Question bank — Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations

True — a shock is adiabatic (no heat crosses the thin front), and $T_0$ tracks total enthalpy, which energy conservation keeps constant through any adiabatic process, reversible or not.

False — a shock is irreversible, entropy jumps up, and $P_0$ always drops across it even though $T_0$ does not.

True — isentropic means reversible and adiabatic, so energy conservation fixes $T_0$ and the entropy-being-constant fixes $P_0$ and ${\rho }_0$ as well.

False — a co-moving thermometer reads the static values; stagnation values are what the gas would reach if brought isentropically to rest, an imagined state.

True — with $V=0$ the gas is already at rest, so there is no kinetic energy to convert and $T_0=T$, $P_0=P$, ${\rho }_0=\rho$.

True — both grow with the same base $1+\frac{\gamma -1}{2}{M}^2$, but $P_0/P$ carries the exponent $\gamma /(\gamma -1)=3.5$ (for air) versus $1$ for $T_0/T$, so pressure climbs far more steeply.

False — it is exactly the $M\to 0$ limit of the compressible relation; it is correct for low speeds and only becomes noticeably inaccurate once $M\gtrsim 0.3$ where the $M^4$ compressibility term matters.

True — no heat crosses the wall so total enthalpy (hence $T_0$) is fixed, but friction is irreversible, raises entropy, and bleeds off $P_0$.

True — isentropically stopping compresses the gas the most, and any real (irreversible) stop gives less recovery, so ${\rho }_0$ is the ceiling.

Bernoulli is only the incompressible limit; at $M=2$ it badly underestimates $P_0$ (roughly half the true value), so you must use the full isentropic $P_0/P=(1+\frac{\gamma -1}{2}{M}^2{)}^{\gamma /(\gamma -1)}$.

It must be $c_p$, not $c_v$, because the steady-flow energy equation uses enthalpy $h=c_{p}T$, not internal energy $u=c_{v}T$.

Adiabatic alone only guarantees $T_0$ (energy); $P_0$ needs the process to also be reversible, and a shock is strongly irreversible, so $P_0$ drops.

The local Mach number uses the static temperature at that point, because the sound speed the fluid actually experiences depends on its static $T$, not the imagined stagnation $T_0$.

It only needs the flow to be adiabatic; it holds even with friction or through shocks, unlike the pressure and density relations which require isentropy.

They differ by the temperature ratio via the ideal gas law: $P_0/P=({\rho }_0/\rho )(T_0/T)$, so the exponents on the base are $\gamma /(\gamma -1)$ and $1/(\gamma -1)$ respectively, not equal.

Missing the sound-speed substitution: $\frac{1}{2}\gamma P{M}^2=\frac{1}{2}\gamma P{V}^2/(\gamma RT)=\frac{1}{2}(P/RT)V^2=\frac{1}{2}\rho V^2$, so the $\gamma$ and one factor of $P/RT$ combine into $\rho$, not $P$.

Temperature reflects total energy, and energy is conserved in any adiabatic stop; pressure reflects the ordering of that energy, which depends on how you stop the gas, so it needs the reversibility (isentropy) assumption.

An isentropic stop wastes no kinetic energy to entropy, so it recovers the most pressure possible; any real, irreversible deceleration recovers less.

Because $P_0$ can only stay the same or fall, and it falls exactly when irreversibility (friction, shocks, mixing) occurs, so a drop in $P_0$ directly measures lost useful energy.

Without heat loss keeps energy in the gas (fixing $T_0$); gentle/frictionless keeps entropy constant (fixing $P_0$ and ${\rho }_0$) — together they define a single, well-defined reference state.

Bernoulli is the mechanical-energy statement for a nearly-incompressible fluid, and at $M\to 0$ density changes vanish, so the compressible relation's leading term collapses exactly to $P_0-P=\frac{1}{2}\rho V^2$.

Speeding the gas up converts thermal/pressure energy into bulk motion, dropping its static pressure and temperature; the ideal gas law then forces the density down, so $\rho /{\rho }_0$ falls to about $0.63$ at $M=1$.

Each factor $1+\frac{\gamma -1}{2}{M}^2\to 1$, so $T_0/T$, $P_0/P$, and ${\rho }_0/\rho$ all approach $1$ — static and stagnation states merge.

Yes, as long as the flow is adiabatic; the relation is a pure algebraic consequence of energy conservation and holds for any Mach number, subsonic or supersonic.

No, the ratio depends on $\gamma$ through the $\frac{\gamma -1}{2}$ factor, so a different $\gamma$ gives a different temperature ratio even at identical Mach number.

No — the shock already destroyed some $P_0$ irreversibly, and the later isentropic stage cannot recover it, so the final $P_0$ is below the upstream value while $T_0$ is unchanged.

Yes — if density truly cannot change then stopping the fluid leaves $\rho$ untouched, and the compressible ${\rho }_0/\rho$ relation smoothly returns $1$ in the $M\to 0$ limit consistent with that.

Around $M\approx 0.3$, where the $M^4$ correction term first exceeds a few percent; below that Bernoulli is a good approximation, above it use the full relation.