Typical Isp values — solid (~260s), LOX - RP1 (~311s), LOX - LH2 (~450s), ion engines (~3000s): Exercises

3.3.5 · D4 · Physics › Rocket Propulsion › Typical Isp values — solid (~260s), LOX - RP1 (~311s), LOX -

Pehle, is page par use hone wale symbols — formula block se pehle inhe zaroor padho:

Definition Is page par use hone wale Symbols

$g_{0} = 9.81 m/s^{2}$ — ek fixed unit-conversion constant (local gravity NAHI). Yeh sirf isliye exist karta hai taaki $I_{s p}$ seconds mein aaye.
$I_{s p}$ — specific impulse, seconds mein: engine + propellant ke liye ek efficiency score.
$v_{e}$ — exhaust velocity, m/s mein: jali hui gas jis physical speed se bahar nikalti hai.
$\overset{m}{˙}$ — propellant mass flow rate, kg/s mein: engine har second kitna mass bahar phenkta hai.
$F$ — thrust, newtons (N) mein: engine jo push force banata hai.
$m_{0}$ / $m_{f}$ — start (wet) mass aur final (dry) mass kg mein; unka ratio $m_{0} / m_{f}$ mass ratio hai.
$T_{c}$ — combustion-chamber temperature kelvin (K) mein: burning gas nozzle se bahar nikalne se pehle kitna garam hai. Zyada garam gas zyada tez chalti hai.
$M$ — exhaust gas ka mean molar mass g/mol (ya kg/mol) mein: average mein har exhaust molecule kitna bhaari hai. Halke molecules same heat ke liye zyada tez chalte hain.

Ab formulas, jinmein har symbol upar already define ho chuka hai:

Neeche ki figure exactly usi idea ka "wiring diagram" hai: yeh center mein master relation $v_{e} = I_{s p} g_{0}$ dikhata hai aur thrust, mass-flow, aur rocket-equation forms ki taraf teen arrows bahar jaate hain. Jab kisi problem mein stuck ho, dekho tumhe teen boxes mein se kaun sa chahiye, phir center tak arrow follow karo to dekho kya substitute karna hai.

Figure — Typical Isp values — solid (~260s), LOX - RP1 (~311s), LOX - LH2 (~450s), ion engines (~3000s)

Reference numbers (inhe yaad karo — L1 inhi par lean karta hai):

Engine	Propellant	$I_{s p}$ (s)	$v_{e} = I_{s p} g_{0}$ (m/s)
Solid	APCP	~260	~2550
Kerolox	LOX/RP-1	~311	~3050
Hydrolox	LOX/LH2	~450	~4410
Ion	Xe/Kr	~3000	~29 400

Do problems (Q7, Q12) neeche di gayi chemical scaling law bhi use karte hain. Yahan kyun yeh hold karta hai, sketch kiya gaya hai:

L1 — Recognition

Recall Solution Q1

Hum kya karte hain: bas benchmark values ki ladder padhte hain. LOX/LH2 = hydrolox ≈ 450 s. Listed chaar families mein (solid 260, kerolox 311, hydrolox 450, ion 3000) yeh chemical engines mein highest hai, lekin ion engine se kaafi neeche hai. Kyun: chemical rockets ke liye exhaust velocity $v_{e} \propto T_{c} / M$ scale karti hai (upar sketch ki gayi scaling law, jahan $T_{c}$ chamber temperature hai aur $M$ exhaust molar mass hai). Hydrolox exhaust mostly water aur leftover H $_{2}$ hai — kisi bhi chemical propellant ka sabse low molar mass $M$ — isliye ise sabse high chemical $v_{e}$ milta hai, hence highest chemical $I_{s p}$ . Answer: $\approx 450$ s, chemicals mein highest.

Recall Solution Q2

Hum kya karte hain: seconds mein $I_{s p}$ exhaust velocity divided by $g_{0}$ hai, toh multiply back karo. Yeh tool kyun: definition $I_{s p} = v_{e} / g_{0}$ literally "exhaust speed measured in units of $g_{0}$ " hai. Physical speed recover karne ke liye us operation ko undo karne ke liye hum $g_{0}$ se multiply karte hain. $v_{e} = I_{s p} g_{0} = 3000 \times 9.81 = 29, 430 m/s \approx 29.4 km/s$ Answer: $\approx 29, 400$ m/s.

L2 — Application

Recall Solution Q3

Hum kya karte hain: thrust per second pheka gaya momentum hai, $F = \overset{m}{˙} v_{e}$ , aur $v_{e} = I_{s p} g_{0}$ . Yeh tool kyun: Newton's 3rd law se har kilogram $v_{e}$ speed se jaate waqt $\overset{m}{˙} v_{e}$ momentum se push back karta hai; wahi force hai. Dekho Thrust and Mass Flow Rate. $F = \overset{m}{˙} I_{s p} g_{0} = 500 \times 311 \times 9.81 = 1, 525, 455 N \approx 1.53 MN$ Answer: $\approx 1.53 \times 1 0^{6}$ N.

Recall Solution Q4

Hum kya karte hain: $F = \overset{m}{˙} I_{s p} g_{0}$ ko $\overset{m}{˙}$ ke liye rearrange karo. Kyun: hame force aur $I_{s p}$ pata hai; sirf yeh unknown hai ki har second kitna mass bahar nikalna chahiye. $\overset{m}{˙} = \frac{F}{I _{s p} g _{0}} = \frac{12 \times 1 0 ^{6}}{260 \times 9.81} = \frac{12 \times 1 0 ^{6}}{2550.6} \approx 4705 kg/s$ Answer: $\approx 4705$ kg/s — har second lagbhag 5 tonnes propellant. Isliye boosters ~2 minutes mein khatam ho jaate hain.

Recall Solution Q5

Hum kya karte hain: wahi $F = \overset{m}{˙} I_{s p} g_{0}$ , ab bahut chhote $\overset{m}{˙}$ ke saath. $F = 5 \times 1 0^{- 6} \times 3000 \times 9.81 = 0.147 N$ Answer: $\approx 0.15$ N — kuch coins ke weight jaisa. Bahut bada $I_{s p}$ , negligible thrust. Dekho Ion and Electric Propulsion.

L3 — Analysis

Recall Solution Q6

Hum kya karte hain: Tsiolkovsky Rocket Equation $Δ v = I_{s p} g_{0} ln (m_{0} / m_{f})$ ko mass ratio ke liye invert karo. Logarithm/exponential kyun: rocket equation kehta hai ki velocity gained mass ratio ke logarithm ke proportional hai (har kg burn karna pehle se kam matter karta hai, kyunki tum abhi-bhi unburned fuel carry kar rahe ho). Ratio ko log se bahar nikalne ke liye hum iska inverse, exponential, apply karte hain. $\frac{m _{0}}{m _{f}} = e^{Δ v / (I_{s p} g_{0})}$ Solid: $\frac{3000}{260 \times 9.81} = \frac{3000}{2550.6} = 1.1762 \Rightarrow e^{1.1762} = 3.242$ . Propellant fraction $= 1 - \frac{1}{3.242} = 0.691 = 69.1%$ . Ion: $\frac{3000}{3000 \times 9.81} = \frac{3000}{29430} = 0.10194 \Rightarrow e^{0.10194} = 1.1073$ . Propellant fraction $= 1 - \frac{1}{1.1073} = 0.0969 = 9.69%$ . Answer: solid ko mass ratio $3.24$ chahiye (69% propellant); ion ko $1.11$ chahiye (sirf 9.7% propellant) — ion same kaam ~7× kam propellant mass se karta hai.

Recall Solution Q7

Hum kya karte hain: $T_{c}$ equal hone par, temperature cancel ho jaata hai aur sirf square root ke andar molar-mass ratio bachta hai. Square root kyun: har molecule ka hot gas thermal energy $T_{c}$ ke saath scale karta hai, lekin woh energy kinetic energy $\frac{1}{2} M v^{2}$ ban jaati hai, isliye $v \propto T_{c} / M$ — same energy ke liye heavier molecules slower chalte hain. $\frac{v _{e, A}}{v _{e, B}} = \frac{T _{c} / M _{A}}{T _{c} / M _{B}} = \frac{M _{B}}{M _{A}} = \frac{40}{18} = 2.222 = 1.491$ Answer: $\approx 1.49$ — light-exhaust engine ~49% faster hai. Yahi reason hai hydrolox (light H $_{2}$ O/H $_{2}$ exhaust) heavier-exhaust chemistries ko beat karta hai.

L4 — Synthesis

Recall Solution Q8

Hum kya karte hain: har engine ke liye $m_{0} = m_{f} e^{Δ v / (I_{s p} g_{0})}$ solve karo. Kyun: fixed payload aur target $Δ v$ diye jaane par, rocket equation exactly fix karta hai ki tumhe kitna propellant (hence total mass) carry karna hoga. (a) Kerolox: exponent $= \frac{9000}{311 \times 9.81} = \frac{9000}{3050.9} = 2.9500$ . $m_{0} = 20000 \times e^{2.9500} = 20000 \times 19.106 = 382, 120 kg$ . (b) Hydrolox: exponent $= \frac{9000}{450 \times 9.81} = \frac{9000}{4414.5} = 2.0388$ . $m_{0} = 20000 \times e^{2.0388} = 20000 \times 7.6813 = 153, 626 kg$ . Comment: hydrolox vehicle $\approx 154$ t vs $\approx 382$ t hai — identical payload aur $Δ v$ ke liye liftoff mass half se bhi kam. Kyunki $m_{0}$ $1/ I_{s p}$ par exponentially depend karta hai, $I_{s p}$ mein ek modest 45% jump mass ko do se zyada kam kar deta hai. Yeh exponential leverage hi reason hai kyun engineers $I_{s p}$ ke har second ke liye ladte hain. Staging and Mass Ratio se compare karo. Answer: (a) $\approx 3.82 \times 1 0^{5}$ kg; (b) $\approx 1.54 \times 1 0^{5}$ kg.

Recall Solution Q9

Hum kya karte hain: har engine ke liye $m_{0} = m_{f} e^{Δ v / (I_{s p} g_{0})}$ nikalao, phir $m_{f}$ minus karo. Hydrolox: exponent $= \frac{6000}{450 \times 9.81} = \frac{6000}{4414.5} = 1.3592$ . $m_{0} = 1000 \times e^{1.3592} = 1000 \times 3.8931 = 3893.1$ kg. Propellant $= 3893.1 - 1000 = 2893.1$ kg. Ion: exponent $= \frac{6000}{3000 \times 9.81} = \frac{6000}{29430} = 0.20388$ . $m_{0} = 1000 \times e^{0.20388} = 1000 \times 1.2262 = 1226.2$ kg. Propellant $= 1226.2 - 1000 = 226.2$ kg. Answer: hydrolox $\approx 2893$ kg propellant; ion $\approx 226$ kg — ion stage ko lagbhag 1/13th propellant chahiye. (Catch yeh hai: ~0.15 N thrust par ion burn months leta hai, jabki hydrolox minutes mein karta hai — high $I_{s p}$ time ko propellant ke liye trade karta hai.)

L5 — Mastery

Recall Solution Q10

(a) $Δ v$ : $m_{0} = m_{f} + propellant = 800 + 600 = 1400$ kg. $Δ v = I_{s p} g_{0} ln \frac{m _{0}}{m _{f}} = 3000 \times 9.81 \times ln \frac{1400}{800}$ $ln (1.75) = 0.55962$ , isliye $Δ v = 29430 \times 0.55962 = 16, 470 m/s \approx 16.5 km/s$ . Itna bada kyun: enormous $v_{e} = 29.4$ km/s matlab yeh hai ki 1.75 ka ek modest mass ratio bhi ~16 km/s deliver karta hai — is mass ratio par ek chemical rocket ke liye impossible. (b) $\overset{m}{˙}$ : $F = \overset{m}{˙} I_{s p} g_{0}$ ko invert karo: $\overset{m}{˙} = \frac{F}{I _{s p} g _{0}} = \frac{0.15}{29430} = 5.097 \times 1 0^{- 6} kg/s \approx 5.1 mg/s .$ (c) Burn time: propellant $\div$ flow rate: $t = \frac{600}{5.097 \times 1 0 ^{- 6}} = 1.1772 \times 1 0^{8} s = \frac{1.1772 \times 1 0 ^{8}}{86400} \approx 1362 days \approx 3.7 years .$ Answer: (a) $\approx 1.65 \times 1 0^{4}$ m/s; (b) $\approx 5.1 \times 1 0^{- 6}$ kg/s; (c) $\approx 1362$ days (~3.7 yr). Yeh ion-engine bargain ek problem mein hai: chhote propellant se spectacular $Δ v$ , saalo ki patient thrust ki kimat par.

Recall Solution Q11

Hum kya karte hain: same $ln (m_{0} / m_{f})$ ke saath, $Δ v = I_{s p} g_{0} ln 4$ purely $I_{s p}$ ke saath scale karta hai. Kyun: mass ratio hold karne se exponential remove ho jaata hai; $Δ v$ tab $I_{s p}$ par linearly depend karta hai, isliye unka ratio sirf $I_{s p, Y} / I_{s p, X}$ hai. $ln 4 = 1.38629$ . $Δ v_{X} = 260 \times 9.81 \times 1.38629 = 2550.6 \times 1.38629 = 3535.9 m/s$ . $Δ v_{Y} = 450 \times 9.81 \times 1.38629 = 4414.5 \times 1.38629 = 6120.0 m/s$ . Difference $= 6120.0 - 3535.9 = 2584.1 m/s$ . Ratio $= \frac{6120.0}{3535.9} = 1.731 = \frac{450}{260}$ . ✓ Answer: Y $\approx 2584$ m/s zyada gain karta hai; ratio $\approx 1.73$ (exactly $450/260$ ).

Recall Solution Q12

Hum kya karte hain: kyunki $v_{e} \propto T_{c} / M$ aur $I_{s p} = v_{e} / g_{0}$ , do engines ke $I_{s p}$ ka ratio unke $T_{c} / M$ values ke ratio ke barabar hai. Ratio kyun lete hain: jab hum ek engine ko doosre se divide karte hain toh dono $g_{0}$ aur unknown proportionality constant cancel ho jaate hain, isliye humhe constant ki value ki kabhi zaroorat nahi — sirf do $T_{c} / M$ numbers ki. $\frac{I _{s p, exp}}{450} = \frac{T _{c, exp} / M _{exp}}{T _{c, ref} / M _{ref}} = \frac{3600/22}{3400/12}$ Inner fraction: $\frac{3600/22}{3400/12} = \frac{163.636}{283.333} = 0.57754$ . Square root $= 0.75996$ . $I_{s p, exp} = 450 \times 0.75996 = 341.98 s \approx 342 s .$ Hotter kyun nahi jeet ta: experimental engine ka chamber 200 K hotter hai (3600 vs 3400 K), jo akele $v_{e}$ sirf $3600/3400 = 1.029$ (~3%) raise karta. Lekin uske exhaust molecules lagbhag do gune bhaari hain ( $22$ vs $12$ g/mol), jo $v_{e}$ ko $12/22 = 0.739$ (~26%) se cut karta hai. Ek square root ke andar, molar-mass penalty temperature gain ko swamp kar deta hai, isliye $I_{s p}$ girkar ~342 s ho jaata hai — lighter-exhaust hydrolox se neeche. Answer: $\approx 342$ s. Low exhaust molar mass high chamber temperature ko beat karta hai — yahi Exhaust Velocity and Nozzle Design ka core lesson hai.