CEA (Chemical Equilibrium with Applications) — using NASA-CEA tool to compute Isp, Tc, products: Worked examples

We reuse three tools from the parent note, so keep them on screen:

Here $R_{u} = 8.314 J/(mol\cdotK)$ is the universal gas constant (energy per mole per kelvin), $γ$ is the ratio of specific heats $c_{p} / c_{v}$ (how "springy" the gas is when squeezed — near $1.2$ for hot combustion products), $M$ is the average molar mass of the exhaust in kg/mol, $p_{c}$ is the chamber pressure (the pressure of the hot gas inside the combustion chamber, upstream of the nozzle), $p_{e}$ is the exit pressure (the gas pressure right at the nozzle mouth), and $p_{e} / p_{c}$ is therefore the exit-to-chamber pressure ratio — a number between $0$ and $1$ that says how far the gas has been allowed to expand.

The scenario matrix

Every CEA-flavoured problem falls into one of these cells. The examples below each carry a tag like (A2) so you can see the whole grid is covered.

Cell	Case class	What makes it different	Example
A1	Stoichiometric mixture	Hottest flame, $O / F$ at the balance point	Ex 1
A2	Fuel-rich mixture	Lower $T_{c}$ but lower $M$ → higher $I_{s p}$	Ex 2
A3	Oxidizer-rich (lean)	Heavy exhaust, wasted oxidizer	Ex 3
B1	Degenerate: no reaction (cold)	$T_{c} = T_{in}$ , nothing burns	Ex 4
C1	Limiting: vacuum expansion	$p_{e} / p_{c} \to 0$ , max $v_{e}$	Ex 5
C2	Limiting: sea-level expansion	$p_{e} = p_{atm}$ , back-pressure bites	Ex 5
D1	Frozen vs equilibrium flow	Recombination bonus in the nozzle	Ex 6
E1	Sensitivity to $M$ (sign of change)	Which knob moves $I_{s p}$ more	Ex 7
F1	Real-world word problem	$I_{s p} \to$ Tsiolkovsky Rocket Equation	Ex 8

Figure — CEA (Chemical Equilibrium with Applications) — using NASA-CEA tool to compute Isp, Tc, products

Worked examples

Ex 1 — Stoichiometric $H_{2} / O_{2}$ (A1)

Worked example Compute

$I_{s p}$ at the exact stoichiometric mixture. Reaction $2 H_{2} + O_{2} \to 2 H_{2} O$ . Take (from a CEA run, ignoring dissociation) $T_{c} = 3500$ K, exhaust essentially pure $H_{2} O$ so $M = 0.018$ kg/mol, $γ = 1.2$ , vacuum expansion $p_{e} / p_{c} = 0.001$ .

Forecast: hottest flame here — do you think this gives the highest $I_{s p}$ ? (Hold that thought.)

Find the mass $O / F$ ratio. Two moles $H_{2}$ (mass $2 \times 2 = 4$ g) burn with one mole $O_{2}$ (mass $32$ g), so $O / F = 32/4 = 8$ . Why this step? This anchors "stoichiometric = 8" so the fuel-rich case (Ex 2) has a reference.
Bracket term. $(\frac{γ - 1}{γ}) = \frac{0.2}{1.2} = 0.1667$ , so $0.00 1^{0.1667} = 0.3162$ , bracket $= 1 - 0.3162 = 0.6838$ . Why this step? This is the fraction of thermal energy the nozzle can extract.
Prefactor. $\frac{2 γ}{γ - 1} = 12$ ; $\frac{R _{u} T _{c}}{M} = \frac{8.314 \times 3500}{0.018} = 1.617 \times 1 0^{6}$ . Why this step? Separates the "how much energy per kg" ( $R_{u} T_{c} / M$ ) from the "how efficiently we release it" (the bracket).
Exit velocity. $v_{e} = 12 \times 1.617 \times 1 0^{6} \times 0.6838 = 1.327 \times 1 0^{7} = 3642$ m/s. Why this step? This is the payoff line — multiplying prefactor, energy-per-kg, and bracket together and taking the root turns the thermodynamic numbers into an actual exhaust speed.
Specific impulse. $I_{s p} = 3642/9.81 = 371$ s. Why this step? Dividing $v_{e}$ by $g_{0}$ converts exhaust speed into the size-independent efficiency figure (seconds) that CEA reports and that engineers compare.

Verify: Units: $J/kg = m^{2} / s^{2} = m/s$ ✓. The answer ( $\approx 371$ s) is lower than the fuel-rich Example 2 below — so hottest flame did not win. Your forecast, if it said "yes," just got corrected.

Ex 2 — Fuel-rich $H_{2} / O_{2}$ (A2)

Worked example Same engine, now run fuel-rich at

$O / F = 6$ . CEA gives a cooler flame $T_{c} = 3300$ K but a lighter exhaust: leftover $H_{2}$ drags the average down to $M = 0.013$ kg/mol. Same $γ = 1.2$ , $p_{e} / p_{c} = 0.001$ .

Forecast: $T_{c}$ dropped $200$ K (bad) but $M$ dropped from $18$ to $13$ g/mol (good). Net — up or down?

Prefactor. $\frac{R _{u} T _{c}}{M} = \frac{8.314 \times 3300}{0.013} = 2.111 \times 1 0^{6}$ . Why this step? Compare to Ex 1's $1.617 \times 1 0^{6}$ : even with a colder flame this is larger, because dividing by a smaller $M$ dominates.
Bracket is unchanged ( $0.6838$ ) — same pressures and $γ$ . Why this step? Isolates the effect: only $T_{c}$ and $M$ changed, so any difference is pure $T_{c} / M$ .
Exit velocity. $v_{e} = 12 \times 2.111 \times 1 0^{6} \times 0.6838 = 1.733 \times 1 0^{7} = 4162$ m/s.
Specific impulse. $I_{s p} = 4162/9.81 = 424$ s.

Verify: $424 > 371$ — fuel-rich beats stoichiometric by $53$ s. Sanity check the ratio directly: $\frac{I _{s p, 2}}{I _{s p, 1}} = \frac{3300/0.013}{3500/0.018} = \frac{2.111 \times 1 0 ^{6}}{1.617 \times 1 0 ^{6}} = 1.142$ , and $371 \times 1.142 = 424$ ✓. This is exactly the peak-shifting shown by the orange curve in the figure.

Ex 3 — Oxidizer-rich (lean) (A3)

Worked example Push the other way:

$O / F = 10$ (too much oxygen). CEA gives $T_{c} = 3400$ K (still hot near stoichiometric) but the extra $O_{2}$ and heavier species raise $M = 0.021$ kg/mol. Same $γ$ , same $p_{e} / p_{c} = 0.001$ .

Forecast: heavier exhaust than stoichiometric — better or worse than Ex 1?

Prefactor. $\frac{R _{u} T _{c}}{M} = \frac{8.314 \times 3400}{0.021} = 1.346 \times 1 0^{6}$ . Why this step? Smaller than Ex 1's $1.617 \times 1 0^{6}$ — the heavy $M$ wins the wrong way.
Bracket term. Same pressures and $γ$ as Ex 1, so $0.00 1^{0.1667} = 0.3162$ and bracket $= 1 - 0.3162 = 0.6838$ . Why this step? We must restate it because the exit-velocity formula multiplies by this bracket; keeping it explicit matches every other example and prevents silently reusing a number.
Exit velocity. $v_{e} = 12 \times 1.346 \times 1 0^{6} \times 0.6838 = 1.105 \times 1 0^{7} = 3324$ m/s. Why this step? Feeds prefactor and bracket into $v_{e}$ ; this is where the heavy- $M$ penalty finally shows up as a lower speed.
Specific impulse. $I_{s p} = 3324/9.81 = 339$ s.

Verify: $339 < 371 < 424$ . Ordering is lean $<$ stoichiometric $<$ fuel-rich, confirming the peak lies on the fuel-rich side. Unburned oxidizer is dead weight — it adds $M$ without adding heat. ✓

Ex 4 — Degenerate case: nothing burns (B1)

Worked example Feed the chamber a mixture that cannot react (e.g. pure

$N_{2}$ , or fuel with no oxidizer at all), injected at $T_{in} = 300$ K.

Forecast: what is $T_{c}$ ? What is $I_{s p}$ if you still "expand" it?

Energy balance with zero reaction. Write the adiabatic constant-pressure balance $\sum_{i} n_{i} h_{i} (T_{c}) = H_{reactants} (T_{in})$ , where $n_{i}$ is the mole number of species $i$ , $h_{i} (T)$ is the molar enthalpy of species $i$ at temperature $T$ (energy per mole, including its formation enthalpy), and $H_{reactants}$ is the total enthalpy of the injected mixture at inlet temperature $T_{in}$ . With no reaction the same species appear on both sides, so no formation-enthalpy difference is released, and the balance forces $T_{c} = T_{in} = 300$ K. Why this step? This is the "zero input" edge case — combustion enthalpy is the only thing that raises temperature.
Expansion of cold gas. Take $N_{2}$ : $M = 0.028$ kg/mol, $γ = 1.4$ , same $p_{e} / p_{c} = 0.001$ . Exponent $\frac{γ - 1}{γ} = \frac{0.4}{1.4} = 0.2857$ , so $0.00 1^{0.2857} = 0.1389$ and bracket $= 1 - 0.1389 = 0.8611$ ; prefactor $\frac{2 γ}{γ - 1} = 7$ ; $\frac{R _{u} T _{c}}{M} = \frac{8.314 \times 300}{0.028} = 8.908 \times 1 0^{4}$ . $v_{e} = 7 \times 8.908 \times 1 0^{4} \times 0.8611 = 5.370 \times 1 0^{5} = 733$ m/s.
Specific impulse. $I_{s p} = 733/9.81 = 74.7$ s.

Verify: A cold gas thruster gives only $\sim 75$ s — that's real (nitrogen cold-gas thrusters sit around $70$ – $80$ s). The lesson: with no chemistry, $I_{s p}$ collapses because $T_{c}$ never leaves room temperature. ✓ This is why we bother with Adiabatic Flame Temperature at all.

Ex 5 — Two limiting expansions: vacuum vs sea level (C1, C2)

Worked example Take the fuel-rich chamber of Ex 2 (

$T_{c} = 3300$ K, $M = 0.013$ , $γ = 1.2$ , $p_{c} = 70$ bar). Compute $I_{s p}$ for (C1) vacuum $p_{e} \to 0$ and (C2) sea level $p_{e} = 1.013$ bar.

Forecast: which is bigger, and by how much roughly?

Vacuum limit (C1). As $p_{e} / p_{c} \to 0$ , the term $(p_{e} / p_{c})^{0.1667} \to 0$ , bracket $\to 1$ (its maximum). $v_{e} = 12 \times 2.111 \times 1 0^{6} \times 1 = 2.533 \times 1 0^{7} = 5033$ m/s, so $I_{s p} = 5033/9.81 = 513$ s. Why this step? Vacuum is the best possible expansion — no back-pressure fights the exhaust, so every bit of thermal energy converts.
Sea-level (C2). $p_{e} / p_{c} = 1.013/70 = 0.01447$ . Then $0.0144 7^{0.1667} = 0.4841$ , bracket $= 0.5159$ . $v_{e} = 12 \times 2.111 \times 1 0^{6} \times 0.5159 = 1.307 \times 1 0^{7} = 3615$ m/s, so $I_{s p} = 3615/9.81 = 369$ s. Why this step? Higher exit pressure means the gas leaves before fully expanding — less energy extracted.

Verify: Vacuum ( $513$ s) $>$ sea level ( $369$ s), a $\sim 144$ s gap — exactly why upper stages use big vacuum nozzles and first stages don't. Note the earlier Ex 2 value ( $424$ s at $p_{e} / p_{c} = 0.001$ ) sits between these two limits ✓, because $0.001$ is between $0$ and $0.0145$ .

Ex 6 — Frozen vs equilibrium flow (D1)

Worked example Ex 2 case, but compare CEA's two nozzle modes.

Frozen: composition locked at the throat → the hand formula's $γ = 1.2$ , $M = 0.013$ hold all the way → $I_{s p} \approx 424$ s (that's Ex 2). Equilibrium: as the gas cools, $H + OH \to H_{2} O$ recombines, dumping extra heat downstream. Model this crudely as an effective $T_{c}$ raised by $8%$ (the recombination bonus): $T_{c}^{eff} = 3300 \times 1.08 = 3564$ K.

Forecast: equilibrium higher or lower? By roughly how many seconds?

Recompute prefactor with the bonus. $\frac{R _{u} T _{c}^{eff}}{M} = \frac{8.314 \times 3564}{0.013} = 2.279 \times 1 0^{6}$ . Why this step? Recombination refills the enthalpy pool (Dissociation at High Temperature run in reverse), which the scaling law reads as a higher effective $T_{c}$ .
Exit velocity. $v_{e} = 12 \times 2.279 \times 1 0^{6} \times 0.6838 = 1.870 \times 1 0^{7} = 4325$ m/s.
Specific impulse. $I_{s p} = 4325/9.81 = 441$ s.

Verify: Equilibrium ( $441$ s) $>$ frozen ( $424$ s), and it lands near the parent note's stated CEA value $\sim 450$ s ✓. Real engines sit between frozen and equilibrium because reactions have finite speed. Ordering: $frozen < real < equilibrium$ .

Ex 7 — Sensitivity: which knob matters more? (E1)

Worked example Starting from Ex 2, you can either (a) raise

$T_{c}$ by $10%$ (better preheating) or (b) drop $M$ by $10%$ (more fuel-rich). Both feed $T_{c} / M$ . Which gives more $I_{s p}$ , and are they equal?

Forecast: since $I_{s p} \propto T_{c} / M$ , guess whether $+ 10%$ on $T_{c}$ and $- 10%$ on $M$ move $I_{s p}$ by the same amount.

Option (a): factor $1.10 = 1.0488$ → $I_{s p} = 424 \times 1.0488 = 444.7$ s. Why this step? $T_{c}$ is in the numerator under the square root, so a $10%$ rise gives a $1.1$ boost.
Option (b): factor $1/0.90 = 1.1111 = 1.0541$ → $I_{s p} = 424 \times 1.0541 = 446.9$ s. Why this step? Cutting $M$ by $10%$ divides by $0.90$ , and $1/0.90 = 1.111 > 1.10$ , so the mass cut is slightly stronger.

Verify: Option (b) wins narrowly ( $446.9 > 444.7$ ). The asymmetry is real: a $- 10%$ change in a denominator ( $\times \frac{1}{0.9}$ ) is a bigger factor than $+ 10%$ in a numerator. This is why chasing low $M$ (light $H_{2}$ exhaust) is such a powerful lever. ✓

Ex 8 — Real-world word problem (F1)

Worked example A vacuum upper stage has

$I_{s p} = 450$ s (from a CEA run) and mass ratio $\frac{m _{0}}{m _{f}} = 4$ (start mass over dry mass). What $Δ v$ does it deliver? Use the Tsiolkovsky Rocket Equation: $Δ v = v_{e} ln \frac{m _{0}}{m _{f}}$ , where $v_{e} = I_{s p} g_{0}$ .

Forecast: for $I_{s p} = 450$ s and a $4 \times$ mass ratio, will $Δ v$ clear low-Earth-orbit speed ( $\sim 7.8$ km/s)?

Convert $I_{s p}$ to effective exhaust velocity. $v_{e} = 450 \times 9.81 = 4414.5$ m/s. Why this step? The rocket equation is written in terms of a velocity, not seconds; multiplying $I_{s p}$ by $g_{0}$ just undoes the definition $I_{s p} = v_{e} / g_{0}$ , recovering the exhaust speed the equation actually needs.
Take the logarithm of the mass ratio. $ln \frac{m _{0}}{m _{f}} = ln 4 = 1.3863$ . Why this step? The $ln$ appears because the rocket lightens as it burns — each kilogram of fuel is spent against an ever-smaller remaining mass, and summing that differential effect integrates to a natural logarithm.
Apply Tsiolkovsky. $Δ v = v_{e} ln \frac{m _{0}}{m _{f}} = 4414.5 \times 1.3863 = 6120$ m/s. Why this step? This multiplies the "how fast we throw mass" ( $v_{e}$ ) by the "how much mass fraction we throw" ( $ln$ term) to get the total velocity change the stage can produce.
Interpret against the target. $6120$ m/s $= 6.12$ km/s $< 7.8$ km/s. Why this step? Comparing the delivered $Δ v$ to the mission requirement is the whole point — it tells you whether this single stage can reach orbit.

Verify: $Δ v \approx 6.12$ km/s — not enough for LEO on its own ( $< 7.8$ km/s), so you'd stage or raise the mass ratio. Units: m/s $\times$ (dimensionless log) = m/s ✓. The whole CEA chain — Gibbs minimization $\to T_{c}$ and $M$ $\to I_{s p}$ $\to Δ v$ — closes here. ✓

Recall Self-check

Why does peak $I_{s p}$ sit fuel-rich of stoichiometric? ::: Fuel-rich lowers $M$ (leftover light $H_{2}$ ) faster than it lowers $T_{c}$ , and $I_{s p} \propto T_{c} / M$ . In the degenerate no-reaction case, what is $T_{c}$ ? ::: It equals the inlet temperature $T_{in}$ — no combustion enthalpy is released. Which nozzle expansion gives the largest $v_{e}$ , vacuum or sea level? ::: Vacuum: as $p_{e} / p_{c} \to 0$ the bracket $\to 1$ , its maximum. Order frozen, equilibrium, and real-engine $I_{s p}$ . ::: frozen $<$ real $<$ equilibrium (recombination adds heat downstream). A $10%$ cut in $M$ vs a $10%$ rise in $T_{c}$ — which boosts $I_{s p}$ more? ::: The $M$ cut, because $1/0.9 = 1.111 > 1.10$ .

Related vault notes: Gibbs Free Energy and Equilibrium · Chemical Equilibrium and Kp · Hess's Law and Enthalpy of Formation · Rocket Nozzle and Thrust Equation · back to parent the CEA topic note.