Properties

Although Euler first presented these equations in 1755, many fundamental questions or concepts about them remain unanswered.

In three space dimensions, in certain simplified scenarios, the Euler equations produce singularities.^[6]

Smooth solutions of the free (in the sense of without source term: g=0) equations satisfy the conservation of specific kinetic energy:

$${\displaystyle {\partial \over \partial t}\left({\frac {1}{2}}u^{2}\right)+\nabla \cdot \left(u^{2}\mathbf {u} +w\mathbf {u} \right)=0}$$

In the one-dimensional case without the source term (both pressure gradient and external force), the momentum equation becomes the inviscid Burgers equation:

$${\displaystyle {\partial u \over \partial t}+u{\partial u \over \partial x}=0}$$

This model equation gives many insights into Euler equations.

Nondimensionalisation

In order to make the equations dimensionless, a characteristic length ${\displaystyle r_{0}}$, and a characteristic velocity ${\displaystyle u_{0}}$, need to be defined. These should be chosen such that the dimensionless variables are all of order one. The following dimensionless variables are thus obtained:

$${\displaystyle {\begin{aligned}u^{*}&\equiv {\frac {u}{u_{0}}},&r^{*}&\equiv {\frac {r}{r_{0}}},\\[5pt]t^{*}&\equiv {\frac {u_{0}}{r_{0}}}t,&p^{*}&\equiv {\frac {w}{u_{0}^{2}}},\\[5pt]\nabla ^{*}&\equiv r_{0}\nabla \end{aligned}}}$$

and of the field unit vector:

$${\displaystyle {\hat {\mathbf {g} }}\equiv {\frac {\mathbf {g} }{g}},}$$

Substitution of these inversed relations in Euler equations, defining the Froude number, yields (omitting the * at apix):

Euler equations in the Froude limit (no external field) are named free equations and are conservative. The limit of high Froude numbers (low external field) is thus notable and can be studied with perturbation theory.

Conservation form

The conservation form emphasizes the mathematical properties of Euler equations, and especially the contracted form is often the most convenient one for computational fluid dynamics simulations. Computationally, there are some advantages in using the conserved variables. This gives rise to a large class of numerical methods called conservative methods.^[1]

The free Euler equations are conservative, in the sense they are equivalent to a conservation equation:

$${\displaystyle {\frac {\partial \mathbf {y} }{\partial t}}+\nabla \cdot \mathbf {F} ={\mathbf {0} },}$$

or simply in Einstein notation:

$${\displaystyle {\frac {\partial y_{j}}{\partial t}}+{\frac {\partial f_{ij}}{\partial r_{i}}}=0_{i},}$$

where the conservation quantity ${\displaystyle \mathbf {y} }$ in this case is a vector, and ${\displaystyle \mathbf {F} }$ is a flux matrix. This can be simply proved.

Demonstration of the conservation form

First, the following identities hold:

$${\displaystyle \nabla \cdot (w\mathbf {I} )=\mathbf {I} \cdot \nabla w+w\nabla \cdot \mathbf {I} =\nabla w}$$

$${\displaystyle \mathbf {u} \cdot \nabla \cdot \mathbf {u} =\nabla \cdot (\mathbf {u} \otimes \mathbf {u} )}$$

where ${\displaystyle \otimes }$ denotes the outer product. The same identities expressed in Einstein notation are:

$${\displaystyle \partial _{i}\left(w\delta _{ij}\right)=\delta _{ij}\partial _{i}w+w\partial _{i}\delta _{ij}=\delta _{ij}\partial _{i}w=\partial _{j}w}$$

$${\displaystyle u_{j}\partial _{i}u_{i}=\partial _{i}\left(u_{i}u_{j}\right)}$$

where I is the identity matrix with dimension N and δ_ij its general element, the Kroenecker delta.

Thanks to these vector identities, the incompressible Euler equations with constant and uniform density and without external field can be put in the so-called conservation (or Eulerian) differential form, with vector notation:

$${\displaystyle \left\{{\begin{aligned}{\partial \mathbf {u} \over \partial t}+\nabla \cdot \left(\mathbf {u} \otimes \mathbf {u} +w\mathbf {I} \right)&=\mathbf {0} \\{\partial 0 \over \partial t}+\nabla \cdot \mathbf {u} &=0,\end{aligned}}\right.}$$

or with Einstein notation:

$${\displaystyle \left\{{\begin{aligned}\partial _{t}u_{j}+\partial _{i}\left(u_{i}u_{j}+w\delta _{ij}\right)&=0\\\partial _{t}0+\partial _{j}u_{j}&=0,\end{aligned}}\right.}$$

Then incompressible Euler equations with uniform density have conservation variables:

$${\displaystyle \mathbf {y} ={\begin{pmatrix}\mathbf {u} \\0\end{pmatrix}};\qquad \mathbf {F} ={\begin{pmatrix}\mathbf {u} \otimes \mathbf {u} +w\mathbf {I} \\\mathbf {u} \end{pmatrix}}.}$$

Note that in the second component u is by itself a vector, with length N, so y has length N+1 and F has size N(N+1). In 3D for example y has length 4, I has size 3×3 and F has size 4×3, so the explicit forms are:

$${\displaystyle {\mathbf {y} }={\begin{pmatrix}u_{1}\\u_{2}\\u_{3}\\0\end{pmatrix}};\quad {\mathbf {F} }={\begin{pmatrix}u_{1}^{2}+w&u_{1}u_{2}&u_{1}u_{3}\\u_{2}u_{1}&u_{2}^{2}+w&u_{2}u_{3}\\u_{3}u_{1}&u_{3}u_{2}&u_{3}^{2}+w\\u_{1}&u_{2}&u_{3}\end{pmatrix}}.}$$

At last Euler equations can be recast into the particular equation:

Spatial dimensions

For certain problems, especially when used to analyze compressible flow in a duct or in case the flow is cylindrically or spherically symmetric, the one-dimensional Euler equations are a useful first approximation. Generally, the Euler equations are solved by Riemann's method of characteristics. This involves finding curves in plane of independent variables (i.e., ${\displaystyle x}$ and ${\displaystyle t}$) along which partial differential equations (PDEs) degenerate into ordinary differential equations (ODEs). Numerical solutions of the Euler equations rely heavily on the method of characteristics.

Conservation form

The incompressible Euler equations in the Froude limit are equivalent to a single conservation equation with conserved quantity and associated flux respectively:

$${\displaystyle \mathbf {y} ={\begin{pmatrix}\rho \\\rho \mathbf {u} \\0\end{pmatrix}};\qquad {\mathbf {F} }={\begin{pmatrix}\rho \mathbf {u} \\\rho \mathbf {u} \otimes \mathbf {u} +p\mathbf {I} \\\mathbf {u} \end{pmatrix}}.}$$

Here ${\displaystyle \mathbf {y} }$ has length ${\displaystyle N+2}$ and ${\displaystyle \mathbf {F} }$ has size ${\displaystyle (N+2)N}$.^{[lower-alpha 1]} In general (not only in the Froude limit) Euler equations are expressible as:

$${\displaystyle {\frac {\partial }{\partial t}}{\begin{pmatrix}\rho \\\rho \mathbf {u} \\0\end{pmatrix}}+\nabla \cdot {\begin{pmatrix}\rho \mathbf {u} \\\rho \mathbf {u} \otimes \mathbf {u} +p\mathbf {I} \\\mathbf {u} \end{pmatrix}}={\begin{pmatrix}0\\\rho \mathbf {g} \\0\end{pmatrix}}}$$

Conservation variables

The variables for the equations in conservation form are not yet optimised. In fact we could define:

$${\displaystyle {\mathbf {y} }={\begin{pmatrix}\rho \\\mathbf {j} \\0\end{pmatrix}};\qquad {\mathbf {F} }={\begin{pmatrix}\mathbf {j} \\{\frac {1}{\rho }}\,\mathbf {j} \otimes \mathbf {j} +p\mathbf {I} \\{\frac {\mathbf {j} }{\rho }}\end{pmatrix}}.}$$

where ${\displaystyle \mathbf {j} =\rho \mathbf {u} }$ is the momentum density, a conservation variable.

where ${\displaystyle \mathbf {f} =\rho \mathbf {g} }$ is the force density, a conservation variable.

Incompressible constraint (revisited)

Coming back to the incompressible case, it now becomes apparent that the incompressible constraint typical of the former cases actually is a particular form valid for incompressible flows of the energy equation, and not of the mass equation. In particular, the incompressible constraint corresponds to the following very simple energy equation:

$${\displaystyle {\frac {De}{Dt}}=0}$$

Thus for an incompressible inviscid fluid the specific internal energy is constant along the flow lines, also in a time-dependent flow. The pressure in an incompressible flow acts like a Lagrange multiplier, being the multiplier of the incompressible constraint in the energy equation, and consequently in incompressible flows it has no thermodynamic meaning. In fact, thermodynamics is typical of compressible flows and degenerates in incompressible flows.^[7]

Basing on the mass conservation equation, one can put this equation in the conservation form:

$${\displaystyle {\partial \rho e \over \partial t}+\nabla \cdot (\rho e\mathbf {u} )=0}$$

meaning that for an incompressible inviscid nonconductive flow a continuity equation holds for the internal energy.

Enthalpy conservation

Since by definition the specific enthalpy is:

$${\displaystyle h=e+{\frac {p}{\rho }}}$$

The material derivative of the specific internal energy can be expressed as:

$${\displaystyle {De \over Dt}={Dh \over Dt}-{\frac {1}{\rho }}\left({Dp \over Dt}-{\frac {p}{\rho }}{D\rho \over Dt}\right)}$$

Then by substituting the momentum equation in this expression, one obtains:

$${\displaystyle {De \over Dt}={Dh \over Dt}-{\frac {1}{\rho }}\left(p\nabla \cdot \mathbf {u} +{Dp \over Dt}\right)}$$

And by substituting the latter in the energy equation, one obtains that the enthalpy expression for the Euler energy equation:

$${\displaystyle {Dh \over Dt}={\frac {1}{\rho }}{Dp \over Dt}}$$

In a reference frame moving with an inviscid and nonconductive flow, the variation of enthalpy directly corresponds to a variation of pressure.

Thermodynamics of ideal fluids

In thermodynamics the independent variables are the specific volume, and the specific entropy, while the specific energy is a function of state of these two variables.

Deduction of the form valid for thermodynamic systems

Considering the first equation, variable must be changed from density to specific volume. By definition:

$${\displaystyle v\equiv {\frac {1}{\rho }}}$$

Thus the following identities hold:

$${\displaystyle \nabla \rho =\nabla \left({\frac {1}{v}}\right)=-{\frac {1}{v^{2}}}\nabla v}$$

$${\displaystyle {\frac {\partial \rho }{\partial t}}={\frac {\partial }{\partial t}}\left({\frac {1}{v}}\right)=-{\frac {1}{v^{2}}}{\frac {\partial v}{\partial t}}}$$

Then by substituting these expressions in the mass conservation equation:

$${\displaystyle -{\frac {\mathbf {u} }{v^{2}}}\cdot \nabla v-{\frac {1}{v^{2}}}{\frac {\partial v}{\partial t}}=-{\frac {1}{v}}\nabla \cdot \mathbf {u} }$$

And by multiplication:

$${\displaystyle {\partial v \over \partial t}+\mathbf {u} \cdot \nabla v=v\nabla \cdot \mathbf {u} }$$

This equation is the only belonging to general continuum equations, so only this equation have the same form for example also in Navier-Stokes equations.

On the other hand, the pressure in thermodynamics is the opposite of the partial derivative of the specific internal energy with respect to the specific volume:

$${\displaystyle p(v,s)=-{\partial e(v,s) \over \partial v}}$$

since the internal energy in thermodynamics is a function of the two variables aforementioned, the pressure gradient contained into the momentum equation should be explicited as:

$${\displaystyle -\nabla p(v,s)=-{\frac {\partial p}{\partial v}}\nabla v-{\frac {\partial p}{\partial s}}\nabla s={\frac {\partial ^{2}e}{\partial v^{2}}}\nabla v+{\frac {\partial ^{2}e}{\partial v\partial s}}\nabla s}$$

It is convenient for brevity to switch the notation for the second order derivatives:

$${\displaystyle -\nabla p(v,s)=e_{vv}\nabla v+e_{vs}\nabla s}$$

Finally, the energy equation:

$${\displaystyle {De \over Dt}=-pv\nabla \cdot \mathbf {u} }$$

can be further simplified in convective form by changing variable from specific energy to the specific entropy: in fact the first law of thermodynamics in local form can be written:

$${\displaystyle {De \over Dt}=T{Ds \over Dt}-p{Dv \over Dt}}$$

by substituting the material derivative of the internal energy, the energy equation becomes:

$${\displaystyle T{Ds \over Dt}+{\frac {p}{\rho ^{2}}}\left({D\rho \over Dt}+\rho \nabla \cdot \mathbf {u} \right)=0}$$

now the term between parenthesis is identically zero according to the conservation of mass, then the Euler energy equation becomes simply:

$${\displaystyle {Ds \over Dt}=0}$$

For a thermodynamic fluid, the compressible Euler equations are consequently best written as:

where:

• ${\displaystyle v}$ is the specific volume
• ${\displaystyle \mathbf {u} }$ is the flow velocity vector
• ${\displaystyle s}$ is the specific entropy

In the general case and not only in the incompressible case, the energy equation means that for an inviscid thermodynamic fluid the specific entropy is constant along the flow lines, also in a time-dependent flow. Basing on the mass conservation equation, one can put this equation in the conservation form:^[8]

$${\displaystyle {\partial \rho s \over \partial t}+\nabla \cdot (\rho s\mathbf {u} )=0}$$

meaning that for an inviscid nonconductive flow a continuity equation holds for the entropy.

On the other hand, the two second-order partial derivatives of the specific internal energy in the momentum equation require the specification of the fundamental equation of state of the material considered, i.e. of the specific internal energy as function of the two variables specific volume and specific entropy:

$${\displaystyle e=e(v,s)}$$

The fundamental equation of state contains all the thermodynamic information about the system (Callen, 1985),^[9] exactly like the couple of a thermal equation of state together with a caloric equation of state.

Conservation form

The Euler equations in the Froude limit are equivalent to a single conservation equation with conserved quantity and associated flux respectively:

$${\displaystyle \mathbf {y} ={\begin{pmatrix}\rho \\\mathbf {j} \\E^{t}\end{pmatrix}};\qquad {\mathbf {F} }={\begin{pmatrix}\mathbf {j} \\{\frac {1}{\rho }}\mathbf {j} \otimes \mathbf {j} +p\mathbf {I} \\\left(E^{t}+p\right){\frac {1}{\rho }}\mathbf {j} \end{pmatrix}}.}$$

where:

• ${\displaystyle \mathbf {j} =\rho \mathbf {u} }$ is the momentum density, a conservation variable.
• ${\textstyle E^{t}=\rho e+{\frac {1}{2}}\rho u^{2}}$ is the total energy density (total energy per unit volume).

Here ${\displaystyle \mathbf {y} }$ has length N + 2 and ${\displaystyle \mathbf {F} }$ has size N(N + 2).^{[lower-alpha 2]} In general (not only in the Froude limit) Euler equations are expressible as:

where ${\displaystyle \mathbf {f} =\rho \mathbf {g} }$ is the force density, a conservation variable.

We remark that also the Euler equation even when conservative (no external field, Froude limit) have no Riemann invariants in general.^[10] Some further assumptions are required

However, we already mentioned that for a thermodynamic fluid the equation for the total energy density is equivalent to the conservation equation:

$${\displaystyle {\partial \over \partial t}(\rho s)+\nabla \cdot (\rho s\mathbf {u} )=0}$$

Then the conservation equations in the case of a thermodynamic fluid are more simply expressed as:

where ${\displaystyle S=\rho s}$ is the entropy density, a thermodynamic conservation variable.

Another possible form for the energy equation, being particularly useful for isobarics, is:

$${\displaystyle {\frac {\partial H^{t}}{\partial t}}+\nabla \cdot \left(H^{t}\mathbf {u} \right)=\mathbf {u} \cdot \mathbf {f} -{\frac {\partial p}{\partial t}}}$$

where ${\textstyle H^{t}=E^{t}+p=\rho e+p+{\frac {1}{2}}\rho u^{2}}$ is the total enthalpy density.

Characteristic equations

The compressible Euler equations can be decoupled into a set of N+2 wave equations that describes sound in Eulerian continuum if they are expressed in characteristic variables instead of conserved variables.

In fact the tensor A is always diagonalizable. If the eigenvalues (the case of Euler equations) are all real the system is defined hyperbolic, and physically eigenvalues represent the speeds of propagation of information.^[11] If they are all distinguished, the system is defined strictly hyperbolic (it will be proved to be the case of one-dimensional Euler equations). Furthermore, diagonalisation of compressible Euler equation is easier when the energy equation is expressed in the variable entropy (i.e. with equations for thermodynamic fluids) than in other energy variables. This will become clear by considering the 1D case.

If ${\displaystyle \mathbf {p} _{i}}$ is the right eigenvector of the matrix ${\displaystyle \mathbf {A} }$ corresponding to the eigenvalue ${\displaystyle \lambda _{i}}$, by building the projection matrix:

$${\displaystyle \mathbf {P} =\left[\mathbf {p} _{1},\mathbf {p} _{2},...,\mathbf {p} _{n}\right]}$$

One can finally find the characteristic variables as:

$${\displaystyle \mathbf {w} =\mathbf {P} ^{-1}\mathbf {y} ,}$$

Since A is constant, multiplying the original 1-D equation in flux-Jacobian form with P⁻¹ yields the characteristic equations:^[12]

$${\displaystyle {\frac {\partial w_{i}}{\partial t}}+\lambda _{j}{\frac {\partial w_{i}}{\partial r_{j}}}=0_{i}}$$

The original equations have been decoupled into N+2 characteristic equations each describing a simple wave, with the eigenvalues being the wave speeds. The variables w_i are called the characteristic variables and are a subset of the conservative variables. The solution of the initial value problem in terms of characteristic variables is finally very simple. In one spatial dimension it is:

$${\displaystyle w_{i}(x,t)=w_{i}\left(x-\lambda _{i}t,0\right)}$$

Then the solution in terms of the original conservative variables is obtained by transforming back:

$${\displaystyle \mathbf {y} =\mathbf {P} \mathbf {w} ,}$$

this computation can be explicited as the linear combination of the eigenvectors:

$${\displaystyle \mathbf {y} (x,t)=\sum _{i=1}^{m}w_{i}\left(x-\lambda _{i}t,0\right)\mathbf {p} _{i},}$$

Now it becomes apparent that the characteristic variables act as weights in the linear combination of the jacobian eigenvectors. The solution can be seen as superposition of waves, each of which is advected independently without change in shape. Each i-th wave has shape w_ip_i and speed of propagation λ_i. In the following we show a very simple example of this solution procedure.

Waves in 1D inviscid, nonconductive thermodynamic fluid

If one considers Euler equations for a thermodynamic fluid with the two further assumptions of one spatial dimension and free (no external field: g = 0):

$${\displaystyle \left\{{\begin{aligned}{\partial v \over \partial t}+u{\partial v \over \partial x}-v{\partial u \over \partial x}&=0\\[1.2ex]{\partial u \over \partial t}+u{\partial u \over \partial x}-e_{vv}v{\partial v \over \partial x}-e_{vs}v{\partial s \over \partial x}&=0\\[1.2ex]{\partial s \over \partial t}+u{\partial s \over \partial x}&=0\end{aligned}}\right.}$$

If one defines the vector of variables:

$${\displaystyle \mathbf {y} ={\begin{pmatrix}v\\u\\s\end{pmatrix}}}$$

recalling that ${\displaystyle v}$ is the specific volume, ${\displaystyle u}$ the flow speed, ${\displaystyle s}$ the specific entropy, the corresponding jacobian matrix is:

$${\displaystyle {\mathbf {A} }={\begin{pmatrix}u&-v&0\\-e_{vv}v&u&-e_{vs}v\\0&0&u\end{pmatrix}}.}$$

At first one must find the eigenvalues of this matrix by solving the characteristic equation:

$${\displaystyle \det(\mathbf {A} (\mathbf {y} )-\lambda (\mathbf {y} )\mathbf {I} )=0}$$

that is explicitly:

$${\displaystyle \det {\begin{bmatrix}u-\lambda &-v&0\\-e_{vv}v&u-\lambda &-e_{vs}v\\0&0&u-\lambda \end{bmatrix}}=0}$$

This determinant is very simple: the fastest computation starts on the last row, since it has the highest number of zero elements.

$${\displaystyle (u-\lambda )\det {\begin{bmatrix}u-\lambda &-v\\-e_{vv}v&u-\lambda \end{bmatrix}}=0}$$

Now by computing the determinant 2×2:

$${\displaystyle (u-\lambda )\left((u-\lambda )^{2}-e_{vv}v^{2}\right)=0}$$

by defining the parameter:

$${\displaystyle a(v,s)\equiv v{\sqrt {e_{vv}}}}$$

or equivalently in mechanical variables, as:

$${\displaystyle a(\rho ,p)\equiv {\sqrt {\partial p \over \partial \rho }}}$$

This parameter is always real according to the second law of thermodynamics. In fact the second law of thermodynamics can be expressed by several postulates. The most elementary of them in mathematical terms is the statement of convexity of the fundamental equation of state, i.e. the hessian matrix of the specific energy expressed as function of specific volume and specific entropy:

$${\displaystyle {\begin{pmatrix}e_{vv}&e_{vs}\\e_{vs}&e_{ss}\end{pmatrix}}}$$

is defined positive. This statement corresponds to the two conditions:

$${\displaystyle \left\{{\begin{aligned}e_{vv}&>0\\[1.2ex]e_{vv}e_{ss}-e_{vs}^{2}&>0\end{aligned}}\right.}$$

The first condition is the one ensuring the parameter a is defined real.

The characteristic equation finally results:

$${\displaystyle (u-\lambda )\left((u-\lambda )^{2}-a^{2}\right)=0}$$

That has three real solutions:

$${\displaystyle \lambda _{1}(v,u,s)=u-a(v,s)\quad \lambda _{2}(u)=u,\quad \lambda _{3}(v,u,s)=u+a(v,s)}$$

Then the matrix has three real eigenvalues all distinguished: the 1D Euler equations are a strictly hyperbolic system.

At this point one should determine the three eigenvectors: each one is obtained by substituting one eigenvalue in the eigenvalue equation and then solving it. By substituting the first eigenvalue λ₁ one obtains:

$${\displaystyle {\begin{pmatrix}a&-v&0\\-e_{vv}v&a&-e_{vs}v\\0&0&a\end{pmatrix}}{\begin{pmatrix}v_{1}\\u_{1}\\s_{1}\end{pmatrix}}=0}$$

Basing on the third equation that simply has solution s₁=0, the system reduces to:

$${\displaystyle {\begin{pmatrix}a&-v\\-a^{2}/v&a\end{pmatrix}}{\begin{pmatrix}v_{1}\\u_{1}\end{pmatrix}}=0}$$

The two equations are redundant as usual, then the eigenvector is defined with a multiplying constant. We choose as right eigenvector:

$${\displaystyle \mathbf {p} _{1}={\begin{pmatrix}v\\a\\0\end{pmatrix}}}$$

The other two eigenvectors can be found with analogous procedure as:

$${\displaystyle \mathbf {p} _{2}={\begin{pmatrix}e_{vs}\\0\\-\left({\frac {a}{v}}\right)^{2}\end{pmatrix}},\qquad \mathbf {p} _{3}={\begin{pmatrix}v\\-a\\0\end{pmatrix}}}$$

Then the projection matrix can be built:

$${\displaystyle \mathbf {P} (v,u,s)=(\mathbf {p} _{1},\mathbf {p} _{2},\mathbf {p} _{3})={\begin{pmatrix}v&e_{vs}&v\\a&0&-a\\0&-\left({\frac {a}{v}}\right)^{2}&0\end{pmatrix}}}$$

Finally it becomes apparent that the real parameter a previously defined is the speed of propagation of the information characteristic of the hyperbolic system made of Euler equations, i.e. it is the wave speed. It remains to be shown that the sound speed corresponds to the particular case of an isentropic transformation:

$${\displaystyle a_{s}\equiv {\sqrt {\left({\partial p \over \partial \rho }\right)_{s}}}}$$

Compressibility and sound speed

Sound speed is defined as the wavespeed of an isentropic transformation:

$${\displaystyle a_{s}(\rho ,p)\equiv {\sqrt {\left({\partial p \over \partial \rho }\right)_{s}}}}$$

by the definition of the isoentropic compressibility:

$${\displaystyle K_{s}(\rho ,p)\equiv {\frac {1}{\rho }}\left({\partial p \over \partial \rho }\right)_{s}}$$

the soundspeed results always the square root of ratio between the isentropic compressibility and the density:

$${\displaystyle a_{s}\equiv {\sqrt {\frac {K_{s}}{\rho }}}}$$

Ideal gas

The sound speed in an ideal gas depends only on its temperature:

$${\displaystyle a_{s}(T)={\sqrt {\gamma {\frac {T}{m}}}}}$$

Deduction of the form valid for ideal gases

In an ideal gas the isoentropic transformation is described by the Poisson's law:

$${\displaystyle d\left(p\rho ^{-\gamma }\right)_{s}=0}$$

where γ is the heat capacity ratio, a constant for the material. By explicitating the differentials:

$${\displaystyle \rho ^{-\gamma }(dp)_{s}+\gamma p\rho ^{-\gamma -1}(d\rho )_{s}=0}$$

and by dividing for ρ^−γ dρ:

$${\displaystyle \left({\partial p \over \partial \rho }\right)_{s}=\gamma p\rho }$$

Then by substitution in the general definitions for an ideal gas the isentropic compressibility is simply proportional to the pressure:

$${\displaystyle K_{s}(p)=\gamma p}$$

and the sound speed results (Newton–Laplace law):

$${\displaystyle a_{s}(\rho ,p)={\sqrt {\gamma {\frac {p}{\rho }}}}}$$

Notably, for an ideal gas the ideal gas law holds, that in mathematical form is simply:

$${\displaystyle p=nT}$$

where n is the number density, and T is the absolute temperature, provided it is measured in energetic units (i.e. in joules) through multiplication with the Boltzmann constant. Since the mass density is proportional to the number density through the average molecular mass m of the material:

$${\displaystyle \rho =mn}$$

The ideal gas law can be recast into the formula:

$${\displaystyle {\frac {p}{\rho }}={\frac {T}{m}}}$$

By substituting this ratio in the Newton–Laplace law, the expression of the sound speed into an ideal gas as function of temperature is finally achieved.

Since the specific enthalpy in an ideal gas is proportional to its temperature:

$${\displaystyle h=c_{p}T={\frac {\gamma }{\gamma -1}}{\frac {T}{m}}}$$

the sound speed in an ideal gas can also be made dependent only on its specific enthalpy:

$${\displaystyle a_{s}(h)={\sqrt {(\gamma -1)h}}}$$

Incompressible case and Lamb's form

The vector calculus identity of the cross product of a curl holds:

$${\displaystyle \mathbf {v\ \times } \left(\mathbf {\nabla \times F} \right)=\nabla _{F}\left(\mathbf {v\cdot F} \right)-\mathbf {v\cdot \nabla } \mathbf {F} \ ,}$$

where the Feynman subscript notation ${\displaystyle \nabla _{F}}$ is used, which means the subscripted gradient operates only on the factor ${\displaystyle \mathbf {F} }$.

Lamb in his famous classical book Hydrodynamics (1895), still in print, used this identity to change the convective term of the flow velocity in rotational form:^[13]

$${\displaystyle \mathbf {u} \cdot \nabla \mathbf {u} ={\frac {1}{2}}\nabla \left(u^{2}\right)+(\nabla \times \mathbf {u} )\times \mathbf {u} }$$

the Euler momentum equation in Lamb's form becomes:

$${\displaystyle {\frac {\partial \mathbf {u} }{\partial t}}+{\frac {1}{2}}\nabla \left(u^{2}\right)+(\nabla \times \mathbf {u} )\times \mathbf {u} +{\frac {\nabla p}{\rho }}=\mathbf {g} ={\frac {\partial \mathbf {u} }{\partial t}}+{\frac {1}{2}}\nabla \left(u^{2}\right)-\mathbf {u} \times (\nabla \times \mathbf {u} )+{\frac {\nabla p}{\rho }}}$$

Now, basing on the other identity:

$${\displaystyle \nabla \left({\frac {p}{\rho }}\right)={\frac {\nabla p}{\rho }}-{\frac {p}{\rho ^{2}}}\nabla \rho }$$

the Euler momentum equation assumes a form that is optimal to demonstrate Bernoulli's theorem for steady flows:

$${\displaystyle \nabla \left({\frac {1}{2}}u^{2}+{\frac {p}{\rho }}\right)-\mathbf {g} =-{\frac {p}{\rho ^{2}}}\nabla \rho +\mathbf {u} \times (\nabla \times \mathbf {u} )-{\frac {\partial \mathbf {u} }{\partial t}}}$$

In fact, in case of an external conservative field, by defining its potential φ:

$${\displaystyle \nabla \left({\frac {1}{2}}u^{2}+\phi +{\frac {p}{\rho }}\right)=-{\frac {p}{\rho ^{2}}}\nabla \rho +\mathbf {u} \times (\nabla \times \mathbf {u} )-{\frac {\partial \mathbf {u} }{\partial t}}}$$

In case of a steady flow the time derivative of the flow velocity disappears, so the momentum equation becomes:

$${\displaystyle \nabla \left({\frac {1}{2}}u^{2}+\phi +{\frac {p}{\rho }}\right)=-{\frac {p}{\rho ^{2}}}\nabla \rho +\mathbf {u} \times (\nabla \times \mathbf {u} )}$$

And by projecting the momentum equation on the flow direction, i.e. along a streamline, the cross product disappears because its result is always perpendicular to the velocity:

$${\displaystyle \mathbf {u} \cdot \nabla \left({\frac {1}{2}}u^{2}+\phi +{\frac {p}{\rho }}\right)=-{\frac {p}{\rho ^{2}}}\mathbf {u} \cdot \nabla \rho }$$

In the steady incompressible case the mass equation is simply:

$${\displaystyle \mathbf {u} \cdot \nabla \rho =0,}$$

that is the mass conservation for a steady incompressible flow states that the density along a streamline is constant. Then the Euler momentum equation in the steady incompressible case becomes:

$${\displaystyle \mathbf {u} \cdot \nabla \left({\frac {1}{2}}u^{2}+\phi +{\frac {p}{\rho }}\right)=0}$$

The convenience of defining the total head for an inviscid liquid flow is now apparent:

$${\displaystyle b_{l}\equiv {\frac {1}{2}}u^{2}+\phi +{\frac {p}{\rho }},}$$

which may be simply written as:

$${\displaystyle \mathbf {u} \cdot \nabla b_{l}=0}$$

That is, the momentum balance for a steady inviscid and incompressible flow in an external conservative field states that the total head along a streamline is constant.

Compressible case

In the most general steady (compressible) case the mass equation in conservation form is:

$${\displaystyle \nabla \cdot \mathbf {j} =\rho \nabla \cdot \mathbf {u} +\mathbf {u} \cdot \nabla \rho =0.}$$

Therefore, the previous expression is rather

$${\displaystyle \mathbf {u} \cdot \nabla \left({\frac {1}{2}}u^{2}+\phi +{\frac {p}{\rho }}\right)={\frac {p}{\rho }}\nabla \cdot \mathbf {u} }$$

The right-hand side appears on the energy equation in convective form, which on the steady state reads:

$${\displaystyle \mathbf {u} \cdot \nabla e=-{\frac {p}{\rho }}\nabla \cdot \mathbf {u} }$$

The energy equation therefore becomes:

$${\displaystyle \mathbf {u} \cdot \nabla \left(e+{\frac {p}{\rho }}+{\frac {1}{2}}u^{2}+\phi \right)=0,}$$

so that the internal specific energy now features in the head.

Since the external field potential is usually small compared to the other terms, it is convenient to group the latter ones in the total enthalpy:

$${\displaystyle h^{t}\equiv e+{\frac {p}{\rho }}+{\frac {1}{2}}u^{2}}$$

and the Bernoulli invariant for an inviscid gas flow is:

$${\displaystyle b_{g}\equiv h^{t}+\phi =b_{l}+e,}$$

which can be written as:

$${\displaystyle \mathbf {u} \cdot \nabla b_{g}=0}$$

That is, the energy balance for a steady inviscid flow in an external conservative field states that the sum of the total enthalpy and the external potential is constant along a streamline.

In the usual case of small potential field, simply:

$${\displaystyle \mathbf {u} \cdot \nabla h^{t}\sim 0}$$

Friedmann form and Crocco form

By substituting the pressure gradient with the entropy and enthalpy gradient, according to the first law of thermodynamics in the enthalpy form:

$${\displaystyle v\nabla p=-T\nabla s+\nabla h}$$

in the convective form of Euler momentum equation, one arrives to:

$${\displaystyle {\frac {D\mathbf {u} }{Dt}}=T\nabla \,s-\nabla \,h}$$

Friedmann deduced this equation for the particular case of a perfect gas and published it in 1922.^[14] However, this equation is general for an inviscid nonconductive fluid and no equation of state is implicit in it.

On the other hand, by substituting the enthalpy form of the first law of thermodynamics in the rotational form of Euler momentum equation, one obtains:

$${\displaystyle {\frac {\partial \mathbf {u} }{\partial t}}+{\frac {1}{2}}\nabla \left(u^{2}\right)+(\nabla \times \mathbf {u} )\times \mathbf {u} +{\frac {\nabla p}{\rho }}=\mathbf {g} }$$

and by defining the specific total enthalpy:

$${\displaystyle h^{t}=h+{\frac {1}{2}}u^{2}}$$

one arrives to the Crocco–Vazsonyi form^[15] (Crocco, 1937) of the Euler momentum equation:

$${\displaystyle {\frac {\partial \mathbf {u} }{\partial t}}+(\nabla \times \mathbf {u} )\times \mathbf {u} -T\nabla s+\nabla h^{t}=\mathbf {g} }$$

In the steady case the two variables entropy and total enthalpy are particularly useful since Euler equations can be recast into the Crocco's form:

$${\displaystyle \left\{{\begin{aligned}\mathbf {u} \times \nabla \times \mathbf {u} +T\nabla s-\nabla h^{t}&=\mathbf {g} \\\mathbf {u} \cdot \nabla s&=0\\\mathbf {u} \cdot \nabla h^{t}&=0\end{aligned}}\right.}$$

Finally if the flow is also isothermal:

$${\displaystyle T\nabla s=\nabla (Ts)}$$

by defining the specific total Gibbs free energy:

$${\displaystyle g^{t}\equiv h^{t}+Ts}$$

the Crocco's form can be reduced to:

$${\displaystyle \left\{{\begin{aligned}\mathbf {u} \times \nabla \times \mathbf {u} -\nabla g^{t}&=\mathbf {g} \\\mathbf {u} \cdot \nabla g^{t}&=0\end{aligned}}\right.}$$

From these relationships one deduces that the specific total free energy is uniform in a steady, irrotational, isothermal, isoentropic, inviscid flow.

Rankine–Hugoniot equations

Starting from the simplest case, one consider a steady free conservation equation in conservation form in the space domain:

$${\displaystyle \nabla \cdot \mathbf {F} =\mathbf {0} }$$

where in general F is the flux matrix. By integrating this local equation over a fixed volume V_m, it becomes:

$${\displaystyle \int _{V_{m}}\nabla \cdot \mathbf {F} \,dV=\mathbf {0} .}$$

Then, basing on the divergence theorem, we can transform this integral in a boundary integral of the flux:

$${\displaystyle \oint _{\partial V_{m}}\mathbf {F} \,ds=\mathbf {0} .}$$

This global form simply states that there is no net flux of a conserved quantity passing through a region in the case steady and without source. In 1D the volume reduces to an interval, its boundary being its extrema, then the divergence theorem reduces to the fundamental theorem of calculus:

$${\displaystyle \int _{x_{m}}^{x_{m+1}}\mathbf {F} (x')\,dx'=\mathbf {0} ,}$$

that is the simple finite difference equation, known as the jump relation:

$${\displaystyle \Delta \mathbf {F} =\mathbf {0} .}$$

That can be made explicit as:

$${\displaystyle \mathbf {F} _{m+1}-\mathbf {F} _{m}=\mathbf {0} }$$

where the notation employed is:

$${\displaystyle \mathbf {F} _{m}=\mathbf {F} (x_{m}).}$$

Or, if one performs an indefinite integral:

$${\displaystyle \mathbf {F} -\mathbf {F} _{0}=\mathbf {0} .}$$

On the other hand, a transient conservation equation:

$${\displaystyle {\partial y \over \partial t}+\nabla \cdot \mathbf {F} =\mathbf {0} }$$

brings to a jump relation:

$${\displaystyle {\frac {dx}{dt}}\,\Delta u=\Delta \mathbf {F} .}$$

For one-dimensional Euler equations the conservation variables and the flux are the vectors:

$${\displaystyle \mathbf {y} ={\begin{pmatrix}{\frac {1}{v}}\\j\\E^{t}\end{pmatrix}},}$$

$${\displaystyle \mathbf {F} ={\begin{pmatrix}j\\vj^{2}+p\\vj\left(E^{t}+p\right)\end{pmatrix}},}$$

where:

• ${\displaystyle v}$ is the specific volume,
• ${\displaystyle j}$ is the mass flux.

In the one dimensional case the correspondent jump relations, called the Rankine–Hugoniot equations, are:<^[16]

$${\displaystyle \left\{{\begin{aligned}{\frac {dx}{dt}}\Delta \left({\frac {1}{v}}\right)&=\Delta j\\[1.2ex]{\frac {dx}{dt}}\Delta j&=\Delta (vj^{2}+p)\\[1.2ex]{\frac {dx}{dt}}\Delta E^{t}&=\Delta (jv(E^{t}+p)).\end{aligned}}\right.}$$

In the steady one dimensional case the become simply:

$${\displaystyle \left\{{\begin{aligned}\Delta j&=0\\[1.2ex]\Delta \left(vj^{2}+p\right)&=0\\[1.2ex]\Delta \left(j\left({\frac {E^{t}}{\rho }}+{\frac {p}{\rho }}\right)\right)&=0.\end{aligned}}\right.}$$

Thanks to the mass difference equation, the energy difference equation can be simplified without any restriction:

$${\displaystyle \left\{{\begin{aligned}\Delta j&=0\\[1.2ex]\Delta \left(vj^{2}+p\right)&=0\\[1.2ex]\Delta h^{t}&=0\end{aligned}}\right.,}$$

where ${\displaystyle h^{t}}$ is the specific total enthalpy.

These are the usually expressed in the convective variables:

$${\displaystyle \left\{{\begin{aligned}\Delta j&=0\\[1.2ex]\Delta \left({\frac {u^{2}}{v}}+p\right)&=0\\[1.2ex]\Delta \left(e+{\frac {1}{2}}u^{2}+pv\right)&=0,\end{aligned}}\right.}$$

where:

• ${\displaystyle u}$ is the flow speed
• ${\displaystyle e}$ is the specific internal energy.

The energy equation is an integral form of the Bernoulli equation in the compressible case. The former mass and momentum equations by substitution lead to the Rayleigh equation:

$${\displaystyle {\frac {\Delta p}{\Delta v}}=-{\frac {u_{0}^{2}}{v_{0}}}.}$$

Since the second term is a constant, the Rayleigh equation always describes a simple line in the pressure volume plane not dependent of any equation of state, i.e. the Rayleigh line. By substitution in the Rankine–Hugoniot equations, that can be also made explicit as:

$${\displaystyle \left\{{\begin{aligned}\rho u&=\rho _{0}u_{0}\\[1.2ex]\rho u^{2}+p&=\rho _{0}u_{0}^{2}+p_{0}\\[1.2ex]e+{\frac {1}{2}}u^{2}+{\frac {p}{\rho }}&=e_{0}+{\frac {1}{2}}u_{0}^{2}+{\frac {p_{0}}{\rho _{0}}}.\end{aligned}}\right.}$$

One can also obtain the kinetic equation and to the Hugoniot equation. The analytical passages are not shown here for brevity.

These are respectively:

$${\displaystyle \left\{{\begin{aligned}u^{2}(v,p)&=u_{0}^{2}+(p-p_{0})(v_{0}+v)\\[1.2ex]e(v,p)&=e_{0}+{\tfrac {1}{2}}(p+p_{0})(v_{0}-v).\end{aligned}}\right.}$$

The Hugoniot equation, coupled with the fundamental equation of state of the material:

$${\displaystyle e=e(v,p)}$$

describes in general in the pressure volume plane a curve passing by the conditions (v₀, p₀), i.e. the Hugoniot curve, whose shape strongly depends on the type of material considered.

It is also customary to define a Hugoniot function:^[17]

$${\displaystyle {\mathfrak {h}}(v,s)\equiv e(v,s)-e_{0}+{\tfrac {1}{2}}(p(v,s)+p_{0})(v-v_{0})}$$

allowing to quantify deviations from the Hugoniot equation, similarly to the previous definition of the hydraulic head, useful for the deviations from the Bernoulli equation.

Finite volume form

On the other hand, by integrating a generic conservation equation:

$${\displaystyle {\frac {\partial \mathbf {y} }{\partial t}}+\nabla \cdot \mathbf {F} =\mathbf {s} }$$

on a fixed volume V_m, and then basing on the divergence theorem, it becomes:

$${\displaystyle {\frac {d}{dt}}\int _{V_{m}}\mathbf {y} dV+\oint _{\partial V_{m}}\mathbf {F} \cdot {\hat {n}}ds=\mathbf {S} .}$$

By integrating this equation also over a time interval:

$${\displaystyle \int _{V_{m}}\mathbf {y} (\mathbf {r} ,t_{n+1})\,dV-\int _{V_{m}}\mathbf {y} (\mathbf {r} ,t_{n})\,dV+\int _{t_{n}}^{t_{n+1}}\oint _{\partial V_{m}}\mathbf {F} \cdot {\hat {n}}\,ds\,dt=\mathbf {0} .}$$

Now by defining the node conserved quantity:

$${\displaystyle \mathbf {y} _{m,n}\equiv {\frac {1}{V_{m}}}\int _{V_{m}}\mathbf {y} (\mathbf {r} ,t_{n})\,dV,}$$

we deduce the finite volume form:

$${\displaystyle \mathbf {y} _{m,n+1}=\mathbf {y} _{m,n}-{\frac {1}{V_{m}}}\int _{t_{n}}^{t_{n+1}}\oint _{\partial V_{m}}\mathbf {F} \cdot {\hat {n}}\,ds\,dt.}$$

In particular, for Euler equations, once the conserved quantities have been determined, the convective variables are deduced by back substitution:

$${\displaystyle {\begin{cases}\displaystyle \mathbf {u} _{m,n}={\frac {\mathbf {j} _{m,n}}{\rho _{m,n}}}\\[1.2ex]\displaystyle e_{m,n}={\frac {E_{m,n}^{t}}{\rho _{m,n}}}-{\frac {1}{2}}u_{m,n}^{2}.\end{cases}}}$$

Then the explicit finite volume expressions of the original convective variables are:<^[18]