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

3.3.5 · D5 · Physics › Rocket Propulsion › Typical Isp values — solid (~260s), LOX - RP1 (~311s), LOX -

Is bank mein the parent topic ke ideas aur bhi sharp ho jaate hain. Agar koi line chubhe toh ye prerequisites dobara padh lo: Tsiolkovsky Rocket Equation, Thrust and Mass Flow Rate, Exhaust Velocity and Nozzle Design, Ion and Electric Propulsion, Combustion Chamber Temperature, aur Staging and Mass Ratio.

Symbol kit — traps se pehle yeh padh lo

Neeche ke traps mein letters ka ek chhota sa set baar baar aata hai. Yahan har ek plain words mein diya hai, taaki koi line surprise na kare. Figure dekho: ye ek cartoon hai ek rocket ka jo gas peechhe phenk raha hai.

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

Definition Symbols, ek ek karke

== $F$ (thrust)== — rocket par forward push, newtons (N) mein measure hota hai. Figure mein yeh pale-yellow arrow hai jo forward (upar) point kar raha hai.
== $\overset{m}{˙}$ (mass flow rate)== — har second mein kitne kilograms propellant nozzle se nikalta hai, kg/s mein. $m$ ke upar chhota dot bas "rate of, per second" ka matlab hai. Figure mein yeh chalk-blue gas stream ki width hai.
== $v_{e}$ (exhaust velocity)== — woh gas kitni tez nikalti hai, m/s mein. Pink arrow jo backward (neeche) point kar raha hai.
== $g_{0}$ == — ek fixed number, $9.81 m/s^{2}$ . Yeh woh gravity NAHI hai jahan tum ho; isse ek printed conversion factor samjho, jo neeche explain hai.
== $I_{s p}$ (specific impulse)== — engine ka efficiency score, seconds mein. Define hai $I_{s p} = v_{e} / g_{0}$ ke roop mein.
== $m_{0}$ == — rocket ki starting (wet) mass: structure + payload + saara propellant, kg mein.
== $m_{f}$ == — burn ke baad rocket ki final (dry) mass: propellant khatam hone ke baad jo kuch bhi bachta hai, kg mein. Toh $m_{0} > m_{f}$ hamesha, aur $m_{0} - m_{f}$ woh propellant hai jo jal gaya.
== $Δ v$ == — burn se rocket ki speed mein jo kul change aata hai, m/s mein. Greek $Δ$ ("delta") ka bas matlab hai "change in."
== $T_{c}$ == — chamber temperature, kelvin mein: gas nikalne se pehle kitna hot hai.
== $M$ == — exhaust ka molar mass: exhaust molecules ke ek "mole" (ek fixed count) ki mass, kg/mol mein. Chhota $M$ = halke molecules jaise H $_{2}$ ; bada $M$ = bhaari molecules.
== $γ$ (gamma) aur $R$ == — gas ki ek property ( $γ$ , ye kitni springy hai) aur universal gas constant ( $R$ ); dono neeche diye full $v_{e}$ formula mein aate hain.

Teen formulas, aur YE kyon aisa dikhte hain

Tum exactly teen relations use karoge. Yahan se har ek kahan se aata hai, simple words mein, taaki beginner bhi samajh sake.

Formula 3. Hot gas ki exhaust velocity — FULL version

Traps mein use hone wala shortcut $v_{e} \propto T_{c} / M$ asal nozzle result ka ek simplification hai: $v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M} [1 - (\frac{p _{e}}{p _{c}})^{\frac{γ - 1}{γ}}]$ Piece by piece padhte hain: zyada hot gas ( $T_{c}$ up) aur halke molecules ( $M$ down) dono $v_{e}$ badhate hain — yahi hai $T_{c} / M$ core. $γ$ terms batate hain ki jab gas expand hoti hai toh uski heat kitni efficiently motion mein convert hoti hai. $p_{e} / p_{c}$ wala bracket (nozzle exit pressure over chamber pressure) kehta hai: gas ko jitna zyada expand karke pressure drop karne do, utni zyada uski energy speed ban jaati hai. Toh nozzle expansion ratio bhi matter karta hai — bada bell zyada $v_{e}$ nikalta hai. Jab hum $v_{e} \propto T_{c} / M$ likhte hain toh hum $γ$ , $R$ , aur pressure bracket ko freeze karke sirf un do dials ko dekh rahe hain jo propellants ke beech sabse zyada differ karte hain. (Zyada detail: Exhaust Velocity and Nozzle Design.)

True or false — justify karo

Chand par bheji gayi rocket ka

I_{s p}

zyada hoga kyunki Moon gravity weak hai.

False.

I_{s p} = v_{e} / g_{0}

mein

g_{0}

ek fixed defined constant hai (

9.81 m/s^{2}

), ek unit-conversion factor — local gravitational field nahi.

I_{s p}

engine aur propellant ki property hai, toh Moon par bhi wahi rahega.

Engine ka

I_{s p}

double karo (same mass ratio) toh achievable

Δ v

bhi double ho jaata hai.

True. Kyunki

Δ v = I_{s p} g_{0} ln (m_{0} / m_{f})

hai aur log term mass ratio se fixed hai,

Δ v

linearly

I_{s p}

ke saath scale karta hai.

Higher

I_{s p}

wala engine hamesha zyada thrust produce karta hai.

False. Thrust hai

F = \overset{m}{˙} v_{e}

; yeh depend karta hai ki tum kitna mass phenko. Ion engines ka

v_{e}

bahut bada hota hai lekin

\overset{m}{˙}

tiny hota hai, toh unka thrust sirf millinewtons hota hai.

Solid rocket motor ek kharab engine hai kyunki uska

I_{s p}

sirf ~260 s hai.

False. Kam

I_{s p}

ka matlab sirf kam efficiency-per-kilogram hai; solids enormous thrust deliver karte hain, simple, storable aur reliable hain — liftoff ke liye ideal boosters jahan brute force matter karta hai.

I_{s p}

seconds mein aur

v_{e}

m/s mein bilkul same physical information carry karte hain.

True. Ye

v_{e} = I_{s p} g_{0}

se locked hain, toh ek jaanna doosra deta hai; seconds bas exhaust velocity ka units-cancelling rescaling hai.

Hydrolox (LOX/LH2) kerolox se zyada mainly isliye better hai kyunki woh hotter jalta hai.

False (mostly). Kyunki

v_{e} \propto T_{c} / M

hai, hydrolox kaafi cases mein kerolox se cooler jalta hai — woh jeet ta hai kyunki uska exhaust (H

_{2}

O plus excess H

_{2}

) bahut low molar mass

M

ka hota hai, aur chhota

M

v_{e}

badhata hai.

Agar chamber temperature hamesha badhate raho, toh chemical rocket ka

I_{s p}

ion engine se zyada ho sakta hai.

False in practice. Materials melt ho jaate hain, toh

T_{c}

capped hai, aur

M

hydrogen-rich exhaust se neeche nahi ja sakta — saath milke ye chemical

I_{s p}

ko ~450 s ke paas cap karte hain, ion engines ke ~3000 s se bahut kam.

Do engines mein same

\overset{m}{˙}

hai lekin alag

I_{s p}

— dono same thrust produce karte hain.

False.

F = \overset{m}{˙} I_{s p} g_{0}

hai, toh equal mass flow par higher-

I_{s p}

engine proportionally zyada thrust produce karta hai. (Ion engines weak hain kyunki unka

\overset{m}{˙}

tiny hai, unka

I_{s p}

nahi.)

Rocket ka

I_{s p}

sea level aur vacuum mein same hota hai.

False. Kyunki real thrust mein pressure term

(p_{e} - p_{a}) A_{e}

hota hai, sea level par ambient air pressure

p_{a}

effective thrust aur isliye

I_{s p}

kam karta hai; vacuum mein

p_{a} \to 0

aur

I_{s p}

apni higher vacuum value tak badh jaata hai.

Error dhundho

"Kyunki

I_{s p} = F / (\overset{m}{˙} g_{0})

aur

g_{0}

gravity hai, toh vacuum mein test kiya gaya engine jahan koi gravity nahi hai, uska

I_{s p}

infinite hoga."

Galat:

g_{0}

ek defined constant

9.81 m/s^{2}

hai, formula mein hamesha present hai chahe actual gravitational environment kuch bhi ho. Kuch bhi infinite nahi jaata.

v_{e} \propto T_{c} \cdot M

, toh bhaari exhaust molecules faster exhaust dete hain."

Relation hai

v_{e} \propto T_{c} / M

— molar mass

M

denominator mein hai. Bhaari exhaust ka matlab hai lower

v_{e}

, yahi exact reason hai ki heavy solid-motor exhaust low

I_{s p}

deta hai.

"Ion engine ka 29 km/s exhaust matlab yeh heavy payloads ko launch pad se utha sakta hai."

High exhaust speed akele kuch nahi uthata; thrust

F = \overset{m}{˙} v_{e}

woh hai jo weight se ladhta hai, aur ion engine ka

\overset{m}{˙}

itna chhota hai ki

F

fraction of a newton hota hai — liftoff thrust ke kahin paas nahi.

"LOX/RP-1 ka

I_{s p}

311 s hai, toh uski exhaust velocity 311 m/s hai."

Tum

g_{0}

se multiply karna bhool gaye. Exhaust velocity hai

v_{e} = I_{s p} g_{0} = 311 \times 9.81 \approx 3050 m/s

, roughly 3 km/s.

"Kyunki

Δ v = I_{s p} g_{0} ln (m_{0} / m_{f})

hai, double thrust wala engine double

Δ v

deta hai."

Thrust

Δ v

formula mein appear nahi karta — sirf

I_{s p}

aur mass ratio appear karte hain. Thrust set karta hai ki tum

Δ v

kitni fast gain karte ho, total achievable nahi.

I_{s p}

thrust divided by mass flow rate hai, toh uske units already seconds hain."

Bilkul nahi —

F / \overset{m}{˙}

ke units m/s hain (woh

v_{e}

hai). Tumhe weight flow rate

\overset{m}{˙} g_{0}

se divide karna hoga; extra

g_{0}

(m/s²) hi woh hai jo plain seconds mein cancel down karta hai.

"Hum LOX/LH2 fuel-rich accidentally run karte hain kyunki mixing imperfect hai."

Galat — yeh deliberate hai. Excess H

_{2}

exhaust molar mass

M

kam karta hai (

v_{e}

badhata hai) aur engine protect karne ke liye heat absorb karta hai; fuel-rich mixture ek engineered

I_{s p}

optimization hai, carelessness nahi.

Why questions

I_{s p}

deliberately weight flow rate use karke define kyon kiya gaya, mass flow rate ki jagah?

\overset{m}{˙} g_{0}

se divide karne par time ke siwa har unit cancel ho jaata hai, metric ya imperial dono systems mein same number milta hai — ek historical convenience taaki dono systems ke engineers engines directly compare kar sakein.

Thrust ko weight flow rate se divide karne par seconds kyun milte hain, meters nahi?

F

N = kg·m/s² mein hai, aur

\overset{m}{˙} g_{0}

hai (kg/s)(m/s²) = kg·m/s³; ratio mein

1/ (1/ s) = s

bachta hai, toh mass, length aur ek time factor sab cancel ho jaate hain.

Rocket equation mein simple product ki jagah logarithm kyon hai?

Kyunki rocket jalte waqt mass lose karta hai, toh exhaust ka baad ka har kilogram ek aur halke vehicle ko accelerate karta hai; un contributions ko add karna jab mass

m_{0}

m_{f}

tak girta hai,

ln (m_{0} / m_{f})

deta hai.

Ion engines us ~450 s ceiling se kaise bachte hain jo saare chemical rockets ko trap karti hai?

Chemical engines molecular bonds ki fixed energy se limited hain (jo

T_{c}

set karta hai) aur achievable exhaust

M

se. Ion engines ek bahari source se electrical energy add karke ions accelerate karte hain, toh woh combustion chemistry se bilkul bound nahi hain.

Deep-space cruising ke liye high

I_{s p}

kyun prized hai lekin launch ke liye nahi?

Space mein time hota hai, toh ek weak but efficient thrust slowly large

Δ v

build karta hai thode propellant se. Launch par tumhe gravity ko turant beat karna hota hai, jiski liye large thrust chahiye — low-

I_{s p}

, high-

\overset{m}{˙}

chemical engines favor hote hain.

Same

Δ v

maneuver ion engine se bahut kam propellant kyun leta hai?

Mass ratio hai

m_{0} / m_{f} = e^{Δ v / (I_{s p} g_{0})}

; bada

I_{s p}

exponent ghata deta hai, toh exponential mass ratio 1 ke karib aa jaata hai, matlab sirf ek chhota propellant fraction spend hota hai.

Longer, wider nozzle bell same propellant ke saath

v_{e}

kyon badhata hai?

Full formula ka pressure bracket

[1 - (p_{e} / p_{c})^{(γ - 1) / γ}]

badhta hai jab gas lower exit pressure

p_{e}

tak expand hoti hai; bada bell zyada expansion allow karta hai, zyada thermal energy exhaust speed mein convert hoti hai.

I_{s p}

ordering solid < kerolox < hydrolox physically kyon expected hai?

Sab

v_{e} \propto T_{c} / M

follow karte hain: solids mein heavy metal-oxide exhaust hota hai (bada

M

), kerolox mein moderate-

M

_{2}

_{2}

O, aur hydrolox mein sabse halka exhaust (H

_{2}

O + excess H

_{2}

), toh

v_{e}

aur isliye

I_{s p}

usi order mein climb karta hai.

Edge cases

Agar

\overset{m}{˙} \to 0

ho (engine barely propellant trickle kar raha hai), toh thrust aur

I_{s p}

ka kya hoga?

Thrust

F = \overset{m}{˙} v_{e} \to 0

(woh vanish ho jaata hai), lekin

I_{s p} = v_{e} / g_{0}

unchanged rehta hai kyunki woh depend nahi karta kitna mass flow ho — sirf exhaust speed par. Yahi ion-engine limit hai: near-zero thrust, high

I_{s p}

Agar engine ki exhaust velocity exactly

g_{0} \times 1 s

ho, toh uska

I_{s p}

kya hoga?

Exactly 1 second, kyunki

I_{s p} = v_{e} / g_{0}

; yeh dikhata hai ki seconds mein

I_{s p}

literally hai "tumhari exhaust speed

g_{0}

ke kitne multiples hai."

Maan lo ek chemical engine same

T_{c}

par pure atomic hydrogen exhaust kar sake —

I_{s p}

ka kya hoga?

M

apni lowest possible value tak drop kar jaata hai, toh

v_{e} \propto T_{c} / M

badhta hai,

I_{s p}

ko chemical range ki extreme upper edge ki taraf push karta hai — isi liye hydrogen-rich exhaust itni mehnat se chase ki jaati hai.

Do maneuvers ko same

Δ v

chahiye, ek solid stage se aur ek hydrolox se. Agar dono ki dry mass identical ho toh kaun zyada propellant leta hai, aur kyon?

Solid stage (lower

I_{s p}

) ko bada mass ratio

e^{Δ v / (I_{s p} g_{0})}

chahiye, toh woh same

Δ v

reach karne ke liye substantially zyada propellant jalata hai.

Fixed

Δ v

ke liye jab

I_{s p} \to \infty

ho toh mass ratio

m_{0} / m_{f}

ka limiting behaviour kya hai?

Exponent

Δ v / (I_{s p} g_{0}) \to 0

ho jaata hai, toh

m_{0} / m_{f} \to e^{0} = 1

— infinitely efficient engine ko practically koi propellant nahi chahiye us

Δ v

ke liye.

Kya thrust kabhi negative (backflow) ja sakta hai, aur iska physically kya matlab hoga?

Haan pressure sense mein: agar sea level par nozzle over-expanded ho, toh exit pressure

p_{e}

ambient

p_{a}

se neeche gir jaati hai,

(p_{e} - p_{a}) A_{e}

term negative ho jaata hai aur net thrust reduce ho jaata hai; severe cases mein flow nozzle wall se separate ho jaati hai, isi liye sea-level nozzles vacuum wale se chhote rakhe jaate hain.

Ek booster Earth par fire hota hai aur ek identical unit Mars par. Uska

I_{s p}

(spec sheet par printed) alag hoga?

g_{0}

-based spec value same hai, lekin effective

I_{s p}

thoda differ karta hai kyunki Mars ka thin atmosphere (

p_{a}

low) exhaust ko Earth se kam fight karta hai — yeh ek pressure effect hai, gravity effect nahi.

Agar chamber temperature

T_{c}

zero ki taraf gire (thanda, barely-reacting mix), toh

v_{e}

ka kya hoga?

Kyunki

v_{e} \propto T_{c} / M

hai,

T_{c} \to 0

ke saath

v_{e} \to 0

; koi thermal energy expand karne ke liye nahi, gas slowly nikalti hai aur

I_{s p}

collapse ho jaata hai — isi liye hot chamber maintain karna essential hai.

Recall Har trap ka one-line summary

Efficiency ( $I_{s p}$ ) force nahi hai ( $F = \overset{m}{˙} v_{e}$ ); $g_{0}$ ek constant hai local gravity nahi; exhaust speed full nozzle formula follow karta hai lekin halke exhaust ( $M$ down) aur hot chamber ( $T_{c}$ up) dominate karte hain; aur ambient air pressure — gravity nahi — woh hai jo $I_{s p}$ ko altitude ke saath shift karta hai.