# 11.5 Normal Shock

Fig. 11.11 A shock wave inside a tube, but it can also be viewed as a one–dimensional shock wave.

In this section the relationships between the two sides of normal shock are presented. In this discussion, the flow is assumed to be in a steady state, and the thickness of the shock is assumed to be very small. A shock can occur in at least two different mechanisms. The first is when a large difference (above a small minimum value) between the two sides of a membrane, and when the membrane bursts (see the discussion about the shock tube). Of course, the shock travels from the high pressure to the low pressure side. The second is when many sound waves "run into'' each other and accumulate (some refer to it as "coalescing'') into a large difference, which is the shock wave. In fact, the sound wave can be viewed as an extremely weak shock. In the speed of sound analysis, it was assumed the medium is continuous, without any abrupt changes. This assumption is no longer valid in the case of a shock. Here, the relationship for a perfect gas is constructed. In Figure 11.11 a control volume for this analysis is shown, and the gas flows from left to right. The conditions, to the left and to the right of the shock, are assumed to be uniform. The conditions to the right of the shock wave are uniform, but different from the left side. The transition in the shock is abrupt and in a very narrow width. Therefore, the increase of the entropy is fundamental to the phenomenon and the understanding of it. It is further assumed that there is no friction or heat loss at the shock (because the heat transfer is negligible due to the fact that it occurs on a relatively small surface). It is customary in this field to denote $$x$$ as the upstream condition and $$y$$ as the downstream condition. The mass flow rate is constant from the two sides of the shock and therefore the mass balance is reduced to

$\rho_{x} \,U_{x} = \rho _{y}\, U_{y} \label{shock:eq:mass}$

In a shock wave, the momentum is the quantity that remains constant because there are no external forces. Thus, it can be written that

$P_{x} - P_{y} = \left(\rho_{x} {U_{y}}^2- \rho_{y}\, {U_{x}}^2 \right) \label{shock:eq:momentum}$ The process is adiabatic, or nearly adiabatic, and therefore the energy equation can be written as

$C_{p} \,T_{x} + \dfrac{{U_{x}}^{2} }{ 2} = C_{p}\, T_{y} + \dfrac{{U_{y}}^{2} }{ 2} \label{shock:eq:energy}$ The equation of state for perfect gas reads

$P = \rho\, R\, T \label{eq:shock:state}$ If the conditions upstream are known, then there are four unknown conditions downstream. A system of four unknowns and four equations is solvable. Nevertheless, one can note that there are two solutions because of the quadratic of equation (3). These two possible solutions refer to the direction of the flow. Physics dictates that there is only one possible solution. One cannot deduce the direction of the flow from the pressure on both sides of the shock wave. The only tool that brings us to the direction of the flow is the second law of thermodynamics. This law dictates the direction of the flow, and as it will be shown, the gas flows from a supersonic flow to a subsonic flow. Mathematically, the second law is expressed by the entropy. For the adiabatic process, the entropy must increase. In mathematical terms, it can be written as follows:

$s_{y} - s_{x} > 0 \label{eq:shock:entropy}$ Note that the greater–equal signs were not used. The reason is that the process is irreversible, and therefore no equality can exist. Mathematically, the parameters are $$P, T, U,$$ and $$\rho$$, which are needed to be solved. For ideal gas, equation (5) is

$\ln \left(\dfrac{T_y }{ T_x} \right) - \left(k - 1\right)\, \dfrac{P_y }{ P_x} > 0 \label{shock:eq:entropyIdeal}$ It can also be noticed that entropy, $$s$$, can be expressed as a function of the other parameters. These equations can be viewed as two different subsets of equations. The first set is the energy, continuity, and state equations, and the second set is the momentum, continuity, and state equations. The solution of every set of these equations produces one additional degree of freedom, which will produce a range of possible solutions. Thus, one can have a whole range of solutions. In the first case, the energy equation is used, producing various resistance to the flow. This case is called Fanno flow, and Section 11.8 deals extensively with this topic. Instead of solving all the equations that were presented, one can solve only four (4) equations (including the second law), which will require additional parameters. If the energy, continuity, and state equations are solved for the arbitrary value of the $$T_y$$, a parabola in the $$T−s$$ diagram will be obtained. On the other hand, when the momentum equation is solved instead of the energy equation, the degree of freedom is now energy, i.e., the energy amount "added'' to the shock. This situation is similar to a frictionless flow with the addition of heat, and this flow is known as Rayleigh flow. This flow is dealt with in greater detail in Section ??. Since the shock has no heat transfer (a special case of Rayleigh flow) and there isn't essentially any momentum transfer (a special case of Fanno flow), the intersection of these two curves is what really happened in the shock. The entropy increases from point $$x$$ to point $$y$$.