Effective exhaust velocity c = v_e + (P_e − P_a)A_e - ṁ: Worked examples

3.3.8 · D3 · Physics › Rocket Propulsion › Effective exhaust velocity c = v_e + (P_e − P_a)A_e - ṁ

Scenario matrix

Numbers touch karne se pehle, yeh naam lete hain ki "har scenario" ka matlab kya hai. Hamare formula mein paanch inputs hain — $v_{e}, P_{e}, P_{a}, A_{e}, \overset{m}{˙}$ — aur ek output $c$ . Interesting behaviour almost poori tarah se pressure difference ke sign se drive hoti hai

$Δ P \equiv P_{e} - P_{a}$

(exit gas pressure minus bahari hawa ki pressure) aur isse ki koi input extreme value hit kare (zero, bahut bada, ya designed-for match). Yeh poora grid hai.

Cell	Scenario class	Trigger	Kya expect karte hain
A	Optimally expanded	$P_{e} = P_{a}$ ( $Δ P = 0$ )	$c = v_{e}$ exactly
B	Underexpanded	$P_{e} > P_{a}$ ( $Δ P > 0$ )	$c > v_{e}$ (bonus thrust)
C	Overexpanded	$P_{e} < P_{a}$ ( $Δ P < 0$ )	$c < v_{e}$ (thrust penalty)
D	Vacuum limit	$P_{a} \to 0$	us engine ke liye sabse bada $c$
E	Ascent (variable $P_{a}$ )	$P_{a}$ altitude ke saath girta hai	climb ke dauran $c$ badhta hai
F	Degenerate area	$A_{e} \to 0$	pressure term khatam, $c \to v_{e}$
G	Solve-backwards	$c$ diya hai, $v_{e}$ ya $A_{e}$ dhundhna hai	definition ko rearrange karo
H	Real-world word problem	data prose mein chhupi hai	translate → same formula
I	Exam twist / units trap	kPa vs Pa, gauge vs absolute	units ki consistency
J	Degenerate $\overset{m}{˙}$	$\overset{m}{˙} \to 0$ ya $\overset{m}{˙} \to \infty$	$c \to \infty$ ya $c \to v_{e}$
K	Degenerate $v_{e}$	$v_{e} \to 0$	sirf pressure term se $c$

Neeche har worked example us cell(s) se tagged hai jo wo cover karta hai. Milke wo poora grid fill karte hain.

Scenario map padhna (Figure 1)

Grid ko sirf padhne ki jagah dekhne ke liye, neeche diya quadrant map dekho. Har region ko exactly aise padho:

Horizontal axis pressure difference hai $Δ P = P_{e} - P_{a}$ : left mein negative, centre line par zero, right mein positive.
Vertical axis velocity mein resulting correction hai, $\frac{Δ P A _{e}}{m ˙}$ — yaani $c$ bare gas speed $v_{e}$ se kitna upar ya neeche hai.
Pale-yellow horizontal line beech mein $c = v_{e}$ hai: "no pressure help" baseline.
Blue region (right) wahan hai jahan $Δ P > 0$ → correction positive → $c > v_{e}$ : yeh underexpanded hai (cell B).
Pink region (left) wahan hai jahan $Δ P < 0$ → correction negative → $c < v_{e}$ : yeh overexpanded hai (cell C).
Centre dot jahan sloped line baseline ko cross karti hai woh matched hai (cell A), $Δ P = 0$ , $c = v_{e}$ .
Seedhi sloped line $\frac{A _{e}}{m ˙} Δ P$ hai; iska steepness $A_{e} / \overset{m}{˙}$ se set hoti hai — bada exit area ya chhota mass flow ise upar tilt karta hai, correction ko amplify karta hai (yahi cells F aur J disguise mein hain).

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

Neeche, teen nozzle mouths dekho: har ek $Δ P$ ke ek sign ke saath ek arrow dikhata hai jis taraf leftover pressure dhakelta hai — bahar (blue, bonus), kuch nahi (yellow, matched), ya andar (pink, penalty).

Har cell ko cover karne wale examples

Worked example Example 1 — Cell A (optimally expanded,

$Δ P = 0$ ) Ek nozzle $\overset{m}{˙} = 400$ kg/s, $v_{e} = 3000$ m/s par run karta hai. Ise exactly us altitude par fire kiya jaata hai jahan uska exit pressure hawa se match karta hai: $P_{e} = P_{a} = 45$ kPa, $A_{e} = 1.5 m^{2}$ ke saath. $c$ nikalo.

Forecast: pressure term vanish ho jaani chahiye, toh guess karo $c = 3000$ m/s exactly.

$Δ P = P_{e} - P_{a} = 45000 - 45000 = 0$ Pa compute karo. Yeh step kyun? Poora doosra term $Δ P A_{e} / \overset{m}{˙}$ hai; agar $Δ P = 0$ toh area aur mass flow irrelevant ho jaate hain.
$c = v_{e} + \frac{0 \times 1.5}{400} = 3000 + 0 = 3000$ m/s. Yeh step kyun? Kuch bhi times zero zero hota hai — nozzle apne design altitude par hai.

Verify: doosre term ke units hain $\frac{Pa \cdot m ^{2}}{kg/s} = \frac{N}{kg/s} = \frac{kg\cdotm/s ^{2}}{kg/s} = m/s$ ✓. Answer forecast se match karta hai: $c = v_{e}$ precisely jab matched.

Worked example Example 2 — Cell B (underexpanded,

$Δ P > 0$ ) Same engine jaise Ex. 1 mein ( $\overset{m}{˙} = 400$ kg/s, $v_{e} = 3000$ m/s, $A_{e} = 1.5 m^{2}$ ) lekin zyada upar fly kiya gaya jahan $P_{a} = 20$ kPa jabki $P_{e} = 45$ kPa. $c$ nikalo.

Forecast: ab $P_{e} > P_{a}$ , toh $Δ P > 0$ → $c$ 3000 m/s se upar hona chahiye.

$Δ P = 45000 - 20000 = 25000$ Pa (positive — underexpanded). Yeh step kyun? Positive $Δ P$ ka matlab hai gas ka mouth bahari hawa se abhi bhi zyada squeeze ho raha hai; woh leftover push help karta hai.
Pressure term $= \frac{25000 \times 1.5}{400} = \frac{37500}{400} = 93.75$ m/s. Yeh step kyun? Pressure force $Δ P A_{e}$ ko $\overset{m}{˙}$ se divide karo taaki force ko velocity boost mein convert kar sako (yahi $c$ ki definition hai).
$c = 3000 + 93.75 = 3093.75$ m/s. Yeh step kyun? Positive boost ko bare gas speed mein add karo — yahi poora formula assemble hua.

Verify: $c > v_{e}$ as forecast ✓. Correction ka sign $+$ hai, "Underexpanded → Up" mnemonic ke consistent.

Worked example Example 3 — Cell C (overexpanded,

$Δ P < 0$ ) Same engine, ab sea level par fire kiya: $P_{a} = 101.3$ kPa, $P_{e} = 45$ kPa.

Forecast: $P_{e} < P_{a}$ , toh $Δ P < 0$ → correction negative hai, $c < 3000$ m/s.

$Δ P = 45000 - 101300 = - 56300$ Pa (negative — overexpanded). Yeh step kyun? Bahari hawa ab exhaust se zyada squeeze kar rahi hai, isliye woh nozzle mein wapas push karti hai.
Pressure term $= \frac{- 56300 \times 1.5}{400} = \frac{- 84450}{400} = - 211.125$ m/s. Yeh step kyun? Wahi "force ÷ mass-flow → velocity" conversion pehle jaisi; minus sign carry through hota hai kyunki net push ab andar point karta hai.
$c = 3000 - 211.125 = 2788.875$ m/s. Yeh step kyun? (Negative) correction ko $v_{e}$ mein add karo — negative number add karna subtract karta hai, $c$ bare gas speed se neeche aata hai.

Verify: $c < v_{e}$ as forecast ✓. Yeh sea-level penalty hai; wohi engine (Ex. 2) upar zyada accha karta hai. Physically yeh over-expansion flow separation bhi cause kar sakta hai — Nozzle Expansion (Under/Over/Optimal) dekho.

Worked example Example 4 — Cell D + E (vacuum limit aur ascent trend)

Ex. 1–3 ka engine lo ( $v_{e} = 3000$ , $A_{e} = 1.5$ , $\overset{m}{˙} = 400$ , $P_{e} = 45$ kPa) aur teen altitudes par $c$ tabulate karo: sea level ( $P_{a} = 101.3$ kPa), high altitude ( $P_{a} = 20$ kPa), aur vacuum ( $P_{a} = 0$ ).

Forecast: jaise hum climb karte hain, $P_{a}$ shrink hota hai, toh $- P_{a} A_{e} / \overset{m}{˙}$ kam negative hota jaata hai → $c$ strictly increase hona chahiye, vacuum mein sabse bada.

Sea level ( $P_{a} = 101.3$ kPa): Ex. 3 ka arithmetic reuse karo, $c = 2788.875$ m/s. Yeh step kyun? Ex. 3 ne already is exact engine ko sea-level pressure par evaluate kiya tha; re-derive karna wahi $Δ P = - 56300$ Pa computation repeat karna hoga, isliye ise trend ke low anchor ke roop mein quote karte hain.
High altitude ( $P_{a} = 20$ kPa): Ex. 2 ka arithmetic reuse karo, $c = 3093.75$ m/s. Yeh step kyun? Ex. 2 same engine hai $P_{a} = 20$ kPa par; ise quote karne se hum $Δ P = 25000$ Pa step dobara kiye bina ek axis par teen altitudes line up kar sakte hain.
Vacuum ( $P_{a} = 0$ ): $Δ P = 45000 - 0 = 45000$ Pa, $c = 3000 + \frac{45000 \times 1.5}{400} = 3000 + 168.75 = 3168.75 m/s .$ Yeh step kyun? $P_{a} = 0$ set karne se saari back-push hat jaati hai; poori exit pressure ab help karti hai — is geometry ke liye theoretical maximum $c$ .

Verify: $2788.875 < 3093.75 < 3168.75$ — monotonically rising as forecast ✓. Yahi reason hai vacuum $I_{s p}$ zyada quote hota hai; Atmospheric Pressure vs Altitude dekho. Rising staircase neeche drawn hai.

Ascent staircase padhna (Figure 3)

Kaise padhen: horizontal axis teen altitudes list karta hai (sea level → high altitude → vacuum), left se right jaisi rocket climb karti hai. Vertical axis $c$ m/s mein hai. Blue dots jo ek line se jude hain woh teen values hain jo abhi compute hui; dekho kaise woh step upward karte hain jaise hum right jaate hain — har step $P_{a}$ ke shrinkage ka hai. Pink dashed line bare gas speed $v_{e} = 3000$ m/s mark karti hai: notice karo ki dots iske neeche se shuru hote hain (overexpanded penalty), ise cross karte hain, aur iske upar khatam hote hain (vacuum bonus). Kisi bhi dot aur pink line ke beech ka vertical gap precisely us altitude ke liye pressure correction $Δ P A_{e} / \overset{m}{˙}$ hai.

Worked example Example 5 — Cell F (degenerate area

$A_{e} \to 0$ ) Ek idealised "no-nozzle" thruster: same gas speed $v_{e} = 3000$ m/s, $\overset{m}{˙} = 400$ kg/s, lekin exit area zero ki taraf shrink karte hain, $A_{e} \to 0$ . Sea level par $c$ ka kya hoga ( $P_{a} = 101.3$ kPa, $P_{e} = 45$ kPa)?

Forecast: agar $A_{e} = 0$ toh pressure push karne ke liye koi ring nahi hai, toh correction khatam ho jaani chahiye aur $c \to v_{e} = 3000$ .

Pressure term $= \frac{( 45000 - 101300 ) \cdot A _{e}}{400}$ . Yeh step kyun? $A_{e}$ poori pressure force ko multiply karta hai; koi area nahi means koi force nahi.
Jaise $A_{e} \to 0$ yeh term $\to 0$ hota hai, toh $c \to 3000$ m/s. Yeh step kyun? Shrinking factor poore product ko zero ki taraf khichta hai, sirf $v_{e}$ bachta hai.

Verify: limit $Δ P$ ke sign se independent hai — chahe under- ya over-expanded ho, zero area correction ko khatam kar deta hai ✓. Sanity: bina expanding bell ke plain gas jet mein pure momentum thrust hi hota hai.

Worked example Example 6 — Cell J (degenerate

$\overset{m}{˙}$ : throttle to nothing, ya open wide) Same underexpanded numbers jaise Ex. 2 mein ( $v_{e} = 3000$ , $Δ P = 25000$ Pa, $A_{e} = 1.5 m^{2}$ ). Mass flow $\overset{m}{˙}$ ke baare mein do limiting questions poochho.

Forecast: $\overset{m}{˙}$ pressure term ke denominator mein hai. Tiny $\overset{m}{˙}$ term ko blow up karna chahiye ( $c \to \infty$ ); huge $\overset{m}{˙}$ ise nothing mein shrink karna chahiye ( $c \to v_{e}$ ).

Correction likhte hain $\frac{Δ P A _{e}}{m ˙} = \frac{25000 \times 1.5}{m ˙} = \frac{37500}{m ˙}$ . Yeh step kyun? Denominator mein $\overset{m}{˙}$ ko isolate karne se dono limits obvious ho jaate hain.
Jaise $\overset{m}{˙} \to 0^{+}$ : $\frac{37500}{m ˙} \to + \infty$ , toh $c \to \infty$ . Yeh step kyun? Ek fixed positive force ko ever-smaller mass flow se divide karna matlab har kilogram pressure push ka unbounded share carry karta hai — "effective speed" per kg diverge karta hai. (Practically yeh idealisation hai: zero propellant leave karne ke saath pressure thrust nahi ho sakti.)
Jaise $\overset{m}{˙} \to \infty$ : $\frac{37500}{m ˙} \to 0$ , toh $c \to v_{e} = 3000$ m/s. Yeh step kyun? Huge flow fixed pressure force ko itne kilograms mein dilute kar deta hai ki per-kg boost vanish ho jaata hai — sirf raw gas speed bachti hai.

Verify: dono limits sensible range bracket karte hain; koi bhi real $\overset{m}{˙}$ underexpanded flow ke liye strictly $v_{e}$ aur $\infty$ ke beech $c$ deta hai ✓. Symbolically $lim_{\overset{m}{˙} \to \infty} c = v_{e}$ aur $lim_{\overset{m}{˙} \to 0^{+}} c = \infty$ .

Worked example Example 7 — Cell K (degenerate

$v_{e} \to 0$ ) Ek hypothetical "cold-gas" case jahan gas barely move karta hai, $v_{e} \to 0$ , lekin abhi bhi ambient se zyada squeeeze ho raha hai: $P_{e} = 60$ kPa, $P_{a} = 20$ kPa, $A_{e} = 0.5 m^{2}$ , $\overset{m}{˙} = 100$ kg/s. $c$ kya hai?

Forecast: $v_{e} = 0$ ke saath momentum half gone hai; $c$ sirf pressure term ke barabar hona chahiye — purely pressure-driven effective speed, phir bhi positive kyunki $Δ P > 0$ .

$c = v_{e} + \frac{Δ P A _{e}}{m ˙}$ mein $v_{e} = 0$ set karo, $c = \frac{Δ P A _{e}}{m ˙}$ bachta hai. Yeh step kyun? Momentum term drop karne se isolate hota hai ki pressure akela kya contribute karta hai.
$Δ P = 60000 - 20000 = 40000$ Pa. Yeh step kyun? Sirf difference enter karta hai; yeh positive hai toh leftover push bahar ki taraf hai.
$c = \frac{40000 \times 0.5}{100} = \frac{20000}{100} = 200$ m/s. Yeh step kyun? Wahi force-÷-mass-flow conversion; koi gas speed add karne ke liye nahi hai, toh yeh hi poora $c$ hai.

Verify: $c = 200$ m/s $> 0$ , purely pressure thrust se as forecast ✓. Agar yahan $Δ P$ negative hota, $c$ negative aata — ek "reverse" effective velocity, yaani net drag thrust se zyada.

Worked example Example 8 — Cell G (

$v_{e}$ ke liye backwards solve karo) Ek test stand $c = 4000$ m/s measure karta hai $\overset{m}{˙} = 100$ kg/s, $P_{e} = 50$ kPa, $P_{a} = 20$ kPa, $A_{e} = 0.8 m^{2}$ ke saath. Raw gas exit velocity $v_{e}$ nikalo.

Forecast: $Δ P > 0$ (underexpanded), toh measured $c$ pressure boost se inflated hai — true $v_{e}$ 4000 m/s se neeche hona chahiye.

Definition rearrange karo: $v_{e} = c - \frac{( P _{e} - P _{a} ) A _{e}}{m ˙}$ . Yeh step kyun? $c$ mein pehle se pressure boost bundle hai; use strip karne ke liye subtract karte hain.
Pressure term $= \frac{( 50000 - 20000 ) \times 0.8}{100} = \frac{24000}{100} = 240$ m/s. Yeh step kyun? Woh boost compute karo jo measured $c$ mein baked in tha.
$v_{e} = 4000 - 240 = 3760$ m/s. Yeh step kyun? Bare gas speed reveal karne ke liye woh boost remove karo.

Verify: $v_{e} < c$ as forecast ✓. Wapas plug karo: $3760 + 240 = 4000 = c$ ✓.

Worked example Example 9 — Cell G (

$A_{e}$ ke liye backwards solve karo) Ek designer chahta hai nozzle $c = 3200$ m/s hit kare ek altitude par $P_{a} = 30$ kPa ke saath, $v_{e} = 3050$ m/s, $P_{e} = 60$ kPa, $\overset{m}{˙} = 250$ kg/s use karke. Kaun sa exit area $A_{e}$ chahiye?

Forecast: humein $3200 - 3050 = 150$ m/s ka positive boost chahiye, aur $Δ P > 0$ hai, toh ek sensible positive area exist hona chahiye.

$c = v_{e} + \frac{Δ P A _{e}}{m ˙}$ se, $A_{e} = \frac{( c - v _{e} ) m ˙}{Δ P}$ solve karo. Yeh step kyun? Sirf $A_{e}$ unknown hai; ise algebraically isolate karo.
$Δ P = 60000 - 30000 = 30000$ Pa; needed boost $c - v_{e} = 150$ m/s. Yeh step kyun? Divide karne se pehle ratio ke dono ingredients SI mein hone chahiye.
$A_{e} = \frac{150 \times 250}{30000} = \frac{37500}{30000} = 1.25 m^{2}$ . Yeh step kyun? Required boost times mass-flow ko pressure difference se divide karo area pane ke liye.

Verify: wapas rakho: $3050 + \frac{30000 \times 1.25}{250} = 3050 + 150 = 3200 = c$ ✓. Units: $\frac{( m/s ) ( kg/s )}{Pa} = \frac{kg\cdotm/s ^{2}}{N/m ^{2}} = \frac{N}{N/m ^{2}} = m^{2}$ ✓.

Worked example Example 10 — Cell H (real-world word problem)

"Ek launch engine har 6 seconds mein 1.8 tonnes propellant jalata hai. Uska exhaust 2.9 km/s par nikalta hai. Liftoff ke waqt nozzle ka 1.1 m² ka mouth 90 kPa par gas carry karta hai jabki bahar sea-level hawa 101.3 kPa par hai. Effective exhaust velocity m/s mein batao."

Forecast: exhaust 90 kPa par hai, hawa 101.3 kPa par → $Δ P < 0$ → overexpanded → $c$ thoda 2900 m/s se neeche.

Prose convert karo. Mass flow: $\overset{m}{˙} = \frac{1.8 t}{6 s} = \frac{1800 kg}{6 s} = 300$ kg/s. Exit speed: $v_{e} = 2.9 km/s = 2900$ m/s. Yeh step kyun? Formula ko SI base units chahiye: kg, s, m, Pa. Tonnes aur km/s traps hain.
$Δ P = 90000 - 101300 = - 11300$ Pa. Yeh step kyun? Sign pehle fix karo; negative matlab overexpanded, forecast se match karta hai.
Pressure term $= \frac{- 11300 \times 1.1}{300} = \frac{- 12430}{300} = - 41.4 \overline{3}$ m/s. Yeh step kyun? Force-÷-mass-flow conversion, minus sign carry hua.
$c = 2900 - 41.433 \dots = 2858.57$ m/s (2 dp tak). Yeh step kyun? Chhoti negative correction ko bare gas speed mein add karo.

Verify: $c < v_{e}$ as forecast, aur sirf thode margin se kyunki $Δ P$ chhota hai ✓. Rounded answer $\approx 2858.6$ m/s. Units ki sanity: kg, s, m, Pa sab SI hain, toh output genuinely m/s hai.

Worked example Example 11 — Cell I (exam twist: gauge vs absolute pressure trap)

"Ek engine $v_{e} = 3200$ m/s, $A_{e} = 0.5 m^{2}$ , $\overset{m}{˙} = 120$ kg/s deta hai. Ek gauge exit pressure $+ 15$ kPa gauge read karta hai jabki local absolute air pressure $85$ kPa hai. $c$ nikalo."

Forecast: "+15 kPa gauge" trap hai — gauge pressure already "surrounding air se upar" ka matlab hai. Toh exit $85$ kPa bahar se $15$ kPa upar hai: $Δ P = + 15$ kPa directly. $c$ thoda 3200 se upar hona chahiye.

Gauge reading interpret karo. Gauge pressure = absolute minus ambient, toh $+ 15$ kPa gauge reading ka matlab hai $P_{e} - P_{a} = 15$ kPa jo exactly $Δ P$ pehle se hai. Koi absolute conversion needed nahi — sirf difference formula mein enter karta hai. Yeh step kyun? Formula $P_{e} - P_{a}$ use karta hai; agar gauge already woh difference report karta hai, use as-is use karo. (Agar problem ne do absolute values diye hote, hum unhe subtract karte.)
$Δ P = 15 kPa = 15000$ Pa. Yeh step kyun? SI consistency ke liye kPa ko Pa mein convert karo divide karne se pehle — doosra common exam trap (Pa formula mein kPa mix karna).
Pressure term $= \frac{15000 \times 0.5}{120} = \frac{7500}{120} = 62.5$ m/s. Yeh step kyun? Standard force-÷-mass-flow conversion.
$c = 3200 + 62.5 = 3262.5$ m/s. Yeh step kyun? Positive boost ko bare gas speed mein add karo.

Verify: $c > v_{e}$ as forecast ✓. Trap se bachao: agar hum galti se $85$ kPa ambient ko gauge reading ke upar add karte, toh $Δ P = 100$ kPa use karte aur answer $\approx 3617$ m/s tak blow up ho jaata — galat. Sirf difference $Δ P$ kabhi bhi enter karta hai; Nozzle Expansion (Under/Over/Optimal) dekho.

Recall

Recall Har situation kaunsa cell hit karti hai?

Sea-level firing $P_{e} < P_{a}$ ke saath ::: Cell C — overexpanded, $c < v_{e}$ . Same engine space mein climb karte huye ::: Cells D aur E — $P_{a} \to 0$ , $c$ apne max tak badhta hai. Ek gauge pehle se $P_{e} - P_{a}$ read kar raha hai ::: Cell I — directly $Δ P$ ke roop mein use karo, koi absolute conversion nahi. $A_{e}$ ko zero ki taraf shrink karna ::: Cell F — pressure term khatam, $c \to v_{e}$ $Δ P$ sign se independent. $\overset{m}{˙}$ ko zero ki taraf throttle karna ::: Cell J — pressure term blow up karta hai, $c \to \infty$ (idealisation). $\overset{m}{˙}$ ko bahut wide open karna ::: Cell J — pressure term zero tak dilute hota hai, $c \to v_{e}$ . Gas barely moving, $v_{e} \to 0$ ::: Cell K — $c$ sirf pressure term ke barabar. $c$ diya hai, raw gas speed dhundhna ::: Cell G — $v_{e} = c - Δ P A_{e} / \overset{m}{˙}$ .

Connections

Effective exhaust velocity $c$ (parent) — woh formula jise yeh examples exercise karte hain.
Specific Impulse — kisi bhi $c$ ko yahan $g_{0}$ se divide karo $I_{s p}$ pane ke liye.
Tsiolkovsky Rocket Equation — inhe $c$ values $Δ v = c ln (m_{0} / m_{f})$ mein feed karo.
Nozzle Expansion (Under/Over/Optimal) — cells A/B/C ko naam deta hai.
Atmospheric Pressure vs Altitude — cells D aur E drive karta hai.
Conservation of Momentum — $v_{e}$ ke peeche momentum-thrust half.