Chamber-to-exit relation - all quantities as f(M_e, γ): Exercises

Before we start, one shared piece of shorthand that appears everywhere. We keep meeting the group of symbols $1 + \frac{γ - 1}{2} M_{e}^{2} .$ Read it aloud as "one plus, gamma-minus-one over two, times Mach squared." It is the single recurring factor that shows up (to different powers) in temperature, pressure, and density. Give it a nickname: call it $Φ$ (the Greek letter "phi", just a name for this bundle).

Figure — Chamber-to-exit relation - all quantities as f(M_e, γ)

The picture above shows $Φ$ climbing as $M_{e}$ climbs — and because the exponents on $Φ$ are all negative, the ratios slide downhill. Keep this shape in your head; every problem is just reading a point off one of these curves.

Level 1 — Recognition

Can you pick the right formula and plug in? No traps yet, just fluency.

Problem L1.1. A gas has $γ = 1.4$ and flows at exit Mach number $M_{e} = 2.0$ . Compute the factor $Φ$ and the temperature ratio $T_{e} / T_{0}$ .

Recall Solution L1.1

$Φ = 1 + \frac{γ - 1}{2} M_{e}^{2} = 1 + \frac{0.4}{2} (2.0)^{2} = 1 + 0.2 \cdot 4 = 1 + 0.8 = 1.8.$

Temperature ratio is just $Φ^{- 1}$ : $\frac{T _{e}}{T _{0}} = \frac{1}{1.8} = 0.556.$ So the exit gas is a bit over half the chamber temperature (in kelvin). What we did: substituted into the definition of $Φ$ , then used the shortest relation. Why: temperature only needs $Φ^{- 1}$ , no exponent gymnastics.

Problem L1.2. For the same $γ = 1.4$ , $M_{e} = 2.0$ , find the pressure ratio $P_{e} / P_{0}$ .

Recall Solution L1.2

The exponent is $- \frac{γ}{γ - 1} = - \frac{1.4}{0.4} = - 3.5.$ $\frac{P _{e}}{P _{0}} = Φ^{- 3.5} = 1. 8^{- 3.5} = 0.1278.$ The exit static pressure is about 12.8% of chamber pressure. Why the exponent: the isentropic link $P / P_{0} = (T / T_{0})^{γ / (γ - 1)}$ turns the $Φ^{- 1}$ of temperature into $Φ^{- γ / (γ - 1)}$ for pressure.

Problem L1.3. Same conditions again. Find the density ratio $ρ_{e} / ρ_{0}$ , and verify it equals (pressure ratio) $\div$ (temperature ratio).

Recall Solution L1.3

Exponent: $- \frac{1}{γ - 1} = - \frac{1}{0.4} = - 2.5.$ $\frac{ρ _{e}}{ρ _{0}} = Φ^{- 2.5} = 1. 8^{- 2.5} = 0.230.$ Check via ideal gas law $ρ \propto P / T$ : $\frac{P _{e} / P _{0}}{T _{e} / T _{0}} = \frac{0.1278}{0.556} = 0.230. ✓$ Why the check works: density carries the exponent $- γ / (γ - 1) - (- 1) = - 1/ (γ - 1)$ , exactly the difference of the pressure and temperature exponents.

Level 2 — Application

Now attach real chamber numbers and units.

Problem L2.1. A rocket has $T_{0} = 3000 K$ , $P_{0} = 50 bar$ , $γ = 1.3$ , $M_{e} = 2.5$ , and specific gas constant $R = 400 J/(kg\cdotK)$ . Find $T_{e}$ , $P_{e}$ , and $ρ_{0}$ (chamber density).

Recall Solution L2.1

$Φ = 1 + \frac{0.3}{2} (2.5)^{2} = 1 + 0.15 \cdot 6.25 = 1 + 0.9375 = 1.9375.$

Temperature: $T_{e} = T_{0} Φ^{- 1} = \frac{3000}{1.9375} = 1548.4 K .$

Pressure: exponent $- γ / (γ - 1) = - 1.3/0.3 = - 4.3333$ . $P_{e} = P_{0} Φ^{- 4.3333} = 50 \cdot 1.937 5^{- 4.3333} = 50 \cdot 0.05612 = 2.806 bar .$

Chamber density from ideal gas law $ρ_{0} = P_{0} / (R T_{0})$ , with $R$ the specific gas constant. Convert $P_{0}$ : $50 bar = 50 \times 1 0^{5} Pa = 5 \times 1 0^{6} Pa$ . $ρ_{0} = \frac{5 \times 1 0 ^{6}}{400 \cdot 3000} = \frac{5 \times 1 0 ^{6}}{1.2 \times 1 0 ^{6}} = 4.167 kg/m^{3} .$ Why convert units first: the gas-law formula needs SI pascals, not bar; mixing them silently scales your density by $1 0^{5}$ .

Problem L2.2. For the rocket in L2.1, find the exit velocity $V_{e}$ .

Recall Solution L2.2

Use $V_{e} = M_{e} \frac{γ R T _{0}}{Φ}$ . Why this formula: $V_{e} = M_{e} a_{e}$ by the definition of Mach number, and $a_{e} = γ R T_{e}$ is the local speed of sound; substituting $T_{e} = T_{0} /Φ$ (our temperature relation) folds the exit temperature into chamber quantities, so we never need to compute $a_{e}$ separately. $γ R T_{0} = 1.3 \cdot 400 \cdot 3000 = 1.56 \times 1 0^{6} .$ $\frac{γ R T _{0}}{Φ} = \frac{1.56 \times 1 0 ^{6}}{1.9375} = 8.052 \times 1 0^{5} .$ $V_{e} = 2.5 \cdot 8.052 \times 1 0^{5} = 2.5 \cdot 897.3 = 2243 m/s .$ What the square root is: $γ R T_{e}$ is the local speed of sound $a_{e}$ ; multiplying by $M_{e}$ turns "how many sound-speeds fast" into an actual speed in m/s.

Problem L2.3. Same rocket. What fraction of the chamber's thermal energy per kilogram has been converted into kinetic energy at the exit? Take $c_{p} = γ R / (γ - 1)$ .

Recall Solution L2.3

Kinetic energy per kg at exit: $\frac{1}{2} V_{e}^{2} = \frac{1}{2} (2243)^{2} = 2.516 \times 1 0^{6} J/kg .$

Chamber thermal energy per kg (enthalpy) $h_{0} = c_{p} T_{0}$ : $c_{p} = \frac{1.3 \cdot 400}{0.3} = 1733.3 J/(kg\cdotK), h_{0} = 1733.3 \cdot 3000 = 5.200 \times 1 0^{6} J/kg .$ Fraction $= \frac{2.516 \times 1 0 ^{6}}{5.200 \times 1 0 ^{6}} = 0.484 \approx 48.4%.$ Sanity: this must equal $1 - T_{e} / T_{0} = 1 - 1/1.9375 = 0.484$ , because energy conservation says the enthalpy lost becomes the kinetic energy gained. It matches. ✓

Level 3 — Analysis

Compare, differentiate, reason about trends.

Problem L3.1. Two nozzles run the same $M_{e} = 3.0$ but different gases: nozzle A has $γ = 1.20$ , nozzle B has $γ = 1.25$ . Which has the hotter exit (larger $T_{e} / T_{0}$ ), and by what factor do their temperature ratios differ?

Recall Solution L3.1

$Φ_{A} = 1 + \frac{0.20}{2} (9) = 1 + 0.9 = 1.9, T_{e} / T_{0}_{A} = 1/1.9 = 0.5263.$ $Φ_{B} = 1 + \frac{0.25}{2} (9) = 1 + 1.125 = 2.125, T_{e} / T_{0}_{B} = 1/2.125 = 0.4706.$

Nozzle A (lower $γ$ ) is hotter at exit. Ratio of ratios: $\frac{0.5263}{0.4706} = 1.118.$ A's exit temperature ratio is about 12% higher. Why lower $γ$ stays hotter: a smaller $γ - 1$ shrinks the kinetic term $\frac{γ - 1}{2} M_{e}^{2}$ , so less thermal energy is drained into motion — consistent with the parent's remark that H₂-rich (low $γ$ ) exhausts are favoured.

Problem L3.2. For a fixed $γ = 1.4$ , how sensitive is $P_{e} / P_{0}$ to $M_{e}$ near $M_{e} = 2$ ? Estimate the fractional change in $P_{e} / P_{0}$ for a $1%$ increase in $M_{e}$ , using the derivative.

Recall Solution L3.2

Write $f (M) = Φ^{- n}$ with $n = γ / (γ - 1) = 3.5$ and $Φ = 1 + 0.2 M^{2}$ .

Why a derivative here: we want "how much does output wiggle when input wiggles a little" — that is exactly the job of the derivative $df / d M$ , the local slope.

$\frac{df}{d M} = - n Φ^{- n - 1} \cdot \frac{d Φ}{d M} = - n Φ^{- n - 1} \cdot (0.4 M) .$ Fractional sensitivity $\frac{d ln f}{d ln M} = \frac{M}{f} \frac{df}{d M} = - n \cdot \frac{0.4 M ^{2}}{Φ} .$ At $M = 2$ : $Φ = 1.8$ , so $\frac{d l n f}{d l n M} = - 3.5 \cdot \frac{0.4 \cdot 4}{1.8} = - 3.5 \cdot \frac{1.6}{1.8} = - 3.111.$ So a $+ 1%$ change in $M_{e}$ gives about $- 3.11%$ change in $P_{e} / P_{0}$ . Reading it: pressure ratio is highly leveraged on Mach number — small Mach gains buy large pressure drops (which is what makes thrust).

Problem L3.3. Show that $ρ_{e} / ρ_{0}$ can be written as $(T_{e} / T_{0})^{1/ (γ - 1)}$ , and check numerically for $γ = 1.4$ , $M_{e} = 2$ .

Recall Solution L3.3

Density ratio $= Φ^{- 1/ (γ - 1)}$ and temperature ratio $= Φ^{- 1}$ . So $(\frac{T _{e}}{T _{0}})^{1/ (γ - 1)} = (Φ^{- 1})^{1/ (γ - 1)} = Φ^{- 1/ (γ - 1)} = \frac{ρ _{e}}{ρ _{0}} . ✓$ Numbers ( $γ = 1.4$ , so $1/ (γ - 1) = 2.5$ ): $T_{e} / T_{0} = 1/1.8 = 0.5556$ ; $0.555 6^{2.5} = 0.2299$ , matching L1.3's $0.230$ . Why this is the isentropic $ρ \propto T^{1/ (γ - 1)}$ : it is the density-temperature form of "adiabatic, reversible" — the same physics, wearing different variables.

Level 4 — Synthesis

Combine multiple relations, including geometry.

First we need the symbol $A^{*}$ (read "A-star") and the area–Mach relation itself.

Formula Area–Mach relation (and where it comes from)

$\frac{A _{e}}{A ^{*}} = \frac{1}{M _{e}} [\frac{2Φ}{γ + 1}]^{\frac{γ + 1}{2 ( γ - 1 )}}$ Why this form. Mass conservation says $\overset{m}{˙} = ρ A V$ is the same at every station, so $\frac{A _{e}}{A ^{*}} = \frac{ρ ^{*} V ^{*}}{ρ _{e} V _{e}}$ — a smaller $ρ V$ needs a bigger area. Now write $ρ V = ρ M a = ρM γ R T$ . Feeding in our ratios $ρ / ρ_{0} = Φ^{- 1/ (γ - 1)}$ and $T / T_{0} = Φ^{- 1}$ (so $T \propto Φ^{- 1/2}$ ), the density and speed-of-sound powers of $Φ$ combine into the single bracket exponent $\frac{γ + 1}{2 ( γ - 1 )}$ ; the throat is evaluated at $M = 1$ (where its own $Φ$ equals $\frac{γ + 1}{2}$ ), and that is what produces the constant $\frac{2}{γ + 1}$ inside the bracket. The lone $1/ M_{e}$ out front is the $V^{*} / V_{e}$ Mach-ratio piece.

Two branches — pick the right one. For a given $A_{e} / A^{*} > 1$ there are two Mach numbers that satisfy this equation: one subsonic ( $M_{e} < 1$ ) and one supersonic ( $M_{e} > 1$ ). Geometry alone cannot decide; the flow "chooses" based on the pressure it is expanding into. For a rocket we want a supersonic jet, so throughout L4–L5 we always take the == $M_{e} > 1$ branch==. (The subsonic branch describes the same bell run as an ordinary diffuser instead of an accelerating nozzle.)

Problem L4.1. A nozzle must produce $M_{e} = 3.0$ with $γ = 1.4$ . Find the required area ratio $A_{e} / A^{*}$ .

Recall Solution L4.1

$\frac{A _{e}}{A ^{*}} = \frac{1}{M _{e}} [\frac{2Φ}{γ + 1}]^{\frac{γ + 1}{2 ( γ - 1 )}} .$ $Φ = 1 + 0.2 (9) = 2.8.$ Exponent $\frac{γ + 1}{2 ( γ - 1 )} = \frac{2.4}{0.8} = 3.0.$ $\frac{2Φ}{γ + 1} = \frac{2 \cdot 2.8}{2.4} = \frac{5.6}{2.4} = 2.3333.$ $\frac{A _{e}}{A ^{*}} = \frac{1}{3.0} \cdot 2.333 3^{3} = \frac{1}{3.0} \cdot 12.70 = 4.235.$ The diverging cone must open to about 4.2× the throat area (taking the supersonic branch, since we want $M_{e} = 3 > 1$ ). What the figure shows: below, the throat (where $M = 1$ ) is the pinch; area grows to reach $M_{e} = 3$ .

Problem L4.2. Take a full engine: $T_{0} = 3400 K$ , $P_{0} = 60 bar$ , $γ = 1.25$ , specific gas constant $R = 480 J/(kg\cdotK)$ , $M_{e} = 3.2$ . Find $V_{e}$ , and using $A_{e} / A^{*}$ report all in one pass.

Recall Solution L4.2

$Φ = 1 + \frac{0.25}{2} (3.2)^{2} = 1 + 0.125 \cdot 10.24 = 1 + 1.28 = 2.28.$

Exit velocity from $V_{e} = M_{e} \frac{γ R T _{0}}{Φ}$ . Why this formula again: it is just $V_{e} = M_{e} a_{e}$ with $a_{e} = γ R T_{e}$ and $T_{e} = T_{0} /Φ$ substituted, so a single expression in chamber quantities does the whole job. $γ R T_{0} = 1.25 \cdot 480 \cdot 3400 = 2.04 \times 1 0^{6} .$ $V_{e} = 3.2 \frac{2.04 \times 1 0 ^{6}}{2.28} = 3.2 8.947 \times 1 0^{5} = 3.2 \cdot 945.9 = 3027 m/s .$

Area ratio: exponent $\frac{γ + 1}{2 ( γ - 1 )} = \frac{2.25}{0.5} = 4.5.$ $\frac{2Φ}{γ + 1} = \frac{2 \cdot 2.28}{2.25} = \frac{4.56}{2.25} = 2.0267.$ $\frac{A _{e}}{A ^{*}} = \frac{1}{3.2} \cdot 2.026 7^{4.5} = \frac{1}{3.2} \cdot 25.10 = 7.844.$ One-pass summary: $V_{e} \approx 3027 m/s$ , area ratio $\approx 7.84$ (supersonic branch). Higher $M_{e}$ needed a much wider exit than L4.1's $M_{e} = 3$ case — the exponent amplifies it.

Problem L4.3. For L4.2, if the throat area is $A^{*} = 0.010 m^{2}$ , find the exit area $A_{e}$ and the exit-to-throat diameter ratio (assume circular).

Recall Solution L4.3

$A_{e} = 7.844 \cdot 0.010 = 0.07844 m^{2} .$

Diameter ratio: since $A \propto d^{2}$ , $\frac{d _{e}}{d ^{*}} = A_{e} / A^{*} = 7.844 = 2.801.$ So the exit diameter is about 2.8× the throat diameter. Why the square root: area scales with diameter squared, so a 7.8× area increase is only a 2.8× diameter increase — nozzles look less dramatic than the area ratio suggests.

Level 5 — Mastery

Design and inverse problems — go from a target to the required inputs.

Problem L5.1. Inverse pressure. A designer needs $P_{e} / P_{0} = 0.010$ with $γ = 1.3$ . What exit Mach number achieves this?

Recall Solution L5.1

We invert $P_{e} / P_{0} = Φ^{- γ / (γ - 1)}$ . Exponent $n = 1.3/0.3 = 4.3333$ . $Φ = (\frac{P _{e}}{P _{0}})^{- 1/ n} = 0.01 0^{- 1/4.3333} = 0.01 0^{- 0.23077} = 2.899.$ Then $Φ = 1 + \frac{0.3}{2} M_{e}^{2} = 1 + 0.15 M_{e}^{2}$ , so $M_{e}^{2} = \frac{Φ - 1}{0.15} = \frac{1.899}{0.15} = 12.66, M_{e} = 3.559.$ What we did: ran every formula backwards — take the log-style root to recover $Φ$ , then solve the quadratic-in- $M_{e}$ piece. Why this is the design workflow: in practice you specify a target expansion (pressure ratio set by altitude) and back out the Mach number, then the area ratio.

Problem L5.2. Altitude matching. Continuing L5.1 ( $γ = 1.3$ , $M_{e} = 3.559$ ), find the area ratio $A_{e} / A^{*}$ this design demands.

Recall Solution L5.2

$Φ = 2.899$ (from L5.1). Exponent $\frac{γ + 1}{2 ( γ - 1 )} = \frac{2.3}{0.6} = 3.8333.$ $\frac{2Φ}{γ + 1} = \frac{2 \cdot 2.899}{2.3} = \frac{5.798}{2.3} = 2.5209.$ $\frac{A _{e}}{A ^{*}} = \frac{1}{3.559} \cdot 2.520 9^{3.8333} = \frac{1}{3.559} \cdot 33.86 = 9.514.$ The nozzle needs an expansion ratio of about 9.5 (supersonic branch) — a large, high-altitude bell. Link: deeper vacuum (lower ambient $P_{e}$ target) always pushes $M_{e}$ up and $A_{e} / A^{*}$ up.

Problem L5.3. Consistency proof. Show that at the throat ( $M = 1$ ), the area-Mach formula gives $A / A^{*} = 1$ for any $γ$ , and verify numerically at $γ = 1.4$ and $γ = 1.2$ .

Recall Solution L5.3

Set $M_{e} = 1$ : then $Φ = 1 + \frac{γ - 1}{2} = \frac{γ + 1}{2} .$ So $\frac{2Φ}{γ + 1} = \frac{2}{γ + 1} \cdot \frac{γ + 1}{2} = 1.$ Anything to any power is still $1$ , and $1/ M_{e} = 1/1 = 1$ , giving $A / A^{*} = 1$ . Why it must: the throat is the $M = 1$ station by definition, so its area equals $A^{*}$ trivially. Numerically: $γ = 1.4 \Rightarrow \frac{2 \cdot 1.2}{2.4} = 1$ , exponent $3$ , result $1$ ; $γ = 1.2 \Rightarrow \frac{2 \cdot 1.1}{2.2} = 1$ , result $1$ . ✓ This self-consistency is a good exam sanity check — and it is the single point where the subsonic and supersonic branches meet.

Recall One-line self-test

All exit ratios are powers of a single factor $Φ$ — name it and its three exponents. ::: $Φ = 1 + \frac{γ - 1}{2} M_{e}^{2}$ ; temperature $Φ^{- 1}$ , pressure $Φ^{- γ / (γ - 1)}$ , density $Φ^{- 1/ (γ - 1)}$ . Why must $Φ \geq 1$ ? ::: Because $M_{e}^{2} \geq 0$ and $γ > 1$ , so $\frac{γ - 1}{2} M_{e}^{2} \geq 0$ and $Φ = 1 + (that) \geq 1$ . For a given $A_{e} / A^{*} > 1$ , how many Mach numbers solve the area relation, and which do we pick? ::: Two — one subsonic, one supersonic; for a rocket jet we pick the supersonic ( $M_{e} > 1$ ) branch.