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

The vocabulary — every symbol defined before use

Deriving the exponents 1, 2.5, 3.5 right here (no external note needed)

Formula Step-by-step: why

$T$ gets exponent 1, $ρ$ gets 2.5, $P$ gets 3.5 (a) Temperature — exponent 1. The energy equation is the temperature statement, so directly $\frac{T _{0}}{T} = B = B^{1} .$ (b) Density and pressure — from two laws we combine. An isentropic (reversible, adiabatic) perfect gas obeys $P v^{γ} = const$ , where $v = 1/ ρ$ is the volume of one kilogram. It also obeys the ideal-gas law $P v = R T$ . We eliminate $v$ :

From ideal gas: $v = \frac{R T}{P}$ . Put this into $P v^{γ} = const$ : $P (\frac{R T}{P})^{γ} = const \Rightarrow P^{1 - γ} T^{γ} = const \Rightarrow P \propto T^{\frac{γ}{γ - 1}} .$ Comparing the moving state to the stopped state (same constant), $\frac{P _{0}}{P} = (\frac{T _{0}}{T})^{γ / (γ - 1)} = B^{γ / (γ - 1)}$ . For air $\frac{γ}{γ - 1} = \frac{1.4}{0.4} = 3.5$ .
Since $ρ = \frac{P}{R T} \propto \frac{T ^{γ / (γ - 1)}}{T} = T^{\frac{1}{γ - 1}}$ , we get $\frac{ρ _{0}}{ρ} = B^{1/ (γ - 1)}$ . For air $\frac{1}{γ - 1} = \frac{1}{0.4} = 2.5$ .

The one extra $γ$ in the pressure numerator is exactly why pressure carries the biggest exponent and therefore drops fastest. (Also in Isentropic Process Relations for Perfect Gas.)

Figure — reading the three curves

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

Intuition How to read this figure to solve problems directly

The horizontal axis is the Mach number $M$ ; the vertical axis is any of the three local/stagnation ratios. All three curves start at 1 when $M = 0$ (no motion ⇒ nothing spent ⇒ local $=$ stagnation) and fall monotonically — none turns back up, because $B = 1 + 0.2 M^{2}$ only grows with $M$ , and each ratio is a decreasing power of $B$ . To use it as a table: pick your $M$ on the bottom axis, go straight up, and read where you hit each curve. The red curve ( $P / P_{0}$ ) always sits lowest (exponent 3.5, plunges fastest); the dashed density curve is in the middle (2.5); the solid black temperature curve is on top (exponent 1, drops slowest) — this is the ordering you'll prove in L1.1. The vertical dotted line at $M = 1$ marks the sonic / critical point where the red curve reads $P / P_{0} = 0.528$ (that value governs choking in L3.2). As $M$ grows large all three curves flatten toward 0. Every exercise below is literally reading, or inverting, one point off these curves.

Level 1 — Recognition

L1.1

State whether each ratio is $\geq 1$ or $\leq 1$ for $M > 0$ , and order the three local-over-stagnation ratios from smallest to largest. Also state their limits as $M \to 0$ and $M \to \infty$ .

Recall Solution

Every local property is what's left after some of the "bank account" is spent as motion, so local $<$ stagnation: $\frac{T}{T _{0}}, \frac{P}{P _{0}}, \frac{ρ}{ρ _{0}}$ are all $\leq 1$ . Their reciprocals equal $B^{1}, B^{2.5}, B^{3.5}$ with $B > 1$ ; a bigger exponent means a bigger reciprocal, hence a smaller ratio. So $\frac{P}{P _{0}} < \frac{ρ}{ρ _{0}} < \frac{T}{T _{0}} .$ Pressure has the biggest exponent (3.5) so it falls fastest. Limits: as $M \to 0$ , $B \to 1$ and all three ratios $\to 1$ (nothing spent). As $M \to \infty$ , $B \to \infty$ and all three $\to 0$ — see the figure's curves flattening onto the axis.

L1.2

For air at $M = 0.5$ , compute the base factor $B$ and then $T / T_{0}$ .

Recall Solution

$B = 1 + 0.2 (0.5)^{2} = 1 + 0.2 (0.25) = 1 + 0.05 = 1.05$ . $\frac{T}{T _{0}} = \frac{1}{B} = \frac{1}{1.05} = 0.9524$ . Matches the parent table row $M = 0.5$ . ✓

Level 2 — Application

L2.1

Air enters a duct with $T_{0} = 320$ K, $P_{0} = 600$ kPa. At a point where $M = 0.8$ , find $T$ , $P$ , and $ρ / ρ_{0}$ .

Recall Solution

$B = 1 + 0.2 (0.8)^{2} = 1 + 0.2 (0.64) = 1.128$ .

$T = T_{0} / B = 320/1.128 = 283.7$ K.
$P = P_{0} / B^{3.5} = 600/ (1.128)^{3.5} = 600/1.5243 = 393.6$ kPa.
$ρ / ρ_{0} = B^{- 2.5} = (1.128)^{- 2.5} = 0.7400$ . Note $B^{- 2.5} = 1/ B^{2.5}$ — the negative exponent is inversion, nothing more.

L2.2

A wind tunnel supplies air at $M = 1.5$ . Given the local static temperature is $T = 200$ K, what is the stagnation temperature $T_{0}$ ?

Recall Solution

$B = 1 + 0.2 (1.5)^{2} = 1 + 0.2 (2.25) = 1.45$ . $T_{0} = B \cdot T = 1.45 \times 200 = 290$ K. (Here we multiply, because $T_{0} / T = B$ so $T_{0} = B T$ .)

L2.3

For $M = 2.0$ air, verify the Feynman consistency check numerically: show $\frac{P _{0} / P}{ρ _{0} / ρ} = \frac{T _{0}}{T}$ .

Recall Solution

$B = 1 + 0.2 (4) = 1.8$ . $P_{0} / P = 1. 8^{3.5} = 7.824$ , $ρ_{0} / ρ = 1. 8^{2.5} = 4.347$ , $T_{0} / T = 1.8$ . $\frac{7.824}{4.347} = 1.800 = B$ . ✓ This is just the ideal-gas law $P / ρ \propto T$ reappearing — and it works precisely because the pressure exponent $\frac{γ}{γ - 1}$ minus the density exponent $\frac{1}{γ - 1}$ equals $\frac{γ - 1}{γ - 1} = 1$ , the temperature exponent.

Level 3 — Analysis

L3.1 (inverse problem)

At a nozzle station the measured static pressure is $40%$ of the reservoir pressure: $P / P_{0} = 0.40$ . Find the Mach number (air).

Recall Solution

$\frac{P _{0}}{P} = \frac{1}{0.40} = 2.5 = B^{3.5}$ . Undo the exponent by raising to $1/3.5$ : $B = 2. 5^{1/3.5} = 2. 5^{0.2857} = 1.30931$ . Recover the speed using the kinetic-energy relation $0.2 M^{2} = B - 1$ (remember: $B - 1$ is the spent kinetic energy in Mach units): $0.2 M^{2} = 0.30931 \Rightarrow M^{2} = 1.54657 \Rightarrow M = 1.2436$ .

L3.2 (choking check)

The reservoir feeding a converging nozzle is at $P_{0} = 250$ kPa. The back pressure is $150$ kPa. Is the nozzle choked?

Recall Solution

Critical ratio for air: $P^{*} / P_{0} = B^{- 3.5}$ at $M = 1$ , where $B = 1.2$ , giving $P^{*} / P_{0} = 1. 2^{- 3.5} = 0.5283$ . Critical exit static pressure $= 0.5283 \times 250 = 132.1$ kPa. Back pressure $150$ kPa $> 132.1$ kPa, so the exit pressure need not fall to $P^{*}$ ; the flow reaches only $M < 1$ . Not choked. (See Converging-Diverging Nozzle & Choking.)

L3.3 (density from two measurements)

At a point $ρ / ρ_{0} = 0.50$ . Find $M$ and then the corresponding $T / T_{0}$ (air).

Recall Solution

$\frac{ρ _{0}}{ρ} = \frac{1}{0.50} = 2 = B^{2.5}$ . Undo the exponent by raising to $1/2.5$ : $B = 2^{1/2.5} = 2^{0.4} = 1.31951$ . Recover the speed with the same kinetic-energy relation $0.2 M^{2} = B - 1$ : $0.2 M^{2} = 0.31951 \Rightarrow M^{2} = 1.59754 \Rightarrow M = 1.2640$ . $T / T_{0} = 1/ B = 1/1.31951 = 0.7579$ .

Level 4 — Synthesis

L4.1 (mass-flow-ish, combining ratios)

Air flows with $M = 1.2$ , static $T = 250$ K, static $P = 80$ kPa. Compute the local speed $V$ , then $T_{0}$ and $P_{0}$ . Use the specific gas constant $R = 287 J\cdotkg^{- 1} K^{- 1}$ (for air) and the speed of sound $a = γ R T$ .

Recall Solution

What is $R$ ? The specific gas constant is the universal constant $8.314 J\cdotmol^{- 1} K^{- 1}$ divided by air's molar mass ( $0.02897 kg/mol$ ), giving $287 J\cdotkg^{- 1} K^{- 1}$ . Its units make $γ R T$ come out as $(m/s)^{2}$ , so $a = γ R T$ is a speed (details in Speed of Sound and Mach Number). $a = 1.4 \times 287 \times 250 = 100450 = 316.9$ m/s. $V = M a = 1.2 \times 316.9 = 380.3$ m/s. $B = 1 + 0.2 (1.2)^{2} = 1 + 0.288 = 1.288$ . $T_{0} = B T = 1.288 \times 250 = 322.0$ K. $P_{0} = P B^{3.5} = 80 \times (1.288)^{3.5} = 80 \times 2.4303 = 194.4$ kPa.

L4.2 (two stations, isentropic)

In an isentropic nozzle, station 1 has $M_{1} = 0.3$ , station 2 has $M_{2} = 1.8$ . The reservoir feeds both. Find the pressure ratio $P_{2} / P_{1}$ between the two stations.

Recall Solution

Since $P_{0}$ is the same (isentropic, shock-free) and $P / P_{0} = B^{- 3.5}$ , $\frac{P _{2}}{P _{1}} = \frac{P _{0} B _{2}^{- 3.5}}{P _{0} B _{1}^{- 3.5}} = \frac{B _{2}^{- 3.5}}{B _{1}^{- 3.5}} = (\frac{B _{1}}{B _{2}})^{3.5} .$ $B_{1} = 1 + 0.2 (0.09) = 1.018$ , $B_{2} = 1 + 0.2 (3.24) = 1.648$ . $\frac{B _{1}}{B _{2}} = \frac{1.018}{1.648} = 0.6177$ . $P_{2} / P_{1} = 0.617 7^{3.5} = 0.1855$ .

Level 5 — Mastery

L5.1 (across a shock — the forbidden move done right)

Air with $M_{1} = 2.0$ , $P_{0, 1} = 500$ kPa passes through a normal shock. Downstream $M_{2} = 0.5774$ and the stagnation-pressure ratio across the shock is $P_{0, 2} / P_{0, 1} = 0.7209$ (from Normal Shock Relations). A probe downstream reads static pressure. Find the downstream static pressure $P_{2}$ .

Recall Solution

Key idea: the isentropic table works within each side separately, but you must use the correct (post-shock) $P_{0}$ downstream. $P_{0, 2} = 0.7209 \times 500 = 360.45$ kPa. Downstream base factor $B_{2} = 1 + 0.2 (0.5774)^{2} = 1 + 0.2 (0.33339) = 1.06668$ . $P_{2} = P_{0, 2} \cdot B_{2}^{- 3.5} = 360.45 \times (1.06668)^{- 3.5} = 360.45 \times 0.79371 = 286.06$ kPa.

L5.2 (variable $γ$ )

Recompute the sonic (critical) static pressure ratio $P^{*} / P_{0}$ — again $*$ denotes the value at $M = 1$ — for a diatomic-vs-monatomic comparison: (a) $γ = 1.4$ (air), (b) $γ = 1.667$ (argon). Which chokes at a higher $P^{*} / P_{0}$ ?

Recall Solution

At $M = 1$ : $B = 1 + \frac{γ - 1}{2}$ and $P^{*} / P_{0} = B^{- γ / (γ - 1)}$ . The exponent here is still the pressure exponent $\frac{γ}{γ - 1}$ from the isentropic law — it changes because $γ$ changes. (a) $γ = 1.4$ : $B = 1.2$ , exponent $= 3.5$ , $P^{*} / P_{0} = 1. 2^{- 3.5} = 0.5283$ . (b) $γ = 1.667$ : $B = 1 + 0.3335 = 1.3335$ , exponent $= γ / (γ - 1) = 1.667/0.667 = 2.4993$ , $P^{*} / P_{0} = 1.333 5^{- 2.4993} = 0.4871$ . Air ( $γ = 1.4$ ) chokes at the higher ratio $0.528$ ; the monatomic gas must drop to $\approx 0.487$ . (Uses Isentropic Process Relations for Perfect Gas.)

L5.3 (full inverse synthesis)

A Pitot-static setup in subsonic air reads $P_{0} = 105$ kPa, static $P = 101$ kPa, static $T = 288$ K. Find $M$ and the flow speed $V$ . Use $R = 287 J\cdotkg^{- 1} K^{- 1}$ as above.

Recall Solution

$\frac{P _{0}}{P} = \frac{105}{101} = 1.03960 = B^{3.5}$ . Undo the exponent: $B = 1.0396 0^{1/3.5} = 1.0396 0^{0.2857} = 1.01116$ . Recover speed via the kinetic-energy relation $0.2 M^{2} = B - 1 = 0.01116$ : $M^{2} = 0.05581 \Rightarrow M = 0.2362$ . $a = 1.4 \times 287 \times 288 = 115718 = 340.2$ m/s. $V = M a = 0.2362 \times 340.2 = 80.35$ m/s. (Subsonic, so the plain isentropic Pitot formula is valid — no shock in front of the probe.)

Connections

Parent: Isentropic flow tables
Speed of Sound and Mach Number — $a = γ R T$ and $R$ used in L4.1, L5.3.
Stagnation Properties and Energy Equation — why $T_{0} = B T$ and where $\frac{γ - 1}{2} M^{2}$ is born.
Normal Shock Relations — supplies $M_{2}$ and $P_{0, 2} / P_{0, 1}$ in L5.1.
Converging-Diverging Nozzle & Choking — critical ratio in L3.2, L5.2.
Area-Mach Relation A/A* — same stagnation framework.
Isentropic Process Relations for Perfect Gas — supplies the exponents $\frac{γ}{γ - 1}$ , $\frac{1}{γ - 1}$ .