Hypergolic propellants — N2O4 - UDMH, MMH: Worked examples

3.3.50 · D3 · Physics › Rocket Propulsion › Hypergolic propellants — N2O4 - UDMH, MMH

Yahan sab kuch teen imported facts par tikaa hai. Main unhe dobara state karta hoon taaki koi bhi symbol bina wajah ka na lage.

Formula Teen tools (dobara state kiye, taaki kuch bhi assumed na ho)

(1) Specific impulse se exhaust speed — dekho Specific Impulse: $v_{e} = I_{s p} g_{0}, g_{0} = 9.81 m/s^{2}$ Yahan $I_{s p}$ ("specific impulse", seconds mein) sirf exhaust speed ki ek repackaging hai; constant $g_{0}$ se multiply karne par ye metres per second mein convert ho jaata hai. $v_{e}$ woh speed hai jis par jaali gas nozzle se nikalti hai.

(2) Tsiolkovsky — dekho Tsiolkovsky Rocket Equation: $Δ v = v_{e} ln \frac{m _{0}}{m _{f}}$ $m_{0}$ = burn se pehle mass (dry rocket + saara propellant), $m_{f}$ = baad mein mass (sirf dry rocket, plus jo bacha). Symbol $ln$ natural logarithm hai — woh sawaal ki " $e$ ko kaunsi power par raise karein ki ye number mile?". Ye isliye aata hai kyunki velocity barabar fractional steps mein build hoti hai jab mass phenkaa jaata hai, aur barabar fractional steps ko jodna exactly wahi hai jo logarithm karta hai.

(3) Mixture ratio bookkeeping: $O / F = \frac{m _{o x}}{m _{f u e l}}, m_{p r o p} = m_{o x} + m_{f u e l}$ $O / F$ ("oh over eff") simply kitne kilograms oxidizer per kilogram fuel hai.

Scenario matrix

Is topic mein jo bhi problem aa sakti hai woh in cells mein se kisi ek mein aati hai. Neeche ke worked examples mein se har ek ke saath woh cell tagged hai jo use cover karta hai.

#	Cell (case class)	Tricky kyun hai	Covered by
A	Ordinary Δv — moderate mass ratio	plug-and-chug baseline	Ex 1
B	Large mass ratio limit ( $m_{0} / m_{f} ≫ 1$ )	$ln$ dheere badhta hai — diminishing returns	Ex 2
C	Small mass ratio limit ( $m_{0} / m_{f} \to 1$ )	$ln (1 + x) \approx x$ — tiny burns	Ex 3
D	Degenerate / zero input	$Δ v = 0$ jab kuch nahi jalta	Ex 3 (part b)
E	Mixture-ratio bookkeeping — tank split solve karo	total mass ko $O / F$ se split karo	Ex 4
F	Ignition-delay ratio (the exponential)	ratio prefactor cancel kar deta hai	Ex 5
G	Real-world word problem (RCS pulse train)	bahut saare tiny burns, Reaction Control Systems (RCS)	Ex 6
H	Exam twist — "hotter but heavier" $v_{e} \propto T_{c} / M$	temperature badha par $I_{s p}$ ghata	Ex 7
I	Cross-compare with cryogenic	same Δv, different $I_{s p}$	Ex 8

Figure — Hypergolic propellants — N2O4 - UDMH, MMH

Upar ki figure is page ka map hai: $ln$ curve ka flat middle part woh jagah hai jahan ordinary rockets rehte hain (cell A), far right "diminishing returns" ki tail hai (cell B), aur $m_{0} / m_{f} = 1$ ke paas wala region tiny-burn limit hai (cells C, D). Baar baar isko dekhte rehna.

Ex 1 — Cell A · ordinary case

Worked example Ek moderate kick burn

Ek probe stage: $m_{0} = 1200 kg$ , usne N₂O₄/MMH ka $700 kg$ jaala, $I_{s p} = 320 s$ . $Δ v$ nikalo.

Forecast: Compute karne se pehle andaaza lagao — mass ratio $1200/500 = 2.4$ hai. Kya tumhe lagta hai $Δ v$ , $v_{e}$ se bada hoga ya chhota? (Kyunki $ln 2.4 \approx 0.88 < 1$ , expect karo chhota $v_{e}$ se.)

Step 1. $v_{e} = I_{s p} g_{0} = 320 \times 9.81 = 3139 m/s$ . Ye step kyun? Tool (1): $I_{s p}$ seconds mein Tsiolkovsky mein useless hai jab tak m/s mein convert na ho.

Step 2. $m_{f} = m_{0} - m_{p r o p} = 1200 - 700 = 500 kg$ . Ye step kyun? Tsiolkovsky ko final mass chahiye, aur jaala hua propellant chala jaata hai.

Step 3. $Δ v = v_{e} ln \frac{m _{0}}{m _{f}} = 3139 ln \frac{1200}{500} = 3139 \times 0.8755 = 2748 m/s$ . Ye step kyun? Tool (2): fractional mass phenkaa gaya, $ln$ se sum kiya.

Verify: Units: $m/s \times (dimensionless log) = m/s$ ✓. Sanity: $2748 < 3139 = v_{e}$ exactly jaise forecast kiya tha, kyunki $ln 2.4 < 1$ . ✓

Ex 2 — Cell B · large-ratio limit

Worked example Zyada propellant se kyun nahi cheat kar sakte

Same stage, same $I_{s p} = 320 s$ , lekin ab dry mass sirf $m_{f} = 120 kg$ hai aur $m_{0} = 1200 kg$ (mass ratio $= 10$ ). Phir koi tanks double kar deta hai: $m_{0} = 2400 kg$ , phir bhi $m_{f} = 120$ (ratio $= 20$ ). Dono $Δ v$ compare karo.

Forecast: Tumne propellant double kiya. Kya $Δ v$ bhi double hua? Andaaza lagao.

Step 1. Ratio 10: $Δ v = 3139 ln 10 = 3139 \times 2.3026 = 7228 m/s$ . Ye step kyun? Comparison ke liye baseline.

Step 2. Ratio 20: $Δ v = 3139 ln 20 = 3139 \times 2.9957 = 9403 m/s$ . Ye step kyun? Naye ratio ke saath phir Tool (2).

Step 3. Doubling se gain: $9403 - 7228 = 2175 m/s$ — sirf $30%$ increase $100%$ propellant increase ke liye. Ye step kyun? Yahi "diminishing returns" ka sabak hai: $ln$ is tarah badhta hai $ln (2 x) = ln x + ln 2$ , toh doubling karne se ek fixed $3139 ln 2 \approx 2176 m/s$ add hota hai chahe kahin se bhi shuru karo.

Verify: Predicted fixed add-on $= 3139 ln 2 = 3139 \times 0.6931 = 2176 m/s$ , jo upar ke $2175$ se match karta hai (rounding). ✓ Figure mein far-right flattening dekho — woh flat slope exactly yahi effect hai.

Ex 3 — Cells C & D · tiny burns aur zero case

Worked example Near-1 limit aur degenerate zero

(a) Ek RCS thruster ek $500 kg$ spacecraft ko nudge karta hai, sirf $2 kg$ N₂O₄/MMH expel karke, $I_{s p} = 290 s$ . $Δ v$ nikalo. (b) Agar thruster fire hi na kare aur $0 kg$ jaale toh $Δ v$ kya hoga?

Forecast: (a) ke liye mass ratio $500/498 = 1.004$ hai — barely 1 se zyada. Andaaza: exact $ln$ ya shortcut $ln (1 + x) \approx x$ yahan theek rahega?

Step 1 (a). $v_{e} = 290 \times 9.81 = 2844.9 m/s$ . Ye step kyun? Tool (1).

Step 2 (a). $m_{f} = 500 - 2 = 498 kg$ , ratio $= 500/498 = 1.004016$ . Ye step kyun? Log argument set up karo.

Step 3 (a). $Δ v = 2844.9 ln (1.004016) = 2844.9 \times 0.0040080 = 11.40 m/s$ . Ye step kyun? Tool (2). Note karo $ln (1 + x) \approx x$ ke saath $x = m_{p r o p} / m_{f} = 2/498 = 0.004016$ deta hai $2844.9 \times 0.004016 = 11.43$ — sirf $0.3%$ andar. Tiny burns ke liye log bilkul skip kar sakte ho.

Step 4 (b). $0$ kg jaao $\Rightarrow m_{f} = m_{0} = 500 \Rightarrow ln (500/500) = ln 1 = 0 \Rightarrow Δ v = 0$ . Ye step kyun? Degenerate case: $ln 1 = 0$ kyunki $e^{0} = 1$ . Koi mass nahi phenkaa, koi push nahi — equation exactly zero deta hai, koi error nahi. Ye figure ka left edge hai jahan curve axis ko touch karta hai.

Verify: (a) units m/s ✓; approximation $x = 0.004016$ exact log se 3 significant figures tak match karta hai ✓. (b) $ln 1 = 0$ exact hai ✓.

Ex 4 — Cell E · do tanks size karna

Ex 5 — Cell F · ignition-delay ratio

Worked example Achha hypergolic vs sluggish pair

Parent note ka delay law hai $τ_{i g} \propto e^{E_{a} / R T_{0}}$ (dekho Arrhenius Rate Law). Ek achhe pair mein $E_{a} / R T_{0} = 3$ hai; ek sluggish pair mein same $T_{0}, ρ, Δ H$ par $E_{a} / R T_{0} = 8$ hai. Sluggish delay kitne guna zyada lamba hai?

Forecast: Exponent mein difference $8 - 3 = 5$ hai. Andaaza lagao $e^{5}$ kitna bada hai — 100 se bada ya chhota?

Step 1. Prefactor ( $ρ c_{p} R T_{0}^{2} /Δ H A [F] [O] E_{a}$ wala lump) dono ke liye same hai, toh ratio mein cancel ho jaata hai. Ye step kyun? Jab do cheezein sirf exponent mein alag hoon, toh ratio lo taaki messy prefactor gayab ho jaaye.

Step 2. $\frac{τ _{s l o w}}{τ _{g oo d}} = \frac{e ^{8}}{e ^{3}} = e^{8 - 3} = e^{5}$ . Ye step kyun? $e^{a} / e^{b} = e^{a - b}$ — exponential ki defining property.

Step 3. $e^{5} = 148.4 \approx 150 \times$ . Ye step kyun? Evaluate karo.

Verify: Parent ke numbers se cross-check: $e^{3} = 20.09$ , $e^{8} = 2981$ , aur $2981/20.09 = 148.4$ ✓ — ratio route aur direct route dono agree karte hain. $150 \times$ zyada lamba delay propellant pool hone deta hai → hard start ka risk. ✓

Ex 6 — Cell G · real-world RCS pulse train

Worked example Bahut saare chhote nudges jodna

Ek Reaction Control Systems (RCS) thruster docking ke dauran 40 short pulses fire karta hai, har baar $600 kg$ spacecraft se $0.5 kg$ N₂O₄/MMH expel karta hai, $I_{s p} = 280 s$ . Total $Δ v$ nikalo.

Forecast: Total propellant $= 40 \times 0.5 = 20 kg$ . Kya ek bade burn mein karna ya 40 chhote mein karna answer change karega? (Trick: nahi — Tsiolkovsky sirf start aur end mass ki parwah karta hai.)

Step 1. $v_{e} = 280 \times 9.81 = 2746.8 m/s$ . Ye step kyun? Tool (1).

Step 2. Total propellant $= 40 \times 0.5 = 20 kg$ , toh $m_{f} = 600 - 20 = 580 kg$ . Ye step kyun? Individual pulses telescope ho jaate hain: $ln \frac{m _{0}}{m _{1}} + ln \frac{m _{1}}{m _{2}} + \dots = ln \frac{m _{0}}{m _{f}}$ . Sirf endpoints bachte hain.

Step 3. $Δ v = 2746.8 ln \frac{600}{580} = 2746.8 \times 0.033902 = 93.1 m/s$ . Ye step kyun? Endpoints par Tool (2).

Verify: Ek $20$ -kg burn se compare karo: identical setup, identical $93.1 m/s$ ✓ — confirm karta hai ki pulse count Δv ke liye irrelevant hai (ye precision ke liye matter karta hai, total impulse ke liye nahi). Units m/s ✓.

Ex 7 — Cell H · exam twist ("hotter but heavier")

Worked example Chamber temperature upar, par I_sp neeche

Ek student propose karta hai ki N₂O₄/MMH engine ko zyada oxidizer-rich chalao taaki chamber temperature $T_{c} = 3000 K$ se $3300 K$ tak badhe, lekin isse mean exhaust molecular mass $M = 20 g/mol$ se $M = 25 g/mol$ ho jaata hai. Kya $v_{e}$ (aur thus $I_{s p}$ ) upar jaayega ya neeche? $v_{e} \propto T_{c} / M$ use karo (dekho Combustion Thermodynamics).

Forecast: Hotter lagta hai faster. Compute karne se pehle direction guess karo.

Step 1. Ratio banao $\frac{v _{e, new}}{v _{e, old}} = \frac{T _{c, new} / M _{new}}{T _{c, old} / M _{old}}$ . Ye step kyun? Sirf combination $T_{c} / M$ matter karta hai; ratio har unknown constant cancel kar deta hai.

Step 2. $T _{c} / M$ : old $= 3000/20 = 150$ ; new $= 3300/25 = 132$ . Ye step kyun? Har $T_{c} / M$ compute karo.

Step 3. Ratio $= 132/150 = 0.88 = 0.938$ . Ye step kyun? $v_{e}$ $93.8%$ tak gir jaata hai — hotter chamber ke bawajood exhaust speed mein $6.2%$ ka loss.

Verify: $150 > 132$ toh ratio $< 1$ ✓, counter-intuitive result confirm karta hai: bhaari products extra heat ko beat kar dete hain. Isliye optimum $O / F$ thoda fuel-rich hota hai — parent ka mistake box, ab quantified. ✓

Ex 8 — Cell I · cryogenic se cross-compare

Worked example Same Δv, kise kam propellant chahiye?

$Δ v = 4000 m/s$ achieve karne ke liye $500 kg$ dry stage ke saath, N₂O₄/MMH ( $I_{s p} = 320 s$ ) aur ek LOX/LH₂ stage ( $I_{s p} = 450 s$ ) compare karo. Har ek ko kitna propellant chahiye?

Forecast: Higher- $I_{s p}$ cryogenic stage ko kam propellant chahiye hoga. Kitna kam — aadha, ya thoda kam?

Step 1. Tsiolkovsky invert karo: $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}}$ , toh $m_{0} = m_{f} e^{Δ v / (I_{s p} g_{0})}$ . Ye step kyun? Humein Δv pata hai aur mass chahiye; rearrange karo, $ln$ aur $e$ ko inverse operations ki tarah use karke.

Step 2 (hypergolic). $v_{e} = 320 \times 9.81 = 3139.2$ ; exponent $= 4000/3139.2 = 1.2742$ ; $m_{0} = 500 e^{1.2742} = 500 \times 3.576 = 1788 kg$ . Propellant $= 1788 - 500 = 1288 kg$ . Ye step kyun? Plug in karo.

Step 3 (cryogenic). $v_{e} = 450 \times 9.81 = 4414.5$ ; exponent $= 4000/4414.5 = 0.9061$ ; $m_{0} = 500 e^{0.9061} = 500 \times 2.4747 = 1237 kg$ . Propellant $= 1237 - 500 = 737 kg$ . Ye step kyun? Same recipe, higher $v_{e}$ .

Verify: Cryogenic ko $737$ kg chahiye vs hypergolic ko $1288$ kg — lagbhag $43%$ kam ✓, iske higher $I_{s p}$ ke saath consistent. Lekin yaad rakho: cryogenics boil off ho jaate hain aur saalon tak wait nahi kar sakte — yahi poora wajah hai kyun Solid Rocket Propellants aur hypergolics exist karte hain. ✓

Recall Kaun sa cell kaun sa tha? (self-test)

"Propellant double karne se sirf ek fixed 2176 m/s add hota hai" — kaun sa cell? ::: Cell B, the large-ratio $ln$ limit (Ex 2). "0 kg jaalne par Δv = 0 kyunki ln 1 = 0" — kaun sa cell? ::: Cell D, the degenerate/zero input (Ex 3b). "Hotter chamber par lower I_sp" — kaun sa cell aur kyun? ::: Cell H; kyunki $v_{e} \propto T_{c} / M$ aur bhaari products jeet jaate hain (Ex 7). "40 pulses same Δv dete hain jaise ek burn" — kyun? ::: Logs telescope ho jaate hain; sirf start aur end mass matter karti hai (Ex 6, cell G).

Connections

Tsiolkovsky Rocket Equation — Ex 1, 2, 3, 6, 8 ke peeche ka engine.
Specific Impulse — har example mein $I_{s p}$ ko $v_{e}$ mein convert karna.
Reaction Control Systems (RCS) — pulse-train problem, Ex 6.
Arrhenius Rate Law — ignition-delay exponential, Ex 5.
Combustion Thermodynamics — $T_{c} / M$ trade-off, Ex 7.
Cryogenic Propellants — LOX-LH2 — cross-comparison, Ex 8.
Solid Rocket Propellants — doosra storable option.