Hypergolic propellants — N2O4 - UDMH, MMH: Exercises

3.3.50 · D4 · Physics › Rocket Propulsion › Hypergolic propellants — N2O4 - UDMH, MMH

Constants jo poori notes mein use hote hain:

$g_{0} = 9.81 m/s^{2}$ (standard gravity — woh number jo Specific Impulse ko seconds mein exhaust speed mein convert karta hai).
$R = 8.314 J mol^{- 1} K^{- 1}$ (gas constant — jab bhi temperature aur energy ki trade hoti hai tab yeh aata hai).

Level 1 — Recognition

Recall Solution L1.1

Answer: (c) Hypergolic. Ek specific pair: oxidizer $N_{2} O_{4}$ (NTO), fuel MMH ( $CH_{3} -NH-NH_{2}$ ). Baaki kyun nahi? Cryogens mahino mein boil ho jaate hain — 8 saal storage ke baad bekar. Solids ko command par band aur dobara start nahi kiya ja sakta. Sirf hypergolics hi saalon tak storable hote hain aur contact par reliably restart karte hain.

Recall Solution L1.2

Answer: $N_{2}$ (nitrogen gas). Nitrogen ka triple bond $N \equiv N$ banana bahut zyada exothermic hota hai — yeh sabse strong chemical bonds mein se ek hai. Poori reaction is liye "downhill" drive hoti hai kyunki loose nitrogen atoms us ultra-stable molecule mein lock ho jaate hain. (Dekho Combustion Thermodynamics.)

Recall Solution L1.3

Answer: activation energy $E_{a}$ (woh energy hill jise ek collision ko react karne se pehle charhna padta hai). Chhota $E_{a}$ → exponent $E_{a} / R T_{0} \to 0$ → $e^{0} = 1$ → bahut chhota delay. Hypergolics ko chemically is tarah engineer kiya jaata hai ki $E_{a} \approx 0$ ho: reaction mein almost koi hill nahi hoti, isliye jaise hi liquids touch karte hain, woh fire ho jaata hai.

Level 2 — Application

Recall Solution L2.1

Formula: $v_{e} = I_{s p} g_{0}$ . Yeh specific impulse ki definition hai — yeh exhaust speed hai jo "seconds of gravity" mein measure hoti hai. $v_{e} = 290 \times 9.81 = 2844.9 m/s \approx 2845 m/s .$

Recall Solution L2.2

Step 1 — oxidizer mass. Definition se $O / F = m_{o x} / m_{f u e l}$ , isliye $m_{o x} = (O / F) \times m_{f u e l} = 1.65 \times 420 = 693 kg .$ Step 2 — total. $m_{p r o p} = m_{f u e l} + m_{o x} = 420 + 693 = 1113 kg .$ Kyun matter karta hai: tank ko dono liquids ke liye size karna hota hai; yahaan oxidizer fuel se zyada bhaari hai.

Recall Solution L2.3

Step 1: $v_{e} = I_{s p} g_{0} = 315 \times 9.81 = 3090.15 m/s$ . Step 2: final mass $m_{f} = 1500 - 900 = 600 kg$ . Step 3: Tsiolkovsky Rocket Equation: $Δ v = v_{e} ln \frac{m _{0}}{m _{f}} = 3090.15 ln \frac{1500}{600} = 3090.15 \times 0.9163 = 2831.5 m/s .$ Log kyun? Exhaust ka har kilogram ek halke hote rocket ko push karta hai, isliye gains compound hote hain — $Δ v$ amount se nahi, ratio se set hoti hai.

Level 3 — Analysis

Recall Solution L3.1

Ratio lo taaki saare messy prefactors cancel ho jaayein: $\frac{τ _{slug}}{τ _{good}} = \frac{e ^{7.0}}{e ^{2.5}} = e^{7.0 - 2.5} = e^{4.5} = 90.0.$ Interpretation: exponent mein 4.5 ki rise delay ko 90× multiply kar deti hai. Us extra time mein, unburned propellant chamber mein pool ho jaata hai — hard start ka seed. Isliye chemists sabse chhota possible $E_{a}$ dhoondthe hain (link: Arrhenius Rate Law).

Recall Solution L3.2

$\frac{v _{e}^{hyper}}{v _{e}^{cryo}} = \frac{T _{c}^{hyper} / M ^{hyper}}{T _{c}^{cryo} / M ^{cryo}} = \frac{3000/20}{3500/13} = \frac{150}{269.2} = 0.5572 = 0.7465.$ Toh hypergolic exhaust approximately 75% utni fast hai — real $I_{s p}$ ratio ( $\sim$ 320 s vs $\sim$ 450 s) se match karta hai. Kaunsa factor dominate karta hai? Molar mass. Bhale hi cryogens thoda hi zyada hot hain, unka exhaust kaafi halka hota hai ( $M = 13$ vs $20$ ). Halke molecules same thermal energy ke liye tezi se nikalte hain. Neeche figure dekho.

Figure — Hypergolic propellants — N2O4 - UDMH, MMH

Recall Solution L3.3

Same $T_{c}$ hai, isliye $v_{e} \propto 1/ M$ : $\frac{v _{e, B}}{v _{e, A}} = \frac{M _{A}}{M _{B}} = \frac{21}{19} = 1.1053 = 1.0513.$ Point B lagbhag 5.1% se jeet jaata hai. Thoda fuel-rich chalane se halke products bante hain (zyada $H_{2}$ , kam $CO_{2}$ ), jo equal temperature par bhi exhaust speed badhata hai. Isliye real engines ek aisa $O / F$ chunte hain jo $T_{c}$ maximize karne wali value se thoda neeche hota hai.

Level 4 — Synthesis

Recall Solution L4.1

Step 1 — exhaust speed. $v_{e} = 320 \times 9.81 = 3139.2 m/s$ . Step 2 — required mass ratio. Tsiolkovsky ko invert karo $Δ v = v_{e} ln (m_{0} / m_{f})$ : $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{2500/3139.2} = e^{0.7964} = 2.2176.$ Step 3 — total mass. Yahaan $m_{f} = m_{dry} = 640 kg$ (saara propellant kharcha ho jaata hai), isliye $m_{0} = 2.2176 \times 640 = 1419.3 kg, m_{p r o p} = m_{0} - m_{f} = 779.3 kg .$ Step 4 — $O / F$ se split karo. $m_{o x} = 1.9 m_{f u e l}$ aur $m_{f u e l} + m_{o x} = m_{p r o p}$ ke saath: $m_{f u e l} (1 + 1.9) = 779.3 \Rightarrow m_{f u e l} = \frac{779.3}{2.9} = 268.7 kg,$ $m_{o x} = 1.9 \times 268.7 = 510.6 kg .$ Check: $268.7 + 510.6 = 779.3 kg .$ ✓

Recall Solution L4.2

Step 1: $v_{e} = 285 \times 9.81 = 2795.85 m/s$ . Step 2: $m_{0} = 940 + 60 = 1000 kg$ , $m_{f} = 940 kg$ . Step 3: $Δ v = 2795.85 ln (1000/940) = 2795.85 \times 0.06188 = 173.0 m/s .$ Bonus — pulses ki sankhya: $60/0.25 = 240$ pulses. Kyunki propellant hypergolic hai, un 240 restarts mein se har ek bina igniter ke reliable hai — exactly isliye RCS is pair ko use karta hai.

Level 5 — Mastery

Recall Solution L5.1

Exponent explicitly likhte hain: $\frac{E _{a}}{R T _{0}}$ . (a) $E_{a} \to 0$ : exponent $\to 0$ , isliye $e^{0} = 1$ . Exponential factor bilkul gayab ho jaata hai aur $τ_{i g}$ sirf prefactor tak collapse ho jaata hai — physically possible sabse chhota delay. Yahi design target hai. (b) $T_{0} \to 0^{+}$ : denominator $R T_{0} \to 0$ , isliye $E_{a} / R T_{0} \to + \infty$ , aur $e^{+ \infty} \to \infty$ . Delay diverge ho jaata hai — propellant light hone se maana kar deta hai. Physically: ek cold-soaked thruster (deep-space shadow, koi heaters nahi) ignite hone mein fail ho sakta hai ya jab finally hota hai toh violent hard start produce karta hai. Exactly isliye hypergolic tanks aur lines par heaters lagate hain.

Recall Solution L5.2

$τ_{i g} = C e^{E_{a} / R T_{0}}$ ko $E_{a}$ ke respect mein differentiate karo ( $T_{0}$ constant maano): $\frac{d τ _{i g}}{d E _{a}} = C e^{E_{a} / R T_{0}} \cdot \frac{1}{R T _{0}} = τ_{i g} \cdot \frac{1}{R T _{0}} .$ Isliye fractional sensitivity hai $\frac{1}{τ _{i g}} \frac{d τ _{i g}}{d E _{a}} = \frac{1}{R T _{0}} .$ Interpretation: $E_{a}$ ko $R T_{0}$ joules per mole badhane par delay $e$ (factor of 2.718) se multiply ho jaata hai. $T_{0} = 300 K$ par, $R T_{0} = 8.314 \times 300 = 2494.2 J/mol \approx 2.49 kJ/mol$ . Toh har extra $\sim$ 2.5 kJ/mol activation energy ignition delay ko nearly triple kar deti hai — ek razor-thin margin, isliye hypergolic chemistry itni carefully tune hoti hai.

Recall Solution L5.3

$a = 1$ fix karo aur har element conserve karo.

Carbon: $2 a = c \Rightarrow c = 2.$
Hydrogen: $8 a = 2 d \Rightarrow d = 4.$
Oxygen: $4 b = 2 c + d = 4 + 4 = 8 \Rightarrow b = 2.$
Nitrogen: $2 a + 2 b = 2 e \Rightarrow 2 + 4 = 2 e \Rightarrow e = 3.$ $C_{2} H_{8} N_{2} + 2 N_{2} O_{4} \to 2 CO_{2} + 4 H_{2} O + 3 N_{2}$ Sanity check — dono taraf atoms gino: C: $2 = 2$ ✓; H: $8 = 8$ ✓; N: $2 + 4 = 6$ , right side $3 \times 2 = 6$ ✓; O: $2 \times 4 = 8$ , right side $2 \times 2 + 4 = 8$ ✓. Saara nitrogen stable $N_{2}$ ke roop mein nikalta hai, jaise thermodynamics demand karta hai.

Connections

Tsiolkovsky Rocket Equation — L2–L4 mein use hone wala $Δ v = v_{e} ln (m_{0} / m_{f})$ .
Specific Impulse — jahaan $v_{e} = I_{s p} g_{0}$ aata hai.
Arrhenius Rate Law — ignition delay ke peeche $e^{- E_{a} / R T}$ .
Cryogenic Propellants — LOX-LH2 — L3.2 mein high- $I_{s p}$ comparison.
Reaction Control Systems (RCS) — L4.2 ka restart-budget context.
Combustion Thermodynamics — kyun N₂ energy carry karta hai.
Parent: 3.3.50 Hypergolic propellants — N2O4 - UDMH, MMH (Hinglish)