Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations: Worked examples

This page is a drill ground. The parent note Stagnation quantities gave you the three master formulas. Here we hunt down every kind of situation those formulas can be thrown at — every regime of speed, every degenerate input, every trap — and work each one from scratch.

Before we start, let us re-anchor the tools so no symbol is used unearned.

Naming that repeated lump $B$ is the single most useful habit for these problems: compute $B$ once from $M$ , then raise it to whichever power you need.

Before diving into individual cases, look at the master map below. It plots all three ratios against Mach number on one canvas — every worked example is just reading a vertical slice off this picture. Notice how the pressure curve (steep power) rockets away from the temperature curve (gentle) as $M$ grows: that single visual fact explains most of the "surprises" in the examples.

Figure — Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations

The scenario matrix

Every problem this topic can throw is one (or a blend) of these case classes. The examples below are tagged with the cell they hit; together they fill the whole grid.

#	Case class	What is tricky about it	Example
A	Given $M$ , find $T_{0}, P_{0}, ρ_{0}$ (forward, supersonic)	Big exponents amplify errors	Ex 1
B	Given $P_{0} / P$ , find $M$ and $V$ (inverse, subsonic)	Must invert the power	Ex 2
C	Degenerate: $M = 0$ (gas already at rest)	Stagnation = static	Ex 3
D	Limiting: $M \to 0$ small but nonzero	Bernoulli vs full formula	Ex 3
E	Sonic point $M = 1$ (throat / reference)	Critical ratios, universal numbers	Ex 4
F	Hypersonic $M ≫ 1$	Enormous heating; asymptotics	Ex 5
G	$T_{0}$ conserved but $P_{0}$ lost (across a shock)	Adiabatic ≠ isentropic	Ex 6
H	Real-world word problem (aircraft nose heating)	Translate words → $M$ → $T_{0}$	Ex 7
I	Exam twist: non-air gas $γ \neq = 1.4$ , unit trap	Different $γ$ , kPa vs Pa, $R$	Ex 8

We attack them in order.

Example 1 — Cell A: forward, supersonic

Worked example Air at

$M = 3$ , $T = 220 K$ , $P = 20 kPa$ , $γ = 1.4$ . Find $T_{0}$ , $P_{0}$ , $ρ_{0}$ . Forecast: $T_{0}$ rises by a factor of a few; $P_{0}$ rises by a large factor because of the $3.5$ power. Guess: is $P_{0}$ closer to $2 \times$ , $10 \times$ , or $30 \times$ the static pressure? (Write your guess before reading on.)

Step 1 — Compute the base factor $B$ . $B = 1 + \frac{γ - 1}{2} M^{2} = 1 + \frac{0.4}{2} (3)^{2} = 1 + 0.2 \times 9 = 2.8$ Why this step? Every relation is a power of $B$ , so we compute it once and reuse it — this is the whole efficiency trick.

Step 2 — Temperature. $T_{0} = B T = 2.8 \times 220 = 616 K$ . Why this step? $T_{0} / T = B^{1}$ ; temperature uses the smallest exponent, so it climbs gently.

Step 3 — Pressure. $P_{0} = P B^{γ / (γ - 1)} = 20 \times (2.8)^{3.5}$ . $(2.8)^{3.5} \approx 36.73$ , so $P_{0} \approx 734.6 kPa$ . Why this step? Exponent $γ / (γ - 1) = 1.4/0.4 = 3.5$ — the steep one — so pressure explodes. (Guessing " $30 \times$ " was right, actually $\approx 37 \times$ .)

Step 4 — Density. $ρ_{0} = ρ B^{1/ (γ - 1)}$ . First get static density from the ideal gas law $ρ = P / (R T)$ with $R = 287 J/kg\cdotK$ : $ρ = \frac{20000}{287 \times 220} = 0.3168 kg/m^{3}$ Then $ρ_{0} = 0.3168 \times (2.8)^{2.5}$ , and $(2.8)^{2.5} \approx 13.12$ , so $ρ_{0} \approx 4.157 kg/m^{3}$ . Why this step? Exponent $1/ (γ - 1) = 2.5$ . Note $P$ was in Pa (20000) so density comes out in SI.

Verify: The ideal-gas consistency check must hold: $\frac{P _{0}}{P} = \frac{ρ _{0}}{ρ} \cdot \frac{T _{0}}{T}$ . $36.73 = ? 13.12 \times 2.8 = 36.74$ ✓ (rounding). All three are mutually consistent — a powerful self-check.

The three answers are exactly the three heights you would read off the master figure above at $M = 3$ : temperature at $2.8$ , density at $13.1$ , pressure at $36.7$ — the pressure curve towering over the others.

Example 2 — Cell B: inverse, subsonic

Worked example A subsonic

Pitot tube gives $P_{0} = 150 kPa$ , static $P = 120 kPa$ , static $T = 260 K$ , air. Find $M$ and $V$ . Forecast: $P_{0} / P = 1.25$ — modest, so $M$ should be somewhere in the $0.5$ – $0.7$ range (a $25%$ pressure rise is a fair chunk but well below the $\sim 89%$ rise you get at $M = 1$ ). Guess a number now.

Step 1 — Get the base factor by inverting the power. $B = (\frac{P _{0}}{P})^{(γ - 1) / γ} = (1.25)^{0.4/1.4} = (1.25)^{0.2857} \approx 1.0658$ Why this step? Pressure gives us $B^{3.5}$ ; to peel off the exponent we raise to the reciprocal $1/3.5 = (γ - 1) / γ$ . This is the "arctan undoing tan" move — we invert the operation that built the formula.

Step 2 — Solve $B$ for $M$ . $1 + 0.2 M^{2} = 1.0658 \Rightarrow M^{2} = \frac{0.0658}{0.2} = 0.329 \Rightarrow M = 0.574$ Why this step? $B = 1 + \frac{γ - 1}{2} M^{2}$ with $\frac{γ - 1}{2} = 0.2$ ; algebra isolates $M$ .

Step 3 — Convert Mach to speed. Local sound speed $a = γ R T = 1.4 \times 287 \times 260$ : $a = 104, 468 \approx 323.2 m/s, V = M a = 0.574 \times 323.2 \approx 185.5 m/s$ Why this step? Mach is dimensionless; to get a real speed we multiply by the local $a$ — see Speed of sound a = √(γRT).

Verify: Rebuild $P_{0} / P$ from $M = 0.574$ : $B = 1 + 0.2 (0.329) = 1.0658$ , $B^{3.5} = 1.065 8^{3.5} \approx 1.250$ ✓. Also $M = 0.574 < 1$ , so "subsonic" assumption holds and the isentropic Pitot formula is legitimate.

Example 3 — Cells C & D: rest and near-rest (degenerate + limit)

The figure below is the "zoom-in" of the master map near $M = 0$ : it overlays the exact pressure curve with the straight Bernoulli line so you can see how they kiss at the origin and only peel apart past $M \approx 0.3$ . That visual is the whole content of this example.

Worked example (a) Gas sits still in a tank at

$T = 300 K$ , $P = 101 kPa$ . What are $T_{0}, P_{0}$ ? (b) A slow duct flow has $V = 30 m/s$ , $T = 300 K$ , $P = 101 kPa$ , $ρ = 1.173 kg/m^{3}$ . Compare full $P_{0}$ vs Bernoulli. Forecast: (a) If nothing moves, stagnation must equal static — the difference is stored in speed, and there is none. (b) At $30 m/s$ ( $M \approx 0.086$ ) Bernoulli should be almost perfect. Guess the error: below $1%$ ?

Part (a) — the $M = 0$ degenerate case (Cell C). Step 1. $M = 0 \Rightarrow B = 1 + 0.2 (0) = 1$ . Step 2. $T_{0} = 1 \times 300 = 300 K$ , $P_{0} = 101 \times 1^{3.5} = 101 kPa$ . Why this step? With no bulk motion there is no ordered kinetic energy to dump — stagnation is static. This is the sanity boundary of the whole theory, and it is the left edge of the figure where all curves meet at $1$ .

Part (b) — the small- $M$ limit (Cell D). Step 1 — get $M$ . $a = 1.4 \times 287 \times 300 = 347.2 m/s$ , $M = 30/347.2 = 0.0864$ . Step 2 — full compressible $P_{0}$ . $B = 1 + 0.2 (0.0864)^{2} = 1.001494$ , $P_{0} = 101 \times B^{3.5} = 101 \times 1.005238 = 101.529 kPa$ . So $P_{0} - P = 0.529 kPa$ . Why this step? The exact relation, no approximation — this is a point on the solid curve of the figure. Step 3 — Bernoulli Bernoulli's equation (incompressible limit): $P_{0} - P = \frac{1}{2} ρ V^{2} = \frac{1}{2} (1.173) (30)^{2} = 527.9 Pa = 0.5279 kPa$ . Why this step? This is the $M \to 0$ limit derived in the parent note — the dashed straight line — and it should nearly match at this tiny $M$ .

Verify: Full gives $529 Pa$ , Bernoulli gives $528 Pa$ — a difference of $\approx 0.2%$ , exactly as forecast for such low $M$ . Bernoulli is safe here; it fails only when $M ≳ 0.3$ .

Example 4 — Cell E: the sonic reference $M = 1$

Worked example For air (

$γ = 1.4$ ), compute the critical ratios $T_{0} / T^{*}$ , $P_{0} / P^{*}$ , $ρ_{0} / ρ^{*}$ at $M = 1$ . These are universal constants you should recognise. Forecast: At $M = 1$ the base factor is $1.2$ . The pressure ratio will be the famous " $1.893$ " you see tabulated. Do you remember its reciprocal (static/total ≈ 0.528)?

Step 1 — base factor. $B = 1 + 0.2 (1)^{2} = 1.2$ . Why this step? $M = 1$ is the reference point (star $^{*}$ ) used all over Isentropic flow relations and nozzle theory.

Step 2 — the three ratios. $\frac{T _{0}}{T ^{*}} = 1.2, \frac{P _{0}}{P ^{*}} = 1. 2^{3.5} \approx 1.8929, \frac{ρ _{0}}{ρ ^{*}} = 1. 2^{2.5} \approx 1.5774$ Why this step? Same $B$ , three exponents — the payoff of naming the lump.

Step 3 — read them as fractions of stagnation. Static/total at sonic: $T^{*} / T_{0} = 0.8333$ , $P^{*} / P_{0} = 0.5283$ , $ρ^{*} / ρ_{0} = 0.6339$ . Why this step? Interpretation: to reach the throat of a nozzle, a gas must drop to $\approx 53%$ of its reservoir pressure. That " $0.528$ " is the choking criterion.

Verify: Consistency: $P_{0} / P^{*} = ? (ρ_{0} / ρ^{*}) (T_{0} / T^{*}) = 1.5774 \times 1.2 = 1.8929$ ✓.

Example 5 — Cell F: hypersonic $M ≫ 1$

Worked example A re-entry probe flies at

$M = 8$ through air at $T = 250 K$ , $γ = 1.4$ . Estimate $T_{0}$ . Why is this dangerous? Forecast: At $M = 8$ the $M^{2} = 64$ term dominates. $T_{0}$ will be enormous — thousands of kelvin. Guess an order of magnitude.

Step 1 — base factor. $B = 1 + 0.2 (8)^{2} = 1 + 12.8 = 13.8$ . Why this step? The " $1$ " is now negligible next to $12.8$ — hypersonic asymptotics: $B \to \frac{γ - 1}{2} M^{2}$ .

Step 2 — stagnation temperature. $T_{0} = B T = 13.8 \times 250 = 3450 K$ . Why this step? Uses only the adiabatic relation — no assumption of smooth deceleration needed for $T_{0}$ (Cell G reminds us why).

Step 3 — physical meaning. $3450 K$ exceeds the melting point of most metals ( $\sim 1800 K$ ). This is why re-entry vehicles need heat shields: the air brought to rest at the nose dumps colossal kinetic energy into heat. Why this step? Connects the number to engineering reality.

Verify: Asymptotic check — $\frac{γ - 1}{2} M^{2} = 0.2 \times 64 = 12.8$ , and $B / (0.2 M^{2}) = 13.8/12.8 = 1.078$ , i.e. the " $+ 1$ " contributes only $\approx 7%$ at $M = 8$ , confirming the high- $M$ approximation $T_{0} \approx \frac{γ - 1}{2} M^{2} T$ is reasonable. Also $3450 K > 1800 K$ , so "dangerous" is justified.

Example 6 — Cell G: $T_{0}$ survives, $P_{0}$ is lost (across a shock)

The figure below is the heart of this example. Read it left-to-right: orange arrow = the flow crossing the violet dashed shock line. The magenta bar runs dead flat across the shock — that is $T_{0} = 540 K$ conserved. The navy bar steps down at the shock — that is $P_{0}$ dropping from $391$ to $283 kPa$ . The vertical dotted drop and the violet " $27.7%$ loss" label mark exactly where the second law bites. Keep your eye on those two bars as you follow the numbers.

Worked example Air enters a

normal shock at $M_{1} = 2$ , static $T_{1} = 300 K$ , $P_{1} = 50 kPa$ . Downstream (from shock tables) $M_{2} = 0.5774$ , $P_{2} = 225 kPa$ , $T_{2} = 506.25 K$ . Show $T_{0}$ is unchanged but $P_{0}$ drops. Forecast: The shock is adiabatic (no heat crosses it) but violently irreversible. So $T_{0, 1} = T_{0, 2}$ exactly, while $P_{0, 2} < P_{0, 1}$ . Guess the pressure loss: $10%$ ? $30%$ ?

Step 1 — upstream stagnation. $B_{1} = 1 + 0.2 (4) = 1.8$ . $T_{0, 1} = 1.8 \times 300 = 540 K$ ; $P_{0, 1} = 50 \times 1. 8^{3.5} = 50 \times 7.824 = 391.2 kPa$ . Why this step? Standard forward computation on the upstream state — the left end of both bars in the figure.

Step 2 — downstream stagnation temperature. $B_{2} = 1 + 0.2 (0.5774)^{2} = 1.0667$ . $T_{0, 2} = 1.0667 \times 506.25 = 540.0 K$ . Why this step? We test conservation of $T_{0}$ using the downstream static values and Mach — this is the right end of the flat magenta bar.

Step 3 — downstream stagnation pressure. $P_{0, 2} = P_{2} B_{2}^{3.5} = 225 \times 1.066 7^{3.5} = 225 \times 1.2569 = 282.8 kPa$ . Why this step? Same isentropic relation applies within the smooth flow downstream of the shock, but referenced to the post-shock state — the lower step of the navy bar.

Step 4 — the loss. $\frac{P _{0, 2}}{P _{0, 1}} = \frac{282.8}{391.2} = 0.723$ — a $27.7%$ loss of total pressure. Why this step? This ratio is the entropy penalty of the shock; energy ( $T_{0}$ ) is conserved, order ( $P_{0}$ ) is not. "Temp is Tame, Press is Power" — and Power is the thing you can lose.

Verify: $T_{0, 1} = T_{0, 2} = 540 K$ ✓ (conserved to the digit). $P_{0, 2} / P_{0, 1} = 0.723 < 1$ ✓ (a genuine loss, as the second law demands for any irreversible adiabatic process).

Example 7 — Cell H: real-world word problem

Worked example A jet cruises at

$V = 250 m/s$ through air at $T = 223 K$ (about $11 km$ altitude). What temperature does a sensor mounted at the stagnation point on the nose read? (There the air is brought fully to rest.) Forecast: The nose sensor reads $T_{0}$ , hotter than the $- 5 0^{\circ} C$ ambient. By how much — a few degrees, or tens of degrees?

Step 1 — translate words to $M$ . The nose stagnation point is where $V \to 0$ , so we need the free-stream Mach. $a = 1.4 \times 287 \times 223 = 89, 577 = 299.3 m/s$ ; $M = 250/299.3 = 0.8353$ . Why this step? "Sensor at stagnation point" is the physical realisation of decelerate to rest — the sensor reads $T_{0}$ .

Step 2 — base factor. $B = 1 + 0.2 (0.8353)^{2} = 1 + 0.1396 = 1.1396$ . Why this step? $B$ is the single quantity that turns a Mach number into every stagnation ratio; we build it once before reaching for the temperature relation.

Step 3 — stagnation temperature. $T_{0} = 1.1396 \times 223 = 254.1 K$ . Why this step? Only the adiabatic $T_{0}$ relation is needed — friction at the nose does not change $T_{0}$ .

Verify: Cross-check via the direct energy form $T_{0} = T + \frac{V ^{2}}{2 c _{p}}$ with $c_{p} = \frac{γ R}{γ - 1} = \frac{1.4 \times 287}{0.4} = 1004.5 J/kg\cdotK$ : $T_{0} = 223 + \frac{25 0 ^{2}}{2 \times 1004.5} = 223 + 31.1 = 254.1 K$ ✓. The nose runs $\approx 31 K$ warmer than ambient — "ram heating", real and measurable.

Example 8 — Cell I: exam twist (non-air gas + unit trap)

Worked example Helium (

$γ = 5/3$ , $R = 2077 J/kg\cdotK$ ) flows at $M = 1.5$ , static $T = 400 K$ , static $P = 80000 Pa$ . Find $T_{0}$ and $P_{0}$ . Trap: do not use $γ = 1.4$ and watch the pressure units. Forecast: Helium's $γ = 5/3 \approx 1.667$ is larger, so the coefficient $\frac{γ - 1}{2} = 0.333$ is bigger — $T_{0}$ rises faster with $M$ than air does. Guess whether $T_{0} / T$ here exceeds the air value ( $1.45$ ) at the same $M$ .

Step 1 — coefficient. $\frac{γ - 1}{2} = \frac{5/3 - 1}{2} = \frac{2/3}{2} = \frac{1}{3} = 0.3333$ . Why this step? Using $0.2$ (air) here would be the classic exam mistake. The coefficient is $γ$ -dependent, so it must be rebuilt for every new gas.

Step 2 — base factor. $B = 1 + 0.3333 (1.5)^{2} = 1 + 0.3333 \times 2.25 = 1.75$ . Why this step? Same " $B$ once" habit — but now with helium's coefficient baked in, not air's.

Step 3 — temperature. $T_{0} = 1.75 \times 400 = 700 K$ . Why this step? $T_{0} / T = 1.75 > 1.45$ — confirmed: heavier $γ$ heats faster, as forecast.

Step 4 — pressure exponent. $\frac{γ}{γ - 1} = \frac{5/3}{2/3} = 2.5$ . So $P_{0} = P B^{2.5} = 80, 000 \times 1.7 5^{2.5}$ . $1.7 5^{2.5} \approx 4.051$ , giving $P_{0} \approx 324, 100 Pa \approx 324.1 kPa$ . Why this step? The exponent $γ / (γ - 1) = 2.5$ differs from air's $3.5$ ; using $3.5$ would be wrong. Pressure came out in Pa because we kept $P$ in Pa throughout.

Verify: Consistency $P_{0} / P = (ρ_{0} / ρ) (T_{0} / T)$ . Density exponent $\frac{1}{γ - 1} = 1.5$ , so $ρ_{0} / ρ = 1.7 5^{1.5} = 2.315$ . Then $2.315 \times 1.75 = 4.051 = 1.7 5^{2.5}$ ✓. The three relations remain mutually consistent for any $γ$ .

Recall

Recall One-line reflexes from this matrix

Base factor definition ::: $B = 1 + \frac{γ - 1}{2} M^{2}$ ; then $T_{0} / T = B$ , $P_{0} / P = B^{γ / (γ - 1)}$ , $ρ_{0} / ρ = B^{1/ (γ - 1)}$ . At $M = 0$ what are $T_{0}, P_{0}$ ? ::: Equal to static $T, P$ — no motion, nothing stored. Critical pressure ratio $P^{*} / P_{0}$ for air? ::: $\approx 0.528$ (the choking value). Across a normal shock, which total quantity is conserved? ::: $T_{0}$ (adiabatic); $P_{0}$ drops (irreversible). Same $M$ : does helium ( $γ = 5/3$ ) or air ( $γ = 1.4$ ) get a bigger $T_{0} / T$ ? ::: Helium, because $\frac{γ - 1}{2}$ is larger. When may you use Bernoulli for $P_{0}$ ? ::: Only $M ≲ 0.3$ (error stays small); above that use the full power law.