Exercises — Effective exhaust velocity c = v_e + (P_e − P_a)A_e - ṁ

(a) $v_e$ — the true velocity of gas at the nozzle exit. (b) $P_e$ (exit static pressure) and $P_a$ (ambient pressure) — both absolute, since only their difference is physically meaningful and mixing gauge/absolute corrupts that difference. (c) $A_e$ — the nozzle exit area, in ${\text{m}}^2$. (d) $c$ is the effective exhaust velocity: the single make-believe speed such that $F=\dot{m}c$ gives the total thrust (momentum plus pressure).

Optimal expansion means $P_e=P_a$, so $(P_e-P_a)=0$. The whole pressure term is $0$. Therefore $c=v_e.$ All thrust is momentum thrust; there is no orange piece in our bar picture.

Pressure term: $\frac{(P_e-P_a)A_e}{\dot{m}}=\frac{(90000-50000)(1.5)}{250}=\frac{60000}{250}=240\text{m/s}.$ Since $P_e>P_a$, this is underexpanded → positive correction → $c>v_e$. $c=2900+240=3140\text{m/s}.$

$\frac{(90000-101300)(1.5)}{250}=\frac{(-11300)(1.5)}{250}=\frac{-16950}{250}=-67.8\text{m/s}.$ $P_e<P_a$ → overexpanded → ambient air pushes back → negative correction. $c=2900-67.8=2832.2\text{m/s}.$

$F=\dot{m}c=250\times 3140=785000\text{N}=785\text{kN}.$

Rearrange the definition (subtract the orange piece back off): $v_e=c-\frac{(P_e-P_a)A_e}{\dot{m}}=4200-\frac{(40000)(0.9)}{120}=4200-\frac{36000}{120}=4200-300=3900\text{m/s}.$ The measured $c$ already contained a $+300$ m/s pressure boost, so the true gas speed is lower.

The pressure term equals $c-v_e=3070-3000=70\text{m/s}$. So $\frac{(P_e-P_a)A_e}{\dot{m}}=70\Rightarrow A_e=\frac{70\dot{m}}{P_e-P_a}=\frac{70\times 180}{35000}=\frac{12600}{35000}=0.36{\text{m}}^2.$

Break-even: $c=v_e$ requires the pressure term $=0$, i.e. $P_e-P_a=0$, so $P_a=P_e=65\text{kPa}.$ This is exactly the optimal expansion altitude. For $+120$ m/s: $\frac{(P_e-P_a)A_e}{\dot{m}}=120\Rightarrow P_e-P_a=\frac{120\dot{m}}{A_e}=\frac{120\times 200}{1.2}=20000\text{Pa}.$ $P_a=65000-20000=45000\text{Pa}=45\text{kPa}.$

(a) Sea level: $\frac{(80000-101300)(1.6)}{300}=\frac{(-21300)(1.6)}{300}=\frac{-34080}{300}=-113.6\text{m/s}.$ $c=2950-113.6=2836.4\text{m/s},F=300\times 2836.4=850920\text{N}\approx 851\text{kN}.$ (b) Vacuum: $\frac{(80000)(1.6)}{300}=\frac{128000}{300}=426.67\text{m/s}.$ $c=2950+426.67=3376.67\text{m/s},F=300\times 3376.67=1013000\text{N}\approx 1013\text{kN}.$ (c) Rise in $c$: $3376.67-2836.4=540.27\text{m/s}$. Rise in thrust: $1013000-850920=162080\text{N}\approx 162\text{kN}$. The same hardware gains $\approx 540$ m/s of effective velocity climbing into space — the ambient back-push disappears.

(a) $I_{sp}=\frac{3376.67}{9.81}=344.2\text{s}.$ (b) $\Delta v=3376.67\times \ln 4=3376.67\times 1.386294=4681.2\text{m/s}.$ The pressure boost that raised $c$ in vacuum flows straight through into higher $I_{sp}$ and higher $\Delta v$ — see Specific Impulse.

Vacuum design: required pressure term $=c-v_e=3500-3100=400\text{m/s}$. With $P_a=0$: $\frac{P_e{A}_e}{\dot{m}}=400\Rightarrow A_e=\frac{400\dot{m}}{P_e}=\frac{400\times 160}{55000}=\frac{64000}{55000}=1.1636{\text{m}}^2.$ Sea-level check with this $A_e=1.1636{\text{m}}^2$: $\frac{(55000-101300)(1.1636)}{160}=\frac{(-46300)(1.1636)}{160}=\frac{-53876}{160}=-336.7\text{m/s}.$ $c_{\text{sea}}=3100-336.7=2763.3\text{m/s}.$ So the nozzle designed for space loses $\approx 737$ m/s of effective velocity at liftoff — a real engineering trade-off (a vacuum-optimised nozzle is badly overexpanded at sea level).

(a) Liftoff, $P_a=101.3>P_e=40$: overexpanded, $(P_e-P_a)<0$ → correction negative → $c<v_e$. Thrust penalty; risk of flow separation. (b) $P_a=40=P_e$: optimal expansion, correction $=0$, $c=v_e$. This altitude is where the geometry is optimal. (c) High vacuum, $P_a\to 0<P_e$: underexpanded, $(P_e-P_a)>0$ → correction positive → $c>v_e$. Maximum $c$. Across the mission $c$ rises monotonically as $P_a$ falls, passing through $c=v_e$ exactly at the design point $P_a=P_e$.

Target thrust: $F_A={\dot{m}}_A{c}_A=200\times 3000=600000\text{N}$. For Engine B, $F_B={\dot{m}}_B{v}_e+(P_e-P_a)A_e$. Set equal: $600000=150\times 3400+(30000)A_e=510000+30000A_e.$ $30000A_e=90000\Rightarrow A_e=3.0{\text{m}}^2.$ (Check: $c_B=F_B/{\dot{m}}_B=600000/150=4000\text{m/s}$, consistent with $v_e+(P_e-P_a)A_e/\dot{m}=3400+90000/150=3400+600=4000$.)

Collapse the term first ::: Every "solve for X" reduces to the single number $c-v_e=(P_e-P_a)A_e/\dot{m}$. Sign lives where ::: Inside $(P_e-P_a)$ — compute it with its sign before dividing. $c$ vs altitude ::: Rises as $P_a$ falls; equals $v_e$ exactly at $P_a=P_e$ (optimal). Chain forward ::: $I_{sp}=c/g_0$ and $\Delta v=c\ln (m_0/m_f)$ inherit every change in $c$.