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Aerospace
Stagnation-pressure retention through a normal shock
Fraction of the total pressure that survives a stationary adiabatic normal shock.
- Ref
- AER-41
- Rev
- v1
- Field
- Aerospace
- Result
- pt2/pt1
A shock is irreversible: it raises entropy and so lowers the stagnation pressure, while stagnation temperature stays constant in an adiabatic shock. The retained fraction is pt2/pt1 = [(γ+1)M1²/(2+(γ−1)M1²)]^(γ/(γ−1)) × [(γ+1)/(2γM1²−γ+1)]^(1/(γ−1)). It is 1 at M1 = 1, 0.72 at Mach 2, 0.33 at Mach 3 and 0.06 at Mach 5 for air. Inlets use several weak shocks to reduce this loss.
Fraction of the total pressure that survives a stationary adiabatic normal shock. It is below one for any M1 > 1 and drops steeply as M1 rises. No shaft work is done, so stagnation temperature is conserved, and the same γ applies on both sides.
Calculator
- Total-pressure loss, 1 − pt2/pt10.2791
- Static pressure ratio p₂/p₁4.5
- Entropy rise in units of R, −ln(pt2/pt1)0.3273
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 keeps 72.09 % of the stagnation pressure and loses 27.91 % of it, which would be a thrust loss in an engine inlet.
- The entropy generated is 0.3273 R, or 93.93 J/(kg·K) for air.
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 | 0.02965 | 1 | 0.9703 | +10% | −12.9 % |
| γ Heat-capacity ratio | 1.1 to 1.67 | 0.6483 | 0.7634 | 0.1152 | +10% | +3.32 % |
Sweep of Upstream Mach number
| Upstream Mach number | Total-pressure ratio |
|---|---|
| 1.01 | 1 |
| 1.634 | 0.8824 |
| 2.258 | 0.6022 |
| 2.881 | 0.3635 |
| 3.505 | 0.212 |
| 4.129 | 0.1245 |
| 4.753 | 0.0749 |
| 5.376 | 0.04641 |
| 6 | 0.02965 |
Sweep one variable
Variables and how each one contributes
Total-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 +3.32 %, from 0.7209 to 0.7448.
Upstream Mach number
The upstream Mach number, above 1. Lowers the retained fraction steeply as the shock gets stronger. Raising it by 10 % (to 2.2) moves the result by −12.9 %, from 0.7209 to 0.6281.
Worked example with the values above
- First factor: [(γ + 1)M1²/(2 + (γ − 1)M1²)]^(γ/(γ−1)) = 30.9664.
- Second factor: [(γ + 1)/(2γM1² − γ + 1)]^(1/(γ−1)) = 0.0232792.
- pt2/pt1 = 30.9664 × 0.0232792 = 0.720874.
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