Normal-shock relations

Rewriting the continuity equation (Eqn. \ref{eq:governing:cont})

$$\frac{{\rho }_2}{{\rho }_1}=\frac{u_1}{u_2}=\frac{u_1^2}{u_1{u}_2}=\{{a^{*}}^2=u_1{u}_2\}=\frac{u_1^2}{{a^{*}}^2}={M_1^{*}}^2$$

Eqn. \ref{eq:MachStar} in Eqn. \ref{eq:Normal:density:a} gives

$$\frac{{\rho }_2}{{\rho }_1}=\frac{(\gamma +1)M_1^2}{2+(\gamma -1)M_1^2}$$

To get a corresponding relation for the pressure ratio over the shock, we go back to the momentum equation (Eqn. \ref{eq:governing:mom})

$$p_2-p_1={\rho }_1{u}_1^2-{\rho }_2{u}_2^2=\{{\rho }_1{u}_1={\rho }_2{u}_1\}={\rho }_1{u}_1(u_1-u_2)={\rho }_1{u}_1^2(1-\frac{u_2}{u_1})$$

$$\frac{p_2-p_1}{p_1}=\frac{{\rho }_1{u}_1^2}{p_1}(1-\frac{u_2}{u_1})=\{a_1=\sqrt{\frac{\gamma p_1}{{\rho }_1}}\}=\gamma \frac{u_1^2}{a_1^2}(1-\frac{u_2}{u_1})=\gamma M_1^2(1-\frac{u_2}{u_1})$$

$$\frac{p_2}{p_1}-1=\gamma M_1^2(1-\frac{u_2}{u_1})=\{\frac{u_2}{u_1}=\frac{{\rho }_1}{{\rho }_2}\}=\gamma M_1^2(1-\frac{2+(\gamma -1)M_1^2}{(\gamma +1)M_1^2})$$

$$\frac{p_2}{p_1}=1+\frac{2\gamma }{\gamma +1}(M_1^2-1)$$

Figure~\ref{fig:shock:pressure:ratio} shows that the pressure must increase over the shock due to the fact that, based on the discussion above, the upstream Mach number must be greater than one and thus the shock is a discontinuous compression process.

The temperature ratio over the shock can be obtained using the already derived relations for pressure ratio and density ratio together with the equation of state $p=\rho RT$

$$\frac{T_2}{T_1}=\left(\frac{p_2}{p_1}\right)\left(\frac{{\rho }_1}{{\rho }_2}\right)$$

$$\frac{T_2}{T_1}=[1+\frac{2\gamma }{\gamma +1}(M_1^2-1)]\left[\frac{(\gamma +1)M_1^2}{2+(\gamma -1)M_1^2}\right]$$

Figure~\ref{fig:normal:shock:relations} below shows how different flow properties change over a normal shock as a function of upstream Mach number.

Now, one question remains. How come that we by analyzing the control volume using the upstream and downstream states get the normal shock relations. There is no way that the governing equations could have known about the fact that we assumed that there would be a shock inside of the control volume, or is it? The answer is that we have assumed that there will be a change in flow properties from upstream to downstream. We have further assumed that the flow is adiabatic (we are using the adiabatic energy equation) so there is no heat exchange. We are, however, allowing for irreversibilities in the flow. The only way to accomplish a change in flow properties under those constraints is a formation of a normal shock (a discontinuity in flow properties - a sudden flow compression) between station 1 and station 2.

The Hugoniot equation

The Hugoniot equation is an alternative normal shock relation based on thermodynamic quantities only. It is derived from the governing equations and relates the change in energy to the change in pressure and specific volume. The starting point of the derivation of the Hugoniot equation is the governing equations (Eqns~\ref{eq:governing:cont} - \ref{eq:governing:energy}).

The continuity equation is rewritten and inserted into the momentum equation

$$u_1=\left(\frac{{\rho }_2}{{\rho }_1}\right)u_2$$

Replace $u_1$ in Eqn. \ref{eq:governing:mom} using Eqn. \ref{eq:governing:cont:b}

$${\rho }_1{\left(\frac{{\rho }_2}{{\rho }_1}\right)}^2{u}_2^2+p_1={\rho }_2{u}_2^2+p_2$$

$$u_2^2({\rho }_1{\left(\frac{{\rho }_2}{{\rho }_1}\right)}^2-{\rho }_2)=(p_2-p_1)$$

$$u_2^2\left(\left(\frac{{\rho }_2}{{\rho }_1}\right)({\rho }_2-{\rho }_1)\right)=(p_2-p_1)$$

$$u_2^2=\left(\frac{{\rho }_1}{{\rho }_2}\right)\frac{p_2-p_1}{{\rho }_2-{\rho }_1}$$

Eqn. \ref{eq:governing:cont:b} and \ref{eq:governing:mom:b} gives

$$u_1^2=\left(\frac{{\rho }_2}{{\rho }_1}\right)\frac{p_2-p_1}{{\rho }_2-{\rho }_1}$$

Eqn. \ref{eq:governing:mom:b} and Eqn. \ref{eq:governing:mom:c} inserted in the energy equation (Eqn. \ref{eq:governing:energy}) gives

$$h_1+\frac{1}{2}\left(\frac{{\rho }_2}{{\rho }_1}\right)\left(\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\right)=h_2+\frac{1}{2}\left(\frac{{\rho }_1}{{\rho }_2}\right)\left(\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\right)$$

$$h_2-h_1=\frac{p_2-p_1}{2}[\left(\frac{{\rho }_2}{{\rho }_1}\right)\left(\frac{1}{{\rho }_2-{\rho }_1}\right)-\left(\frac{{\rho }_1}{{\rho }_2}\right)\left(\frac{1}{{\rho }_2-{\rho }_1}\right)]$$

$$h_2-h_1=\frac{p_2-p_1}{2}\left[\frac{{\rho }_2^2-{\rho }_1^2}{{\rho }_1{\rho }_2({\rho }_2-{\rho }_1)}\right]=\frac{p_2-p_1}{2}\left[\frac{{\rho }_2+{\rho }_1}{{\rho }_1{\rho }_2}\right]$$

$$h_2-h_1=\frac{p_2-p_1}{2}(\frac{1}{{\rho }_1}+\frac{1}{{\rho }_2})$$

Now, replacing the enthalpies with internal energies using $h=e+p/\rho$ gives

$$e_2-e_1=\frac{p_1}{{\rho }_1}-\frac{p_2}{{\rho }_2}+\frac{p_2-p_1}{2}(\frac{1}{{\rho }_1}+\frac{1}{{\rho }_2})$$

which after some rewriting becomes the Hugoniot equation

$$e_2-e_1=\frac{p_2+p_1}{2}(\frac{1}{{\rho }_1}-\frac{1}{{\rho }_2})=\frac{p_2+p_1}{2}({\nu }_1-{\nu }_2)$$

To give an idea about how the normal shock relates to an isentropic compression (a flow compression process without losses) the change in flow density as a function of pressure ratio is compared in Figure~\ref{fig:normal:shock:compression:vs:isentropic}. One can see that the normal-shock compression is more effective but less efficient than the corresponding isentropic process.

Introducing $C$ as the massflow per unit area (which is a constant)

$${\rho }_1{u}_1={\rho }_2{u}_2=C$$

Inserted into the momentum equation this gives

$$p_1+\frac{C^2}{{\rho }_1}=p_2+\frac{C^2}{{\rho }_2}$$

or

$$\frac{p_2-p_1}{{\nu }_2-{\nu }_1}=-C^2$$

which implies that all possible solutions to the governing equations must be located on a line in $p\nu$-space (the so-called Rayleigh line). If we add the Hugoniot relation to this we will find that there are two possible solutions, the upstream condition and the condition downstream of the normal shock and the flow cannot be in any of the intermediate stages. The normal-process is a so-called wave solution to the governing equations where the flow state must jump directly from one flow state to another without passing the intermediate conditions. If we add heat or friction to the problem we will instead get continuous solutions as we will see in the following sections. Figures \ref{fig:shock:pv} and \ref{fig:shock:Ts} shows a normal shock process in a $p\nu$- and $Ts$-diagram, respectively. Note that the flow passes the characteristic conditions as it is going through the shock, which means that the flow goes from supersonic to subsonic.