Lesson 4 of 14

Propellant Performance

Where exhaust velocity comes from: heat of combustion, flame temperature, dissociation and molecular mass — why every engine runs fuel-rich, why hydrogen's cooler flame wins, and how chemists compute the numbers in Sutton's tables.

6 min3 live simulations
you are here
The question

Kerosene burning in oxygen reaches about 3700 K. Hydrogen burning in oxygen, at the mixture ratios engines actually use, reaches only 3000–3600 K. Yet the hydrogen engine's exhaust leaves 30% faster. How can a cooler flame make a faster jet?

Two ingredients: heat and lightness

Lesson 2 gave us the exhaust velocity of an ideal nozzle. Strip it to its essentials and one ratio remains:

ve  ∝  TcM\cvel{v_e} \;\propto\; \sqrt{\frac{\cflame{T_c}}{\mathfrak{M}}}

The chamber temperature Tc\cflame{T_c} is how much energy each kilogram of gas holds. The molar mass M\mathfrak{M} is how heavy the molecules carrying it are. A propellant is good if it releases a lot of heat and leaves light molecules to carry it away.

Why the square root of T/MT/\mathfrak M? Because temperature is the average kinetic energy of the molecules, 12mv2ˉ∝T\tfrac12 m\bar{v^2} \propto T. At the same temperature, a hydrogen molecule (mass 2) jiggles three times faster than a water molecule (mass 18). A nozzle's job is to turn that random jiggling into an ordered jet, and the jet can never go faster than the molecules could.

WatchThe molecule race
H₂ (M = 2)H₂O (M = 18)CO (M = 28)CO₂ (M = 44)
3000 K
Hydrogen exhaust√(T/M) = 15.9
Kerosene exhaust√(T/M) = 10.9
Exhaust velocity ratio1.45×hydrogen / kerosene
Challenges0/2
  • Switch to real chamber temperatures: does kerosene’s hotter flame win the race?
  • Find a temperature at which hydrogen exhaust still beats kerosene exhaust at 3700 K.
Two boxes of hot exhaust gas. Every molecule has the same average kinetic energy at a given temperature, ½mv² ∝ T — so light molecules move fast and heavy ones slowly. A nozzle turns this random motion into an ordered jet, and the jet can be no faster than the molecules were. That is why vₑ ∝ √(T/M).
Estimate first
What is the average speed of the water molecules in a 3000 K rocket chamber, bouncing around at random?
101001,00010,000100,000
Drag along the line to place your guess — each tick is ten times the last.

Burning at the right mixture

The Mixture ratio rOxidizer-to-fuel mass flow ratio, O/F. Engines run fuel-rich of stoichiometric because the lighter exhaust gives more Isp. r=m˙ox/m˙fr = \dot m_{ox}/\dot m_{f} is the rocket engineer's version of the equivalence ratio. For hydrogen and oxygen the stoichiometric ratio, where every hydrogen atom finds an oxygen, is 7.94. Combustion there is hottest. So why does no hydrogen engine run at 7.94?

ExploreThe hottest flame is not the best propellant
2467.99.911.91000200030004000mixture ratio O/F (by mass)chamber temperature (K)300 s350 ssea-level Isp (orange)stoichiometricpeak Isphottestmolar mass (dashed)
6.00 (φ 1.32)
Chamber temperature3595 K
Molar mass12.8 g/mol
c*2355 m/s
Isp sea level / vacuum382 / 459 svacuum with ε = 60
The highest-Isp practical combination — but hydrogen is 14 times less dense than water and boils at 20 K.
Challenges0/3
  • Find the peak-Isp mixture ratio for hydrogen.
  • For kerosene, find how far the Isp peak sits from the temperature peak.
  • Explore all four propellant pairs (1/4).
Chamber temperature (red), exhaust molar mass (purple) and sea-level Isp (orange) against mixture ratio O/F, at 68 bar with optimum expansion — the condition of Sutton's tables. Exhaust velocity goes like √(T/M): the peak of Isp is where temperature and lightness strike the best bargain.

Follow the purple dashed line. Adding fuel beyond stoichiometric leaves unburned hydrogen in the exhaust. It carries no chemical energy away — it has not burned — but it is extraordinarily light, and it drags the average molar mass down faster than it drags the temperature down. For a while T/M\sqrt{T/\mathfrak M} keeps rising. The Isp peak for hydrogen sits near r≈4r \approx 4–5, far rich of stoichiometric; for kerosene the effect is smaller but still there, because the unburned fuel turns into light CO and H₂ rather than heavy CO₂.

Real engines don't even run at the Isp peak. The RS-25 runs at r=6r = 6: slightly less Isp, but the extra oxygen is dense, so the hydrogen tank — the biggest and lightest-built structure on the vehicle — gets smaller. Engine designers optimize the vehicle, not the engine.

Check your intuition
Why does running fuel-rich also help the engine survive?

Dissociation: the ceiling on flame temperature

If you simply divide the heat of combustion by the heat capacity of the products, oxygen flames should reach 4500 K and more. Real ones stop around 3500–3900 K. The reason is DissociationAt chamber temperatures H₂O and CO₂ partly break into H, O and OH, storing energy that caps the flame temperature.: at those temperatures, water and carbon dioxide partly break apart into H, O, OH and CO. Breaking bonds absorbs energy, so the harder you try to heat the gas, the more of the extra energy is soaked up by pulling molecules apart.

Two consequences matter for rockets:

  • Pressure helps. Dissociation increases the number of molecules; high pressure pushes back against it (Le Chatelier). Higher chamber pressure means a slightly hotter flame — one of many reasons engines have crept from 20 bar to 300 bar.
  • Some energy can come back. As the gas expands and cools in the nozzle, the fragments recombine, releasing their stored energy as extra velocity — if the chemistry is fast enough.
CompareFrozen or shifting: does the chemistry keep up?
020406080100240260280300320340360380nozzle area ratio εvacuum Isp (s)shifting equilibriumfrozen flow
68 bar
Higher pressure suppresses dissociation
Dissociation19%of the ideal flame temperature lost
Recombination bonus+2.4%shifting over frozen, ε = 100
Chamber temperature3801 K
Real nozzles sit between the two: recombination keeps up near the throat, where the gas is dense and slow, then freezes as it thins out and accelerates. Measured performance usually lands between the curves.
In the chamber some of the water and CO₂ is dissociated, storing energy. If the gas recombines as it cools in the nozzle (shifting equilibrium), that energy comes back as velocity. If the flow is too fast for the chemistry (frozen), it leaves with the exhaust. Vacuum Isp against nozzle area ratio.

Sutton calls the two limits Shifting equilibriumAssuming the exhaust keeps re-reacting (recombining) as it cools in the nozzle, returning dissociation energy as velocity. Frozen flow assumes it cannot. (composition keeps up with the falling temperature) and frozen flow (composition fixed at the throat). The truth lies in between, and the gap — a few percent of Isp — is one of the "real nozzle" losses of Lesson 2.

How the tables are made

The numbers in Sutton's performance tables come from programs like NASA's CEA, which find the chemical equilibrium composition by minimizing the Gibbs free energy of dozens of species at fixed pressure and enthalpy, then expand that mixture isentropically through the nozzle. The model inside this course is a much simpler cousin — product composition from bookkeeping rules, flame temperature from an energy balance, dissociation as a soft ceiling — but it is calibrated to land close to the tables:

Propellant (O/F)TcT_c (K)M\mathfrak Mc∗c^* (m/s)Isp (s)*
LOX / LH₂ (4.0)~3000~10~2390~390
LOX / CH₄ (3.2)~3550~20~1840~310
LOX / RP-1 (2.6)~3700~22~1800~300
NTO / MMH (2.2)~3400~21.5~1720~289
*68 bar chamber, optimum expansion to sea level, shifting equilibrium: the reference condition of Sutton's table 5-5. In vacuum with a large nozzle, add 15–20%.
Going deeper
The energy balance behind the flame temperature

The adiabatic flame temperature comes from conservation of enthalpy: the enthalpy of the reactants (including their heats of formation, which for liquid propellants include vaporization) equals the enthalpy of the products at the flame temperature,

∑reactantsnjΔhf,j∘=∑productsni[Δhf,i∘+∫298Tccp,i dT].\sum_{reactants} n_j \Delta h^\circ_{f,j} = \sum_{products} n_i\left[\Delta h^\circ_{f,i} + \int_{298}^{T_c} c_{p,i}\,dT\right].

The catch is that the product composition nin_i itself depends on TcT_c through dissociation, so the problem is solved iteratively: guess TcT_c, find the equilibrium composition, check the energy balance, adjust. The same equilibrium solution also gives the effective γ\gamma and M\mathfrak{M}, and from them c∗c^* and CFC_F.

Recap
What to carry forward
  • Exhaust velocity scales as Tc/M\sqrt{T_c/\mathfrak M}: heat and lightness both matter.
  • The mixture ratio for peak Isp is fuel-rich of stoichiometric, because excess fuel makes the exhaust lighter faster than it makes it cooler. Real engines trade a little Isp for denser propellant.
  • Dissociation caps flame temperatures near 3500–3900 K; higher chamber pressure pushes the cap up slightly.
  • Recombination in the nozzle returns some energy: shifting equilibrium is the upper limit, frozen flow the lower.
  • Performance tables come from Gibbs-energy minimization plus isentropic expansion.

Next: how to feed these propellants into a chamber at hundreds of kilograms per second. The anatomy of a liquid rocket engine.

Finished this lesson?Mark it complete to track your progress through the course (saved in this browser).
Up next · Lesson 5Liquid Engines & PropellantsWhy pump when you can push?Keep going →