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Specific entropy increase across an adiabatic shock
Entropy generated per unit mass of gas passing through an adiabatic shock, from the two stagnation pressures for an ideal gas at equal stagnation temperature.
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- Δs [J/(kg·K)]
In an adiabatic shock the stagnation temperature is unchanged, so the entropy rise of an ideal gas depends only on the total pressures: Δs = −R ln(pt2/pt1). It is positive, as the second law requires, and it measures the work that has been made unavailable. A total-pressure ratio of 0.8 gives Δs = 0.223 R, or 64 J/(kg·K) in air.
Entropy generated per unit mass of gas passing through an adiabatic shock, from the two stagnation pressures for an ideal gas at equal stagnation temperature. It is positive, since pt2 ≤ pt1, and measures the irreversibility of the shock.
Calculator
- Total-pressure ratio pt2/pt10.8
- Total-pressure loss, 1 − pt2/pt10.2
- Entropy rise in units of R, Δs/R0.2231
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 generates 64.042 J/(kg·K) of entropy, 0.2231 times R, from a 20 % loss of total pressure.
- At the stagnation temperature of the flow this corresponds to 19,210 J/kg of lost work at 300 K, the energy that cannot be recovered.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J/(kg·K)) | Local step | Change in result |
|---|---|---|---|---|---|---|
| pt2 Downstream total pressure | 20,000 to 100,000 Pa | 0 | 461.9 | 461.9 | +10% | −42.7 % |
| pt1 Upstream total pressure | 80,000 to 300,000 Pa | 0 | 379.3 | 379.3 | +10% | +42.7 % |
| R Gas constant | 200 to 500 J/(kg·K) | 44.63 | 111.6 | 66.94 | +10% | +10 % |
Sweep of Downstream total pressure
| Downstream total pressure (Pa) | Entropy increase (J/(kg·K)) |
|---|---|
| 20,000 | 461.9 |
| 30,000 | 345.5 |
| 40,000 | 263 |
| 50,000 | 198.9 |
| 60,000 | 146.6 |
| 70,000 | 102.4 |
| 80,000 | 64.04 |
| 90,000 | 30.24 |
| 100,000 | 0 |
Sweep one variable
Variables and how each one contributes
Entropy increase
What the equation gives you. Everything else feeds it.
Gas constant
The specific gas constant, in J/(kg·K). Scales the entropy increase in direct proportion. Raising it by 10 % (to 315.7) moves the result by +10 %, from 64.04 to 70.45 J/(kg·K).
Upstream total pressure
The upstream total pressure, in Pa. Raises the pressure ratio loss and so the entropy rise. Raising it by 10 % (to 110,000) moves the result by +42.7 %, from 64.04 to 91.4 J/(kg·K).
Downstream total pressure
The downstream total pressure, in Pa. A lower downstream value means more loss and more entropy generated. Raising it by 10 % (to 88,000) moves the result by −42.7 %, from 64.04 to 36.69 J/(kg·K).
Worked example with the values above
- Pressure ratio: pt2/pt1 = 80,000 / 100,000 = 0.8.
- Logarithm: ln = −0.223144.
- Δs = −287 × (−0.223144) = 64.0422 J/(kg·K).
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
- Putting static pressures in place of total pressures.
- Using the formula for a non-adiabatic process, where stagnation temperature changes.
- Entering pt2 larger than pt1, which would make the entropy change negative.
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