1.1.8.1. The Euler Equations in Vector Form

1.1.8.1.1. Euler Equations in Strong Conservative Form in 1D (Scalar Notation)

1.1.8.1.1.1. Conservation of Mass

\[{\partial \rho \over \partial t} + {\partial \over \partial x}(\rho u) = 0\]

\(\rho u\) is conserved across the shock

1.1.8.1.1.2. Conservation of Momentum

\[{\partial \over \partial t}(\rho u) + {\partial \over \partial x}(\rho u^2 + p) = 0\]

\(u\) is the transported quantity

\(\rho u\) is conserved across the shock

\(\rho u^2 + p\) is also conserved across the shock

1.1.8.1.1.3. Conservation of Energy

We also need to work with the energy equation:

\[e_T(x,t) = e(x,t) + {u^2 \over 2}(x,t)\]

where:

\(e_T\) = specific total energy

\(e(x,t)\) = internal energy = heat added \(q\) - work done \(p \over \rho\)

\({u^2 \over 2}(x,t)\) = kinetic energy

Enthalpy change is the amount of heat content added to the system (by the shock) at constant pressure - is this valid if the pressure is not constant? - possibly, as we are computing total enthalpy:

\[h_T(x,t) = e_T + {p \over \rho} = h + {u^2 \over 2}\]

where:

\(h_T\) = specific total enthalpy

\(e_T\) = specific total energy

\(h\) = enthalpy \(h = e + {p \over \rho}\)

../_images/enthalpy.png

Giving the Conservation of Energy:

\[{\partial \over \partial t}(\rho e_T) + {\partial \over \partial x}(\rho u e_T + pu) = {\partial \over \partial t}(\rho e_T) + {\partial \over \partial x}(\rho u h_T) = 0\]

Interpretation:

  • Specific total energy \(e_T\) is the transported quantity

  • The terms \(\rho e_T\) and \(\rho u e_T + pu\) are conserved across the shock

  • \(\rho u h_T\) is also conserved across the shock (by the definition of enthalpy) Question: do we solve for enthalpy or energy or both?

About the source term:

  • We have included the source term with the convection term in strong conservative form

  • The source term is the net rate of work done by the pressure (simple thought - Pressure x Area x Velocity = Force x Distance / Time)

In 1D the independent variable \(u\) is a scalar i.e. it only has one value or component at one point.

1.1.8.1.2. Euler Equations in Strong Conservative Form in 1D (Vector Notation)

1.1.8.1.2.1. Solution Vector \(\mathbf{U}\)

Define a column vector of conserved variables:

\[\begin{split}\mathbf{U} = \begin{bmatrix} \rho \\ \rho u \\ \rho e_T \end{bmatrix}\end{split}\]

1.1.8.1.2.2. Flux Vector \(\mathbf{F}\)

We can also define a flux vector:

\[\begin{split}\mathbf{F} = \begin{bmatrix} \rho u \\ \rho u^2 + p \\ \rho u e_T + pu \end{bmatrix} = \begin{bmatrix} \rho u \\ \rho u^2 + p \\ \rho u h_T \end{bmatrix}\end{split}\]

Refer to the components of \(\mathbf{F}\) as \(F_1\), \(F_2\) and \(F_3\), representing:

\(F_1 = \rho u\) = Mass Flux

\(F_2 = \rho u^2 + p\) = Momentum Flux + Pressure Force

\(F_3 = \rho u e_T + pu\) = Total Energy Flux + Pressure Work

We call \(\mathbf{F}\) a flux vector, but it includes pressure effects

1.1.8.1.2.3. Euler Equations in Vector Notation

\[{\partial \mathbf{U} \over \partial t} + {\partial \mathbf{F} \over \partial x} = 0\]

1.1.8.1.2.4. The Jacobian Matrix: Linearisation of the System

\(\mathbf{F}\) can be written as a function of \(\mathbf{U}\):

\[{\partial \mathbf{F} \over \partial x} = {d \mathbf{F} \over d \mathbf{U}} {\partial \mathbf{U} \over \partial x}\]

Key step in the linearisation: It is assumed that \(\mathbf{F}\) is only a function of \(\mathbf{U}\), so this is the ordinary derivative.

However, \(\mathbf{F}\) is not just a function of x and \(\mathbf{U}\) is also not just a function of x, so these are both partial.

Where The Jacobian Matrix is:

\[\begin{split}{d \mathbf{F} \over d \mathbf{U}} = \begin{bmatrix} \partial \mathbf{F} \over \partial U_1 & \partial \mathbf{F} \over \partial U_2 & \partial \mathbf{F} \over \partial U_3 \end{bmatrix} = \begin{bmatrix} \partial F_1 \over \partial U_1 & \partial F_1 \over \partial U_2 & \partial F_1 \over \partial U_3 \\ \partial F_2 \over \partial U_1 & \partial F_2 \over \partial U_2 & \partial F_2 \over \partial U_3 \\ \partial F_3 \over \partial U_1 & \partial F_3 \over \partial U_2 & \partial F_3 \over \partial U_3 \end{bmatrix} = \mathbf{A}\end{split}\]

Can now write:

\[{\partial \mathbf{U} \over \partial t} + \mathbf{A} {\partial \mathbf{U} \over \partial x} = 0\]

Can now more easily solve the equations using linear solvers

Note: If \(\mathbf{F} = F_1\) (i.e. F is a scalar)

Then the Jacobian Matrix is just a row vector:

\[{d \mathbf{F} \over d \mathbf{U}} = \begin{bmatrix} \partial F_1 \over \partial U_1 & \partial F_1 \over \partial U_2 & \partial F_1 \over \partial U_3 \end{bmatrix}\]

In other words, the gradient of F, \(\nabla \mathbf{F}\)