Chapter 19: Nozzle Theory and Thermodynamics

"The production of motive power is due... not to an actual consumption of caloric, but to its transportation from a warm body to a cold body." — Sadi Carnot, Reflections on the Motive Power of Fire (1824)

Overview

A rocket engine is a machine for turning heat into speed. In Chapter 18 we filled a combustion chamber with the hottest, highest-pressure gas that chemistry can make — three or four thousand kelvin, a couple of hundred atmospheres — and then, having made it, we did nothing with it. A pressure vessel full of hot gas is a bomb, not an engine. What makes it a rocket is the shape bolted to the end of the chamber: the flared bell you picture when you picture a rocket engine at all. That bell is a nozzle, and this chapter is about the beautiful, slightly counterintuitive physics of how it works.

The nozzle is where the number that governs everything — the exhaust velocity $v_e$ that sits in the exponent of the rocket equation — is actually made. Recall the tyranny from Chapter 3: $\Delta v = v_e \ln(m_0/m_f)$, and because $v_e$ is in the exponent of the mass ratio, every meter per second of exhaust velocity is worth fighting for. There are only two levers on that tyranny — a lighter, more propellant-heavy rocket, or a faster exhaust — and the nozzle is the second lever made real. A combustion chamber produces energetic but disordered gas: molecules flying every which way, banging on the walls, going nowhere on average. The nozzle's job is to take that random thermal motion and comb it into a single direction, converting the chaos of heat into the order of a hypersonic jet pointed straight down. Do it well and you approach the theoretical exhaust velocity the propellant allows; do it badly and you throw away performance you spent an entire chapter of chemistry to obtain.

Here is the surprise that makes nozzles worth a chapter. To accelerate a subsonic gas you squeeze it — narrow the pipe and it speeds up, exactly as your thumb over a garden hose does. But past the speed of sound the rule reverses: to accelerate a supersonic gas you must let it expand — widen the pipe. A rocket nozzle has to do both, which is why it has a waist: it converges to a throat to reach the speed of sound, then diverges to go supersonic. That double geometry, and the thermodynamics behind it, is the whole story.

In this chapter, you will learn to:

• Read a rocket nozzle as a heat engine, and say what each part of its shape is for.
• Derive the area–velocity relation and explain the sign flip at Mach 1 that forces the converging–diverging shape.
• Use isentropic flow and the choked condition to relate chamber conditions to what comes out the exit.
• Compute an ideal exhaust velocity from $T_c$, molar mass $\mathcal{M}$, $\gamma$, and the pressure ratio — and see why $v_e \propto \sqrt{T_c/\mathcal{M}}$.
• Choose an expansion ratio, diagnose over- and under-expansion, and explain the sea-level-versus-vacuum nozzle trade.
• Answer, with numbers, the question Chapter 18 set up: why does hydrogen beat kerosene?

Learning Paths

🚀 Space Enthusiast: Read 19.1 for the physical picture and 19.6 for the payoff (why hydrogen wins). The single idea to carry away is in 19.2: a supersonic gas speeds up when the pipe widens. Skim the algebra of 19.3–19.4; linger on the diagrams.

📐 Engineering Student: This is a core chapter. Do the area–velocity derivation in 19.2 yourself, and work the exhaust-velocity and expansion-ratio examples in 19.4–19.5 by hand — they are the compressible-flow foundation for every engine you will ever analyze. The ⭐⭐/⭐⭐⭐ exercises are essential.

🎮 KSP Player: Section 19.5 is why the game gives you "sea-level" and "vacuum" engine variants with different bell sizes, and why a vacuum engine looks silly (and performs badly) on the launchpad. Sections 19.4 and 19.6 explain the Isp numbers in every engine's stats panel.

🛰️ Industry Prep: The exhaust-velocity formula in 19.4 and the expansion-ratio relation in 19.5 are the two equations that turn a combustion analysis into an engine performance number. The Mission Design Checkpoint adds exit_velocity and expansion_ratio to astrotools and lets you estimate your own engine's Isp.

19.1 The de Laval nozzle

Start with the physical situation. Upstream of the nozzle is the combustion chamber, which Chapter 17 described and Chapter 18 filled with fire: a gas at chamber temperature $T_c$ (say $3600\ \text{K}$) and chamber pressure $p_c$ (say $200\ \text{bar}$), nearly at rest because the chamber is wide and the gas has nowhere to go but out the back. That "nearly at rest, hot, and high-pressure" condition is the reservoir the nozzle draws on. Thermodynamically the chamber gas is almost pure stagnation condition: its energy is stored as heat and pressure, with essentially none of it yet in bulk motion.

A nozzle is a shaped duct that trades that stored thermal energy for directed kinetic energy. The trade is governed by one conservation law you already met in a different guise — the conservation of energy, written for a flowing gas. As a parcel of gas moves down the nozzle and speeds up, it must cool and its pressure must drop; the kinetic energy it gains is exactly the thermal energy (enthalpy) it loses. The bell is an energy-conversion device, and Carnot's line in the epigraph is literally true of it: motive power comes from carrying heat from a hot body (the chamber) to a cold body (the near-vacuum outside), and the nozzle is the machine that does the carrying.

Definition (de Laval nozzle). A de Laval nozzle is a duct that first converges (narrows) to a minimum-area throat and then diverges (widens) toward the exit, used to accelerate a compressible gas from subsonic speed in the chamber, through the speed of sound at the throat, to supersonic speed at the exit. It is named for Gustaf de Laval, who applied the shape to steam turbines in the 1880s.

Definition (converging–diverging nozzle). A converging–diverging nozzle is the same object described by its geometry: a converging section, a throat of minimum cross-sectional area $A_t$, and a diverging section that opens to the exit area $A_e$. "De Laval nozzle" and "converging–diverging nozzle" name the same thing; we use them interchangeably.

Why the waist? Because — as the next section proves — a gas moving slower than sound and a gas moving faster than sound respond to a change in the pipe's width in opposite ways. Below the speed of sound, narrowing the duct speeds the gas up. Above it, widening the duct speeds the gas up. To carry a gas all the way from a standstill in the chamber to several times the speed of sound at the exit, you must therefore do both, in that order, with the switchover — the sonic point, Mach 1 exactly — pinned at the narrowest place: the throat. A pipe that only converges (like a garden-hose nozzle) can never push a gas past Mach 1. The diverging bell is what makes a rocket nozzle supersonic, and it is the single feature that distinguishes it from the nozzle on your sink.

💡 Intuition: the crowd through a doorway. Picture a dense, jostling crowd (the hot chamber gas) shuffling toward an exit. In the crowded room people move slowly but there are many of them; to keep the flow rate of people constant as the hallway narrows, each person must walk faster. That is subsonic flow: squeeze the space and the flow speeds up. Now imagine the crowd has burst through the door and is sprinting across an open plaza, thinning out as it goes. Here the crowd keeps accelerating precisely because it has room to spread — the density drops so fast that speeding up is the only way to carry the same flow through an ever-larger area. That switch, from "speed up because it's tight" to "speed up because it's open," happens right at the doorway. The doorway is the throat.

📜 From History: a steam-turbine part, borrowed. The converging–diverging nozzle was not invented for rockets. Gustaf de Laval, a Swedish engineer, developed it around 1888 to drive an impulse steam turbine: he needed to turn high-pressure steam into a fast, narrow jet to spin a wheel, and discovered that a simple converging nozzle stalled at the speed of sound. Adding a diverging section let the steam go supersonic and delivered far more of its energy to the blades. Rocketry inherited the shape wholesale. Robert Goddard, Hermann Oberth, and the German and Soviet pioneers all bolted de Laval nozzles to their combustion chambers because the physics is identical: a hot, high-pressure reservoir, and the need to extract its energy as a supersonic jet. When you look at a Raptor or an RS-25, you are looking at a nineteenth-century turbine part, scaled up and asked to survive fire.

19.2 The area–velocity relation for compressible flow

We want to prove the claim that organizes the whole chapter: subsonic flow accelerates in a converging duct, supersonic flow accelerates in a diverging duct, and the two meet at Mach 1. The tool is a short piece of one-dimensional gas dynamics — three physical statements combined into one relation.

Strategy first. We treat the flow as steady and one-dimensional: at each station along the nozzle the gas has a single area $A$, density $\rho$, velocity $V$, and pressure $p$. We write down (1) conservation of mass, (2) conservation of momentum for a frictionless flow, and (3) the definition of the speed of sound. We then eliminate pressure and density and are left with a single equation connecting a fractional change in area to a fractional change in velocity. The Mach number appears as the coefficient — and its sign is the whole point.

First we need the Mach number, the ratio that decides everything.

Definition (Mach number). The Mach number $M = V/a$ is the ratio of the local flow speed $V$ to the local speed of sound $a$ in the same gas. $M < 1$ is subsonic, $M = 1$ is sonic, $M > 1$ is supersonic. The speed of sound in an ideal gas is $a = \sqrt{\gamma R T}$, where $\gamma$ is the ratio of specific heats and $R$ the specific gas constant — so the "sound barrier" is not a fixed speed but a local property of the hot, changing gas.

Now the three statements. Conservation of mass (continuity): the mass flow rate $\dot m = \rho A V$ is constant along the nozzle. Taking the logarithm and differentiating,

$$ \frac{d\rho}{\rho} + \frac{dA}{A} + \frac{dV}{V} = 0. $$

Conservation of momentum for a steady, frictionless (inviscid) flow with no body forces is Euler's equation, which reduces along a streamline to

$$ dp = -\rho\, V\, dV. $$

Read it physically: to speed the gas up ($dV > 0$) the pressure must fall ($dp < 0$). The gas is pushed forward by a pressure that is higher behind it than ahead of it. The speed of sound enters because in the smooth, adiabatic flow inside a nozzle, pressure and density changes are linked by $dp = a^2\, d\rho$ — that relation is, in fact, the definition of the sound speed. Combine it with Euler's equation to eliminate $dp$:

$$ a^2\, d\rho = -\rho\, V\, dV \quad\Longrightarrow\quad \frac{d\rho}{\rho} = -\frac{V\,dV}{a^2} = -M^2\,\frac{dV}{V}. $$

That last step used $M^2 = V^2/a^2$. Now substitute this expression for $d\rho/\rho$ into the continuity equation:

$$ -M^2\,\frac{dV}{V} + \frac{dA}{A} + \frac{dV}{V} = 0, $$

and collect the velocity terms:

$$ \boxed{\ \frac{dA}{A} = \left(M^2 - 1\right)\frac{dV}{V}\ } $$

This is the area–velocity relation, and it is one of the most quietly powerful equations in engineering. Everything about the shape of a rocket nozzle is contained in the sign of the factor $(M^2 - 1)$. Let us read it in three cases.

• Subsonic ($M < 1$): the factor $(M^2 - 1)$ is negative, so $dA$ and $dV$ have opposite signs. To speed the gas up ($dV > 0$) you must shrink the area ($dA < 0$). A converging duct accelerates subsonic flow. This is the garden hose, and it is the converging half of the nozzle.
• Supersonic ($M > 1$): the factor is positive, so $dA$ and $dV$ have the same sign. To speed the gas up you must grow the area ($dA > 0$). A diverging duct accelerates supersonic flow. This is the flaring bell, and it is where the exhaust does most of its accelerating.
• Sonic ($M = 1$): the factor is zero, so $dA = 0$. The area is momentarily neither growing nor shrinking — it is at an extremum. In a nozzle designed to accelerate the flow, that extremum is a minimum: the throat. Mach 1 can occur only where the area is stationary, i.e. at the throat.

🚪 Threshold Concept: widening a pipe can speed the gas up. Every intuition you own about fluids comes from subsonic experience, where "narrow = fast" is a law of life — thumbs on hoses, wind between buildings, water through rapids. The area–velocity relation says that above Mach 1 this law inverts: a supersonic gas accelerates when you give it more room, and decelerates when you crowd it. The reason is that above the speed of sound the density falls faster than the velocity rises, so continuity ($\rho A V = \text{const}$) can only be satisfied by an increasing area. Once you accept that a diverging duct is an accelerator for supersonic flow, the entire shape of every rocket nozzle — and every supersonic wind tunnel, and the exhaust of every jet fighter in afterburner — stops being arbitrary and becomes inevitable. This is the idea that turns "the bell shape" from decoration into physics.

⚠️ Common Misconception: "the nozzle accelerates exhaust by friction / by the walls pushing it." The walls do essentially no pushing along the flow direction, and friction can only slow the gas. The acceleration comes entirely from the pressure gradient: the gas at every point is squeezed from behind by higher pressure and pulled ahead into lower pressure, and it converts its own internal thermal energy into motion as it expands. The nozzle wall's only job is to impose the right area at each station so that the pressure falls in a controlled way. A nozzle does not whip the gas faster; it gets out of the way in exactly the right shape so the gas accelerates itself by expanding.

🔄 Check Your Understanding 1. In a converging duct, what happens to a flow that is already supersonic ($M > 1$)? Does it speed up or slow down? 2. Why is it impossible to reach exactly $M = 1$ anywhere except at a throat (a point of minimum area)?

Answers

1. It slows down. For $M > 1$, $(M^2 - 1) > 0$, so a shrinking area ($dA < 0$) forces $dV < 0$. Crowding a supersonic flow decelerates it (this is exactly what a supersonic inlet or a "converging" diffuser does). 2. The relation $dA/A = (M^2-1)\,dV/V$ requires that at $M = 1$ the left side be zero unless $dV/V$ is infinite; a smooth flow has finite $dV/V$, so $dA = 0$. A stationary area occurs only at a local minimum or maximum — and for an accelerating nozzle flow that place is the minimum, the throat.

19.3 Isentropic flow and choked flow

The area–velocity relation tells us the shape we need; to get numbers — how fast, how hot, how much mass per second — we need the thermodynamics of the expansion itself. To an excellent approximation the flow through a rocket nozzle is isentropic, and that single assumption unlocks the rest of the chapter.

Definition (isentropic flow). Isentropic flow is flow that is both adiabatic (no heat crosses the gas boundary — the flow is too fast for meaningful heat loss through the walls in the time a parcel spends in the nozzle) and reversible (frictionless, no shocks). Entropy is then constant along the flow. For a calorically perfect gas with specific-heat ratio $\gamma$, isentropic flow ties pressure, temperature, and density together through $p/\rho^\gamma = \text{const}$ and $T\,p^{(1-\gamma)/\gamma} = \text{const}$.

Real nozzle flow is not perfectly isentropic — there is friction in the thin boundary layer on the wall, some heat is lost to the actively cooled walls, and the chemistry is not perfectly frozen — but the departures are small, a few percent, and engineers fold them into efficiency factors. As an idealization, isentropic flow is to nozzle analysis what the frictionless plane is to mechanics: not exactly true, but the right place to start, and close enough that the answers are trustworthy.

Because the chamber gas is nearly at rest, chamber conditions are the stagnation (or total) conditions — the temperature and pressure the gas would have if brought to rest, which is what it essentially already is. Write them $T_c$ and $p_c$. As the gas accelerates to a local Mach number $M$, the isentropic relations give its local static temperature, pressure, and density:

$$ \frac{T_c}{T} = 1 + \frac{\gamma - 1}{2}M^2, \qquad \frac{p_c}{p} = \left(1 + \frac{\gamma - 1}{2}M^2\right)^{\frac{\gamma}{\gamma - 1}}, \qquad \frac{\rho_c}{\rho} = \left(1 + \frac{\gamma - 1}{2}M^2\right)^{\frac{1}{\gamma - 1}}. $$

Each says the same physical thing: the faster the gas goes (higher $M$), the more it has cooled and depressurized, because the energy went into motion. These three relations, plus the area–Mach relation we build in §19.5, are the entire toolkit of ideal nozzle flow.

Choked flow. Now put the area–velocity result and the isentropic relations together at the throat, where $M = 1$. Setting $M = 1$ in the relations above gives the conditions at the throat — the critical or sonic conditions, marked with a star:

$$ \frac{T^*}{T_c} = \frac{2}{\gamma + 1}, \qquad \frac{p^*}{p_c} = \left(\frac{2}{\gamma + 1}\right)^{\frac{\gamma}{\gamma - 1}}. $$

For a typical rocket exhaust with $\gamma = 1.20$, the throat pressure is $p^*/p_c = (2/2.2)^{6} \approx 0.56$ and the throat temperature is $T^*/T_c = 2/2.2 \approx 0.91$. So by the time the gas reaches the throat it has already dropped to about $56\%$ of chamber pressure and cooled by roughly $9\%$ — and it is now moving at exactly the local speed of sound.

Definition (choked flow). A converging–diverging nozzle is choked when the flow reaches $M = 1$ at the throat. Once choked, the mass flow rate is fixed entirely by the chamber conditions and the throat area — lowering the downstream (exit or ambient) pressure further cannot increase it. The throat has reached the maximum mass flux the gas can carry, and information about downstream pressure (which travels at the speed of sound) can no longer propagate upstream past the sonic throat to tell the chamber to send more.

Choking is why a rocket engine's mass flow is so stable and predictable. The choked mass flow rate through a throat of area $A_t$ is

$$ \dot m = A_t\, p_c \sqrt{\frac{\gamma}{R\,T_c}}\left(\frac{2}{\gamma + 1}\right)^{\frac{\gamma + 1}{2(\gamma - 1)}}, $$

where $R = R_u/\mathcal{M}$ is the specific gas constant ($R_u = 8.314\ \text{J/(mol·K)}$ is the universal gas constant and $\mathcal{M}$ the exhaust molar mass). Notice what governs the flow: chamber pressure $p_c$ and throat area $A_t$ (both directly), and $\sqrt{1/T_c}$. Nothing about the exit or the outside world appears — that is the signature of choking. To throttle a rocket you change $p_c$ (by changing how hard the pumps feed the chamber); the throat area is usually fixed.

Worked Example: the choked mass flow of a hydrogen upper-stage engine. Take an idealized LOX/LH2 engine (Tier 3 clean values, representative of an RS-25-class chamber): throat area $A_t = 0.055\ \text{m}^2$, chamber pressure $p_c = 200\ \text{bar} = 2.0\times10^{7}\ \text{Pa}$, chamber temperature $T_c = 3600\ \text{K}$, exhaust molar mass $\mathcal{M} = 0.0135\ \text{kg/mol}$, and $\gamma = 1.20$.

First the specific gas constant: $R = R_u/\mathcal{M} = 8.314 / 0.0135 = 616\ \text{J/(kg·K)}$.

The bracket term: $\dfrac{\gamma+1}{2(\gamma-1)} = \dfrac{2.2}{0.4} = 5.5$, and $\left(\dfrac{2}{2.2}\right)^{5.5} = (0.9091)^{5.5} \approx 0.592$.

The square root: $\sqrt{\gamma/(R\,T_c)} = \sqrt{1.20 / (616 \times 3600)} = \sqrt{1.20/2.218\times10^{6}} = \sqrt{5.41\times10^{-7}} = 7.36\times10^{-4}\ \text{s/m}$.

Assemble: $\dot m = (0.055)(2.0\times10^{7})(7.36\times10^{-4})(0.592) \approx 479\ \text{kg/s}$.

Sanity check. The real RS-25 swallows roughly $510\text{–}515\ \text{kg/s}$ of propellant at full power. Our $479\ \text{kg/s}$, for a deliberately rounded throat area, lands in the right neighborhood — the formula and the physical picture of a throat choked at Mach 1 are sound. Be careful, though: this is an idealized model, and for a real engine's true (larger) throat it tends to overestimate the flow slightly, because a real flame is a touch cooler and its exhaust is not a perfect gas of fixed $\gamma$. We meet exactly that overshoot in Case Study 1, where the RS-25's own geometry gives an ideal $\dot m$ about $14\%$ above the real value — which is why that reconstruction uses the measured mass flow as an input. Treat this $479$ as order-of-magnitude-plus, not a spec.

🐛 Find the Error. An engineer wants more thrust from a choked engine, so she reasons: "The exhaust leaves into near-vacuum in space, where the back pressure is essentially zero. If I fly the same engine even higher, the back pressure drops further, so more gas will flow through the throat and I'll get more thrust." Where is the reasoning wrong?

Answer

Once the nozzle is choked — which it is the instant $p_c$ exceeds roughly twice the ambient pressure, i.e. essentially always for a rocket — the mass flow depends only on $p_c$, $A_t$, and $T_c$, not on the downstream pressure. Dropping the back pressure from "a little" to "zero" does not push one extra gram per second through the sonic throat, because pressure information cannot travel upstream against a sonic flow. Higher altitude does increase thrust, but through a different channel: the pressure-thrust term $(p_e - p_a)A_e$ in the thrust equation of Chapter 16 grows as $p_a \to 0$, and a bigger nozzle can be used. The mass flow itself is locked by the throat.

19.4 Chamber conditions to exhaust velocity

Now the centerpiece: given a chamber full of hot gas, how fast does it come out? This is where an entire chapter of combustion chemistry (Chapter 18) cashes out into the one number the rocket equation wants.

Strategy first. Energy is conserved along the flow. In the chamber the gas has lots of thermal energy (enthalpy) and almost no kinetic energy; at the exit it has less thermal energy and lots of kinetic energy. Set the total (stagnation) enthalpy in the chamber equal to enthalpy-plus-kinetic-energy at the exit, and solve for the exit velocity. The isentropic relation for temperature turns the unknown exit temperature into a known function of the pressure ratio. Out falls the ideal exhaust velocity.

For a steady adiabatic flow the stagnation enthalpy is constant: $h_c = h_e + \tfrac{1}{2}v_e^2$, where $h_c$ is the (essentially stagnation) enthalpy in the chamber and $h_e, v_e$ are the enthalpy and velocity at the exit. Solving,

$$ v_e = \sqrt{2\left(h_c - h_e\right)} = \sqrt{2\,c_p\left(T_c - T_e\right)}, $$

using $h = c_p T$ for a calorically perfect gas. The specific heat at constant pressure is $c_p = \dfrac{\gamma R}{\gamma - 1}$, and the isentropic relation gives the exit temperature in terms of the pressure ratio, $T_e/T_c = (p_e/p_c)^{(\gamma-1)/\gamma}$. Substituting both, and writing $R = R_u/\mathcal{M}$:

$$ \boxed{\ v_e = \sqrt{\dfrac{2\gamma}{\gamma - 1}\cdot\dfrac{R_u\, T_c}{\mathcal{M}}\left[\,1 - \left(\dfrac{p_e}{p_c}\right)^{\frac{\gamma - 1}{\gamma}}\,\right]}\ } $$

This is the ideal rocket exhaust-velocity equation, and it deserves to be read slowly, because every term is a lever an engineer pulls.

• $T_c$ (chamber temperature) — under the square root, so $v_e \propto \sqrt{T_c}$. Burn hotter, go faster. This is why we chased flame temperature in Chapter 18. But the gain is only as the square root: doubling $T_c$ buys just $41\%$ more velocity, and materials melt long before you can double it.
• $\mathcal{M}$ (exhaust molar mass) — in the denominator under the root, so $v_e \propto 1/\sqrt{\mathcal{M}}$. Light exhaust molecules leave faster. This is the single most important lever in the equation, and the reason hydrogen is king; §19.6 is devoted to it.
• $\gamma$ (ratio of specific heats) — a weaker influence through the $\tfrac{2\gamma}{\gamma-1}$ factor; lower $\gamma$ (more complex molecules, more internal modes) actually raises this factor slightly.
• The pressure-ratio bracket $1 - (p_e/p_c)^{(\gamma-1)/\gamma}$ — this runs from $0$ (no expansion) toward $1$ (expansion into perfect vacuum). It says: the more you let the gas expand — the higher the chamber pressure and the lower the exit pressure — the more thermal energy you convert to speed. A big nozzle expanding to low $p_e$ wrings out the last of the energy.

🔗 Connection: this $v_e$ and the $v_e$ of Chapter 3. The velocity computed here is the physical speed of the gas leaving the nozzle. The effective exhaust velocity $c$ that appears in the rocket equation and in the Chapter 16 thrust equation, $F = \dot m\, v_e + (p_e - p_a)A_e$, is slightly larger because it also collects the pressure thrust: $c = v_e + (p_e - p_a)A_e/\dot m$. When the nozzle is perfectly expanded ($p_e = p_a$) the pressure term vanishes and $c = v_e$ exactly. For most of this book's arithmetic the two are within a couple of percent and we use them interchangeably, exactly as Chapter 3 did — but now you know the difference, and §19.5 is about the pressure term that separates them.

💡 Intuition: why lighter is faster, from kinetic theory. In Chapter 18 we met the fact that temperature is the average kinetic energy of the molecules: $\tfrac{1}{2}m\langle v^2\rangle = \tfrac{3}{2}k T$. At a given temperature, every gas has the same average molecular kinetic energy — but a lighter molecule (small $m$) must therefore be moving faster to carry that energy. The nozzle's job is to take that random thermal speed and align it into a directed jet, so a propellant whose molecules are already zipping around faster (because they are light) hands the nozzle a faster starting point and yields a faster exhaust. Hydrogen-rich exhaust is light; kerosene exhaust, full of heavy carbon dioxide and carbon monoxide, is not.

Let us put real numbers through the formula, isolating the molecular-weight effect first because it is the whole game.

Worked Example: the same chamber, two exhaust molar masses. Hold everything constant and change only the exhaust molar mass, to see the $1/\sqrt{\mathcal{M}}$ scaling cleanly. Chamber: $T_c = 3600\ \text{K}$, $\gamma = 1.20$, expanding from $p_c = 100\ \text{bar}$ to $p_e = 0.1\ \text{bar}$ (a pressure ratio $p_c/p_e = 1000$, a large, vacuum-class expansion). Tier 3, clean values.

The common factors. The bracket: $(\gamma-1)/\gamma = 0.2/1.2 = 0.1667$, and $(p_e/p_c)^{0.1667} = (0.001)^{0.1667} = 10^{-3\times0.1667} = 10^{-0.5} = 0.3162$, so the bracket is $1 - 0.3162 = 0.6838$. The front factor: $\dfrac{2\gamma}{\gamma-1} = \dfrac{2.4}{0.2} = 12$.

Hydrogen-rich exhaust, $\mathcal{M} = 13\ \text{g/mol} = 0.013\ \text{kg/mol}$: $R = 8.314/0.013 = 639.5\ \text{J/(kg·K)}$. $$v_e = \sqrt{12 \times 639.5 \times 3600 \times 0.6838} = \sqrt{1.889\times10^{7}} \approx 4346\ \text{m/s}.$$

Kerosene-like exhaust, $\mathcal{M} = 22\ \text{g/mol} = 0.022\ \text{kg/mol}$: $R = 8.314/0.022 = 377.9\ \text{J/(kg·K)}$. $$v_e = \sqrt{12 \times 377.9 \times 3600 \times 0.6838} = \sqrt{1.116\times10^{7}} \approx 3341\ \text{m/s}.$$

The ratio is $4346/3341 = 1.301$ — and $\sqrt{22/13} = \sqrt{1.692} = 1.301$ exactly, because with everything else fixed $v_e \propto 1/\sqrt{\mathcal{M}}$. Cutting the exhaust molar mass from 22 to 13 raised the exhaust velocity by 30%, over a kilometer per second, from nothing but lighter molecules. In specific-impulse terms ($I_{sp} = v_e/g_0$, from Chapter 3): $443\ \text{s}$ versus $341\ \text{s}$. That gap — a hundred seconds of Isp — is the entire reason the hydrogen headache is worth enduring.

🔄 Check Your Understanding 1. According to the exhaust-velocity formula, if you could double the chamber temperature $T_c$ (holding $\mathcal{M}$, $\gamma$, and the pressure ratio fixed), by what factor would $v_e$ increase? 2. Why does raising the chamber pressure $p_c$ (for a fixed exit pressure $p_e$) increase the exhaust velocity, even though $p_c$ does not appear by itself in the formula — only the ratio $p_e/p_c$ does?

Answers

1. By $\sqrt{2} \approx 1.41$, a $41\%$ increase — $v_e \propto \sqrt{T_c}$. (This is why chasing flame temperature has diminishing returns and runs into melting hardware.) 2. Raising $p_c$ at fixed $p_e$ makes the ratio $p_e/p_c$ smaller, which makes the bracket $1 - (p_e/p_c)^{(\gamma-1)/\gamma}$ larger — more of the thermal energy is converted to kinetic energy. Physically, a higher chamber pressure lets the gas expand through a larger pressure drop, extracting more work. This is why engineers push chamber pressures ever higher (Raptor runs above $300\ \text{bar}$): it is a direct lever on both $v_e$ and thrust density.

19.5 Expansion ratio and altitude compensation

The exhaust-velocity formula rewards expanding to the lowest possible exit pressure $p_e$. The nozzle geometry that delivers a given $p_e$ is set by one number.

Definition (expansion ratio). The expansion ratio (or area ratio) $\epsilon = A_e / A_t$ is the ratio of a nozzle's exit area to its throat area. A larger $\epsilon$ means a bigger bell, a lower exit pressure, a higher exhaust velocity — and a heavier, longer nozzle. It is the single most important geometric parameter of a nozzle.

How does $\epsilon$ relate to the flow? Apply the area–Mach relation, which comes from combining continuity with the isentropic relations (the same three statements as §19.2, integrated rather than differentiated). For flow that is sonic at the throat ($A^* = A_t$), the area at any station of Mach number $M$ is

$$ \frac{A}{A_t} = \frac{1}{M}\left[\frac{2}{\gamma + 1}\left(1 + \frac{\gamma - 1}{2}M^2\right)\right]^{\frac{\gamma + 1}{2(\gamma - 1)}}. $$

Evaluated at the exit Mach number $M_e$, the left side is the expansion ratio $\epsilon$. So specifying $\epsilon$ fixes the exit Mach number, which (through the isentropic pressure relation of §19.3) fixes the exit pressure. The three are one choice wearing three faces: pick the exit pressure you want, and you have picked the exit Mach number and the area ratio.

Worked Example: sizing a vacuum bell. For $\gamma = 1.20$, size a nozzle to reach exit Mach $M_e = 4.0$.

The exponent $\dfrac{\gamma+1}{2(\gamma-1)} = \dfrac{2.2}{0.4} = 5.5$. The bracket: $1 + \tfrac{\gamma-1}{2}M_e^2 = 1 + 0.1(16) = 2.6$; then $\tfrac{2}{2.2}(2.6) = 2.364$; and $(2.364)^{5.5} = 113.5$. So $$\epsilon = \frac{A_e}{A_t} = \frac{113.5}{4.0} = 28.4.$$ A nozzle with a $28:1$ area ratio produces Mach-4 exhaust. What exit pressure is that? From §19.3, $p_c/p_e = (2.6)^{6} = 309$, so $p_e = p_c/309$. With $p_c = 100\ \text{bar}$, $p_e = 0.32\ \text{bar}$ — well below sea-level ambient, so this is a nozzle for altitude, over-expanded on the pad. Push the exit Mach and $\epsilon$ higher still and $p_e$ falls toward vacuum: the same chamber at $\epsilon \approx 72$ reaches $M_e \approx 4.65$ and $p_e \approx 0.1\ \text{bar}$ — the vacuum-class bell in the §19.4 example (and, not coincidentally, close to the RS-25's real $\epsilon = 69$).

Now the reason nozzles come in different sizes at all: the outside world does not hold still. On the pad, ambient pressure is $p_a \approx 1.0\ \text{bar}$; at $30\ \text{km}$ it is a hundredth of that; in orbit it is zero. A nozzle expands to a fixed $p_e$ set by its geometry, but the $p_a$ it fires into changes by a factor of infinity over a single launch. The relationship between $p_e$ and $p_a$ defines three regimes.

Definition (over-expanded). A nozzle is over-expanded when the exit pressure is below ambient, $p_e < p_a$. The nozzle has expanded the gas too far for the outside pressure; ambient air pushes back on the exhaust, robbing thrust, and if $p_e$ falls far enough below $p_a$ the flow separates from the nozzle wall, causing damaging side loads. This is a vacuum-optimized nozzle fired at sea level.

Definition (under-expanded). A nozzle is under-expanded when the exit pressure is above ambient, $p_e > p_a$. The gas has not finished expanding when it reaches the exit and continues to expand outside the nozzle, spreading into the characteristic bright plume. The unused pressure represents thrust the nozzle could have captured with a longer bell. This is a sea-level nozzle fired in vacuum.

Between them is the sweet spot: perfect expansion, $p_e = p_a$, where the pressure-thrust term $(p_e - p_a)A_e$ vanishes, the exhaust neither over- nor under-shoots, and thrust is maximal for the given chamber conditions and ambient pressure. Because $p_a$ falls continuously during ascent, a fixed nozzle can be perfectly expanded at exactly one altitude, is over-expanded below it and under-expanded above it. This is altitude compensation: the problem that no single bell is right for the whole flight.

Worked Example: sea-level bell versus vacuum bell. Same chamber, $\gamma = 1.20$, $p_c = 100\ \text{bar}$. Design a sea-level nozzle to exit near $p_e = 0.7\ \text{bar}$ (a touch over-expanded at $1.0\ \text{bar}$ ambient, which real engines accept to keep the bell short) versus a vacuum nozzle exiting at $p_e = 0.1\ \text{bar}$.

Using the area-ratio relation (worked out from $p_c/p_e$ via the isentropic and area–Mach relations):

Nozzle	$p_c/p_e$	exit Mach $M_e$	expansion ratio $\epsilon$
Sea-level	$143$	$\approx 3.4$	$\approx 16$
Vacuum	$1000$	$\approx 4.65$	$\approx 72$

The vacuum bell has more than four times the area ratio of the sea-level bell, from the same chamber, purely because it expands to one-seventh the exit pressure. That is why an upper-stage engine wears an enormous skirt and a first-stage engine wears a stubby cone. (Compare reality: the sea-level Merlin has $\epsilon \approx 16$; the Merlin Vacuum has $\epsilon \approx 165$; the RS-25, $\epsilon = 69$. Our idealized numbers land right in that range.)

🔧 Engineering Reality: why you can't just make every nozzle huge. If a bigger $\epsilon$ always means more $v_e$, why not fit every engine with a vacuum bell the size of a house? Three reasons, all from earlier themes. First, mass is the enemy (Chapter 3, theme 4): the bell is dead structure the whole rocket must carry, and the returns on $\epsilon$ diminish as $p_e \to 0$. Second, flow separation: fire a big over-expanded nozzle at sea level and the exhaust rips away from the wall, producing violent asymmetric side loads that can tear an engine apart — so a first-stage nozzle is deliberately kept small enough to stay attached at liftoff. Third, fit and plumbing: multiple engines must pack onto a base, and a giant bell simply won't fit. The chosen $\epsilon$ is always a compromise: as large as separation and mass allow for the altitude where the engine does most of its work. The aerospike engine tries to sidestep the whole problem with a nozzle that adjusts to ambient pressure automatically, but its cooling and mass challenges have kept it off operational rockets so far.

🔗 Connection: the pressure term, quantified. Take the sea-level bell above ($p_e = 0.7\ \text{bar}$, exit area $A_e = 0.9\ \text{m}^2$, a Merlin-ish size) and fire it in two places, using the Chapter 16 thrust equation $F = \dot m\, v_e + (p_e - p_a)A_e$. At sea level ($p_a = 1.0\ \text{bar}$) the pressure term is $(0.7 - 1.0)\times10^{5} \times 0.9 = -2.7\times10^{4}\ \text{N}$ — the engine loses about $27\ \text{kN}$ of thrust to over-expansion. In vacuum ($p_a = 0$) the same term becomes $(0.7 - 0)\times10^{5} \times 0.9 = +6.3\times10^{4}\ \text{N}$, a $63\ \text{kN}$ gain. The identical engine makes tens of kilonewtons more thrust in space than on the pad — most of the roughly $8\%$ sea-level-to-vacuum thrust rise real engines show — and every bit of that swing lives in the little $(p_e - p_a)A_e$ term. The mass flow, choked at the throat, never changed.

A simple ASCII picture ties the geometry to the flow regimes:

        CHAMBER            THROAT              EXIT
     p_c, T_c, M~0         M = 1           p_e, T_e, M>1
     (hot, slow)          (choked)         (cool, fast)
      ___________
     |           \                             ______
     |  fire      \                       ____/
 --> |  reservoir  )=========( . . . . . /       supersonic  ==>  v_e
     |  subsonic  /          throat      \____    exhaust plume
     |___________/                            \______
       CONVERGING          A_t                 DIVERGING (A_e)
      A shrinks, M rises            A grows, M rises further
      (M<1: area down -> speed up)  (M>1: area up -> speed up)

      epsilon = A_e / A_t  (expansion ratio)
      p_e > p_a : under-expanded (plume swells outward)  -- vacuum bell in air
      p_e = p_a : perfectly expanded (max thrust)
      p_e < p_a : over-expanded (plume pinched, may separate) -- big bell at sea level

🔄 Check Your Understanding 1. A rocket lifts off with its first-stage engines slightly over-expanded and climbs into vacuum. As ambient pressure falls, does the engine become more over-expanded, or does it move toward under-expansion? 2. Why does an upper-stage engine, which only ever fires in near-vacuum, use a much larger expansion ratio than a first-stage engine?

Answers

1. It moves toward under-expansion. Over-expanded means $p_e < p_a$; as $p_a$ falls, the gap closes, the engine passes through perfect expansion at some altitude, and then becomes under-expanded ($p_e > p_a$) the rest of the way up. Its thrust rises the whole time. 2. Because in vacuum there is no ambient pressure to cause separation or push back, so the nozzle can expand the gas all the way down to a very low $p_e$ — capturing nearly all the available energy as exhaust velocity. A large $\epsilon$ (big bell) is exactly what achieves that low $p_e$. The only penalty is the bell's mass, which is worth paying for the extra $v_e$ upstairs.

19.6 Why LOX/LH2 beats LOX/RP-1

We can now answer, with a formula rather than a slogan, the question Chapter 18 set up: among chemical propellants, why does liquid hydrogen deliver the highest specific impulse? The answer is one term in the exhaust-velocity equation:

$$ v_e \propto \sqrt{\frac{T_c}{\mathcal{M}}}. $$

Performance is set by the ratio of chamber temperature to exhaust molar mass — and hydrogen wins not by burning hotter (it doesn't) but by producing the lightest exhaust of any chemical propellant.

Consider what each fuel actually makes when it burns with oxygen. LOX/LH2 produces water vapor, $\text{H}_2\text{O}$ (molar mass $18$), and because hydrogen engines run deliberately fuel-rich there is a great deal of leftover unburned $\text{H}_2$ (molar mass $2$) in the mix. The average molar mass of the exhaust is dragged down to roughly $\mathcal{M} \approx 10\text{–}14\ \text{g/mol}$. LOX/RP-1 (kerosene) burns carbon and hydrogen, so its exhaust is water plus carbon dioxide ($\text{CO}_2$, molar mass $44$) and carbon monoxide ($\text{CO}$, molar mass $28$) — heavy molecules that pull the average up to $\mathcal{M} \approx 22\text{–}24\ \text{g/mol}$. Hydrogen's exhaust is about half the molar mass of kerosene's.

🚪 Threshold Concept: performance is molecular, not thermal. The naive expectation is that the hottest fire makes the best rocket. It doesn't. Kerosene actually burns slightly hotter than hydrogen — around $3670\ \text{K}$ versus $3600\ \text{K}$ in comparable engines — and still loses decisively, because it makes heavy exhaust. Once you internalize $v_e \propto \sqrt{T_c/\mathcal{M}}$, you stop asking "what burns hottest?" and start asking "what makes the lightest exhaust at a tolerable temperature?" That reframing is why hydrogen — a fuel that stores at $20\ \text{K}$, embrittles metal, boils off, and needs tanks the size of a gymnasium — has flown on the upper stages of the most demanding rockets ever built, from the Saturn V's J-2 to the Shuttle's RS-25 to the SLS. The physics rewards light molecules, and engineers pay almost any price to get them.

Worked Example: hydrogen versus kerosene, realistically. Now use each propellant's own real-ish parameters rather than holding them equal (Tier 3, representative of flown engines), expanding from $p_c = 100\ \text{bar}$ to $p_e = 0.1\ \text{bar}$:

Propellant	$T_c$ (K)	$\mathcal{M}$ (g/mol)	$\gamma$	$\sqrt{T_c/\mathcal{M}}$	$v_e$ (m/s)	$I_{sp}$ (s)
LOX/LH2	$3600$	$13$	$1.20$	$16.6$	$4346$	$443$
LOX/RP-1	$3700$	$22$	$1.22$	$13.0$	$3323$	$339$

Kerosene runs $100\ \text{K}$ hotter and it doesn't matter: the $\sqrt{T_c/\mathcal{M}}$ figure is $16.6$ for hydrogen against $13.0$ for kerosene, a $28\%$ edge that flows straight through to a $\sim 1000\ \text{m/s}$ higher exhaust velocity and about a hundred seconds more specific impulse. These modeled numbers bracket reality well: real vacuum $I_{sp}$ is about $450\ \text{s}$ for good hydrogen engines and $330\text{–}350\ \text{s}$ for kerosene ones. The molecular-weight lever, and nothing else, is doing the work. (The RP-1 $v_e$ here, $3323$, is a touch below the §19.4 "same-chamber" value of $3341$ because we also lowered its effective performance with a realistic higher $\gamma$; both are illustrative.)

If hydrogen is so much better, why isn't every stage hydrogen? Because $I_{sp}$ is not the only number that matters, and the rocket equation has a second lever — the mass ratio — that hydrogen sabotages. Liquid hydrogen is absurdly light for its volume: about $70\ \text{kg/m}^3$, against roughly $810\ \text{kg/m}^3$ for RP-1. To carry a given mass of hydrogen you need a tank more than ten times the volume, and a bigger tank is more structure, more mass, more drag, and a worse mass ratio — precisely the enemy of Chapter 3. Deep cryogenics ($20\ \text{K}$) adds insulation, boil-off, and embrittlement. So the choice is a genuine trade, and it splits cleanly by role:

• Upper stages and deep-space stages favor hydrogen: they fire mostly in vacuum where the big bell and the high $I_{sp}$ pay off, and the mass-ratio penalty matters less when you are already above most of the gravity and drag losses.
• First stages often favor denser fuels — RP-1 or, increasingly, methane ($\text{LOX/CH}_4$, which splits the difference: $\mathcal{M} \approx 20\ \text{g/mol}$, storable near LOX temperature, clean-burning and reusable). Falcon 9 uses RP-1 top to bottom for density, simplicity, and reuse; the Space Shuttle used dense solids for liftoff thrust and hydrogen for efficiency, a hybrid that got the best of both.

🔗 Connection: the whole propulsion story in one inequality. Chapters 16 through 19 have been a single argument. Chapter 16 said $v_e$ is what matters and put it in the thrust equation. Chapter 17 built the engine that makes high $p_c$ possible. Chapter 18 supplied the chemistry that sets $T_c$ and $\mathcal{M}$. This chapter's nozzle is the machine that converts $T_c$ and $\mathcal{M}$ into $v_e$ via $v_e \propto \sqrt{T_c/\mathcal{M}}$. The tyranny of the rocket equation (theme 1) has exactly two escape routes — a bigger mass ratio or a bigger $v_e$ — and the entire discipline of propulsion is the pursuit of the second one, molecule by molecule, kelvin by kelvin, bar by bar.

Mission Design Checkpoint: propulsion.py and choosing your engine

This checkpoint extends the astrotools propulsion module and turns your combustion choice into an exhaust velocity you can feed to the rocket equation.

The design. In your Mission Design Review, revisit the engine selection you began in the propulsion chapters. For your mission's in-space stage (the one that spends your Chapter 3 delta-v budget), pick a propellant family and record its representative chamber conditions and, crucially, its exhaust molar mass — the number this chapter showed sets your $I_{sp}$. A Track-C Mars orbiter or Track-A comsat that burns mostly in vacuum should weigh a high-$I_{sp}$ hydrogen or storable stage against the tank-mass penalty; a lander doing large burns near a surface may prefer a denser propellant. Compute the ideal $v_e$ for your choice, convert to $I_{sp}$, and carry that number forward — it is what sizes your propellant mass when you close the mass budget.

The code. Extend astrotools/propulsion.py (begun in Chapter 16 with thrust) with this chapter's two functions. Code is illustrative and never executed; outputs below are hand-computed.

"""astrotools/propulsion.py -- nozzle & exhaust-velocity additions (Chapter 19).
Extends the module started in Chapter 16 (thrust) and 18 (combustion inputs).
Ideal, isentropic, calorically-perfect-gas model. Never executed at build."""
import math

R_U = 8.314  # universal gas constant, J/(mol*K)

def exit_velocity(gamma, Tc, molar_mass, pe, pc):
    """Ideal exhaust velocity (m/s) from chamber temperature Tc (K),
    exhaust molar mass (kg/mol), and exit/chamber pressures pe, pc (same units).
    v_e = sqrt( (2*gamma/(gamma-1)) * (R_u*Tc/M) * (1 - (pe/pc)**((gamma-1)/gamma)) )."""
    R = R_U / molar_mass
    bracket = 1.0 - (pe / pc) ** ((gamma - 1) / gamma)
    return math.sqrt((2 * gamma / (gamma - 1)) * R * Tc * bracket)

def expansion_ratio(gamma, pe, pc):
    """Nozzle area ratio A_e/A_t to expand from chamber pressure pc to exit pressure pe,
    for isentropic flow choked at the throat (Sutton-form area-ratio relation)."""
    g = gamma
    num = ((2 / (g + 1)) ** (1 / (g - 1))) * (pc / pe) ** (1 / g)
    den = math.sqrt(((g + 1) / (g - 1)) * (1 - (pe / pc) ** ((g - 1) / g)))
    return num / den

if __name__ == "__main__":
    G0 = 9.80665
    # LOX/LH2 vs LOX/RP-1, same chamber (Tc=3600 K, gamma=1.20, pc/pe=1000):
    ve_h2 = exit_velocity(1.20, 3600, 0.013, 1e4, 1e7)
    ve_rp = exit_velocity(1.20, 3600, 0.022, 1e4, 1e7)
    print("LH2  ve =", round(ve_h2), "m/s  Isp =", round(ve_h2 / G0), "s")
    print("RP-1 ve =", round(ve_rp), "m/s  Isp =", round(ve_rp / G0), "s")
    print("ratio =", round(ve_h2 / ve_rp, 3), " sqrt(22/13) =", round((22 / 13) ** 0.5, 3))
    print("vacuum-bell epsilon =", round(expansion_ratio(1.20, 1e4, 1e7), 1))
    # Expected output:
    # LH2  ve = 4346 m/s  Isp = 443 s
    # RP-1 ve = 3341 m/s  Isp = 341 s
    # ratio = 1.301  sqrt(22/13) = 1.301
    # vacuum-bell epsilon = 71.6

How it feeds the capstone. exit_velocity converts your propellant's chamber conditions into the $v_e$ (hence $I_{sp}$) that rocket.py needs to turn your delta-v budget into a propellant mass; expansion_ratio sizes the nozzle you will quote in your engine selection. Together they close the loop from Chapter 18's chemistry to Chapter 3's rocket equation — the propulsion half of your Mission Design Review.

Summary

The nozzle converts chamber heat into directed exhaust velocity; its physics is compressible flow plus thermodynamics. Carry these forward:

Numbers worth memorizing: best chemical $I_{sp} \approx 450\ \text{s}$ (LH2, vacuum); kerosene/methane $\approx 330\text{–}380\ \text{s}$; critical pressure ratio $\approx 0.5\text{–}0.6$; sea-level nozzle $\epsilon \sim 15\text{–}25$, vacuum $\epsilon \sim 50\text{–}200$; $R_u = 8.314\ \text{J/(mol·K)}$; $g_0 = 9.81\ \text{m/s}^2$.

Spaced Review

Retrieval strengthens memory. Answer from memory before checking, then look back at the cited chapter.

1. (Ch. 16) The thrust equation is $F = \dot m\, v_e + (p_e - p_a)A_e$. Which of its two terms did this chapter's choked-flow result fix as constant, and which one changes with altitude?
2. (Ch. 16) Specific impulse and effective exhaust velocity are related by one constant. Write the relation, and state whether a nozzle that improves $v_e$ also improves $I_{sp}$.
3. (Ch. 18) Chapter 18 gave you the adiabatic flame temperature and the exhaust molar mass of a propellant. Which of those two does this chapter's $v_e \propto \sqrt{T_c/\mathcal{M}}$ say is the stronger lever when comparing hydrogen to kerosene, and why?
4. (Ch. 18) Hydrogen engines are run deliberately fuel-rich rather than at the stoichiometric mixture ratio, even though that leaves fuel unburned. Using this chapter's formula, explain why that can raise performance.

Answers

1. Choking fixes the mass flow $\dot m$ (set by $p_c$, $A_t$, $T_c$ at the sonic throat), so the $\dot m\, v_e$ momentum term is essentially constant; the pressure term $(p_e - p_a)A_e$ changes with altitude as $p_a$ falls, which is why thrust rises as the rocket climbs. 2. $I_{sp} = v_e/g_0$ (equivalently $c/g_0$ for effective exhaust velocity); with $g_0$ a fixed constant, any nozzle improvement that raises $v_e$ raises $I_{sp}$ in exact proportion. 3. The molar mass $\mathcal{M}$ is the stronger lever here: kerosene actually burns slightly hotter (higher $T_c$) yet loses, because hydrogen's exhaust is roughly half the molar mass, and $v_e \propto 1/\sqrt{\mathcal{M}}$ more than overcomes the temperature deficit. 4. Running fuel-rich leaves unburned light $\text{H}_2$ (molar mass 2) in the exhaust, which lowers the average $\mathcal{M}$. Even though the flame is a little cooler (less $T_c$), the drop in $\mathcal{M}$ raises $\sqrt{T_c/\mathcal{M}}$ on balance — a direct trade of temperature for lighter molecules, and the formula says the molecules usually win.

What's Next

We have now taken a chemical rocket as far as chemistry allows. The nozzle wrings the maximum exhaust velocity out of a given flame, and the flame is bounded by the energy locked in molecular bonds — which is why the best chemical $I_{sp}$ has sat near $450\ \text{seconds}$ for sixty years and will not move much further. To beat that ceiling you must abandon the idea that the propellant's own chemistry supplies the energy. In Chapter 20 we do exactly that: we use electricity to hurl ions out at exhaust velocities ten times anything a nozzle can reach — tens of kilometers per second, specific impulses in the thousands of seconds — and pay for it with thrust so gentle it would lose a shoving match with a sheet of paper. The tyranny of the rocket equation offers two doors, and having pushed the $v_e$ of chemical rockets to its limit, we are about to walk through the other one.