Nuclear thermal propulsion — NTR Isp ~900 s concept: Question bank

3.3.44 · D5 · Physics › Rocket Propulsion › Nuclear thermal propulsion — NTR Isp ~900 s concept

Ek fact jis par har answer tika hai: $v_{e} \propto \frac{T _{c}}{M}, I_{s p} = \frac{v _{e}}{g _{0}}$ jahan == $v_{e}$ exhaust speed hai, $T_{c}$ chamber (reactor) temperature hai, $M$ ek exhaust molecule ka molar mass hai, aur $g_{0} = 9.81 m/s^{2}$ == standard gravity hai jo sirf ek unit-converter ke roop mein use hoti hai.

True or false — justify karo

NTR chemical rocket se zyada hot hoti hai, aur yahi iski

I_{s p}

zyada hone ki main wajah hai.

False — solid-core NTRs thande chalaate hain (~2700 K vs H₂/O₂ ke liye ~3600 K); asli winner hai pure hydrogen ka tiny molar mass (

M \approx 2

18

) jo root ke andar

1/ M

mein hai.

Chamber temperature ko double karne se exhaust velocity bhi double ho jaati hai.

False —

v_{e} \propto T_{c}

, isliye

T_{c}

double karne se

v_{e}

sirf

2 \approx 1.41

guna hoti hai; temperature ek square root ke neeche hai.

9 guna halka propellant use karne se exhaust bilkul 9 guna fast ho jaati hai.

False —

v_{e} \propto 1/ M

, isliye 9 guna halka hone par

9 = 3 \times

fast milta hai, 9 guna nahi.

I_{s p}

measure karta hai ki rocket kitna force (thrust) produce karta hai.

False —

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

ek efficiency hai (effective exhaust seconds); thrust hai

F = \overset{m}{˙} v_{e}

jo mass flow

\overset{m}{˙}

par bhi depend karta hai.

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

mein jo

g_{0}

hai uska matlab hai ki specific impulse tab badal jaati hai jab rocket Moon par chalao.

False —

g_{0}

ek fixed reference constant (9.81) hai jo isliye choose kiya gaya hai taaki

I_{s p}

seconds mein aaye; ye local gravity nahi hai, isliye

I_{s p}

har jagah same rehti hai.

Dharti ki surface se payload lift karne ke liye NTR ek badhiya choice hai.

False — NTRs ka thrust-to-weight poor hota hai (bhaari reactor, modest

\overset{m}{˙}

); unka high

I_{s p}

deep-space

Δ v

mein kaam aata hai, pad chhodne ke high-thrust kaam mein nahi.

Agar same reactor rakho lekin hydrogen ki jagah steam use karo, toh

I_{s p}

lagbhag same rahegi kyunki heat source nahi badla.

False — same

T_{c}

hai lekin

M

2 se 18 ho jaata hai, isliye

v_{e}

lagbhag

9 = 3 \times

girta hai (~900 s se ~335 s tak); propellant, sirf heat source nahi,

I_{s p}

set karta hai.

Fission chemical bonds se bahut zyada energy per kg release karta hai, isliye NTR principle mein almost koi bhi

I_{s p}

reach kar sakti hai.

Practice mein False — abundant energy se tum choose kar sakte ho light hydrogen, lekin achievable

T_{c}

reactor ke melting (~2700–3000 K) se cap hai, isliye solid-core

I_{s p}

~900 s ke paas top out hoti hai.

Error dhundho

"Flow energy balance ke liye hum

\frac{1}{2} v_{e}^{2} = c_{v} T_{c}

set karte hain kyunki kinetic energy internal thermal energy ke equal hai."

Error

c_{v}

mein hai; ek flowing gas pressure (flow) work bhi karta hai, isliye correct energy currency enthalpy

c_{p} T

hai.

c_{v}

use karne se

v_{e}

γ

factor se underestimate hoti hai.

"Kyunki

v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M}

universal

R = 8.314

use karta hai, isliye hume

M

grams per mole mein plug karna chahiye."

SI mein

M

kg/mol mein chahiye, isliye hydrogen

2 \times 1 0^{- 3}

hai,

2

nahi. Grams use karne se

v_{e}

1000 \approx 31.6 \times

inflate ho jaata hai — ek bakwaas answer.

"Formula

v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M}

already assume karta hai ki nozzle sirf aadha expand karta hai."

Nahi — ye full expansion assume karta hai,

T_{e} \to 0

, taaki saari enthalpy

c_{p} (T_{c} - T_{e})

kinetic energy mein convert ho. Partial expansion se

T_{e} > 0

rehta hai aur smaller

v_{e}

milta hai.

"H₂+O₂ jalane wala chemical rocket high

I_{s p}

isliye deta hai kyunki hydrogen uska fuel hai."

Exhaust water hai (

M \approx 18

), free hydrogen nahi; heavy product molecule exactly yahi wajah hai ki chemical

I_{s p}

sirf ~450 s hai.

"Reactor ko 5000 K tak cranking karna obvious upgrade hai kyunki

v_{e}

T_{c}

ke saath badhta hai."

Physically

v_{e}

5000/2700 \approx 1.36 \times

badhega, lekin solid fuel elements 5000 K se kaafi neeche melt ho jaate hain, isliye limit materials ki hai, equation ki nahi.

Why questions

T / M

ratio, akela temperature nahi, NTR aur chemical ke beech winner kyun decide karta hai?

Kyunki

M

same root ke neeche hai aur hydrogen ka

M

water se 9 guna chhota hai; woh

9 = 3 \times

boost NTR ke thode chhote

T_{c}

ko overwhelm kar deta hai.

Nozzle ke liye enthalpy

h = c_{p} T

"energy per kg" kyun sahi hai, plain internal energy nahi?

Gas khud ko nozzle se push karta hai flow-work

p V

karte hue; enthalpy

=

internal energy

+ p V

dono ko bundle karta hai, isliye ye steady adiabatic flow mein conserved currency hai.

Agar har kg gas mein same thermal energy ho toh halka molecule zyada fast kyun nikalti hai?

Wahi energy zyada, halke molecules mein share hoti hai, isliye har ek kam mass carry karta hai lekin apna kinetic energy ka share rakhne ke liye zyada fast move karta hai — speed

1/ M

ke scale par hai.

~900 s ko "practical ceiling" kyun kaha jaata hai, fundamental physics limit nahi?

Physics higher

T_{c}

par higher

v_{e}

allow karti, lekin solid reactor cores ~2700–3000 K ke paas melt ho jaate hain, isliye engineering materials, equation nahi, wall set karta hai.

NTR ka excellent

I_{s p}

ho sakta hai lekin phir bhi Earth-launch engine ke liye bekar ho sakta hai, kyun?

I_{s p}

exhaust efficiency hai; launch ke liye high thrust-to-weight chahiye, aur bhaari reactor plus modest mass flow NTRs ko low thrust per unit weight deta hai.

Prefactor

\frac{2 γ}{γ - 1}

sirf ek bare 2 ki jagah kyun aata hai?

Ye us fraction ko account karta hai jitni thermal energy directed motion ban jaati hai;

γ = c_{p} / c_{v}

encode karta hai ki gas energy kaise store karta hai, isliye heat-to-motion conversion ki efficiency us par depend karti hai.

Edge cases

Agar nozzle sirf partially expand kare taaki

T_{e}

tiny nahi ho toh

v_{e}

ka kya hoga?

Sirf drop

c_{p} (T_{c} - T_{e})

kinetic energy mein convert hota hai, isliye

v_{e} = 2 c_{p} (T_{c} - T_{e})

chhota hoga; boxed full-expansion formula ek upper bound hai.

M \to

bahut bada (bahut heavy propellant) ki limit mein,

I_{s p}

kya approach karta hai?

v_{e} \propto 1/ M \to 0

, isliye

I_{s p} \to 0

; heavy exhaust slowly throw hoti hai aur rocket temperature ki parwah kiye bina hopelessly inefficient ho jaata hai.

Agar

T_{c}

fixed rakho aur

M

ko ek proton (pure atomic H) ke mass ki taraf shrink karo toh kya hoga?

v_{e}

1/ M

ke roop mein badhta rahega, yahi wajah hai ki atomic (dissociated) hydrogen theoretical dream propellant hai — though ise atomic rakhna aur heat contain karna hard parts hain.

Exhaust velocity formula kya predict karta hai jab

γ \to 1

(bahut saare internal energy modes wali gas)?

Factor

\frac{2 γ}{γ - 1}

blow up ho jaata hai, isliye temperature drop per degree zyada enthalpy available hoti hai — lekin H₂ jaise real light gases

γ \approx 1.4

rakhte hain, prefactor ko ~7 ke paas rakhte hue.

Agar

T_{c}

exhaust temperature

T_{e}

ke equal ho (koi expansion hi nahi), toh

v_{e}

kya hai?

Zero — koi enthalpy drop nahi toh convert karne ke liye koi energy nahi, isliye "rocket" koi directed exhaust produce nahi karta; nozzle se expansion zaroori hai.

Recall Yahan har trap ki ek-line summary

Lagbhag saare NTR misconceptions "nuclear = hot = powerful" ko over-weight karne se aate hain. Corrective mantra yeh hai: $I_{s p}$ efficiency hai ( $v_{e} / g_{0}$ ), $v_{e} \propto T / M$ hai jisme $T$ ek root ke neeche hai aur melting se capped hai, aur light molecule $M$ — temperature nahi — wajah hai ki NTR jeet jaata hai.

Connections

Specific Impulse — wo efficiency metric jis par har trap ghoomta hai.
Tsiolkovsky Rocket Equation — jahan $I_{s p}$ advantage $Δ v$ mein cash out hota hai.
De Laval Nozzle — wo device jiska enthalpy-to- $v_{e}$ conversion in questions mein probe hota hai.
Nuclear Fission — kyun energy light hydrogen par spend karne ke liye "free" hai.
Chemical Rocket Propulsion — ~450 s baseline jisse kai items compare karte hain.
Adiabatic Flow & Enthalpy — $c_{p} T$ -vs- $c_{v} T$ trap ka source.
Nuclear Electric Propulsion — sister concept jo thrust trade karke aur bhi higher $I_{s p}$ leta hai.