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

3.3.8 · Physics › Rocket Propulsion

$c$ KYA hai?

KAISE: First principles se thrust derive karo

Step 1 — Momentum thrust (Newton ka 2nd/3rd law). Rocket ko apna system maano. Time $d t$ mein woh mass $d m = \overset{m}{˙} d t$ ko speed $v_{e}$ pe peeche eject karta hai (rocket ke relative).

Yeh step kyun? Force = momentum ke change ki rate. Gas ko rearward momentum milta hai, toh reaction se rocket ko equal forward momentum milta hai.

$F_{mom} = \frac{d p _{gas}}{d t} = \frac{v _{e} d m}{d t} = \overset{m}{˙} v_{e}$

Step 2 — Pressure thrust (exit pe gas ka force balance). Exit plane consider karo, area $A_{e}$ . Exit ke andar gas $P_{e}$ pressure se bahar push karti hai; bahar, ambient air $P_{a}$ se andar push karti hai.

Yeh step kyun? Pressure, force per unit area hota hai. Exit ring pe net outward push $(P_{e} - P_{a})$ hai jo $A_{e}$ pe act karta hai. Yeh force kisi cheez se cancel nahi hota (wahan koi wall nahi hai), isliye yeh thrust ban jaata hai.

$F_{pres} = (P_{e} - P_{a}) A_{e}$

Step 3 — Dono add karo.

$F = momentum \overset{m}{˙} v_{e} + pressure (P_{e} - P_{a}) A_{e}$

Step 4 — Sab kuch ek velocity $c$ mein absorb karo. Hum $c$ ko define karte hain taaki $F = \overset{m}{˙} c$ . Poori cheez ko $\overset{m}{˙}$ se divide karo:

$c = \frac{F}{m ˙} = v_{e} + \frac{( P _{e} - P _{a} ) A _{e}}{m ˙}$

Yeh zahmat kyun? Kyunki ab har rocket equation ( $F = \overset{m}{˙} c$ , $I_{s p} = c / g_{0}$ , Tsiolkovsky $Δ v = c ln (m_{0} / m_{f})$ ) EK number use karti hai.

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

Special cases (Pehle forecast karo, phir verify)

Common mistakes

Active Recall

Recall Quick self-test (hide karke answer do)

Rocket pressure thrust kyun produce karta hai?
$c = v_{e}$ exactly kab hota hai?
Rocket jaise upar jaata hai, $c$ badhta hai ya ghatat hai? Kyun?
Agar $P_{e} < P_{a}$ ho toh correction ka sign kya hai, aur us condition ko kya kehte hain?

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho ek balloon phula ke chodh do. Hawa peeche rush karti hai aur balloon ko aage dhakka deti hai — yeh hai "speedy gas" push. Ab socho balloon ka munh thoda tanga hai toh andar ki hawa bahar ki hawa se zyada squeeze hai bilkul opening pe — woh extra squeeze balloon ko thoda aur dhakka deti hai. Rocket mein dono dhakke hote hain. Do shoves track karne ki bajaye, scientists pretend karte hain ki gas ONE make-believe speed $c$ pe nikali, jo same total push deti hai. Space mein upar room-air peeche nahi dhakkelti, toh make-believe speed badi ho jaati hai aur rocket zyada push karta hai!

Flashcards

Effective exhaust velocity

c

define karo.

Woh single equivalent speed jo

F = \overset{m}{˙} c

satisfy kare, jisme momentum aur pressure thrust shamil ho:

c = v_{e} + (P_{e} - P_{a}) A_{e} / \overset{m}{˙}

c

define karne se pehle full thrust equation likho.

F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}

c = v_{e}

kab hota hai?

Jab nozzle optimally expanded ho, yani

P_{e} = P_{a}

(pressure term zero ho jaata hai).

Underexpanded (

P_{e} > P_{a}

) hone par pressure term ka sign?

Positive →

c > v_{e}

(extra thrust).

Overexpanded (

P_{e} < P_{a}

) hone par sign?

Negative →

c < v_{e}

(thrust penalty, flow separation ka risk).

Same engine ke liye altitude badhne par

c

kyun badhta hai?

Kyunki

P_{a}

kam hoti hai, jisse

- P_{a} A_{e} / \overset{m}{˙}

kam negative hota hai; exit pressure ki push ko ambient air kam oppose karti hai.

Pressure thrust physically kahan se aati hai?

Net outward force

(P_{e} - P_{a})

jo exit area

A_{e}

pe act karta hai, jise cancel karne ke liye koi wall nahi hai.

c

aur specific impulse ka relation?

I_{s p} = c / g_{0}

, toh

c = I_{s p} g_{0}

c

diya ho toh raw exit velocity

v_{e}

kaise nikalein?

v_{e} = c - (P_{e} - P_{a}) A_{e} / \overset{m}{˙}

Connections

Thrust Equation — is note ka parent; $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ .
Specific Impulse — $I_{s p} = c / g_{0}$ ; effective velocity ek disguise mein.
Tsiolkovsky Rocket Equation — $c$ use karta hai: $Δ v = c ln (m_{0} / m_{f})$ .
Nozzle Expansion (Under/Over/Optimal) — pressure term ka sign set karta hai.
Conservation of Momentum — momentum thrust ki origin.
Atmospheric Pressure vs Altitude — kyun $c$ ascent ke dauran vary karta hai.