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Aerospace
Static pressure jump through a normal shock
Ratio of static pressure behind a normal shock to that ahead of it, for upstream Mach number M1 > 1.
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- AER-39
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- Aerospace
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- p2/p1
Momentum conservation across a normal shock gives the static pressure ratio p2/p1 = 1 + [2γ/(γ+1)](M1² − 1). It rises without limit with M1², reaching 4.5 at Mach 2 and 29 at Mach 5 in air. The pressure rise happens over a few mean free paths, and it is accompanied by a loss of stagnation pressure, as the process is irreversible.
Ratio of static pressure behind a normal shock to that ahead of it, for upstream Mach number M1 > 1. It grows with M1² without limit. Stagnation pressure behaves differently, because the shock is irreversible.
Calculator
- Density ratio ρ₂/ρ₁2.667
- Temperature ratio T₂/T₁1.687
- Pressure rise relative to p₁, (p₂ − p₁)/p₁3.5
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The shock raises the static pressure by a factor of 4.5, so a gas at 100 kPa upstream is at 450 kPa downstream.
- The density rises by a smaller factor, 2.667, so most of the pressure rise goes into heating the gas.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| M1 Upstream Mach number | 1.01 to 6 | 1.023 | 41.83 | 40.81 | +10% | +21.8 % |
| γ Heat-capacity ratio | 1.1 to 1.67 | 4.143 | 4.753 | 0.61 | +10% | +3.06 % |
Sweep of Upstream Mach number
| Upstream Mach number | Static pressure ratio |
|---|---|
| 1.01 | 1.023 |
| 1.634 | 2.947 |
| 2.258 | 5.779 |
| 2.881 | 9.519 |
| 3.505 | 14.17 |
| 4.129 | 19.72 |
| 4.753 | 26.18 |
| 5.376 | 33.55 |
| 6 | 41.83 |
Sweep one variable
Variables and how each one contributes
Static pressure ratio
What the equation gives you. Everything else feeds it.
Heat-capacity ratio
The heat-capacity ratio cp/cv of the gas, about 1.4 for air. A lower γ (hotter, more complex gas) changes the compressible-flow exponents and the sizes of the shock and expansion effects. Raising it by 10 % (to 1.54) moves the result by +3.06 %, from 4.5 to 4.638.
Upstream Mach number
The upstream Mach number, above 1. Raises the pressure ratio with M1², so strong shocks compress enormously. Raising it by 10 % (to 2.2) moves the result by +21.8 %, from 4.5 to 5.48.
Worked example with the values above
- Factor: 2γ/(γ + 1) = 1.16667; M1² − 1 = 3.
- Product: 3.5.
- p2/p1 = 1 + 3.5 = 4.5.
Assumptions and limits
- A steady normal shock in a calorically perfect gas, with the same γ on both sides and no heat addition.
- Upstream Mach number above 1; the downstream flow is subsonic.
Common mistakes
- Using the upstream Mach number on the downstream side.
- Using total pressure where static pressure is needed, or the reverse.
- Applying normal-shock formulas to the full Mach number of an oblique shock.
Statistics and uncertainty
Normal-shock relations are exact for a perfect gas. At high Mach numbers, dissociation and real-gas effects change the pressure and temperature ratios by several per cent, and weak shocks are close to isentropic, so small pressure losses are sensitive to inputs.
Where it is used
- Estimating shock losses in inlets, nozzles and supersonic tunnels.
- Interpreting pitot readings behind a bow shock.
Maths you need
- Rankine–Hugoniot relations: conservation of mass, momentum and energy across a shock
- Stagnation pressure loss: a measure of the irreversibility of the shock
- Entropy: a measure of irreversibility