Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions of M: Worked examples

Everything below uses air, $γ = 1.4$ , so the recurring numbers are:

Here $\frac{γ - 1}{2} = \frac{0.4}{2} = 0.2$ , so the base factor is always $1 + 0.2 M^{2}$ .

Definition Symbols and conventions used on this page (read first)

The subscript $0$ (as in $T_{0}, P_{0}, ρ_{0}$ ) always means the stagnation (total) value — the property the flow would reach if brought to rest isentropically. The plain symbols $T, P, ρ$ are the local (static) values that the moving gas actually has. So $T_{0} / T$ , $P_{0} / P$ , $ρ_{0} / ρ$ are always "stopped-over-flowing" and are $\geq 1$ .
$M = \frac{V}{a}$ is the Mach number: the flow speed $V$ divided by the local speed of sound $a$ . For a perfect gas $a = γ R T$ (see Speed of Sound and Mach Number); we build $a$ from local temperature whenever a problem gives velocity instead of $M$ .
$R$ is the specific gas constant of the fluid (energy per unit mass per kelvin). For air $R = 287 J\cdotkg^{- 1} K^{- 1}$ ; it appears inside $a = γ R T$ and in the ideal-gas law $P = ρR T$ .
A star superscript ( $T^{*}, P^{*}, ρ^{*}$ ) denotes the ==critical values at $M = 1$ == — the "sonic" or "choked" state a nozzle reaches at its throat (see Converging-Diverging Nozzle & Choking).
$s$ is entropy, the thermodynamic bookkeeper of irreversibility. Isentropic means " $s$ = constant"; our three formulas were derived assuming exactly this. When $s$ rises (e.g. across a shock) they no longer connect the two states.
Rounding convention: we carry full precision through the arithmetic and report final answers to 4 significant figures (intermediate values shown to 4 s.f. too). Small last-digit differences between an answer and its verification are pure rounding.

The scenario matrix

Every question this topic can ask falls into one of these cells. The examples below are labelled with the cell they hit.

Cell	What makes it distinct	Example
A. $M = 0$ (degenerate)	flow at rest: all ratios $= 1$ , sanity anchor	Ex 1
B. Subsonic $0 < M < 1$	ratios slightly below 1, "gentle" regime	Ex 2
C. Sonic $M = 1$ (critical)	the choke point, starred quantities $T^{}, P^{}, ρ^{*}$	Ex 3
D. Supersonic $M > 1$	large drops, all three ratios small	Ex 4
E. Inverse: given a ratio → find $M$	invert the formula (root, not power)	Ex 5
F. Limiting $M \to \infty$	asymptotic behaviour, ratios → 0	Ex 6
G. Real-world word problem	Pitot / air-intake, must build $M$ first	Ex 7
H. Exam twist — trap	ratios applied across a shock (illegal)	Ex 8
I. Cross-check between ratios	use $\frac{P _{0} / P}{ρ _{0} / ρ} = \frac{T _{0}}{T}$ to catch errors	Ex 9

Cell A — the degenerate case $M = 0$

Cell C — the sonic / critical case $M = 1$

Worked example Example 3 — Nozzle throat (choked)

A converging nozzle is fed from a reservoir at $T_{0} = 500$ K, $P_{0} = 800$ kPa. At the throat the flow reaches $M = 1$ . Find $T^{*}, P^{*}, ρ^{*}$ (the star superscript denotes the critical, $M = 1$ , values).

Forecast: at $M = 1$ the pressure ratio is the famous $\approx 0.528$ , temperature $\approx 0.833$ .

Base factor $= 1 + 0.2 (1)^{2} = 1.2$ . Why this step? $M = 1$ gives the critical/star condition used in Converging-Diverging Nozzle & Choking.
$T^{*} = T_{0} /1.2 = 500/1.2 = 416.7$ K. Why? $T^{*} / T_{0} = 1/1.2 = 0.8333$ ; divide the stagnation value down.
$P^{*} = P_{0} \cdot 1. 2^{- 3.5} = 800 \times 0.5283 = 422.6$ kPa. Why? $P^{*} / P_{0} = 0.5283$ is the choke threshold — the back pressure must fall this far to reach $M = 1$ .
$ρ^{*} = ρ_{0} \cdot 1. 2^{- 2.5} = 0.6339 ρ_{0}$ . Why? Density exponent 2.5.

Verify: $P^{*} / P_{0} = 0.5283 < ρ^{*} / ρ_{0} = 0.6339 < T^{*} / T_{0} = 0.8333$ , the strict ordering holds, and $0.5283 \approx 0.528$ matches the universal air choke ratio. ✓

Cell D — supersonic flow

Figure 1 below traces all three local-over-stagnation ratios across the whole $M$ axis and pins the four worked cells (A, B, C, D) onto their curves. Each curve is labelled directly and in the legend, so it reads correctly even in grayscale: the pressure curve ( $P / P_{0}$ , exponent 3.5) lies lowest because it drops fastest, the density curve ( $ρ / ρ_{0}$ , exponent 2.5) sits in the middle, and the temperature curve ( $T / T_{0}$ , exponent 1) stays highest — everywhere for $M > 0$ .

Figure — Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions of M

Figure 1 — Local/stagnation ratios versus Mach number, with worked cells A–D marked. Curves are labelled and colour-coded (lavender = $T / T_{0}$ , mint = $ρ / ρ_{0}$ , coral = $P / P_{0}$ ).

Cell E — the inverse problem (given a ratio, find $M$ )

Worked example Example 5 — A Pitot reads

$P / P_{0} = 0.3$ . What is $M$ ? Forecast: $0.3$ is below the sonic value $0.528$ , so the flow must be supersonic ( $M > 1$ ).

Write the pressure formula and flip it: $(1 + 0.2 M^{2})^{3.5} = \frac{P _{0}}{P} = \frac{1}{0.3} = 3.333$ . Why this step? We know the left is the base factor to the 3.5 power; the right is the given ratio inverted.
Undo the power by taking the $3.5$ -th root (raise to $1/3.5$ ): $1 + 0.2 M^{2} = 3.33 3^{1/3.5} = 1.4053$ . Why this step? This is the isentropic version of "arctan undoes tan" — the root undoes the power, recovering the base factor.
Isolate $M^{2}$ : $0.2 M^{2} = 0.4053 \Rightarrow M^{2} = 2.026$ . Why? Subtract the 1, divide by $0.2$ .
$M = 2.026 = 1.424$ . Why? Mach number is positive, so take the positive root.

Verify: $M = 1.424 > 1$ matches the forecast (supersonic, because the ratio undercut $0.528$ ), and feeding $M = 1.424$ forward gives $P / P_{0} = 1.405 3^{- 3.5} = 0.300$ . ✓

Cell F — the limiting behaviour $M \to \infty$

Worked example Example 6 — What do the ratios do as

$M$ grows without bound? Forecast: at huge speed, stopping the flow releases enormous kinetic energy, so local $T, P, ρ$ should all shrink toward zero relative to stagnation.

Base factor $\approx 0.2 M^{2}$ for large $M$ (the "+1" becomes negligible). Why this step? When $0.2 M^{2} ≫ 1$ the constant term is dwarfed.
$\frac{T}{T _{0}} = \frac{1}{1 + 0.2 M ^{2}} \to 0$ like $\frac{5}{M ^{2}}$ . Why? Denominator grows like $M^{2}$ , so the ratio decays as $1/ M^{2}$ .
$\frac{P}{P _{0}} = (1 + 0.2 M^{2})^{- 3.5} \to 0$ like $M^{- 7}$ ; $\frac{ρ}{ρ _{0}} \to 0$ like $M^{- 5}$ . Why? Raising a growing base to a negative power drives it to zero faster the bigger the exponent.

Numeric check at $M = 10$ : base $= 1 + 0.2 (100) = 21$ ; $T / T_{0} = 1/21 = 0.04762$ , $P / P_{0} = 2 1^{- 3.5} = 2.472 \times 1 0^{- 5}$ , $ρ / ρ_{0} = 2 1^{- 2.5} = 5.191 \times 1 0^{- 4}$ .

Verify: all three collapse toward 0, pressure fastest — a hypersonic vehicle's surface pressure is a tiny fraction of what you'd get by stopping the flow, which is why hypersonic stagnation-point heating (from $T_{0}$ ) dominates design. ✓ See Speed of Sound and Mach Number for how $M$ itself scales.

Cell G — real-world word problem

Worked example Example 7 — Pitot tube on an aircraft

A jet cruises where the ambient (static) air is $T = 223$ K, $P = 26.5$ kPa, and it flies at $V = 250$ m/s. The nose Pitot port stops the air isentropically. What total pressure and total temperature does it read? ( $R = 287$ J/kg·K.)

Forecast: subsonic-ish airliner speed, so $M < 1$ ; expect a total pressure maybe 40–50% above static.

Speed of sound: $a = γ R T = 1.4 \times 287 \times 223 = 299.3$ m/s. Why this step? Ratios need $M$ , and $M = V / a$ — we must build $a$ from the local temperature first (see Speed of Sound and Mach Number).
Mach number: $M = V / a = 250/299.3 = 0.8353$ . Why? This single number feeds every ratio.
Base factor $= 1 + 0.2 (0.8353)^{2} = 1 + 0.2 (0.6977) = 1.1395$ . Why? Standard step.
$T_{0} = T \cdot 1.1395 = 223 \times 1.1395 = 254.1$ K. Why? Temperature exponent 1.
$P_{0} = P \cdot 1.139 5^{3.5} = 26.5 \times 1.5793 = 41.85$ kPa. Why? Pressure exponent 3.5; this is what the Pitot gauge reads.

Verify: $M = 0.8353 < 1$ confirms subsonic (forecast ✓), and $P_{0}$ is 58% above static — plausible for high-subsonic flight. Units: $(J/kg\cdotK)(K) = J/kg = m^{2} / s^{2} =$ m/s ✓. The stagnation framework here is exactly Stagnation Properties and Energy Equation.

Cell H — the exam trap (ratios across a shock)

Worked example Example 8 — Applying the tables where they're illegal

A flow at $M_{1} = 2$ has stagnation pressure $P_{0, 1} = 500$ kPa. It passes through a normal shock. A student computes the downstream static pressure by looking up the $M_{2}$ isentropic ratio against the same $P_{0}$ . Why is this wrong, and what's the fix?

Forecast: across a shock entropy $s$ jumps, so $P_{0}$ can't be conserved — the naive lookup overestimates $P_{0, 2}$ .

Recognise the shock is not isentropic: the entropy $s$ (the irreversibility bookkeeper defined at the top) increases. Why this step? Our three formulas were derived assuming an isentropic path ( $s$ constant) from local to stagnation — the derivation breaks the instant entropy rises.
Consequence: $P_{0, 2} < P_{0, 1}$ . For $M_{1} = 2$ (from Normal Shock Relations) the stagnation-pressure ratio is $P_{0, 2} / P_{0, 1} = 0.7209$ . Why? The shock "wastes" available pressure as entropy; only $T_{0}$ survives (energy still conserved).
Correct procedure: use the downstream stagnation pressure. $P_{0, 2} = 0.7209 \times 500 = 360.4$ kPa, and only then apply the isentropic ratio with the downstream Mach number ( $M_{2} = 0.5774$ for $M_{1} = 2$ ). Why? Each side of the shock has its own isentropic bank account.
As a check, $T_{0}$ is unchanged: $T_{0, 2} = T_{0, 1}$ . Why? Stagnation enthalpy $h_{0}$ is conserved through a shock (adiabatic), so $T_{0}$ carries across untouched.

Verify: $P_{0, 2} = 360.4$ kPa $< 500$ kPa (loss ✓), while $T_{0}$ is preserved — precisely the split the parent's third "mistake" callout warns about. Never reuse one $P_{0}$ across a shock.

Cell I — cross-check between the ratios

Worked example Example 9 — Catch your own arithmetic error

For $M = 2$ you computed $P / P_{0} = 0.1278$ and $ρ / ρ_{0} = 0.2301$ . Use the ideal-gas cross-check to confirm they're mutually consistent with $T / T_{0} = 0.5556$ .

Forecast: dividing pressure ratio by density ratio should reproduce the temperature ratio (because $P = ρR T$ ).

Compute $\frac{P / P _{0}}{ρ / ρ _{0}} = \frac{0.1278}{0.2301} = 0.5554$ . Why this step? From the ideal-gas law $P = ρR T$ , dividing stagnation-normalised pressure by stagnation-normalised density leaves $\frac{T}{T _{0}}$ — an internal consistency law any correct triple must obey (the specific gas constant $R$ cancels).
Compare to the direct temperature ratio $T / T_{0} = 1/1.8 = 0.5556$ . Why this step? If these two numbers disagree by more than rounding, one of your three ratios has a mis-typed exponent — this is the fastest self-audit, the parent's "Feynman test".

Verify: $0.5554 \approx 0.5556$ (agree to 3 decimals; the tiny gap is just rounding of the tabulated $0.1278$ and $0.2301$ ). Recomputing exactly from the base factor $1.8$ gives $1. 8^{- 3.5} /1. 8^{- 2.5} = 1. 8^{- 1} = 0.5556$ , matching $T / T_{0}$ perfectly. ✓

Recall Which cell is which? (quick self-test)

$M = 1$ throat conditions live in cell ::: C (sonic / critical, the starred $T^{*}, P^{*}, ρ^{*}$ ) "Given $P / P_{0}$ , find $M$ " is cell ::: E (inverse — take a root, not a power) Reusing one $P_{0}$ before and after a normal shock is cell ::: H (illegal, entropy $s$ rises) As $M \to \infty$ all three local/stagnation ratios go to ::: 0 (cell F)

Connections

Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions of M (index 3.1.7) — the parent formulas every example uses.
Speed of Sound and Mach Number — needed in Ex 6–7 to build $M$ from $V$ via $a = γ R T$ .
Stagnation Properties and Energy Equation — why $T_{0}$ survives (Ex 7, 8).
Normal Shock Relations — supplies $P_{0, 2} / P_{0, 1}$ and $M_{2}$ in Ex 8.
Converging-Diverging Nozzle & Choking — the critical/star values of Ex 3.
Area-Mach Relation A/A* — same stagnation framework, next step up.
Isentropic Process Relations for Perfect Gas — the $P \propto T^{γ / (γ - 1)}$ used everywhere.

Worked examples — Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions of M

The scenario matrix

Cell A — the degenerate case $M = 0$

Cell B — subsonic flow

Cell C — the sonic / critical case $M = 1$

Cell D — supersonic flow

Cell E — the inverse problem (given a ratio, find $M$ )

Cell F — the limiting behaviour $M \to \infty$

Cell G — real-world word problem

Cell H — the exam trap (ratios across a shock)

Cell I — cross-check between the ratios

Connections