13.3 Normal Shock Waves

Normal shock waves are abrupt, irreversible transitions that convert supersonic flow to subsonic flow. They show up whenever supersonic flow hits an obstacle or encounters certain pressure conditions in a duct or nozzle. Getting comfortable with the normal shock relations is essential because they let you predict exactly how pressure, temperature, density, and Mach number change across the shock, which matters for everything from wind tunnel design to jet engine performance.

Characteristics and Formation of Normal Shock Waves

Characteristics of normal shock waves

A normal shock wave is an extremely thin region (on the order of a few mean free paths) where flow properties change almost discontinuously. The wave is oriented perpendicular to the flow direction, which is why it's called "normal."

Across a normal shock, the following always happens:

• Pressure, density, and temperature increase (often dramatically)
• Velocity and Mach number decrease — the flow goes from supersonic ($M_1 > 1$) upstream to subsonic ($M_2 < 1$) downstream
• Total (stagnation) pressure drops — this is the hallmark of an irreversible process
• Total temperature stays constant — the shock is adiabatic (no heat transfer), so $T_{t1} = T_{t2}$

Normal shocks form in situations like converging-diverging nozzles operating off-design, supersonic wind tunnels, and ahead of blunt bodies in supersonic flight. The key requirement is that the upstream flow must be supersonic; you cannot have a normal shock in subsonic flow.

Normal Shock Relations and Flow Properties

Normal shock relations calculations

The normal shock relations come from applying conservation of mass, momentum, and energy across the shock. All downstream properties can be determined from just the upstream Mach number $M_1$ and the specific heat ratio $\gamma$ (typically 1.4 for air).

Pressure ratio:

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

Density ratio:

$\frac{\rho_2}{\rho_1} = \frac{(\gamma + 1)M_1^2}{(\gamma - 1)M_1^2 + 2}$

Temperature ratio:

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

Downstream Mach number:

$M_2^2 = \frac{(\gamma - 1)M_1^2 + 2}{2\gamma M_1^2 - (\gamma - 1)}$

To use these in practice, follow these steps:

1. Identify the upstream Mach number $M_1$ and the value of $\gamma$ for the gas.
2. Plug $M_1$ into the downstream Mach number equation to find $M_2$.
3. Use the pressure, density, and temperature ratio equations to find $p_2$, $\rho_2$, and $T_2$ from the known upstream values.
4. If you need downstream stagnation properties, use isentropic relations on the downstream side with $M_2$.

As a quick check: for $M_1 = 2.0$ and $\gamma = 1.4$, you get $p_2/p_1 = 4.5$, $T_2/T_1 \approx 1.687$, and $M_2 \approx 0.577$. Notice how the pressure nearly quintuples while the Mach number drops well below 1. Normal shock tables in your textbook tabulate these ratios so you don't have to compute them by hand every time, but you should be able to derive and use the equations directly.

Entropy and pressure across shocks

Normal shocks are non-isentropic. Entropy increases across the shock because kinetic energy is irreversibly dissipated through viscous effects and heat conduction within the thin shock layer.

The entropy change can be related to the total pressure ratio:

$\frac{p_{t2}}{p_{t1}} = \left[\frac{(\gamma + 1)M_1^2}{2 + (\gamma - 1)M_1^2}\right]^{\frac{\gamma}{\gamma - 1}} \left[\frac{\gamma + 1}{2\gamma M_1^2 - (\gamma - 1)}\right]^{\frac{1}{\gamma - 1}}$

Since entropy rise and total pressure loss are connected by $\Delta s = -R \ln(p_{t2}/p_{t1})$, a larger total pressure drop means more entropy generation. Both effects grow with shock strength: a shock at $M_1 = 1.2$ causes a modest total pressure loss of a few percent, while a shock at $M_1 = 3.0$ can destroy over 30% of the total pressure.

This matters for real systems because total pressure represents the flow's ability to do useful work. In a ramjet or scramjet, every bit of total pressure lost to shocks is energy that can't be converted to thrust. That's why engineers work hard to replace strong normal shocks with weaker oblique shock systems whenever possible.

Effects of Back Pressure on Normal Shock Waves in Nozzles

Back pressure effects on shocks

Back pressure is the ambient or imposed pressure at the nozzle exit. In a converging-diverging (C-D) nozzle, varying the back pressure controls whether and where a normal shock appears in the diverging section.

Here's how the shock location shifts as you change back pressure (starting from the fully expanded, shock-free design condition):

1. Back pressure equals the design exit pressure — the flow is fully supersonic throughout the diverging section with no shock. This is the ideal operating point.
2. Back pressure is raised above the design value but below a critical threshold — a normal shock forms inside the diverging section. The flow is supersonic upstream of the shock and subsonic downstream. The subsonic flow then decelerates further (pressure rises) to match the back pressure at the exit.
3. Back pressure increases further — the shock moves upstream toward the throat. Because the Mach number in the diverging section is lower closer to the throat, the shock becomes weaker (smaller jumps in $p$, $\rho$, $T$).
4. Back pressure reaches the value corresponding to a shock at the throat — the shock sits right at the throat and is very weak ($M_1$ just above 1). Raising back pressure beyond this point eliminates the shock entirely, and the flow becomes subsonic throughout the diverging section.

The practical takeaway: the presence of a shock in the nozzle reduces exit velocity and therefore thrust compared to the shock-free design condition. Engineers optimize nozzle geometry (specifically the area ratio $A_e/A^*$) and control back pressure to keep the nozzle operating as close to its design point as possible, minimizing shock losses.