Compressible flow and shock waves

Start with gas crossing a standing front

In a supplied ideal-gas model, gas at 100 kPa and 300 K approaches a stationary planar front at Mach 2. Downstream, its pressure is 450 kPa and its temperature is 506.25 K, while its speed is lower. How can the gas become hotter and more compressed without heat being supplied from outside? What has changed, and what is conserved?

We will first distinguish the speed of the gas from the speed of small pressure signals. Then we will balance mass, momentum and energy across this normal shock and check its entropy increase. The characteristic-ray comparison, shock/contact comparison and nozzle graph are the first-reading path. Oblique geometry and the final algebra form an advanced return path.

This is a physical gas-state calculation within declared ideal assumptions, not a measured shock or a finite-thickness shock structure. The published Shock waves, solitons, and nonlinear waves lesson used separate Burgers and KdV scalar models to explain moving jumps and smooth pulses. Here density, pressure, temperature and total energy are coupled by the compressible Euler system; a scalar entropy rule alone does not supply those gas states.

Published preparation includes Units, standards, and physical constants; Algebra, functions, graphs, and logarithms; Conservation of mass, momentum, and energy; and Euler’s equation and Bernoulli’s principle. The gas-state and entropy terms needed here are defined locally. Later thermodynamic lessons provide a deeper return path rather than an unseen prerequisite.

Name the state and its sound speed

Compressibility concerns a fluid’s density response to changes in its thermodynamic state. Write pressure p in Pa, density ρ in kg/m³ and absolute temperature T in K. For the assigned ideal gas, p = ρRT with specific gas constant R = 287 J/(kg·K). We also prescribe constant heat capacities with γ = c_p/c_v = 1.4, c_p − c_v = R, so c_p = 1004.5 and c_v = 717.5 J/(kg·K). These are model coefficients, not an identification or precision measurement of a particular air sample.

Internal energy per mass is e = c_vT and enthalpy per mass is h = e + p/ρ = c_pT, both in J/kg. Enthalpy packages internal energy together with the pressure work carried by flowing gas. This accounts for why the steady energy balance uses h + u²/2, rather than temperature alone.

Small, locally isentropic pressure disturbances have sound speed a defined by a² = (∂p/∂ρ)_s. The subscript s means fixed specific entropy. For our calorically perfect gas, a² = γp/ρ = γRT. This is not the derivative at fixed temperature. At 300 K the assigned sound speed is about 347.188709 m/s, and the upstream Mach-2 gas speed is 694.377419 m/s.

Mach number M is speed relative to the chosen observer divided by local sound speed. For a moving front, use the gas speed relative to that front when applying a shock formula. Changing the observer changes the gas speed and Mach number; it does not change local pressure, density or temperature in a nonrelativistic model.

A low Mach number can support small density changes caused by the motion under suitable conditions, but it does not guarantee uniform density when heating, composition or stratification matters. Conversely, even a small density variation can carry a pressure wave whose travel time matters. Near a sonic region, or across a shock, the constant-density approximation needs particular care.

Follow small signals relative to the gas

In one-dimensional smooth flow, the three characteristic speeds of the ideal-gas Euler system are u − a, u and u + a. These are directions in space–time along which the system carries information. The two acoustic families move at ±a relative to the gas; the middle family carries entropy or contact information with the gas velocity u.

Compare three separate uniform states at the same assigned 300 K and 100 kPa, with downstream Mach numbers 0.5, 1 and 2. At M = 0.5, u − a is negative: a small acoustic signal directed against the gas travels upstream in the laboratory. At M = 1 it is zero. At M = 2 both acoustic directions move downstream in the laboratory, even though one travels against the gas in its own local rest frame.

For these uniform examples a ray from x = 0 follows x = (u ± a)t; the middle family follows x = ut. The diagram shows those information paths, not a plot of pressure amplitude or a set of gas-particle displacements. The three states are alternative comparisons, not three consecutive regions of a solved nozzle.

In a smooth nonlinear compression, characteristic directions can converge because the local velocity and sound speed change with the state. The ideal smooth solution can steepen until an admissible shock description is needed. An expansion instead spreads into a rarefaction fan under the ordinary gas assumptions. The directions are local: the statement about small upstream signals does not forbid a moving shock or a changed global flow pattern from altering an initially supersonic region.

Conserve through a thin shock

Choose a planar front moving at speed S along +x and let w = u − S be the gas velocity relative to it. A control volume moving with that front gives the steady local jump balances ρ₁w₁ = ρ₂w₂ = j, p₁ + ρ₁w₁² = p₂ + ρ₂w₂², and h₁ + w₁²/2 = h₂ + w₂²/2. Here j is mass flow per area, in kg/(m²·s), and states 1 and 2 are uniform limits on the upstream and downstream sides with gas crossing from 1 to 2.

These relations assume no external heat transfer or shaft work across the thin region and negligible potential-energy change. They conserve mass, momentum and energy flux; a pressure rise is part of the momentum exchange. For a normal shock, the incident velocity is normal to the front. In our main example S = 0, so w = u.

The thin jump is an outer-state idealisation. Real shocks contain molecular transport and entropy generation over a finite region; the inviscid jump model does not calculate that region’s thickness. Writing conservation on its two sides does not make the passage reversible or justify an isentropic relation across the front.

Conservation by itself also admits algebraic jumps that are physically inadmissible. The gas crossing an ordinary compression shock must have increasing specific entropy. With the stated direction j > 0, require s₂ − s₁ ≥ 0; a nonzero shock has a strict increase in this perfect-gas model. A reversed expansion jump that violates this condition is replaced by the appropriate rarefaction in a physical Riemann solution.

Calculate the normal-shock states

Solving the jump balances with the assigned gas closure gives p₂/p₁ = 1 + [2γ/(γ + 1)](M₁² − 1), ρ₂/ρ₁ = (γ + 1)M₁²/[(γ − 1)M₁² + 2], and T₂/T₁ = (p₂/p₁)/(ρ₂/ρ₁). The downstream Mach number satisfies M₂² = [1 + (γ − 1)M₁²/2]/[γM₁² − (γ − 1)/2]. Mach numbers in these relations are relative to the normal shock.

For M₁ = 2 and γ = 1.4, the pressure ratio is 4.5, the density ratio is 8/3 and the temperature ratio is 1.6875. Thus p₂ = 450000 Pa, T₂ = 506.25 K and ρ₂ ≈ 3.097174 kg/m³. Continuity gives u₂ = u₁ρ₁/ρ₂ ≈ 260.391532 m/s.

The downstream sound speed rises to about 451.011363 m/s because T rises. Therefore M₂ ≈ 0.577350: a normal compression shock changes the normal flow from supersonic to subsonic in its own rest frame. The changing Mach number involves both gas speed and sound speed; using the upstream a on both sides would give the wrong downstream Mach.

The mass flux is about 806.477839 kg/(m²·s) on each side. The momentum flux p + ρu² is 660000 Pa on each side, and h + u²/2 is 542430 J/kg on each side. In the ideal jump description, the two uniform states meet at x = 0; no collection of intermediate gas states or physical shock thickness is resolved.

Separate static heating from stagnation-pressure loss

Static temperature is the local thermodynamic temperature of the gas. In the shock frame, stagnation temperature T₀ is the temperature reached by hypothetically bringing that state to rest adiabatically without shaft work: T₀ = T + w²/(2c_p). To define stagnation pressure p₀, bring each state to rest isentropically; then p₀ = p(T₀/T)^[γ/(γ − 1)]. These definitions do not require the actual shock passage to be isentropic.

In the main example, T₀ = 540 K on both sides. Lower bulk kinetic energy accompanies higher static enthalpy and temperature, conserving total enthalpy under the stated assumptions. Energy has not disappeared from the gas, and outside heat is not required to explain the static-temperature rise.

Stagnation pressure behaves differently: p₀₁ ≈ 782.444907 kPa and p₀₂ ≈ 564.044081 kPa, so p₀₂/p₀₁ ≈ 0.720874. Static pressure rises while stagnation pressure falls. The latter records an irreversible loss of pressure recovery potential relative to an isentropic deceleration from the upstream state, not a failure of energy conservation.

For a calorically perfect gas, the entropy change is Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁). The assigned normal shock gives Δs ≈ +93.932548 J/(kg·K). At equal T₀ this also equals −R ln(p₀₂/p₀₁). Adiabatic means no heat transferred through the boundary; isentropic means constant specific entropy. A shock is the example that separates those two conditions.

The incompressible Bernoulli expression p + ρu²/2 with constant density cannot replace the gas energy/entropy calculation across a shock. For a moving shock, be equally careful with the observer: h + w²/2 is the conserved shock-frame total enthalpy here; h + u²/2 in a different laboratory frame need not be unchanged across that moving front.

Distinguish a material contact from a shock

A one-dimensional inviscid material contact has the same pressure and normal velocity on both sides, but density and temperature can differ. It travels with the gas normal velocity, S = u. In its own rest frame j = ρ(u − S) = 0: gas does not cross the contact. A shock instead has a nonzero through-going mass flux and changes the gas as it crosses.

For a separate assigned contact, put p = 100000 Pa and u = 100 m/s on both sides, with ρ_L = 1 and ρ_R = 2 kg/m³. The equation of state gives T_L ≈ 348.432056 K and T_R ≈ 174.216028 K. Starting at x = 0, the contact reaches x = 0.200 m at t = 0.002 s. Both the gas velocity and contact speed are 100 m/s; the density step is not an acoustic wave moving at u ± a.

Specific entropy may jump in either spatial direction across this ideal contact. That is not a negative entropy production experienced by crossing gas: no gas crosses the material boundary. Molecular diffusion or heat conduction would alter a real contact’s structure over time; those effects are omitted from this exact inviscid discontinuity.

A shock-tube initial-value problem can contain a shock, a contact and a rarefaction, with different speeds and intervening states. Showing one of these structures does not calculate all of the others. We use independent contact and stationary-shock cases here rather than splice unrelated states into a purported complete shock-tube solution.

Use isentropic relations only on their path

For smooth adiabatic reversible flow of the assigned gas, p/ρ^γ is constant along an isentropic path. With steady no-work flow and negligible potential change, T₀ = T[1 + (γ − 1)M²/2], p₀/p = [1 + (γ − 1)M²/2]^[γ/(γ − 1)] and ρ₀/ρ = [1 + (γ − 1)M²/2]^[1/(γ − 1)]. A subscript zero denotes that path’s rest state, not the state just upstream of any arbitrary wave.

These relations apply separately to isentropic decelerations from either side of our shock, which have different p₀ values. They do not set one common p₀ or one common p/ρ^γ across the entropy-generating shock. A small change or a gently shaped wall also does not by itself prove reversibility: friction, heat transfer or other dissipative processes must be assessed.

A steady two-dimensional supersonic stream turning around an ideal convex corner can expand through a Prandtl–Meyer fan. Pressure and temperature decrease, speed and Mach number increase, and stagnation pressure and temperature remain constant in the isentropic model. A finite compression through an ordinary shock instead generates entropy. Curving compression waves can coalesce into a shock; smooth compression is not automatically a global shock-free solution.

For an advanced return path, define ν_PM(M) = √[(γ + 1)/(γ − 1)] arctan√[(γ − 1)(M² − 1)/(γ + 1)] − arctan√(M² − 1), for M ≥ 1. For this ideal expansion fan the turn magnitude is ν_PM(M₂) − ν_PM(M₁), with angles in radians when evaluated by this formula. This function is an angle, distinct from the kinematic viscosity ν used in earlier lessons. Its use requires the particular supersonic, planar, isentropic corner model; it is not a normal-shock formula.

Accelerate gas through a changing area

Now supply an ideal steady quasi-one-dimensional nozzle. ‘Quasi-one-dimensional’ represents each section by a uniform axial state while area A changes slowly along x; it does not resolve the detailed two-dimensional flow. For smooth isentropic flow without external work, mass flow is ṁ = ρuA and the energy relation is h + u²/2 = constant.

Combining these with dp = a²dρ gives dA/A = (M² − 1)du/u. For a subsonic stream, a decreasing area accompanies increasing speed; for a supersonic stream, an increasing area accompanies increasing speed. Therefore a diverging duct is not automatically a diffuser: the flow branch matters.

At a smooth sonic transition M = 1, a regular solution needs an area extremum in this ideal model. A converging–diverging nozzle can provide a throat followed by a supersonic branch when the reservoir, downstream pressure and entire solution admit it. The existence of a throat does not force every operating condition to be sonic or supersonic.

For fixed reservoir stagnation values p₀, T₀ and gas constants, the isentropic area relation is A/A* = (1/M){[2/(γ + 1)][1 + (γ − 1)M²/2]}^[(γ + 1)/(2(γ − 1))]. A* is the area associated with M = 1 for that mass flow and stagnation state. It is an actual throat area only when that throat is sonic. For area ratios above one there are separate subsonic and supersonic mathematical branches.

Find what choking does and does not limit

Take a large reservoir at p₀ = 200000 Pa and T₀ = 300 K feeding an ideal converging nozzle with throat/exit area A_e = 0.0001 m². Let p_b be the imposed surrounding back pressure. While the exit is subsonic and matches p_b, reducing p_b accelerates the gas and increases its mass flow.

At a fixed area and fixed stagnation state, ṁ = (Ap₀/√T₀)√(γ/R) M[1 + (γ − 1)M²/2]^[−(γ + 1)/(2(γ − 1))] has its maximum at M = 1. The critical exit pressure is p*/p₀ = [2/(γ + 1)]^[γ/(γ − 1)] ≈ 0.528282, or p* ≈ 105656.358 Pa. At this sonic exit, T* = 250 K and ṁ* ≈ 0.0466712 kg/s.

Below that back pressure, the simple converging-nozzle model keeps the exit sonic and its mass flow fixed, with adjustment outside the nozzle. For p_b/p₀ = 0.4, p_b is 80000 Pa but the sonic exit pressure is still about 105656 Pa. Substituting p_b into the subsonic isentropic exit formula after choking would falsely continue the internal branch.

Choking does not mean that the gas stops or that no greater flow is ever possible. The limiting value here depends on area, p₀, T₀ and gas model. Increasing area or reservoir pressure changes it. Nor is it a universal speed cap: a suitably operated diverging section can accelerate a sonic stream to supersonic speed.

At A/A* = 1.6875, our area equation admits M ≈ 0.372244 and M = 2 on different isentropic branches. With the assigned reservoir they have different local pressures, about 181751.6 Pa and 25560.9 Pa. Area ratio alone chooses neither. In a real converging–diverging nozzle the back pressure can select subsonic flow, a shock-containing pattern or a supersonic exit with external adjustment. We have not specified a complete diverging geometry or solved a shock position.

Separate normal and tangential oblique motion

An oblique shock is inclined to the upstream flow. Let β be the angle between the upstream velocity and the shock line, and θ the flow turn through the compression. Resolve the velocity into normal and tangential components relative to the shock. In the ideal thin inviscid jump, the tangential component is unchanged; the normal component obeys the normal-shock balances.

Use M_n1 = M₁ sinβ in the normal-shock formulas, not the full M₁. Downstream, M_n2 = M₂ sin(β − θ), so the full downstream Mach number need not be subsonic even though the normal component becomes subsonic. For a planar perfect-gas compression, tanθ = 2cotβ(M₁²sin²β − 1)/[M₁²(γ + cos2β) + 2].

For a separate assigned case, retain p₁ = 100 kPa, T₁ = 300 K and M₁ = 2 but choose β = 45°. Then M_n1 ≈ 1.414214, θ ≈ 14.743563°, p₂/p₁ ≈ 2.166667, ρ₂/ρ₁ ≈ 1.714286 and M₂ ≈ 1.456324. The full flow remains supersonic; the normal downstream Mach number is about 0.733799. The unchanged tangential speed is about 490.998982 m/s.

For a prescribed corner turn rather than a supplied β, the angle relation can have weak and strong branches or no attached solution. Their admissibility and realised pattern require the geometry and downstream conditions. If the required compression cannot remain attached, a detached, generally curved shock must be analysed; one normal-shock value does not describe every point on a curved front. Nor is a planar wedge calculation identical to the axisymmetric flow around a cone.

Return path: the Euler system and Hugoniot relation

For one-dimensional inviscid flow without external sources, define total energy density E = ρe + ρu²/2, with units J/m³. The conserved state is q = (ρ, ρu, E) and the flux is f = (ρu, ρu² + p, u(E + p)). The system is ∂q/∂t + ∂f/∂x = 0. The energy-flux pressure term is essential: omitting it would fail to account for flow work.

For a front at x = St, integral conservation gives S[q] = [f], where [·] means right minus left. This is the system Rankine–Hugoniot relation. Moving to the front frame and using nonzero mass flux reduces it to the three balances above. It is the multi-component counterpart of the scalar moving-jump rule in the earlier nonlinear-wave lesson, with a different state vector and physical closure.

Write v = 1/ρ for specific volume, not velocity. From j = ρw obtain w = jv. Momentum gives p₂ − p₁ = j²(v₁ − v₂). Combine energy h₂ − h₁ = (w₁² − w₂²)/2 with h = e + pv to eliminate j: e₂ − e₁ = (p₁ + p₂)(v₁ − v₂)/2. This is the thermodynamic Hugoniot relation under the declared no-heat/no-work jump conditions.

In the assigned normal shock, e₂ − e₁ = 147984.375 J/kg; the pressure–specific-volume expression gives the same number. Combining this relation with the calorically perfect closure and momentum balance yields the ratios already used. Conservation also allows the no-jump branch. Selecting a nonzero ordinary compression requires the admissible supersonic-to-subsonic normal branch and an entropy increase; solving the algebra does not remove that gate.

For a smooth isentropic one-dimensional region, the combinations J_± = u ± 2a/(γ − 1) are constant along dx/dt = u ± a, respectively. These are characteristic compatibility statements for this restricted smooth model. Across a shock they are replaced by the jump conditions. Constancy along a characteristic is also different from the Riemann-invariant quantities held across an entire opposite-family simple wave; the direction and meaning should be named before transferring a formula.

Return path: derive the area and choking conditions

For nonzero smooth nozzle flow, differentiate ṁ = ρuA: dρ/ρ + du/u + dA/A = 0. Steady inviscid momentum gives dp + ρu du = 0. Along an isentrope dp = a²dρ, hence dρ/ρ = −M²du/u. Substitution gives dA/A = (M² − 1)du/u. At the sonic point, the relation identifies a necessary regularity condition; it does not alone prove that a globally acceptable transonic solution exists.

For fixed p₀ and T₀, substitute u = M√(γRT), ρ = p/(RT) and the isentropic p/p₀ and T/T₀ relations into ṁ = ρuA. This gives the mass-flow formula above. Its dimensionless Mach-dependent factor is Φ(M) = M(1 + bM²)⁻ᶜ, where b = (γ − 1)/2 and c = (γ + 1)/[2(γ − 1)].

For M > 0, the logarithmic derivative is Φ′/Φ = 1/M − 2bcM/(1 + bM²). Its sign is the sign of 1 − M² after putting the terms over the common positive denominator M(1 + bM²). Thus Φ rises for subsonic M, reaches its maximum at 1 and decreases on the supersonic mathematical branch for fixed area and stagnation state.

A decrease of Φ beyond M = 1 does not mean lowering a converging nozzle’s back pressure sends its exit onto that branch. The nozzle geometry and characteristic/boundary conditions still choose the allowed solution. Evaluating the sonic state gives the critical pressure ratio and choked flow; solving the area relation farther downstream needs the appropriate branch and, if present, shock losses.

Recognise where the simple model ends

Compressible-flow reasoning is useful in nozzles, pneumatic equipment, inlets, exhausts, pressure-wave devices and high-speed aerodynamics. Useful predictions require the actual boundaries and gas properties. A nozzle mass-flow limit is conditional on its upstream state and area; a shock-state ratio is conditional on the incoming state and front orientation.

Heat addition changes the total-enthalpy balance. Wall friction adds momentum exchange and entropy production even when the passage is adiabatic. Constant-area heat-transfer and friction models are often called Rayleigh and Fanno flow; their distinct balances cannot be replaced by the isentropic area equation. The thermodynamic and transport lessons are the return path for those extensions.

At sufficiently high temperature, heat capacities, molecular vibration, dissociation and chemical reactions can change the thermal closure. With rarefaction, a local continuum equilibrium model or an infinitesimally thin front may no longer resolve the relevant physics. Liquids and dense fluids need their own state relations. These are limits of the supplied perfect-gas calculations, not numerical corrections already included in them.

A shock can interact with a viscous boundary layer and turbulent motion, altering separation and unsteadiness. Boundary layers, lift, and drag and Turbulence develop the neighboring transport mechanisms. Compressibility also changes averaging: density-weighted or Favre averages are often used when density fluctuates. A constant-density covariance model is not automatically a compressible closure.

Check the common misreadings

‘Hotter gas proves outside heat was added’ misses kinetic-to-enthalpy conversion. ‘No heat crosses the shock, so entropy stays constant’ confuses adiabatic and reversible flow. ‘Pressure rises, so stagnation pressure must rise’ confuses static state with isentropic recovery potential.

‘The gas is supersonic, so every observer must see it moving faster than sound’ ignores the reference frame. ‘A contact is a shock because density jumps’ ignores its zero relative mass flux and continuous pressure/velocity. ‘Every oblique shock makes the full downstream flow subsonic’ confuses the normal Mach component with the full Mach number.

‘A throat guarantees choking’ omits operating pressure. ‘Choked means no flow’ misses the finite sonic mass flux. ‘A/A* gives one Mach number’ misses the two branches. A defensible answer names the state equation, front or observer, conservation conditions, entropy test and boundary constraints that select the result.

Worked examples

Try it yourself

1. A normal shock moves at laboratory speed S. Which velocity belongs in its normal Mach number and steady jump relations: u or u − S?

   Show the answer and reasoning

   Use the shock-relative velocity w = u − S.

   The planar local jump balances are written in a frame moving with the front. The sound speed belongs to the local thermodynamic state, while gas velocity depends on the observer.

2. In the assigned uniform 300 K state at M = 0.5, what is the laboratory speed of the acoustic family directed against the gas? Does the same family travel upstream in the M = 2 comparison?

   Show the answer and reasoning

   u − a ≈ −173.594355 m/s at M = 0.5; at M = 2 it is +347.188709 m/s and travels downstream in the laboratory.

   The signal travels at −a relative to the gas, so its laboratory speed is u − a. This is a comparison of infinitesimal acoustic information, not a calculation of a finite moving shock.

3. An algebraic expansion jump conserves the fluxes but takes gas from the original downstream thermodynamic state to the original upstream state. Is conservation alone enough to admit it?

   Show the answer and reasoning

   No. In that reversed crossing direction the gas entropy would decrease.

   The original compression has a positive entropy increase. Reversing it changes the sign. An admissible expanding Riemann pattern uses a rarefaction rather than this entropy-violating shock.

4. The assigned contact starts at x = 0 and both gases have u = 100 m/s. Where is it after 2 ms, and what is the gas mass flux through it in its own frame?

   Show the answer and reasoning

   x = 0.200 m, with zero relative mass flux on both sides.

   The contact speed equals u. Multiplying by 0.002 s gives its position; ρ(u − S) = 0 because the material boundary moves with the gas.

5. For γ = 1.4, A/A* = 1.6875 admits M ≈ 0.372244 and M = 2. Does this equation by itself select the operating Mach at the exit of a nozzle?

   Show the answer and reasoning

   No. The branch and full flow require upstream/downstream conditions and geometry.

   The isentropic area relation has a minimum at M = 1 and two branches above one. A shock can also change stagnation pressure and therefore the local area–Mach bookkeeping.

6. For the ideal choked nozzle, double reservoir p₀ while holding T₀, area and gas constants fixed, and retain sufficiently low back pressure. What happens to ṁ*?

   Show the answer and reasoning

   It doubles, from about 0.0466712 to 0.0933423 kg/s.

   The choked mass-flow expression is proportional to p₀ at fixed other inputs. Choking is a conditional limit for a specified upstream state, not an absolute maximum for every operating condition.

7. The supplied oblique case has M_n2 ≈ 0.733799 but full M₂ ≈ 1.456324. Is one of these numbers necessarily wrong?

   Show the answer and reasoning

   No. They refer to the normal component and the full downstream velocity, respectively.

   The tangential velocity is unchanged in the jump model. The full velocity is larger than its normal projection, with M_n2 = M₂ sin(β − θ).

8. Could the same γ = 1.4, constant-c_p model be used without reassessment for strongly heated reacting gas merely because its upstream Mach number is 2?

   Show the answer and reasoning

   No. Mach number alone does not establish the thermal or chemical model’s validity.

   Heat capacities, vibrational excitation, chemical composition and equilibrium can depend on temperature and other conditions. The closure and energy pathways must be checked before transferring the perfect-gas ratios.

What to carry forward

• Compressible flow couples density, momentum and energy through a state/thermal model; Mach number uses local sound speed and a named observer.
• Small acoustic information travels at u ± a in one dimension, while material contact information travels with u.
• A normal shock conserves the stated fluxes while increasing entropy along crossing gas; a zero-thickness jump does not resolve its molecular structure.
• Static temperature and pressure can rise while stagnation temperature stays fixed and stagnation pressure falls.
• A contact has zero relative mass flux and continuous pressure/normal velocity, despite a possible density and entropy difference.
• Isentropic nozzle relations need their own conditions and branch selection; choking fixes a finite mass flow for specified reservoir state and area.
• An oblique shock acts on the normal component, so its full downstream flow can remain supersonic; attachment and branch selection depend on geometry and boundaries.