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

The one formula that runs this whole page

Before any exercise, pin down the exhaust-velocity relation, because almost every problem leans on it. We met it in the parent note; here is a plain-words re-read so no symbol is a stranger.

Intuition WHERE the exponent

$\frac{γ - 1}{γ}$ comes from (why raise the pressure ratio to that power?) This is the one non-obvious step in the whole formula, so let's earn it. For isentropic (loss-free, constant-entropy) flow of an ideal gas, temperature and pressure are locked together by the relation $\frac{T _{e}}{T _{c}} = (\frac{p _{e}}{p _{c}})^{\frac{γ - 1}{γ}} .$ Why this exact power? It falls out of combining two ideal-gas facts: (i) along an isentrope $p V^{γ} = const$ (pressure and volume trade off with exponent $γ$ ), and (ii) the ideal-gas law $p V \propto T$ . Eliminating $V$ between them leaves $T \propto p^{(γ - 1) / γ}$ . So the exponent is not arbitrary — it is precisely the bookkeeping that keeps entropy constant as the gas drops in pressure. Why we need it here: the energy balance says the kinetic energy gained equals the thermal energy lost, $\frac{1}{2} v_{e}^{2} = c_{p} (T_{c} - T_{e})$ . Substituting $T_{e} = T_{c} (p_{e} / p_{c})^{(γ - 1) / γ}$ turns "temperature drop" into "pressure drop," giving the bracket $[1 - (p_{e} / p_{c})^{(γ - 1) / γ}]$ . That is why the pressure ratio is raised to this particular power: it is the isentropic map from pressure-drop to temperature-drop, i.e. from pressure-drop to usable speed.

Definition Where this formula is allowed to be used (domain of validity)

The ideal-nozzle relation assumes steady, adiabatic, isentropic (loss-free) flow of an ideal gas through a converging–diverging nozzle with fixed $γ$ and $M$ . Three edge behaviours you must respect:

Choked flow / critical pressure ratio. Supersonic exhaust only exists once the flow "chokes" at the throat, which needs $p_{e} / p_{c}$ below the critical ratio $(\frac{2}{γ + 1})^{\frac{γ}{γ - 1}}$ . For $γ = 1.2$ this is $\approx 0.564$ . Every exercise here uses ratios far below that ( $0.01, 0.001$ ), so the flow is comfortably choked and supersonic — the formula applies.
$p_{e} / p_{c} \to 0$ (vacuum limit). The bracket saturates toward $1$ ; $v_{e}$ approaches a finite ceiling $\frac{2 γ}{γ - 1} \frac{R _{u} T _{c}}{M}$ . You cannot get infinite speed from finite heat.
Over- vs under-expanded. The formula gives the ideal $v_{e}$ assuming the nozzle exit pressure matches the design $p_{e}$ . If ambient pressure is higher than $p_{e}$ the jet is over-expanded (flow separates, thrust lost); if lower, under-expanded (jet keeps spreading outside, some thrust unrecovered). In vacuum ( $p_{amb} = 0$ ) you always want $p_{e}$ as small as practical.

Look at the figure below: it is the picture behind every $I_{s p}$ number on this page.

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

The horizontal axis is the pressure ratio $p_{e} / p_{c}$ (plotted on a log scale, expansion increasing to the right), and the vertical axis is the energy-cash-in bracket $1 - (p_{e} / p_{c})^{(γ - 1) / γ}$ — the fraction of the gas's thermal energy the nozzle turns into directed speed. The two marked dots are the Level-2 operating points ( $0.01$ and $0.001$ ); notice how the curve is already flattening near $1$ , which is exactly why deeper expansion gives diminishing returns (Exercise 2.3). The dashed marker shows the critical pressure ratio below which the flow is choked.

Level 1 — Recognition

Exercise 1.1

CEA hands you four outputs from a run: $T_{c}$ , a list of $X_{i}$ , $I_{s p}$ , and $c^{*}$ . Match each to its plain-English meaning: (a) chamber flame temperature, (b) the equilibrium "soup" of species as mole fractions, (c) thrust per unit weight-flow (efficiency, in seconds), (d) characteristic velocity (a chamber-only quality number).

Recall Solution

$T_{c} \leftrightarrow$ (a) chamber flame temperature.
$X_{i} \leftrightarrow$ (b) product mole fractions — the equilibrium mixture.
$I_{s p} \leftrightarrow$ (c) thrust per unit weight-flow, in seconds.
$c^{*} \leftrightarrow$ (d) characteristic velocity, a chamber-only figure independent of the nozzle.

Why: $T_{c}$ and $X_{i}$ come out of the chamber (Gibbs minimization + energy balance). $I_{s p}$ needs the nozzle (it depends on $v_{e}$ ). $c^{*}$ isolates chamber quality by dividing out the nozzle.

Exercise 1.2

State, in one line each, the two things CEA holds fixed while it minimizes Gibbs free energy $G$ to find the chamber products.

Recall Solution

First, a reminder of the three symbols in the constraint (built in the parent note):

$n_{i}$ = the number of moles of species $i$ present (the unknowns CEA solves for).
$a_{ij}$ = how many atoms of element $j$ sit inside one molecule of species $i$ (e.g. for $H_{2} O$ , $a = 2$ for H and $1$ for O).
$b_{j}$ = the total moles of element $j$ the propellant supplied to the chamber.

The two things held fixed:

Conservation of each element: $\sum_{i} a_{ij} n_{i} = b_{j}$ for every element $j$ (atoms can't be created or destroyed).
Constant temperature and pressure framing of $G$ (the chamber is treated at fixed $p$ ; the enthalpy balance then pins $T_{c}$ ).

Everything else — how many of each molecule — is free to rearrange to make $G$ smallest.

Level 2 — Application

Exercise 2.1

A run reports $v_{e} = 3600 m/s$ . Find $I_{s p}$ in seconds.

Recall Solution

$I_{s p} = \frac{v _{e}}{g _{0}} = \frac{3600}{9.81} \approx 367 s .$ What we did: divided exhaust speed by standard gravity to convert m/s into the "seconds" efficiency figure.

Exercise 2.2

A methane/oxygen engine gives $T_{c} = 3500 K$ , exhaust $\overset{ˉ}{M} = 22 g/mol = 0.022 kg/mol$ , $γ = 1.15$ , and expands to $p_{e} / p_{c} = 0.01$ . Estimate $v_{e}$ and $I_{s p}$ with the ideal-nozzle formula.

Recall Solution

Build it in three chunks (this is exactly how the parent note's Example 2 does it):

Front factor $\frac{2 γ}{γ - 1} = \frac{2 ( 1.15 )}{0.15} = 15.33$ .
Energy term $\frac{R _{u} T _{c}}{M} = \frac{8.314 \cdot 3500}{0.022} = 1.323 \times 1 0^{6} J/kg$ .
Exponent $\frac{γ - 1}{γ} = \frac{0.15}{1.15} = 0.1304$ . Here we use the isentropic map (built above): raising the pressure ratio to this power converts pressure-drop into the fraction of temperature (hence energy) released. Numerically $0.0 1^{0.1304} = 0.5485$ , so bracket $= 1 - 0.5485 = 0.4515$ .

$v_{e} = 15.33 \cdot 1.323 \times 1 0^{6} \cdot 0.4515 = 9.156 \times 1 0^{6} \approx 3026 m/s .$ $I_{s p} = \frac{3026}{9.81} \approx 308 s .$

Exercise 2.3

Same engine as 2.2 but you now expand deeper, to $p_{e} / p_{c} = 0.001$ (near-vacuum). Recompute the bracket and the new $v_{e}$ . By how much does $v_{e}$ rise?

Recall Solution

Only the bracket changes: $0.00 1^{0.1304} = 0.4062$ , so bracket $= 1 - 0.4062 = 0.5938$ . $v_{e} = 15.33 \cdot 1.323 \times 1 0^{6} \cdot 0.5938 = 1.2043 \times 1 0^{7} \approx 3470 m/s .$ Increase: $3470 - 3026 \approx 444 m/s$ (about $+ 14.7%$ ). Why the gain, and why it still tapers: deeper expansion (smaller $p_{e} / p_{c}$ ) pushes the isentropic temperature drop further, so the bracket climbs from $0.45$ toward $0.59$ . But because the exponent $0.1304$ is small, the bracket approaches its ceiling of $1$ ever more slowly — the flattening curve in the figure above shows the diminishing returns.

Level 3 — Analysis

Exercise 3.1

Two engines have the same $T_{c} = 3300 K$ and same $γ, p_{e} / p_{c}$ . Engine A's exhaust averages $M = 18 g/mol$ (water-rich); engine B is run fuel-rich and averages $M = 13 g/mol$ (leftover $H_{2}$ ). Using $I_{s p} \propto T_{c} / M$ , find the ratio $I_{s p, B} / I_{s p, A}$ .

Recall Solution

Since $T_{c}, γ, p_{e} / p_{c}$ are shared, only $M$ differs, so $\frac{I _{s p, B}}{I _{s p, A}} = \frac{M _{A}}{M _{B}} = \frac{18}{13} = 1.385 \approx 1.177.$ What this says: the lighter exhaust is $\approx 18%$ more efficient at the same temperature. This is the raw driving force behind running rockets fuel-rich.

Exercise 3.2

Now be honest: going fuel-rich also cools the flame. Suppose B's fuel-rich burn drops $T_{c}$ from $3300 K$ to $3000 K$ while $M$ falls $18 \to 13$ . Does B still beat A? Compute the full ratio $(T_{c, B} / M_{B}) / (T_{c, A} / M_{A})$ .

Recall Solution

$\frac{I _{s p, B}}{I _{s p, A}} = \frac{T _{c, B} / M _{B}}{T _{c, A} / M _{A}} = \frac{3000/13}{3300/18} = \frac{230.8}{183.3} = 1.259 \approx 1.122.$ B still wins by $\approx 12%$ . The lesson: the $M$ drop ( $18 \to 13$ , a factor $0.72$ ) beats the $T_{c}$ drop ( $3300 \to 3000$ , a factor $0.91$ ) inside the square root. The peak of $I_{s p}$ therefore sits on the fuel-rich side of stoichiometric — see the curve in the figure below.

The horizontal axis is the mixture ratio $O / F$ (mass basis; moving left means more fuel-rich), and the vertical axis is the $I_{s p}$ proxy $T_{c} / M$ . The blue curve rises as we go fuel-rich and only turns over well past stoichiometric — the pink dashed line marks $O / F = 8$ (stoichiometric, hottest flame but not the best $I_{s p}$ ), and the yellow arrow marks the true peak sitting on the fuel-rich shoulder. This is the picture behind Exercises 3.1, 3.2 and 5.1.

Exercise 3.3

Explain in terms of the master equilibrium condition $μ_{i} = \sum_{j} λ_{j} a_{ij}$ why raising $T_{c}$ produces more dissociated species (H, OH, O) even though it makes the flame hotter and hotter is "good."

Recall Solution

Reminder of the symbols (from the parent note's Lagrange derivation):

$μ_{i}$ = chemical potential of species $i$ — its "escaping tendency," the marginal Gibbs energy of adding one more mole of $i$ .
$λ_{j}$ = the elemental potential (Lagrange multiplier) of element $j$ — a bookkeeping "price per atom" of element $j$ that enforces atom conservation.
$a_{ij}$ = atoms of element $j$ inside species $i$ (same as Exercise 1.2).

The chemical potential of a gas is $μ_{i} = μ_{i}^{\circ} (T) + R T ln (p_{i} / p^{\circ})$ . Dissociation (e.g. $H_{2} O \to OH + H$ ) increases the number of independent gas particles, and the $R T ln$ mixing term rewards spreading into more species — that reward grows with $T$ . So the composition that satisfies $μ_{i} = \sum_{j} λ_{j} a_{ij}$ shifts toward more dissociated fragments as $T$ rises. The catch: dissociation is endothermic, so it eats energy that would otherwise show up as temperature. That is why real $T_{c}$ is capped well below the naive "no-dissociation" adiabatic flame temperature — a direct link to Dissociation at High Temperature and Adiabatic Flame Temperature.

Level 4 — Synthesis

Exercise 4.1 (CEA + Tsiolkovsky)

An upper stage uses the Exercise-2.2 engine but at deep expansion giving $I_{s p} = 430 s$ . Its wet mass is $12000 kg$ and dry mass $3000 kg$ . Using the Tsiolkovsky Rocket Equation $Δ v = v_{e} ln (m_{0} / m_{f})$ with $v_{e} = I_{s p} g_{0}$ , find the achievable $Δ v$ .

Recall Solution

Exhaust velocity: $v_{e} = I_{s p} g_{0} = 430 \cdot 9.81 = 4218.3 m/s$ . Mass ratio: $m_{0} / m_{f} = 12000/3000 = 4$ . $Δ v = v_{e} ln (4) = 4218.3 \times 1.3863 = 5848 m/s .$ How the pieces link: CEA gives $I_{s p}$ → that gives $v_{e}$ → Tsiolkovsky turns $v_{e}$ and the mass ratio into mission $Δ v$ . This is the whole "propulsion bridge" in one line.

Exercise 4.2 (equilibrium vs frozen bonus)

A CEA run reports frozen $I_{s p} = 418 s$ and equilibrium $I_{s p} = 451 s$ for the same case. (a) What physical process accounts for the extra $33$ s? (b) A real engine is measured at $I_{s p} = 440 s$ ; express its performance as a fraction of the way from frozen to equilibrium.

Recall Solution

(a) In equilibrium expansion, exothermic recombination (e.g. $H + OH \to H_{2} O$ ) keeps releasing heat as the gas cools in the nozzle, refilling the enthalpy pool that becomes exhaust kinetic energy. Frozen flow forbids this, so it loses that "recombination bonus." (b) Fraction $= \frac{440 - 418}{451 - 418} = \frac{22}{33} = 0.667$ , i.e. the real engine sits $\approx 67%$ of the way from frozen toward equilibrium — reasonable, since finite reaction rates keep it below the ideal equilibrium bound.

Exercise 4.3 (Hess + energy balance sanity check)

For the constant-pressure adiabatic chamber, $\sum_{i} n_{i} h_{i} (T_{c}) = H_{reactants} (T_{in})$ with $h_{i} = Δ_{f} h_{i}^{\circ} + \int_{T^{\circ}}^{T} c_{p, i} d T$ . Using Hess's Law and Enthalpy of Formation, suppose burning $1$ mol of fuel releases $Δ_{r} H = - 900 kJ$ at $298 K$ , and the products' total heat capacity is a constant $\overset{ˉ}{C}_{p} = 0.30 kJ K^{- 1}$ . Estimate $T_{c}$ if all the released enthalpy heats the products from $298 K$ (ignore dissociation).

Recall Solution

Energy freed heats the products: $- Δ_{r} H = \overset{ˉ}{C}_{p} (T_{c} - 298)$ . $T_{c} = 298 + \frac{900}{0.30} = 298 + 3000 = 3298 K .$ What we did: balanced "heat of reaction released" against "heat needed to warm products." Why the real CEA value is lower: this ignores the endothermic dissociation that steals energy at high $T$ — so a real CEA $T_{c}$ would come out a few hundred K below this idealized $\sim 3300 K$ .

Level 5 — Mastery

Exercise 5.1 (design an $O / F$ scan)

You want peak $I_{s p}$ for $H_{2} / O_{2}$ . You run CEA at $O / F = 4, 6, 8$ (mass) and read off, at fixed $p_{c}$ and expansion:

$O / F$	$T_{c}$ (K)	$\overset{ˉ}{M}$ (g/mol)
4	3000	10
6	3300	13
8	3550	18

Rank them by $I_{s p}$ using the proxy $T_{c} / M$ (same $γ$ , same expansion). Which $O / F$ wins, and what does that tell you about where to place a finer scan?

Recall Solution

Units note: we only need to rank, and every row uses the same units for $M$ , so the g/mol vs kg/mol distinction cancels in the ratio — the proxy numbers below use g/mol purely for compactness. (If you wanted an actual $v_{e}$ you would still convert $M$ to kg/mol, as flagged at Level 2.)

Compute $T_{c} / M$ then its square root (proportional to $I_{s p}$ ):

$O / F = 4 : 3000/10 = 300, 300 = 17.32.$
$O / F = 6 : 3300/13 = 253.8, 253.8 = 15.93.$
$O / F = 8 : 3550/18 = 197.2, 197.2 = 14.04.$

Ranking (best first): $O / F = 4 > 6 > 8$ . The proxy keeps rising as we go more fuel-rich, so the true peak lies at or below $O / F = 4$ — place your finer scan around $O / F = 3 - 5$ . Design lesson: stoichiometric ( $O / F = 8$ ) is the worst of the three for $I_{s p}$ despite the hottest flame — exactly the fuel-rich-wins pattern.

Exercise 5.2 (judge two propellants)

Propellant X: $T_{c} = 3300 K$ , $\overset{ˉ}{M} = 13 g/mol$ . Propellant Y: $T_{c} = 3900 K$ , $\overset{ˉ}{M} = 22 g/mol$ . Same $γ = 1.2$ , same $p_{e} / p_{c} = 0.001$ . Compute each $v_{e}$ (full formula) and decide which gives more $I_{s p}$ . Then state the one-sentence physical reason.

Recall Solution

Shared pieces: $\frac{2 γ}{γ - 1} = \frac{2.4}{0.2} = 12$ ; exponent $\frac{γ - 1}{γ} = \frac{0.2}{1.2} = 0.1667$ ; $0.00 1^{0.1667} = 0.3162$ , bracket $= 1 - 0.3162 = 0.6838$ . (Remember to use $M$ in kg/mol.)

X: $\frac{R _{u} T _{c}}{M} = \frac{8.314 \cdot 3300}{0.013} = 2.111 \times 1 0^{6} J/kg$ . $v_{e} = 12 \cdot 2.111 \times 1 0^{6} \cdot 0.6838 = 1.732 \times 1 0^{7} \approx 4162 m/s \Rightarrow I_{s p} = \frac{4162}{9.81} \approx 424 s .$

Y: $\frac{R _{u} T _{c}}{M} = \frac{8.314 \cdot 3900}{0.022} = 1.474 \times 1 0^{6} J/kg$ . $v_{e} = 12 \cdot 1.474 \times 1 0^{6} \cdot 0.6838 = 1.209 \times 1 0^{7} \approx 3477 m/s \Rightarrow I_{s p} = \frac{3477}{9.81} \approx 354 s .$

X wins ( $424$ s vs $354$ s). Reason (one sentence): X's much lower molar mass ( $13$ vs $22$ ) beats Y's hotter flame, because inside $T_{c} / M$ the mass ratio $13/22 \approx 0.59$ hurts Y more than its temperature advantage $3900/3300 \approx 1.18$ helps.

Exercise 5.3 (close the loop with $Δ v$ )

Take the winning propellant X from 5.2 ( $I_{s p} = 424 s$ ). A stage has mass ratio $m_{0} / m_{f} = 5$ . Find $Δ v$ . Then say how much $Δ v$ you'd lose if a poor nozzle knocked $I_{s p}$ down to $400 s$ .

Recall Solution

$v_{e} = 424 \cdot 9.81 = 4159.4 m/s$ ; $ln 5 = 1.6094$ . $Δ v = 4159.4 \cdot 1.6094 = 6694 m/s .$ At $I_{s p} = 400 s$ : $v_{e} = 3924 m/s$ , $Δ v = 3924 \cdot 1.6094 = 6316 m/s$ . Loss: $6694 - 6316 \approx 378 m/s$ . Takeaway: a $24$ -second $I_{s p}$ slip costs nearly $0.4 km/s$ — often the difference between reaching orbit and not.

Recall Self-test cloze

The efficiency figure CEA outputs, in seconds, is ::: $I_{s p} = v_{e} / g_{0}$ The two effects that trade off to place the $I_{s p}$ peak fuel-rich are ::: rising $T_{c}$ vs rising $M$ (optimize $T_{c} / M$ ) The extra $I_{s p}$ of equilibrium over frozen flow comes from ::: exothermic recombination releasing heat downstream in the nozzle The equation that turns $v_{e}$ and mass ratio into mission $Δ v$ is ::: Tsiolkovsky $Δ v = v_{e} ln (m_{0} / m_{f})$

Related builds: Gibbs Free Energy and Equilibrium, Chemical Equilibrium and Kp, Rocket Nozzle and Thrust Equation, and the parent CEA topic note.