Specific impulse Isp = v_e - g₀ — definition, physical meaning, units: Exercises

3.3.4 · D4 · Physics › Rocket Propulsion › Specific impulse Isp = v_e - g₀ — definition, physical meani

Level 1 — Recognition

Goal: definitions se seedha quantity padho, koi chained steps nahi.

L1.1

Ek engine ka effective exhaust velocity $v_{e} = 4413 m/s$ hai. Uska specific impulse kya hai?

Recall Solution

KYA: seedha $I_{s p} = v_{e} / g_{0}$ lagao. KYUN: definition ek exhaust speed ko size-independent efficiency number mein convert karti hai. $I_{s p} = \frac{4413}{9.80665} \approx 450 s$

L1.2

Inme se kaun seconds mein measure hota hai: (a) thrust, (b) specific impulse, (c) mass-flow rate?

Recall Solution

(b) specific impulse. Thrust newtons (N) mein hota hai, mass-flow kg/s mein. $I_{s p} = \frac{N\cdots}{N} = s$ : newtons cancel ho jaate hain, seconds bach jaate hain.

L1.3

Ek engine ki data sheet Earth par $I_{s p} = 300 s$ quote karti hai. Mars par uska $I_{s p}$ kya hoga (wahan local $g \approx 3.71 m/s^{2}$ )?

Recall Solution

Phir bhi $300 s$ . Formula constant $g_{0} = 9.80665$ use karta hai, local $g$ kabhi nahi. $I_{s p}$ engine ki property hai, planet ki nahi.

Level 2 — Application

Goal: do formulas ki ek clean chain.

L2.1

$I_{s p} = 320 s$ hai. Effective exhaust velocity $v_{e}$ nikalo.

Recall Solution

KYA: definition ulti karo, $v_{e} = I_{s p} g_{0}$ . KYUN: hum efficiency number se wapas ek speed chahte hain. $v_{e} = 320 \times 9.80665 \approx 3138 m/s$

L2.2

Ek engine ka $I_{s p} = 250 s$ aur mass-flow $\overset{m}{˙} = 300 kg/s$ hai. Thrust nikalo.

Recall Solution

KYA: $F = \overset{m}{˙} g_{0} I_{s p}$ use karo (thrust ko dono — efficiency aur flow — chahiye). KYUN: thrust per second eject hone wala momentum hai; $g_{0} I_{s p} = v_{e}$ ejection speed hai. $F = 300 \times 9.80665 \times 250 \approx 7.36 \times 1 0^{5} N \approx 0.74 MN$

L2.3

Ek stage mein $m_{p} = 15000 kg$ propellant hai, $I_{s p} = 340 s$ hai. Total impulse $J$ nikalo.

Recall Solution

KYA: $J = I_{s p} g_{0} m_{p}$ . KYUN: total impulse poora "push-package" hai, jo $I_{s p}$ aur propellant weight $W_{p} = g_{0} m_{p}$ ke product ke barabar hai. $J = 340 \times 9.80665 \times 15000 \approx 5.00 \times 1 0^{7} N\cdots$

Level 3 — Analysis

Goal: reverse-engineer ya compare karo — socho kaunsa unknown solve karna hai.

L3.1

Ek test mein thrust $F = 1.8 \times 1 0^{6} N$ aur mass-flow $\overset{m}{˙} = 420 kg/s$ measure hota hai. Engine ka $I_{s p}$ kya hai?

Recall Solution

KYA: pehle $v_{e} = F / \overset{m}{˙}$ nikalo, phir $I_{s p} = v_{e} / g_{0}$ . KYUN: thrust aur flow se $F = \overset{m}{˙} v_{e}$ ke zariye ejection speed milti hai; phir seconds mein convert karo. $v_{e} = \frac{1.8 \times 1 0 ^{6}}{420} \approx 4285.7 m/s, I_{s p} = \frac{4285.7}{9.80665} \approx 437 s$

L3.2

Engine A: $I_{s p} = 450 s$ , $\overset{m}{˙} = 500 kg/s$ . Engine B: $I_{s p} = 3000 s$ , $\overset{m}{˙} = 0.006 kg/s$ (ek ion thruster). Kaun zyada thrust deta hai, aur kitne factor se?

Recall Solution

Dono ke liye $F = \overset{m}{˙} g_{0} I_{s p}$ calculate karo. $F_{A} = 500 \times 9.80665 \times 450 \approx 2.21 \times 1 0^{6} N$ $F_{B} = 0.006 \times 9.80665 \times 3000 \approx 176.5 N$ A jeet jaata hai factor $\frac{F _{A}}{F _{B}} \approx \frac{2.21 \times 1 0 ^{6}}{176.5} \approx 1.25 \times 1 0^{4}$ se — yani 12,500× zyada thrust, bhale hi B 6.7× zyada efficient hai. Neeche figure dekho: efficiency (height) aur thrust (ek alag combination) same axis nahi hain.

Figure — Specific impulse Isp = v_e - g₀ — definition, physical meaning, units

L3.3

Do engines same $m_{p} = 10000 kg$ propellant jalate hain. Engine X ka $I_{s p} = 280 s$ hai, Engine Y ka $I_{s p} = 350 s$ hai. Y kitna zyada total impulse deta hai?

Recall Solution

$J = I_{s p} g_{0} m_{p}$ , toh $m_{p}$ aur $g_{0}$ fixed hone par, $Δ J = (I_{s p Y} - I_{s pX}) g_{0} m_{p}$ . $Δ J = (350 - 280) \times 9.80665 \times 10000 = 70 \times 9.80665 \times 10000 \approx 6.86 \times 1 0^{6} N\cdots$

Level 4 — Synthesis

Goal: $I_{s p}$ ko ek doosre physics law (Tsiolkovsky, burn time) ke saath combine karo.

L4.1

Ek rocket ka initial mass $m_{0} = 50000 kg$ , final (dry) mass $m_{f} = 20000 kg$ , aur $I_{s p} = 340 s$ hai. Ideal velocity change $Δ v$ nikalo.

Recall Solution

KYA: $v_{e} = I_{s p} g_{0}$ nikalo, phir Tsiolkovsky $Δ v = v_{e} ln (m_{0} / m_{f})$ mein daalo. KYUN: $I_{s p}$ wo exhaust speed set karta hai jise rocket equation mass ratio ke log se multiply karta hai. $v_{e} = 340 \times 9.80665 \approx 3334.3 m/s$ $Δ v = 3334.3 \times ln \frac{50000}{20000} = 3334.3 \times ln (2.5) \approx 3334.3 \times 0.91629 \approx 3055 m/s$

L4.2

L4.1 wala rocket apna propellant $\overset{m}{˙} = 250 kg/s$ par jalata hai. Burn kitne time ka hoga, aur thrust kya hai?

Recall Solution

Jala hua propellant $m_{p} = m_{0} - m_{f} = 30000 kg$ . Burn time: $t_{b} = m_{p} / \overset{m}{˙} = 30000/250 = 120 s$ . Thrust: $F = \overset{m}{˙} g_{0} I_{s p} = 250 \times 9.80665 \times 340 \approx 8.34 \times 1 0^{5} N \approx 0.83 MN$ .

L4.3

Consistency verify karo: L4.2 ke liye total impulse do tarike se calculate karo — $J = F t_{b}$ se aur $J = I_{s p} g_{0} m_{p}$ se — aur dikhao ki dono agree karte hain.

Recall Solution

Tarika 1 (force × time): $J = F t_{b} = 8.3357 \times 1 0^{5} \times 120 \approx 1.0003 \times 1 0^{8} N\cdots$ . Tarika 2 (definition): $J = I_{s p} g_{0} m_{p} = 340 \times 9.80665 \times 30000 \approx 1.0003 \times 1 0^{8} N\cdots$ . Dono match karte hain (koi bhi chhota gap rounding ka hai). KYUN zaroori hai: $F = \overset{m}{˙} g_{0} I_{s p}$ aur $m_{p} = \overset{m}{˙} t_{b}$ , toh $F t_{b} = \overset{m}{˙} g_{0} I_{s p} \cdot (m_{p} / \overset{m}{˙}) = g_{0} I_{s p} m_{p}$ . $\overset{m}{˙}$ cancel ho jaata hai — wohi cancellation jo $I_{s p}$ ko size-independent banati hai.

Level 5 — Mastery

Goal: multi-stage reasoning, design constraint back-solve, ya ek subtle conceptual limit.

L5.1 (design back-solve)

Ek stage ko mass ratio $m_{0} / m_{f} = 3.2$ ki limit mein $Δ v = 4000 m/s$ deliver karna hai. Engine ko minimum kitna $I_{s p}$ chahiye?

Recall Solution

KYA: Tsiolkovsky ko $v_{e}$ ke liye ulta karo, phir $I_{s p} = v_{e} / g_{0}$ . KYUN: mission $Δ v$ fix karta hai aur tankage mass ratio fix karta hai, $I_{s p}$ woh unknown reh jaata hai jo engine team ko hit karna hai. $v_{e} = \frac{Δ v}{l n ( m _{0} / m _{f} )} = \frac{4000}{l n 3.2} = \frac{4000}{1.16315} \approx 3438.8 m/s$ $I_{s p} = \frac{3438.8}{9.80665} \approx 350.7 s$ Toh engine ko $I_{s p} ≳ 351 s$ chahiye — ek accha cryogenic engine.

L5.2 (do engines, same propellant budget)

Ek spacecraft ka fixed propellant budget $m_{p} = 8000 kg$ hai (dry mass $m_{f} = 4000 kg$ ). Ek chemical engine ( $I_{s p} = 330 s$ ) aur ek ion engine ( $I_{s p} = 3200 s$ ) se milne wale $Δ v$ compare karo.

Recall Solution

Dono ka same mass ratio hai $m_{0} / m_{f} = (m_{f} + m_{p}) / m_{f} = 12000/4000 = 3$ . $Δ v_{chem} = 330 \times 9.80665 \times ln 3 \approx 3236.2 \times 1.09861 \approx 3555 m/s$ $Δ v_{ion} = 3200 \times 9.80665 \times ln 3 \approx 31381.3 \times 1.09861 \approx 3.448 \times 1 0^{4} m/s$ Ion engine same propellant se ~9.7× zyada $Δ v$ deta hai — yahi high $I_{s p}$ kharidta hai. (Pakda: L3.2 se, yeh seconds ki bajay mahino mein tiny thrust deliver karta hai.)

L5.3 (limiting case — degenerate input)

Kya hoga $I_{s p}$ us limit mein jahan ek engine bilkul propellant eject nahi karta ( $\overset{m}{˙} \to 0$ ) lekin jo gas woh eject karta, uski exhaust speed $v_{e}$ abhi bhi hai? Aur us limit mein thrust kya hoga?

Recall Solution

$I_{s p} = v_{e} / g_{0}$ unchanged rahega — yeh sirf speed $v_{e}$ par depend karta hai, kitna gas flow karta hai usp par nahi. Ek perfectly efficient engine ka well-defined $I_{s p}$ ho sakta hai bhale hi flow khatam ho jaaye. Lekin thrust $\to 0$ : $F = \overset{m}{˙} v_{e} \to 0 \cdot v_{e} = 0$ . Yahi L3 lesson ka mathematical core hai: efficiency ( $I_{s p}$ ) aur force ( $F$ ) decoupled hain. Degenerate zero-flow limit mein poori efficiency bachti hai aur saari thrust khatam ho jaati hai.

Wrap-up recall

Recall Kaun sa formula kaun se unknown ke liye?

$v_{e}$ diya hai, $I_{s p}$ chahiye? ::: $I_{s p} = v_{e} / g_{0}$ . $I_{s p}$ diya hai, $v_{e}$ chahiye? ::: $v_{e} = I_{s p} g_{0}$ . $I_{s p}$ aur $\overset{m}{˙}$ diye hain, thrust chahiye? ::: $F = \overset{m}{˙} g_{0} I_{s p}$ . $I_{s p}$ aur $m_{p}$ diye hain, total impulse chahiye? ::: $J = I_{s p} g_{0} m_{p}$ . $I_{s p}$ aur mass ratio diye hain, $Δ v$ chahiye? ::: $Δ v = I_{s p} g_{0} ln (m_{0} / m_{f})$ . Kya $I_{s p}$ location ke saath ya $\overset{m}{˙}$ ke saath badalta hai? ::: Nahi — yeh sirf $v_{e}$ par (aur constant $g_{0}$ par) depend karta hai.

Connections

Rocket Propulsion
Tsiolkovsky Rocket Equation
Thrust and Mass Flow Rate
Effective Exhaust Velocity
Impulse-Momentum Theorem
Standard Gravity g0