Solid rocket Isp derivation from grain properties: Question bank

3.3.38 · D5 · Physics › Rocket Propulsion › Solid rocket Isp derivation from grain properties

Shuru karne se pehle, teen "geometry-of-the-hardware" symbols se miliye jinpar neeche ke traps rely karte hain. Figure dekho — yeh sirf motor ke gas path ka ek labelled cross-section hai burning grain se, narrow throat se hote hue, exit tak.

Figure — Solid rocket Isp derivation from grain properties

Woh core split jis par aapko test kiya ja raha hai, ek line mein:

Woh formula jis par traps baar baar point karte hain (parent note se, yahaan show kiya gaya taaki kuch "invisible" na rahe):

True or false — justify karo

Ek firecracker jo diameter mein double hai lekin identical propellant se bana hai, uska

I_{s p}

bhi double hoga.

False. Grain double karne se burning area

A_{b}

scale hota hai, jo

\overset{m}{˙}

aur thrust double karta hai — lekin

I_{s p} = F / (\overset{m}{˙} g_{0})

mein

A_{b}

cancel ho jaata hai; same chemistry ka matlab same exhaust speed, same

I_{s p}

Do motors jinke

I_{s p}

same hain, unka thrust bhi same hona chahiye.

False.

I_{s p}

exhaust speed

c = g_{0} I_{s p}

fix karta hai, lekin

F = \overset{m}{˙} c

phir bhi

\overset{m}{˙} = ρ_{p} A_{b} r

ke saath scale karta hai; ek bada burning surface identical

I_{s p}

par zyada thrust deta hai.

Exhaust molar mass

M

kam karne se

I_{s p}

badhta hai.

True.

v_{e} \propto 1/ M

, isliye halke exhaust molecules (jaise

H_{2} O

H_{2}

) per unit energy zyada tezi se nikalta hai — yahi wajah hai ki hydrogen-rich propellants high score karte hain (dekho Nozzle Isentropic Expansion).

g_{0} = 9.81

I_{s p}

formula mein hai, woh space mein operate karte rocket ke liye chhota ho jaata hai.

False.

g_{0}

ek fixed bookkeeping constant hai jo velocity ko "seconds" mein convert karta hai; yeh local gravitational field nahi hai. Yeh

9.81

hi rehta hai har jagah, deep space mein bhi.

Pressure exponent

n = 1.2

wala propellant stably run kar sakta hai agar aap chamber pressure low rakhein.

False.

n \geq 1

ke saath equilibrium pressure

p_{c} \propto (\cdot)^{1/ (1 - n)}

ka koi stable fixed point nahi hota — koi bhi chhoti pressure rise burn rate ko throat

A_{t}

ke vent karne se tez boost kar deti hai, jo starting pressure ki parwah kiye bina thermal runaway drive karta hai.

Perfectly matched nozzle (

p_{e} = p_{a}

) ke liye,

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

exactly hota hai.

True. Pressure-thrust term

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

(jahan

A_{e}

nozzle exit area hai) zero ho jaata hai jab

p_{e} = p_{a}

, isliye saara thrust momentum thrust hai aur

I_{s p}

v_{e} / g_{0}

mein reduce ho jaata hai.

Nozzle expand karna taaki

p_{e} / p_{c} \to 0

ho,

v_{e}

ko unbounded badha deta hai.

False. Bracket

[1 - (p_{e} / p_{c})^{(γ - 1) / γ}] \to 1

, isliye

v_{e}

\frac{2 γ}{γ - 1} \frac{R T _{c}}{M}

par saturate hota hai — aap sirf finite chamber enthalpy

c_{p} T_{c}

extract kar sakte ho, kabhi zyada nahi (dekho Tsiolkovsky Rocket Equation ki kyun finite

v_{e}

Δ v

ko cap karta hai).

Chamber pressure

p_{c}

badhane se

I_{s p}

bahut zyada seedha badhta hai.

Mostly false.

I_{s p}

p_{c}

par depend karta hai sirf ratio

p_{e} / p_{c}

ke through bracket ke andar — ek weak, logarithmic-jaisi gain.

T_{c}

aur

M

dominate karte hain;

p_{c}

mainly

\overset{m}{˙}

ke through thrust set karta hai.

Error dhundho

"Star-shaped grain cores specific impulse boost karne ke liye use hote hain."

Error: star cores burning area

A_{b}

boost karte hain (hence thrust aur ek flatter thrust profile) — yeh

\overset{m}{˙}

ko affect karta hai,

I_{s p}

ko nahi. Dekho Grain Geometry and Thrust Profiles.

\overset{m}{˙} = ρ_{p} A_{b} r

, isliye agar main density

ρ_{p}

half kar doon toh

I_{s p}

half ho jaayegi."

Error:

ρ_{p}

half karne se

\overset{m}{˙}

(aur thrust) half hota hai, lekin

I_{s p} = F / (\overset{m}{˙} g_{0})

mein —

ρ_{p}

kabhi

v_{e}

mein enter nahi karta, isliye

I_{s p}

untouched rehti hai.

"Kyunki burn rate

r = a p_{c}^{n}

pressure ke saath badhti hai, zyada pressure ka matlab hai propellant ke har kg mein zyada mass burn hota hai."

Error: zyada

p_{c}

recession ko speed up karta hai, isliye zyada mass per second, lekin propellant ka propellant ke har kilogram mein burned mass chemistry se fixed hai.

r

rate change karta hai, efficiency nahi.

"Characteristic velocity

c^{*}

measure karta hai ki exhaust nozzle se kitni tezi se nikalta hai."

Error:

c^{*} = \frac{1}{Γ} R T_{c} / M

chamber-aur-throat efficiency measure karta hai (ki chemistry chamber ko kitni achi tarah pressurize karti hai), nozzle ke upstream; yahaan

Γ

fixed throat factor hai,

R

gas constant hai,

M

molar mass hai. Nozzle expansion quality ek alag thrust-coefficient factor hai. Dekho Characteristic Velocity c-star.

"Thrust find karne ke liye main sirf

\overset{m}{˙}

v_{e}

se multiply kar doon aur kaam ho jaata hai."

Error: yeh sirf momentum term deta hai. Off-design (

p_{e} \neq = p_{a}

) par aapko

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

add karna hoga, jahan

A_{e}

nozzle exit area hai; altitude par

p_{a}

girta hai aur yeh term badhta hai, isliye ise ignore karna high-altitude thrust ko underestimate karta hai.

"Equilibrium chamber pressure sirf chemistry se set hoti hai."

Error:

p_{c} = (ρ_{p} a c^{*} A_{b} / A_{t})^{1/ (1 - n)}

geometric ratio

A_{b} / A_{t}

(burning area over throat area) par bhi depend karta hai — yahi exactly woh jagah hai jahan grain geometry does chamber mein reach karta hai.

I_{s p}

ke units impulse hain, isliye bada

I_{s p}

matlab bada total push."

Error:

I_{s p}

ke units seconds hain (impulse per unit weight). Total impulse

\int F d t

hai;

I_{s p}

efficiency per kilogram hai, decoupled from kitne kilograms aap carry karte ho.

Why questions

Kyun grain geometry final

I_{s p}

formula se cancel ho jaati hai lekin thrust se nahi?

I_{s p} = F / (\overset{m}{˙} g_{0})

aur

F = \overset{m}{˙} v_{e}

; wohi

\overset{m}{˙}

jo grain geometry control karta hai numerator aur denominator dono mein appear karta hai aur cancel ho jaata hai, sirf chemistry-driven

v_{e}

bacha rehta hai. Thrust mein

\overset{m}{˙}

ka ek factor rehta hai, isliye geometry wahan survive karti hai.

Kyun pressure exponent

n

strictly 1 se kam hona chahiye stable motor ke liye?

Stability ke liye pressure feedback (zyada area burn → zyada pressure → tezi burn) ko ek fixed point par converge karna hota hai; exponent

1/ (1 - n)

sirf

n < 1

ke liye finite aur positive rehta hai.

n = 1

par yeh diverge karta hai — self-reinforcing runaway.

Kyun

g_{0}

ko local gravity ki jagah ek fixed constant choose kiya jaata hai?

Taaki

I_{s p}

ek single universal figure of merit ho jo sea level, altitude, aur vacuum mein comparable ho — kisi rocket ki efficiency sirf isliye nahi badalni chahiye kyunki aapne use alag gravity field mein shift kiya.

Kyun halka exhaust (

s ma l l M

) heavy exhaust se

I_{s p}

ke liye better hai?

Same chamber enthalpy kam, halke molecules mein share hoti hai, isliye har ek zyada speed gain karta hai:

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

. Speed, mass nahi, woh cheez hai jisko

I_{s p}

reward karta hai.

Kyun aap nozzle ko infinitely long banake infinite

v_{e}

nahi pa sakte?

Ek lamba nozzle

p_{e} \to 0

drive karta hai, lekin extractable energy finite chamber enthalpy

c_{p} T_{c}

par capped hai; bracket 1 par saturate hota hai. Convert karne ke liye simply koi aur energy nahi hai.

Kyun burn rate law chamber pressure se couple karta hai?

Zyada pressure hot gas aur heat flux ko solid surface par harder push karta hai, pyrolysis aur recession accelerate karta hai — ek empirical response jo Saint-Robert Burn Rate Law mein

r = a p_{c}^{n}

ke roop mein capture hota hai.

Edge cases

Jis instant burning surface area

A_{b} \to 0

(grain almost consumed) ho,

\overset{m}{˙}

ka kya hota hai?

\overset{m}{˙} = ρ_{p} A_{b} r \to 0

, isliye gas production aur thrust zero ki taraf collapse karte hain — yeh natural burnout/tail-off hai, chahe chemistry (hence last gas ka

I_{s p}

) unchanged rahe.

Upar boxed

I_{s p}

formula mein "ideal" case kya hai, aur yeh kab break down hota hai?

Ideal matlab optimum-expanded hai,

p_{e} = p_{a}

, isliye pressure-thrust term

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

vanish hota hai. Yeh off-design altitudes par break down hota hai jahan

p_{e} \neq = p_{a}

aur woh correction wapas aata hai.

Limit

p_{e} / p_{c} \to 1

(nozzle barely expand karta hai) mein,

v_{e}

kya hai?

Bracket

[1 - (p_{e} / p_{c})^{(γ - 1) / γ}] \to 0

, isliye

v_{e} \to 0

: koi expansion nahi matlab koi enthalpy kinetic energy mein convert nahi — essentially ek plugged nozzle.

Jab

n \to 0

, equilibrium chamber pressure kaise behave karta hai?

Exponent

1/ (1 - n) \to 1

, isliye

p_{c}

ρ_{p} a c^{*} A_{b} / A_{t}

ke saath linearly scale karta hai aur burn rate pressure-independent ho jaata hai — ek bahut stable, aasaani se tune hone wala motor.

Agar do motors identical chemistry share karte hain lekin ek ka throat area

A_{t}

double hai, toh kya change hota hai?

Bada

A_{t}

gas ko zyada aasaani se vent karta hai, equilibrium

p_{c}

ko lower karta hai (

A_{b} / A_{t}

factor ke through).

I_{s p}

barely move karta hai (sirf weak

p_{e} / p_{c}

term ke through); thrust level aur burn time shift hote hain.

Same chemistry ke saath perfect vacuum mein fire kiye motor ka

I_{s p}

kya predict karta hai sea level ke comparison mein?

Vacuum

I_{s p}

zyada hota hai kyunki

p_{a} = 0

full

p_{e} A_{e}

pressure thrust add karta hai; underlying

v_{e}

unchanged hai, lekin extra term vacuum performance ko quoted "best-case" number banata hai.