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

3.3.8 · D4 · Physics › Rocket Propulsion › Effective exhaust velocity c = v_e + (P_e − P_a)A_e - ṁ

Shuru karne se pehle, ek picture ki formula kya hai: ek base speed jiske upar ek choti si correction hai.

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

Blue bar raw gas speed $v_{e}$ hai — woh actual speed jis par exhaust nozzle se nikalti hai.
Orange piece jo upar add hota hai woh pressure correction $\frac{( P _{e} - P _{a} ) A _{e}}{m ˙}$ hai.
Dono ka sum (poori height) $c$ hai, woh "as-if" speed jo total thrust reproduce karta hai.

Yeh picture dimag mein rakho: zyaadatar exercises sirf yahi poochti hain ki orange piece banao, hatao, ya reverse karo.

Level 1 — Recognition

Recall Solution 1.1

(a) $v_{e}$ — nozzle exit par gas ki true velocity. (b) $P_{e}$ (exit static pressure) aur $P_{a}$ (ambient pressure) — dono absolute, kyunki sirf unka difference physically meaningful hai aur gauge/absolute mix karna us difference ko corrupt kar deta hai. (c) $A_{e}$ — nozzle exit area, $m^{2}$ mein. (d) $c$ effective exhaust velocity hai: ek single make-believe speed jisse $F = \overset{m}{˙} c$ total thrust deta hai (momentum aur pressure dono).

Recall Solution 1.2

Optimal expansion ka matlab hai $P_{e} = P_{a}$ , toh $(P_{e} - P_{a}) = 0$ . Poora pressure term $0$ ho jaata hai. Isliye $c = v_{e} .$ Saara thrust momentum thrust hai; hamare bar picture mein koi orange piece nahi hai.

Level 2 — Application

Recall Solution 2.1

Pressure term: $\frac{( P _{e} - P _{a} ) A _{e}}{m ˙} = \frac{( 90000 - 50000 ) ( 1.5 )}{250} = \frac{60000}{250} = 240 m/s .$ Kyunki $P_{e} > P_{a}$ , yeh underexpanded hai → positive correction → $c > v_{e}$ . $c = 2900 + 240 = 3140 m/s .$

Recall Solution 2.2

$\frac{( 90000 - 101300 ) ( 1.5 )}{250} = \frac{( - 11300 ) ( 1.5 )}{250} = \frac{- 16950}{250} = - 67.8 m/s .$ $P_{e} < P_{a}$ → overexpanded → ambient air pushback karta hai → negative correction. $c = 2900 - 67.8 = 2832.2 m/s .$

Recall Solution 2.3

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

Level 3 — Analysis

Recall Solution 3.1

Definition ko rearrange karo (orange piece wapas subtract karo): $v_{e} = c - \frac{( P _{e} - P _{a} ) A _{e}}{m ˙} = 4200 - \frac{( 40000 ) ( 0.9 )}{120} = 4200 - \frac{36000}{120} = 4200 - 300 = 3900 m/s .$ Measured $c$ mein pehle se $+ 300$ m/s ka pressure boost tha, isliye true gas speed lower hai.

Recall Solution 3.2

Pressure term equals $c - v_{e} = 3070 - 3000 = 70 m/s$ hai. Toh $\frac{( P _{e} - P _{a} ) A _{e}}{m ˙} = 70 \Rightarrow A_{e} = \frac{70 m ˙}{P _{e} - P _{a}} = \frac{70 \times 180}{35000} = \frac{12600}{35000} = 0.36 m^{2} .$

Recall Solution 3.3

Break-even: $c = v_{e}$ ke liye pressure term $= 0$ chahiye, yaani $P_{e} - P_{a} = 0$ , toh $P_{a} = P_{e} = 65 kPa .$ Yeh exactly optimal expansion altitude hai. $+ 120$ m/s ke liye: $\frac{( P _{e} - P _{a} ) A _{e}}{m ˙} = 120 \Rightarrow P_{e} - P_{a} = \frac{120 m ˙}{A _{e}} = \frac{120 \times 200}{1.2} = 20000 Pa .$ $P_{a} = 65000 - 20000 = 45000 Pa = 45 kPa .$

Level 4 — Synthesis

Recall Solution 4.1

(a) Sea level: $\frac{( 80000 - 101300 ) ( 1.6 )}{300} = \frac{( - 21300 ) ( 1.6 )}{300} = \frac{- 34080}{300} = - 113.6 m/s .$ $c = 2950 - 113.6 = 2836.4 m/s, F = 300 \times 2836.4 = 850920 N \approx 851 kN .$ (b) Vacuum: $\frac{( 80000 ) ( 1.6 )}{300} = \frac{128000}{300} = 426.67 m/s .$ $c = 2950 + 426.67 = 3376.67 m/s, F = 300 \times 3376.67 = 1013000 N \approx 1013 kN .$ (c) $c$ mein rise: $3376.67 - 2836.4 = 540.27 m/s$ . Thrust mein rise: $1013000 - 850920 = 162080 N \approx 162 kN$ . Same hardware space mein jaate waqt effective velocity mein $\approx 540$ m/s gain karta hai — ambient back-push khatam ho jaata hai.

Agla figure is hi engine ka $c$ poori altitude range mein trace karta hai.

Red dashed line $v_{e} = 2950$ m/s hai — gas ki ek fixed property.
Blue curve $c$ hai. Yeh $v_{e}$ se neeche start hoti hai (overexpanded, sea level), $v_{e}$ ko cross karti hai jahan $P_{a} = P_{e} = 80$ kPa (optimal), aur $P_{a} \to 0$ hote hi upar uth jaati hai (vacuum, underexpanded).
Ek crossing point nozzle ka design altitude hai.

Recall Solution 4.2

(a) $I_{s p} = \frac{3376.67}{9.81} = 344.2 s .$ (b) $Δ v = 3376.67 \times ln 4 = 3376.67 \times 1.386294 = 4681.2 m/s .$ Vacuum mein $c$ badhane wala pressure boost seedha higher $I_{s p}$ aur higher $Δ v$ mein flow hota hai — dekho Specific Impulse.

Level 5 — Mastery

Recall Solution 5.1

Vacuum design: required pressure term $= c - v_{e} = 3500 - 3100 = 400 m/s$ hai. $P_{a} = 0$ ke saath: $\frac{P _{e} A _{e}}{m ˙} = 400 \Rightarrow A_{e} = \frac{400 m ˙}{P _{e}} = \frac{400 \times 160}{55000} = \frac{64000}{55000} = 1.1636 m^{2} .$ Sea-level check is $A_{e} = 1.1636 m^{2}$ ke saath: $\frac{( 55000 - 101300 ) ( 1.1636 )}{160} = \frac{( - 46300 ) ( 1.1636 )}{160} = \frac{- 53876}{160} = - 336.7 m/s .$ $c_{sea} = 3100 - 336.7 = 2763.3 m/s .$ Toh space ke liye design kiya gaya nozzle liftoff par effective velocity mein $\approx 737$ m/s khota hai — yeh ek real engineering trade-off hai (vacuum-optimised nozzle sea level par badly overexpanded hota hai).

Recall Solution 5.2

(a) Liftoff, $P_{a} = 101.3 > P_{e} = 40$ : overexpanded, $(P_{e} - P_{a}) < 0$ → correction negative → $c < v_{e}$ . Thrust penalty; flow separation ka risk. (b) $P_{a} = 40 = P_{e}$ : optimal expansion, correction $= 0$ , $c = v_{e}$ . Yeh woh altitude hai jahan geometry optimal hai. (c) High vacuum, $P_{a} \to 0 < P_{e}$ : underexpanded, $(P_{e} - P_{a}) > 0$ → correction positive → $c > v_{e}$ . Maximum $c$ . Mission ke across $c$ monotonically rise karta hai jaise $P_{a}$ girta hai, design point $P_{a} = P_{e}$ par exactly $c = v_{e}$ se guzarta hai.

Recall Solution 5.3

Target thrust: $F_{A} = \overset{m}{˙}_{A} c_{A} = 200 \times 3000 = 600000 N$ . Engine B ke liye, $F_{B} = \overset{m}{˙}_{B} v_{e} + (P_{e} - P_{a}) A_{e}$ . Equal set karo: $600000 = 150 \times 3400 + (30000) A_{e} = 510000 + 30000 A_{e} .$ $30000 A_{e} = 90000 \Rightarrow A_{e} = 3.0 m^{2} .$ (Check: $c_{B} = F_{B} / \overset{m}{˙}_{B} = 600000/150 = 4000 m/s$ , $v_{e} + (P_{e} - P_{a}) A_{e} / \overset{m}{˙} = 3400 + 90000/150 = 3400 + 600 = 4000$ se consistent hai.)

Recall wrap-up

Recall Kya tumne ladder master kiya?

Pehle term collapse karo ::: Har "solve for X" ek single number $c - v_{e} = (P_{e} - P_{a}) A_{e} / \overset{m}{˙}$ tak reduce ho jaata hai. Sign kahan rehta hai ::: $(P_{e} - P_{a})$ ke andar — divide karne se pehle sign ke saath compute karo. $c$ vs altitude ::: $P_{a}$ girne par rise karta hai; exactly $P_{a} = P_{e}$ par $v_{e}$ ke barabar hota hai (optimal). Aage chain karo ::: $I_{s p} = c / g_{0}$ aur $Δ v = c ln (m_{0} / m_{f})$ $c$ mein har change inherit karte hain.

Connections

Parent note (Hinglish) — woh derivation jin exercises ko yeh train karta hai.
Thrust Equation — $F = \overset{m}{˙} v_{e} + (P_{e} - P_{a}) A_{e}$ , Exercises 2.3 aur 5.3 mein use hua.
Specific Impulse — $I_{s p} = c / g_{0}$ , Exercise 4.2(a).
Tsiolkovsky Rocket Equation — $Δ v = c ln (m_{0} / m_{f})$ , Exercise 4.2(b).
Nozzle Expansion (Under/Over/Optimal) — poore mein regime names.
Conservation of Momentum — momentum term ka origin.
Atmospheric Pressure vs Altitude — kyun ascent ke dauran $c$ change hota hai.