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

False. At fixed temperature and pressure the Second Law drives the system to minimum Gibbs free energy $G$, not maximum entropy — entropy maximization is the criterion at fixed internal energy and volume, a different problem.

False. High temperature drives dissociation, so a large fraction sits as ${\text{H}}_2$, $\text{OH}$, $\text{H}$, $\text{O}$, and ${\text{O}}_2$ coexisting in equilibrium — treating it as pure water overpredicts $T_c$ and mis-estimates $M$.

True, but that is the point — leftover cold ${\text{H}}_2$ soaks up heat so $T_c$ drops, yet the average molar mass $M$ drops faster, and since $I_{sp}\propto \sqrt{T_c/M}$ the net effect raises $I_{sp}$.

False. It is one equation per species, coupled through the shared elemental potentials ${\lambda }_j$; together they reproduce the $K_p$ relation for every possible reaction at once.

Almost always true — in equilibrium mode exothermic recombination ($\text{H}+\text{OH}\to {\text{H}}_2\text{O}$) keeps releasing heat as the gas cools in the nozzle, refilling the enthalpy pool that becomes exhaust velocity.

False. From $I_{sp}=v_e/g_0$ the seconds are just thrust per unit weight-flow with the mass-flow cancelled (units $(\text{m/s})/({\text{m/s}}^2)=\text{s}$); it is a size-independent efficiency figure, not a duration.

False. You supply only the elements present (and a thermo library); CEA considers every species in its database containing those elements and lets the Gibbs minimization decide which survive in meaningful amounts.

False. The hot chamber uses constant enthalpy and pressure (HP), while isentropic nozzle expansion uses constant entropy and pressure (SP) — different constraints for different physics.

The conservation constraint fixes total atoms of each element (${\sum }_i{a}_{ij}{n}_i=b_j$), not the individual $n_i$ — the $n_i$ are exactly the unknowns CEA is free to shuffle to minimize $G$.

The correct scaling is $I_{sp}\propto \sqrt{T_c/M}$; ignoring the exhaust molar mass $M$ hides why a cooler but lighter fuel-rich mixture can beat the hottest stoichiometric one.

That term (with $p_i$ the partial pressure and $p^{\circ }=1$ bar the reference) encodes dilution and mixing entropy — dropping it kills dissociation entirely, since it is the very thing that makes a dilute species' escaping tendency low enough to form.

Peak $I_{sp}$ sits on the fuel-rich side, near $O/F\approx 6$, because the extra ${\text{H}}_2$ lowers $M$ faster than it lowers $T_c$; the hottest flame ($O/F=8$) and the best $I_{sp}$ are different operating points.

Both matter — the energy balance ${\sum }_i{n}_i{h}_i(T_c)=H_{\text{reactants}}$ uses $h_i={\Delta }_f{h}_i^{\circ }+\int c_{p,i}dT$ for every product too, exactly Hess's-law bookkeeping applied at $T_c$.

Isentropic means reversible-adiabatic, not chemically frozen — in equilibrium mode the composition keeps re-solving as pressure drops, and that shifting chemistry is where the recombination bonus comes from.

Real engines fall between frozen and equilibrium bounds — reactions are finite-rate, not fully frozen and not instantaneous, so frozen is a lower estimate, not the truth.

Choosing an independent reaction set by hand is arbitrary and error-prone; minimizing $G$ subject to atom conservation automatically enforces all $K_p$ relations simultaneously through the shared ${\lambda }_j$ multipliers.

In $v_e=\sqrt{(2\gamma /(\gamma -1))(R_u{T}_c/M)[1-(p_e/p_c{)}^{(\gamma -1)/\gamma }]}$ the factor $R_u{T}_c/M$ is energy per unit mass; lighter molecules must move faster to carry the same energy, so $v_e\propto 1/\sqrt{M}$.

Lowering $T$ shifts equilibrium, which changes how much enthalpy is released, which changes $T$ — the adiabatic flame temperature is the self-consistent value where composition and energy balance agree.

The released heat re-raises the enthalpy the flow can convert to directed kinetic energy as it keeps expanding, so it feeds velocity — the payoff the nozzle equation turns into thrust.

$I_{sp}$ sets the effective exhaust velocity $v_e=g_0{I}_{sp}$, which is the single propellant parameter in the Tsiolkovsky rocket equation governing how much velocity change a given propellant mass buys.

Because $I_{sp}$ also depends on the exhaust molar mass $M$; a hydrogen-rich mixture has tiny $M$ and wins even at equal flame temperature.

No exothermic chemistry is possible, so $T_c$ stays at the input temperature and the "products" are just the input species — the Gibbs minimization has nothing to rearrange.

No — even with perfectly balanced atoms, the high $T_c$ still dissociates products into $\text{OH}$, $\text{H}$, $\text{O}$, etc., because equilibrium at that temperature favors the fragments; complete combustion is a low-temperature idealization.

The bracket $[1-(p_e/p_c{)}^{(\gamma -1)/\gamma }]\to 0$, so $v_e\to 0$ and $I_{sp}\to 0$ — with no pressure drop the nozzle extracts no directed kinetic energy.

The bracket approaches $1$, giving the maximum ideal $v_e=\sqrt{(2\gamma /(\gamma -1))(R_u{T}_c/M)}$ — a finite ceiling set by chamber conditions, not infinity.

No — eventually $T_c$ collapses because too little oxidizer releases too little heat, so past the peak $I_{sp}$ falls again; the best point is a balance, not an endpoint.

$T_c$ would drop below the reactant inlet temperature, since the products would absorb enthalpy rather than release it — the same conserved-enthalpy equation runs in reverse.

Yes — CEA handles condensed-phase equilibria too, so fuel-rich hydrocarbon runs can deposit solid carbon (soot) and cooler mixtures can form liquid water or condensed oxides; a condensed phase has $\mu$ fixed by its own standard state (no partial-pressure log term) and appears in the results with a special phase flag.

Condensed carbon carries mass but contributes little to gas expansion, so it raises the effective exhaust $M$ and wastes energy locking up atoms in a solid — CEA flags where increasing fuel-richness starts depositing carbon, which is often where practical $I_{sp}$ gains stop.