Nuclear thermal propulsion — NTR Isp ~900 s concept: Worked examples

3.3.44 · D3 · Physics › Rocket Propulsion › Nuclear thermal propulsion — NTR Isp ~900 s concept

Neeche sab kuch ek hi equation pe ride karta hai, toh chalte hain use ek baar pin karte hain aur kabhi re-derive nahi karte:

The scenario matrix

Har NTR exhaust-velocity problem ek inn cells mein se kisi ek mein hoti hai. Neeche jo examples hain woh us cell ke saath tagged hain jo wo cover karte hain.

Cell	Kya vary hota hai	Physics ka stress test	Example
A	Light propellant, moderate $T_{c}$	baseline ~900 s result	Ex 1
B	Heavy propellant, same reactor	$1/ M$ penalty	Ex 2
C	$T_{c}$ badhao	$T$ (weak) gain	Ex 3
D	$T_{c} \to 0$ (degenerate: cold reactor)	limiting behaviour, sanity floor	Ex 4
E	Finite exhaust temp $T_{e} \neq = 0$	non-ideal full formula, koi shortcut nahi	Ex 5
F	Real word problem: Mars trip ke liye $Δ v$	$I_{s p}$ ko Tsiolkovsky Rocket Equation se jodhna	Ex 6
G	Exam twist: tumhe $v_{e}$ diya, $M$ poochha	formula ko invert karna, sign/root ka dhyan	Ex 7
H	Mixed propellant (H₂ + seeded gas)	effective molar mass, weighted average	Ex 8

Yahan jaanbujhkar koi negative-number ya quadrant cases nahi hain — angle problem ke unlike, har physical input ( $T$ , $M$ , $γ$ ) strictly positive hai, aur $v_{e}$ ek magnitude hai (square root, hamesha $\geq 0$ ). Cell D woh jagah hai jahan hum prove karte hain ki formula boundary par bhi theek behave karta hai.

Cell A — the baseline (light gas, moderate heat)

Worked example Ex 1 — Famous 900 s wapas nikalo

Given: pure hydrogen, $M = 2 \times 1 0^{- 3} kg/mol$ , $T_{c} = 2700 K$ , $γ = 1.4$ . Find: $v_{e}$ aur $I_{s p}$ .

Forecast: compute karne se pehle andaza lagao — $I_{s p}$ 400, 900, ya 2000 s ke kareeb hoga? (Poora topic 900 chillata hai, toh wahi predict karo aur dekho ki numbers sath deti hain ya nahi.)

Prefactor $\frac{2 γ}{γ - 1} = \frac{2 ( 1.4 )}{0.4} = 7.0$ . Yeh step kyun? Yeh random heat ko ek direction ki motion mein convert karne ki thermodynamic efficiency bundle karta hai; yeh sirf gas type par depend karta hai, temperature par nahi.
Thermal energy per kg $\frac{R T _{c}}{M} = \frac{8.314 \times 2700}{2 \times 1 0 ^{- 3}} = 1.122 \times 1 0^{7} J/kg$ . Yeh step kyun? Yeh har kilogram hot hydrogen ke andar band raw energy budget hai; denominator mein chhota $M$ hi ise itna bada banata hai.
Combine $v_{e} = 7.0 \times 1.122 \times 1 0^{7} = 7.86 \times 1 0^{7} \approx 8862 m/s$ .
Convert $I_{s p} = \frac{8862}{9.81} \approx 903 s$ .

Verify: root ke andar units hain $J/kg = m^{2} / s^{2}$ , toh root m/s deta hai ✔. Magnitude ~9 km/s — yeh cold air mein sound speed se lagbhag 26 guna hai — hydrogen sach mein bahut tezi se nikalti hai. Forecast of ~900 s match karta hai.

Cell B — $1/ M$ penalty (heavy gas, same reactor)

Worked example Ex 2 — Same 2700 K, lekin steam push karo

Given: water vapour, $M = 18 \times 1 0^{- 3} kg/mol$ , $T_{c} = 2700 K$ , $γ \approx 1.3$ . Find: $v_{e}$ , $I_{s p}$ , aur Ex 1 se ratio.

Forecast: $M$ 9 guna badha. Kyunki $v_{e} \propto 1/ M$ , andaza lagao $v_{e}$ lagbhag $9 = 3$ se giregi. Predict ~3000 m/s.

Prefactor $\frac{2 ( 1.3 )}{0.3} = 8.667$ . Yeh step kyun? Steam mein zyada internal degrees of freedom hain (lower $γ$ ), jo prefactor ko thoda badhata hai — lekin woh gain $M$ loss ke samne kuch bhi nahi.
Energy per kg $\frac{8.314 \times 2700}{18 \times 1 0 ^{- 3}} = 1.247 \times 1 0^{6} J/kg$ — exactly $1/9$ of Ex 1's value.
Combine $v_{e} = 8.667 \times 1.247 \times 1 0^{6} = 1.081 \times 1 0^{7} \approx 3288 m/s$ .
$I_{s p} = 3288/9.81 \approx 335 s$ .

Verify: ratio $v_{e}^{H_{2}} / v_{e}^{steam} = 8862/3288 \approx 2.70$ . Pure- $M$ prediction thi $9 = 3$ ; thoda kam mila kyunki steam ka higher prefactor thoda wapas laya — sanity confirmed. Lesson: heat source nahi badla, phir bhi $I_{s p}$ 903 se girke 335 ho gaya. Propellant ka molar mass sab kuch decide karta hai.

Cell C — weak $T$ gain (reactor ko crank karo)

Worked example Ex 3 — Agar reactor 5000 K hit kar sake toh?

Given: hydrogen jaise Ex 1 mein, lekin $T_{c}$ ko 2700 K se 5000 K tak badhao. Find: naya $I_{s p}$ .

Forecast: temperature almost double ho gayi. Lekin yeh square root ke neeche hai — toh guess karo gain double se kam hoga. Predict maybe +35%.

Scaling shortcut use karo $\frac{v _{e}^{new}}{v _{e}^{old}} = \frac{T _{c}^{new}}{T _{c}^{old}} = \frac{5000}{2700} = 1.852 = 1.361$ . Yeh step kyun? $T_{c}$ ke alawa sab kuch same hai, toh saari constants cancel ho jaati hain — sirf temperature ratio bachta hai, aur woh square root ke roop mein enter karta hai.
Scale up $I_{s p} = 903 \times 1.361 \approx 1229 s$ .

Verify: temperature 85% badhi, lekin $I_{s p}$ sirf 36% bada — square root ne bilkul forecast ke mutabiq ise damp kiya. Reality check: solid reactor fuel elements 2700–3000 K ke paas melt ho jaate hain, toh 5000 K solid core ke liye fiction hai. Isliye ~900 s ek material ceiling hai, physics ki nahi.

Cell D — degenerate limit (cold reactor)

Worked example Ex 4 — Reactor ko cold hone do:

$T_{c} \to 0$ Given: master formula mein $M, γ$ fixed hain, lekin $T_{c}$ 0 K ki taraf shrink ho raha hai. Find: $v_{e}$ aur $I_{s p}$ ka kya hoga, aur kya formula sane rehta hai?

Forecast: koi heat in nahi → convert karne ke liye koi energy nahi → kuch nahi nikalna chahiye. Predict $v_{e} \to 0$ .

Dependence padho $v_{e} = ek fixed positive number \frac{2 γ}{γ - 1} \frac{R}{M} \times T_{c}$ . Yeh step kyun? $T_{c}$ ko isolate karne se dikhta hai ki $v_{e}$ sirf ek constant times $T_{c}$ hai.
Limit lo jaise $T_{c} \to 0^{+}$ , $T_{c} \to 0$ , toh $v_{e} \to 0$ aur $I_{s p} \to 0$ . Yeh step kyun? Yeh confirm karta hai ki formula gracefully degrade karta hai — koi division blow-up nahi, koi negative root nahi, koi imaginary number nahi.
Saare valid inputs ke liye sign check karo: $T_{c} > 0$ , $M > 0$ , aur $γ > 1$ (toh $γ - 1 > 0$ ). Root ke neeche har factor positive hai, toh $v_{e}$ hamesha ek real, non-negative number hai. Yahan koi quadrant/sign case nahi hai — angle ke unlike, exhaust speed ek magnitude hai.

Verify: hydrogen ke liye ek tiny $T_{c} = 1 K$ par: $v_{e} = 7.0 \times \frac{8.314 \times 1}{2 \times 1 0 ^{- 3}} = 2.91 \times 1 0^{4} \approx 171 m/s$ — chhota aur positive, promise ke mutabiq 0 ki taraf ja raha hai. Physically sound: ek cold gas barely dribble karke nikalti hai.

Cell E — non-ideal full formula ( $T_{e} \neq = 0$ )

Boxed formula ne assume kiya tha ki nozzle itna zyada expand karta hai ki exhaust temperature $T_{e} \to 0$ . Real nozzles ek finite $T_{e}$ par ruk jaate hain. Tab tumhe poora enthalpy drop rakhna padta hai, koi clean shortcut nahi.

Worked example Ex 5 — Hydrogen with a realistic finite exhaust temperature

Given: hydrogen, $M = 2 \times 1 0^{- 3}$ , $γ = 1.4$ , $T_{c} = 2700 K$ , lekin nozzle gas ko sirf $T_{e} = 300 K$ tak hi cool kar pata hai (absolute zero tak nahi pahunchta). Find: $v_{e}$ , aur idealized Ex 1 se compare karo.

Forecast: hum enthalpy ka aakhri $300 K$ waste kar rahe hain, jo ki drop ka $\frac{300}{2700} \approx 11%$ hai. Square root ke neeche, guess karo $v_{e}$ us ka lagbhag aadha, ~5% giregi.

Effective temperature drop $T_{c} - T_{e} = 2700 - 300 = 2400 K$ . Yeh step kyun? Sirf enthalpy ka drop kinetic energy banta hai; jo bhi heat nozzle exit par gas mein abhi bhi hai woh energy hai jise tum extract nahi kar paye.
Plug in $v_{e} = 7.0 \times \frac{8.314 \times 2400}{2 \times 1 0 ^{- 3}} = 7.0 \times 9.977 \times 1 0^{6} = 6.984 \times 1 0^{7} \approx 8357 m/s$ .
Compare $\frac{8357}{8862} = 0.943$ → 5.7% loss.

Verify: lost fraction yeh satisfy karta hai $1 - \frac{300}{2700} = 0.889 = 0.943$ ✔ exactly. Forecast of ~5% confirmed. Lesson: boxed $v_{e} \approx 2 γ R T_{c} / [(γ - 1) M]$ ek upper bound hai; real exhaust kuch percent slower hota hai.

Cell F — real mission (word problem)

Worked example Ex 6 — 900 s tumhe Mars burn par kitna de sakta hai?

Given: ek NTR upper stage with $I_{s p} = 903 s$ (Ex 1 se). Yeh fully fuelled mass $m_{0} = 100 tonnes$ se start karta hai aur $m_{f} = 40 tonnes$ dry tak burn karta hai. Find: achievable $Δ v$ Tsiolkovsky Rocket Equation $Δ v = v_{e} ln \frac{m _{0}}{m _{f}}$ se. Phir same mass ratio ke saath $I_{s p} = 450 s$ wale chemical stage se compare karo.

Forecast: $I_{s p}$ chemical se double hua, aur $Δ v$ linearly $v_{e}$ ke saath scale karta hai, toh guess karo NTR lagbhag double $Δ v$ deta hai.

Exhaust speed $v_{e} = I_{s p} g_{0} = 903 \times 9.81 = 8858 m/s$ . Yeh step kyun? Tsiolkovsky ko $v_{e}$ m/s mein chahiye, seconds mein nahi; $I_{s p}$ definition ko undo karne ke liye $g_{0}$ se multiply karo.
Mass-ratio log $ln \frac{100}{40} = ln 2.5 = 0.9163$ . Yeh step kyun? Rocket equation kehta hai $Δ v$ sirf is dimensionless ratio aur $v_{e}$ par depend karta hai — logarithm ki wajah se extra fuel diminishing returns deta hai.
NTR $Δ v = 8858 \times 0.9163 \approx 8117 m/s$ .
Chemical, same ratio $v_{e} = 450 \times 9.81 = 4414.5$ , $Δ v = 4414.5 \times 0.9163 \approx 4045 m/s$ .

Verify: ratio $8117/4045 = 2.007 \approx 2$ ✔ — $Δ v$ double ho gaya, $I_{s p}$ doubling se match karta hai kyunki $m_{0} / m_{f}$ same tha. Trans-Mars injection ko roughly $3.6 km/s$ chahiye; NTR ka $8.1 km/s$ use margin ke saath clear karta hai, jabki same-mass chemical stage ke paas ek se zyada aisi burn ke liye barely kuch hai.

Cell G — exam twist (formula ko invert karo)

Worked example Ex 7 — Exhaust speed diya, molar mass nikalo

Given: ek mystery NTR propellant $v_{e} = 6000 m/s$ produce karta hai $T_{c} = 2700 K$ aur $γ = 1.4$ par. Find: uska molar mass $M$ . Yeh real gas kaun sa hai?

Forecast: $6000$ m/s pure hydrogen ke $8862$ se slow hai, toh $M$ 2 g/mol se bhaari hona chahiye. Guess around 4 g/mol (helium?).

Dono sides square karo $v_{e}^{2} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M}$ . Yeh step kyun? Unknown $M$ ek square root ke andar trapped hai; squaring se woh free ho jaata hai. Hum sirf positive root rakhte hain — $v_{e}$ ek speed hai, toh koi $\pm$ ambiguity nahi.
$M$ ke liye solve karo $M = \frac{2 γ}{γ - 1} \frac{R T _{c}}{v _{e}^{2}} = 7.0 \times \frac{8.314 \times 2700}{600 0 ^{2}} = 7.0 \times \frac{2.245 \times 1 0 ^{4}}{3.6 \times 1 0 ^{7}}$ .
Compute $M = 7.0 \times 6.235 \times 1 0^{- 4} = 4.36 \times 1 0^{- 3} kg/mol = 4.36 g/mol$ .

Verify: $M = 4.36 \times 1 0^{- 3}$ ko forward formula mein wapas plug karo: $v_{e} = 7.0 \times \frac{8.314 \times 2700}{4.36 \times 1 0 ^{- 3}} = 3.6 \times 1 0^{7} = 6000$ m/s ✔. ~4 g/mol ki value helium hai ( $M \approx 4$ ) — ek real (agar wasteful, kyunki He inert aur expensive hai) NTR candidate. Forecast confirmed: H₂ se bhaari, isliye slower.

Cell H — mixed propellant (effective molar mass)

Kabhi kabhi hydrogen mein thoda sa heavy gas "seed" kiya jaata hai (opacity ke liye, taaki reactor radiation use better heat kare). Mixture ek single gas ki tarah behave karta hai jiske paas ek effective molar mass hoti hai — mole-weighted average.

Worked example Ex 8 — 95% hydrogen + 5% xenon by moles

Given: $95%$ H₂ ( $M = 2 g/mol$ ) aur $5%$ xenon ( $M = 131 g/mol$ ) mole fraction se. Reactor $T_{c} = 2700 K$ , $γ = 1.4$ lo. Find: $M_{eff}$ aur $I_{s p}$ , pure hydrogen ke 903 s se compare karke.

Forecast: sirf 5% xenon — negligible lagta hai. Lekin xenon H₂ se 65 guna bhaari hai. Guess karo $M_{eff}$ noticeably drag hoga, maybe ~8 g/mol tak, aur $I_{s p}$ 903 se kaafi neeche gir jayega.

Weighted average $M_{eff} = 0.95 \times 2 + 0.05 \times 131 = 1.90 + 6.55 = 8.45 g/mol = 8.45 \times 1 0^{- 3} kg/mol$ . Yeh step kyun? Kuch bahut bhaare molecules disproportionate mass carry karte hain, toh woh average dominate karte hain — yahi "small heavy contamination" trap hai.
Exhaust speed $v_{e} = 7.0 \times \frac{8.314 \times 2700}{8.45 \times 1 0 ^{- 3}} = 7.0 \times 2.657 \times 1 0^{6} = 1.860 \times 1 0^{7} \approx 4313 m/s$ .
$I_{s p} = 4313/9.81 \approx 440 s$ .

Verify: ratio se cross-check karo $I_{s p} \propto 1/ M_{eff}$ : $903 \times 2/8.45 = 903 \times 0.4866 = 439$ s ✔. Lesson: ek mere 5% heavy seed ne $I_{s p}$ ko 903 se ~440 s tak slash kar diya — seedha chemical-rocket territory mein. Light propellant ki purity sab kuch hai; isliye real NTRs contamination se bachte hain.

Coverage recap

Figure — Nuclear thermal propulsion — NTR Isp ~900 s concept

Upar ka bar chart saare aath examples ko ek $I_{s p}$ axis par rakhta hai taaki tum story ek nazar mein dekh sako: pure hot hydrogen ~900 s ke paas baithta hai, $T$ crank use barely uthata hai, aur koi bhi mass contamination (steam, helium, xenon seed) use buri tarah neeche kheenchti hai. Molar mass, temperature nahi, woh lever hai.

Recall Saare examples mein kaun sa ek variable

$I_{s p}$ ko sabse zyada move kiya? Molar mass $M$ ::: Yeh $1/ M$ ke roop mein enter karta hai aur cases mein 2→131 g/mol span kiya, $I_{s p}$ ko 903 s se 335 s tak swing kiya. Temperature sirf $T$ ke roop mein enter karti hai aur melting se cap hai, toh woh $I_{s p}$ ko bahut kam move karti hai.

Recall Is poore topic mein koi "negative case" ya "quadrant case" kyun nahi hai?

Kyunki har physical input ( $T_{c} > 0$ , $M > 0$ , $γ > 1$ ) strictly positive hai aur $v_{e}$ ek square root hai — ek magnitude. ::: Formula valid inputs ke liye kabhi negative ya imaginary exhaust speed produce nahi kar sakta; sirf boundary case hai $T_{c} \to 0$ jisse $v_{e} \to 0$ milta hai (Cell D).

Connections

Specific Impulse — woh $I_{s p}$ jo har example compute karta hai.
Tsiolkovsky Rocket Equation — Ex 6 mein explicitly use kiya gaya $I_{s p}$ ko $Δ v$ mein cash karne ke liye.
De Laval Nozzle — Cell E ke peeche finite- $T_{e}$ expander.
Adiabatic Flow & Enthalpy — energy balance $\frac{1}{2} v_{e}^{2} = c_{p} (T_{c} - T_{e})$ jo yahan har formula ke neeche hai.
Nuclear Fission — woh heat source jo $T_{c}$ set karta hai.
Chemical Rocket Propulsion — ~450 s baseline jis par Ex 6 aur Ex 8 wapas land karte hain.
Nuclear Electric Propulsion — woh cousin jo thrust trade karta hai even higher $I_{s p}$ ke liye.