Exercises — Thrust coefficient C_F = F - (P_c A - ) — derivation

WHAT we do: plug straight into the definition — no nozzle geometry needed, just three measured numbers. WHY: $C_F$ is defined as thrust divided by the "reference force" $P_c{A}^{\ast }$. This is the fastest way to grade a real test firing. $P_c{A}^{\ast }=5\times 10^6\times 0.008=4.0\times 10^4\text{N}=40000\text{N}$ $C_F=\frac{90000}{40000}=\boxed{2.25}$ A high value — this nozzle is doing a lot of amplifying.

False. $C_F=\frac{F}{P_c{A}^{\ast }}=\frac{\text{N}}{(\text{Pa})({\text{m}}^2)}=\frac{\text{N}}{({\text{N/m}}^2)({\text{m}}^2)}=\frac{\text{N}}{\text{N}}$ — force over force, so it is dimensionless. It is a pure amplification factor.

WHAT we do: break the square root into a constant $K$ times a bracket, then multiply. WHY split it: the constant $K$ depends only on $\gamma$ and is reused across many problems; the bracket carries the expansion. Constant: $\frac{2{\gamma }^2}{\gamma -1}=\frac{2(1.44)}{0.2}=14.4,{\left(\frac{2}{2.2}\right)}^{\frac{2.2}{0.2}}=(0.9091{)}^{11}=0.3505$ $K=14.4\times 0.3505=5.048$ Bracket (exponent $\frac{\gamma -1}{\gamma }=\frac{0.2}{1.2}=0.1667$): $1-(0.02{)}^{0.1667}=1-0.5214=0.4786$ $C_{F,\text{mom}}=\sqrt{5.048\times 0.4786}=\sqrt{2.416}=\boxed{1.554}$

WHAT we do: add the pressure term to the momentum term. WHY: exit pressure $P_e$ differs from ambient $P_a$, so the unbalanced pressure over the exit area contributes. $\frac{P_e-P_a}{P_c}\frac{A_e}{A^{\ast }}=\frac{0.14-0.10}{7}\times 25=\frac{0.04}{7}\times 25=+0.1429$ The term is positive here because $P_e>P_a$ (slightly under-expanded — see Over- and Under-Expanded Nozzles). $C_F=1.554+0.143=\boxed{1.697}$

WHAT we do: only the pressure term changes with altitude; $C_{F,\text{mom}}$ is fixed by geometry. (a) Sea level: $\frac{0.05-0.10}{6}\times 30=\frac{-0.05}{6}\times 30=-0.25\Rightarrow C_F=1.60-0.25=\boxed{1.35}$ (b) Vacuum: $\frac{0.05-0}{6}\times 30=\frac{0.05}{6}\times 30=+0.25\Rightarrow C_F=1.60+0.25=\boxed{1.85}$ WHY the flip: at sea level the outside air ($0.1$ MPa) is stronger than the exhaust exit ($0.05$ MPa), so ambient pressure pushes back on the exit — a penalty. In vacuum there is nothing pushing back, so the same exit pressure becomes a bonus. Same nozzle gains $\Delta C_F=0.50$ just by climbing to space.

What the figure shows: the curve below plots $C_F$ against ambient pressure $P_a$ for this fixed nozzle. The cyan line slopes downward as $P_a$ rises (more outside push-back). The dashed white line marks the constant momentum coefficient $1.60$; where the cyan line crosses it (dotted vertical, at $P_a=P_e=0.05$ MPa) the pressure term is exactly zero. The two amber dots mark our answers: the vacuum bonus (upper-left, $C_F=1.85$) and the sea-level penalty (lower-right, $C_F=1.35$).

WHAT we do: divide the pressure term by the total $C_F$. WHY: this ratio tells us how much of the nozzle's grade is bought by the pressure bonus versus raw momentum — it justifies (or not) the extra weight of a big bell. $\frac{\text{pressure term}}{C_F}=\frac{0.25}{1.85}=0.135=\boxed{13.5\%}$ Momentum thrust dominates (86.5%), but the pressure bonus is why space engines carry giant bells: adding area $A_e/A^{\ast }$ in vacuum only ever helps.

(a) Momentum term. $K=5.048$ (from 2.1, same $\gamma$). Exponent $0.1667$; $P_e/P_c=0.1/7=0.01429$. WHY the bracket: it measures how much of the gas's energy the expansion has converted to speed — a smaller $P_e/P_c$ means a fuller expansion and a bracket closer to 1. $1-(0.01429{)}^{0.1667}=1-0.4926=0.5074,C_{F,\text{mom}}=\sqrt{5.048\times 0.5074}=\sqrt{2.561}=1.600$ (b) Pressure term (vacuum, $P_a=0$). WHY $P_a=0$: in vacuum nothing pushes back on the exit, so the entire exit pressure $P_e$ becomes a thrust bonus. $\frac{P_e-0}{P_c}\frac{A_e}{A^{\ast }}=\frac{0.1}{7}\times 25=0.3571$ (c) Add the two contributions — the definition of the full coefficient is momentum term plus pressure term: $C_F=1.600+0.357=\boxed{1.957}$ (d) $F=C_F{P}_c{A}^{\ast }=1.957\times (7\times 10^6)\times 0.01=\boxed{1.370\times 10^5\text{N}\approx 137\text{kN}}$. WHY (d) works: $C_F$ is by construction the multiplier on the reference force $P_c{A}^{\ast }$. Multiply and you recover thrust — no need to know $T_c$, $R$, or $\dot{m}$. That independence from $T_c$ is the whole reason $C_F$ exists (see Characteristic Velocity c-star for the part that does carry $T_c$).

WHAT we do: eliminate $\dot{m}$ between the two relations. $I_{sp}=\frac{F}{\dot{m}{g}_0}=\frac{C_F{c}^{\ast }\dot{m}}{\dot{m}{g}_0}=\frac{C_F{c}^{\ast }}{g_0}$ $I_{sp}=\frac{1.85\times 1700}{9.81}=\frac{3145}{9.81}=\boxed{320.6\text{s}}$ WHY it factors so cleanly: exhaust speed $=C_F{c}^{\ast }$; dividing by $g_0$ converts it to seconds. This shows the clean split — nozzle score $C_F$ times combustion score $c^{\ast }$ gives performance $I_{sp}$. See Specific Impulse Isp.

WHAT we do: set the bracket $[1-(P_e/P_c{)}^{(\gamma -1)/\gamma }]\to 1$ (since $P_e\to 0$), leaving only $\sqrt{K}$. $C_{F,\max }=\sqrt{K}=\sqrt{5.048}=\boxed{2.247}$ WHY this is a ceiling: even an infinitely long nozzle can only extract the gas's full available energy once. What remains is set purely by $\gamma$ — the gas's own thermodynamic springiness. No geometry beats $\sqrt{K}$; a softer gas (lower $\gamma$) has a higher ceiling because more of its energy is convertible to directed motion.

The argument. Making the nozzle longer (bigger $A_e/A^{\ast }$) lowers $P_e$. This raises the momentum term (bigger bracket) but the pressure term $\frac{P_e-P_a}{P_c}\frac{A_e}{A^{\ast }}$ turns negative once $P_e<P_a$. The two effects trade off, and calculus shows the total $C_F$ peaks exactly when $P_e=P_a$ — the pressure term crosses zero there. Physically: any exit pressure mismatch wastes energy (either un-expanded pressure left over, or over-expansion pushing sideways). Numeric confirmation for 3.1's engine at sea level ($P_a=0.1$ MPa). If we could re-tune the nozzle so $P_e=P_a=0.1$ MPa, the pressure term vanishes and $C_F=C_{F,\text{mom}}$ at that new (larger) matched expansion. With the given fixed geometry $P_e=0.05<P_a$, the engine is over-expanded and loses $0.25$ (from 3.1a), confirming the un-matched case sits below the matched optimum: $C_{F,\text{sea, actual}}=1.35<C_{F,\text{mom, matched}}=\mathrm{1.60.}$

Bracket at $P_e/P_c=0.10$: $1-(0.10{)}^{0.1667}=1-0.6813=0.3187$, so $C_{F,\text{mom}}=\sqrt{5.048\times 0.3187}=\sqrt{1.609}=1.269$. Bracket at $P_e/P_c=0.01$: $1-(0.01{)}^{0.1667}=1-0.4642=0.5358$, so $C_{F,\text{mom}}=\sqrt{5.048\times 0.5358}=\sqrt{2.705}=1.645$. Factor: $1.645/1.269=\boxed{1.296}$ — about a 30% gain in momentum coefficient for a tenfold drop in exit pressure. Diminishing but real; this is why vacuum bells grow so large.

Why is $C_F$ independent of chamber temperature $T_c$? ::: Because the $\sqrt{R{T}_c}$ from mass flow cancels the $\sqrt{R{T}_c}$ from exit velocity when you divide by $P_c{A}^{\ast }$ — $C_F$ sees only ratios ($P_e/P_c$, $P_a/P_c$) and geometry ($A_e/A^{\ast }$), never absolute temperature.