to saying that $$\nabla \cdot \va{v} = 0$$.
The equation for $$\rho$$ enforces conservation of mass
for inhomogeneous fluids, i.e. fluids that are "lumpy",
but where the size of the lumps is conserved by incompressibility.



## References
1.  B. Lautrup,
    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
    CRC Press.