Nuclear thermal propulsion — NTR Isp ~900 s concept: Exercises

3.3.44 · D4 · Physics › Rocket Propulsion › Nuclear thermal propulsion — NTR Isp ~900 s concept

Parent topic: NTR concept note.

Is page mein use hone wale symbols (compute karne se pehle define karo)

Definition Master formula ka har letter, simple shabdon mein

$v_{e}$ — exhaust velocity (m/s): gas nozzle se kitni tezi se nikalti hai. Yahi woh number hai jiske peeche hum bhaagte hain; bada $v_{e}$ = behtar rocket.
$T_{c}$ — combustion-chamber temperature (K): chamber ke andar propellant ka temperature, jab tak woh accelerate nahi hoti. Ek NTR mein reactor gas ko is $T_{c}$ tak garam karta hai. "Hot chamber" matlab bada $T_{c}$ .
$M$ — molar mass of the propellant (kg/mol): gas ke ek mole ki mass. Hydrogen ka $M = 2 \times 1 0^{- 3}$ kg/mol hai (bahut halka); steam ka $18 \times 1 0^{- 3}$ . Chhota $M$ = halka molecule.
$γ$ — ratio of specific heats $γ = c_{p} / c_{v}$ (dimensionless, koi units nahi). Yeh measure karta hai ki gas expand hone par kitni "springy" hai — uski kitni heat directed motion mein convert hoti hai. Ek diatomic gas jaise H₂ ke liye, $γ \approx 1.4$ ; ek triatomic gas jaise steam/CH₄ ke liye, $γ \approx 1.3$ .
$R = 8.314$ J/mol·K — universal gas constant (har gas ke liye same).
$g_{0} = 9.81$ m/s² — standard gravity, sirf $v_{e}$ ko $I_{s p}$ ke "seconds" mein convert karne ke liye use hota hai.

Formula Master formula — aur

$\frac{2 γ}{γ - 1}$ KAHAAN se aata hai $v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M}, I_{s p} = \frac{v _{e}}{g _{0}} .$ Yeh shape kyun hai? Flowing gas ke liye energy conservation se shuru karo (dekho Adiabatic Flow & Enthalpy): woh heat energy jo gas khoती hai jab woh $T_{c}$ se nozzle-exit temperature $T_{e}$ tak thandi hoti hai, kinetic energy mein badal jaati hai, $\frac{1}{2} v_{e}^{2} = c_{p} (T_{c} - T_{e}) .$ Hum enthalpy ka $c_{p}$ use karte hain ( $c_{v}$ nahi) kyunki ek flowing gas apne aap par push-work bhi karti hai. Ab do substitutions:

Full expansion: ek accha nozzle gas ko almost completely thanda kar deta hai, isliye $T_{e} \to 0$ aur $(T_{c} - T_{e}) \to T_{c}$ . Yeh best case hai ( $v_{e}$ par ek upper bound).
$c_{p}$ ko per unit mass mein likho using $c_{p} = \frac{γ}{γ - 1} \frac{R}{M}$ — yahan $γ$ , $R$ , $M$ enter karte hain.

Dono substitute karo: $\frac{1}{2} v_{e}^{2} = \frac{γ}{γ - 1} \frac{R}{M} T_{c}$ . 2 se multiply karo aur root lo → factor $\frac{2 γ}{γ - 1}$ bas " $\frac{1}{2}$ se woh 2 $* * t im es * * s p ec i f i c - h e a t b u n d l e$ \frac{\gamma}{\gamma-1}$" hai." Yeh heat ko straight-line jet mein convert karne ki thermodynamic efficiency ko package karta hai.

Level 1 — Recognition

L1.1 — Formula se scaling padho

Numbers plug kiye bina, batao ki $v_{e}$ kaise change hota hai agar tum (a) chamber temperature $T_{c}$ double karo, (b) molar mass $M$ quarter karo.

Recall Solution

Hum kya use karte hain: fingerprint $v_{e} \propto T_{c} / M$ (formula mein baki sab yahaan constants hain). (a) $T_{c}$ double karna: $v_{e} \to 2 v_{e}$ , yaani $\times 1.41$ . (b) $M$ quarter karna: $v_{e} \to 1/ (1/4) v_{e} = 4 v_{e} = \times 2$ . Square root kyun? $v_{e}$ aata hai $\frac{1}{2} v_{e}^{2} = (energy per kg)$ se, isliye speed us energy ke square root ke roop mein badhti hai — kabhi linearly nahi.

L1.2 — Metric identify karo

Ek design team report karti hai "hamare engine se 900 kN thrust nikalti hai." Kya yeh tumhe $I_{s p}$ batata hai? Ek line mein explain karo.

Recall Solution

Nahi. Thrust $F = \overset{m}{˙} v_{e}$ hai — yeh exhaust speed ko mass flow $\overset{m}{˙}$ ke saath mix karta hai. $I_{s p} = v_{e} / g_{0}$ sirf exhaust speed par depend karta hai. Tum 900 kN bade slow gas throw karke (low $I_{s p}$ ) ya thodi fast gas se (high $I_{s p}$ ) hasil kar sakte ho.

Level 2 — Application

L2.1 — 900 s number reproduce karo

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

Recall Solution

Prefactor: $\frac{2 γ}{γ - 1} = \frac{2 ( 1.4 )}{0.4} = 7.0$ . 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. $v_{e} = 7.0 \times 1.122 \times 1 0^{7} = 7.86 \times 1 0^{7} \approx 8860 m/s .$ $I_{s p} = \frac{8860}{9.81} \approx 903 s . ✓$

L2.2 — Hydrogen ki jagah Methane

Usi 2700 K reactor ko methane ke saath chalao ( $M = 16 \times 1 0^{- 3}$ kg/mol, $γ = 1.3$ ). $I_{s p}$ nikalo.

Recall Solution

Prefactor: $\frac{2 ( 1.3 )}{0.3} = 8.667$ . $\frac{R T _{c}}{M} = \frac{8.314 \times 2700}{16 \times 1 0 ^{- 3}} = 1.403 \times 1 0^{6}$ J/kg. $v_{e} = 8.667 \times 1.403 \times 1 0^{6} = 1.216 \times 1 0^{7} \approx 3487 m/s .$ $I_{s p} = \frac{3487}{9.81} \approx 355 s .$ Hydrogen se itna zyada worse kyun? $M$ 8× bada hai; sirf $1/ M$ factor ek factor $8 \approx 2.8$ ka cost karta hai.

Level 3 — Analysis

L3.1 — Target match karne ke liye temperature

Tum chahte ho ki hydrogen wala ek NTR $I_{s p} = 1000$ s tak pahunche. Melting limits ignore karte hue, kitna chamber temperature $T_{c}$ chahiye? ( $M = 2 \times 1 0^{- 3}$ , $γ = 1.4$ .)

Recall Solution

Target speed: $v_{e} = I_{s p} g_{0} = 1000 \times 9.81 = 9810$ m/s. Formula ko $T_{c}$ ke liye invert karo ( $v_{e}^{2} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M}$ ko $T_{c}$ ke liye solve karo): $T_{c} = \frac{v _{e}^{2} M}{\frac{2 γ}{γ - 1} R} = \frac{( 9810 ) ^{2} \times 2 \times 1 0 ^{- 3}}{7.0 \times 8.314} .$ Numerator: $(9810)^{2} = 9.624 \times 1 0^{7}$ ; $\times 2 \times 1 0^{- 3} = 1.925 \times 1 0^{5}$ . Denominator: $7.0 \times 8.314 = 58.2$ . $T_{c} = \frac{1.925 \times 1 0 ^{5}}{58.2} \approx 3308 K .$ Interpretation: 1000 s ke liye ~3300 K chahiye — yeh solid fuel elements ke melt point (~2900 K) se pehle hi aa jaata hai. Isi liye solid-core NTR ~900 s ke paas ruk jaata hai.

L3.2 — Kaun sa lever zyada strong hai?

903 s baseline se shuru karke, tum ya to $T_{c}$ ko 2700 K se 3200 K tak badha sakte ho, ya effective $M = 3 \times 1 0^{- 3}$ wale hydrogen mix se pure hydrogen $M = 2 \times 1 0^{- 3}$ par switch kar sakte ho. Kaun sa zyada $I_{s p}$ gain deta hai?

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

Recall Solution

Kyunki $v_{e} \propto T_{c} / M$ , hum ratios compare karte hain. Upar ki figure har option ko gray baseline ke against relative exhaust speed ki bar ke roop mein plot karti hai: Temperature lever (orange bar): factor $3200/2700 = 1.185 = 1.089$ → lagbhag $+ 8.9%$ . Mass lever (green bar): factor $3/2 = 1.5 = 1.225$ → lagbhag $+ 22.5%$ . Jaise figure mein green arrow highlight karta hai, green bar orange se zyada upar jaati hai: mass lever jeetta hai — $M$ ko 3 se 2 g/mol drop karna 500 K temperature bump se behtar hai. Yahi parent note ki poori moral hai: halka molecule > garam reactor.

Level 4 — Synthesis

L4.1 — $Δ v$ budget mein feed karo

Ek NTR upper stage ka $I_{s p} = 903$ s hai. Uska wet (fuelled) mass 40 t aur dry mass 15 t hai. Tsiolkovsky Rocket Equation $Δ v = v_{e} ln (m_{0} / m_{f})$ use karke $Δ v$ nikalo. Phir find karo ki same mass ratio se ek chemical stage at $I_{s p} = 450$ s kya $Δ v$ deta hai.

Recall Solution

Effective exhaust speed: $v_{e} = I_{s p} g_{0} = 903 \times 9.81 = 8858$ m/s. Mass ratio: $m_{0} / m_{f} = 40/15 = 2.667$ , isliye $ln (2.667) = 0.9808$ . $Δ v_{N T R} = 8858 \times 0.9808 \approx 8687 m/s .$ Chemical: $v_{e} = 450 \times 9.81 = 4414.5$ m/s, same $ln$ factor: $Δ v_{c h e m} = 4414.5 \times 0.9808 \approx 4330 m/s .$ Payoff: identical hardware mass ratio, lekin NTR almost exactly double $Δ v$ deliver karta hai — kyunki $Δ v$ $v_{e}$ mein linear hai jabki $v_{e}$ roughly double ho gaya.

L4.2 — Same $Δ v$ , kam fuel

Ek mission ko 15 t dry mass ke saath $Δ v = 6000$ m/s chahiye. NTR ( $I_{s p} = 903$ s) aur chemical stage ( $I_{s p} = 450$ s) ke liye required wet mass compare karo.

Recall Solution

Tsiolkovsky invert karo: $m_{0} = m_{f} e^{Δ v / v_{e}}$ . NTR: $v_{e} = 8858$ ; exponent $6000/8858 = 0.6774$ ; $e^{0.6774} = 1.969$ . $m_{0}^{N T R} = 15 \times 1.969 \approx 29.5 t \Rightarrow fuel \approx 14.5 t .$ Chemical: $v_{e} = 4414.5$ ; exponent $6000/4414.5 = 1.359$ ; $e^{1.359} = 3.893$ . $m_{0}^{c h e m} = 15 \times 3.893 \approx 58.4 t \Rightarrow fuel \approx 43.4 t .$ Payoff: chemical stage ko same kaam ke liye ~3× propellant chahiye. High $I_{s p}$ exponentially pay off karta hai $e^{Δ v / v_{e}}$ ke zariye — yahi deep reason hai ki NTRs deep-space transfers ke liye kyun matter karte hain.

Level 5 — Mastery

L5.1 — Material ceiling ek design proof ke roop mein

Scaling use karke prove karo ki ek solid-core NTR jo wall temperature $T_{ma x} = 2900$ K tak limited hai, hydrogen ke saath kabhi $I_{s p} = 1200$ s nahi pahunch sakta ( $M = 2 \times 1 0^{- 3}$ , $γ = 1.4$ ). Phir batao ki tum kis propulsion family mein switch karoge aur kyun.

Recall Solution

$T_{e}$ yaad karo: master-formula derivation mein $T_{e}$ nozzle-exit temperature thi — nozzle ke bilkul end par gas ka temperature, expand aur cool hone ke baad. "Full expansion" matlab nozzle gas ko jitna ho sake utna thanda karta hai, $T_{e} \to 0$ , jo ki sabse zyada heat ko speed mein convert karta hai. Isliye $T_{e} \to 0$ set karna us chamber temperature par koi bhi real nozzle jo sabse bada $v_{e}$ reach kar sakta hai, woh deta hai. Ceiling par best-case $v_{e}$ (full expansion, $T_{c} = T_{ma x} = 2900$ K): $v_{e}^{ma x} = 7.0 \times \frac{8.314 \times 2900}{2 \times 1 0 ^{- 3}} = 7.0 \times 1.206 \times 1 0^{7} = 8.44 \times 1 0^{7} \approx 9186 m/s .$ $I_{s p}^{ma x} = \frac{9186}{9.81} \approx 936 s .$ Yahan tak ki ideal limit mein bhi ceiling ~936 s deta hai, 1200 s se bahut neeche. Behtar karne ke liye tumhe exhaust speed ko ek solid melting wall se decouple karna hoga — propellant ko bina kisi solid ke chhuye garam karo. Yahi Nuclear Electric Propulsion ki logic hai (ionise + electrostatically accelerate; $I_{s p}$ hazaaron seconds ka), jo bahut zyada $I_{s p}$ ke liye thrust trade off karta hai. Gas-core NTR kyun nahi? Ek gas-core reactor fissile fuel ko khud hi hot gas banne deta hai (koi solid limit nahi), bahut zyada $T_{c}$ tak pahunchta hai — lekin yeh bahut kam mature hai. Clean argument: solid core ⇒ $T$ melting se cap ⇒ $I_{s p}$ ~900 s ke paas cap aur yeh unavoidable hai.

L5.2 — Propellant reverse-engineer karo

Ek future gas-core NTR hydrogen use karke $I_{s p} = 1500$ s achieve karta hai ( $M = 2 \times 1 0^{- 3}$ , $γ = 1.4$ ). Iska matlab kya chamber temperature hai, aur feasibility par comment karo.

Recall Solution

$v_{e} = 1500 \times 9.81 = 14715$ m/s. $T_{c} = \frac{v _{e}^{2} M}{\frac{2 γ}{γ - 1} R} = \frac{( 14715 ) ^{2} \times 2 \times 1 0 ^{- 3}}{7.0 \times 8.314} .$ $(14715)^{2} = 2.166 \times 1 0^{8}$ ; $\times 2 \times 1 0^{- 3} = 4.331 \times 1 0^{5}$ ; $/58.2 = 7442$ K. $T_{c} \approx 7440 K .$ Feasibility: ~7400 K kisi bhi solid ke melting point se bahut zyada hai — sirf ek gas-core design (fissioning plasma, koi solid wall contact mein nahi) ise sustain kar sakta hai. Yeh L5.1 ke argument ko confirm karta hai: $I_{s p}$ ko ~1000 s se aage push karna tumhe solid-core architecture bilkul abandon karne par majboor karta hai.

Recall Poori ladder ki ek-line summary

$v_{e} \propto T_{c} / M$ pehchano (L1) → SI units ke saath numbers plug karo (L2) → levers ko ratios ke roop mein compare karo (L3) → Tsiolkovsky ko feed karo jahan high $I_{s p}$ exponentially pay off karta hai (L4) → prove karo ki material ceiling tumhe ~1000 s se aage Nuclear Electric Propulsion ki taraf force karti hai (L5).

Connections

Specific Impulse — har problem yahi optimize karti hai.
Tsiolkovsky Rocket Equation — L4 ka $Δ v$ payoff.
De Laval Nozzle — enthalpy ko $v_{e}$ mein badalne wala expander.
Nuclear Fission — heat source.
Chemical Rocket Propulsion — ~450 s baseline.
Adiabatic Flow & Enthalpy — jahan $\frac{1}{2} v_{e}^{2} = c_{p} Δ T$ aata hai.
Nuclear Electric Propulsion — jahan L5 solid-core ceiling ke baad point karta hai.