3.3.21 · D3 · Physics › Rocket Propulsion › Characteristic velocity c - and its relation to flame tempe
Intuition Yeh page kya hai
Parent note ne tumhe master formula diya. Yahan hum use har tarah ke questions ke against stress-test karte hain jo ek exam ya test stand throw kar sakta hai — forward calculation, backward calculation, limiting values, degenerate inputs, aur ek real-world trap. Agar tum yahan koi case dekh lo, toh tum usse pehle hi survive kar chuke ho.
Numbers touch karne se pehle, chalte hain is page ke har symbol ko naam dete hain taaki koi bhi andha use na ho:
Definition Woh symbols jo hum use karenge (har ek ki physical meaning)
c ∗ — characteristic velocity (m/s): sirf chamber aur propellant ke liye ek quality score.
p 0 — chamber (stagnation) pressure (Pa): combustion chamber ke andar build up hua pressure, gas ke escape hone se pehle.
A t — throat area (m²): nozzle ke sabse narrow point ka cross-section, jahan flow choke hoti hai.
m ˙ — mass flow rate (kg/s): kitne kilograms propellant har second nikalta hai.
M — molecular weight of the exhaust (kg/mol): ek mole gas ki mass; chhota = halke, zippy molecules.
γ — specific heat ratio (dimensionless): gas kitni "springy" hai; combustion gases ke liye ~1.1–1.67.
R u — universal gas constant = 8.314 J/mol⋅K : har gas ke liye same.
T 0 — chamber (flame) temperature (K): jala hua gas kitna hot hai.
Γ — Vandenkerckhove function (dimensionless): saari γ -dependence ka ek bundle.
C F — thrust coefficient (dimensionless): nozzle ka chamber performance ka amplification.
c e — effective exhaust velocity (m/s): exhaust jet ki real average speed, c e = c ∗ C F se milti hai (chamber quality × nozzle amplification).
I s p — specific impulse (s): propellant ke unit weight-flow per thrust; ek fuel-economy score, I s p g 0 = c e .
Ab yaad karo usi quantity ke do roop:
Is topic ke har problem ka ek cell hota hai. Neeche ke examples mein (Cell …) tag hai taaki tum dekh sako ki poora grid cover hua hai.
Cell
Case class
Kya diya → kya poochha
Kaun sa form
A
Forward (chemistry)
T 0 , M , γ → c ∗
theoretical
B
Measured (test stand)
p 0 , A t , m ˙ → c ∗
measured
C
Backward for m ˙
c ∗ , p 0 , A t → m ˙
measured, inverted
D
Backward for T 0
c ∗ , M , γ → T 0
theoretical, inverted
E
Efficiency / two-form clash
measured vs theoretical → η c ∗
dono
F
Limiting γ
γ → 1 aur γ bada → Γ behaviour
sirf Γ
G
Degenerate input
m ˙ → 0 , A t → 0 , T 0 → 0
formula ki sanity
H
Real-world word problem
mixture ratio shift → c ∗ trade
theoretical, ratio
I
Exam twist (chain to I s p )
c ∗ , C F → c e , I s p
forms ko link karo
Ab hum har cell walk karenge.
Worked example Kerosene/LOX chamber
Diya hai T 0 = 3500 K (flame temperature), M = 22 g/mol = 0.022 kg/mol (molecular weight), γ = 1.22 (specific heat ratio), R u = 8.314 J/mol⋅K (universal gas constant). c ∗ nikalo.
Forecast: H 2 / O 2 se bhaari gas (22 vs 16) → expect karo c ∗ parent ke 2143 m/s se kam hoga. Guess ~1800 m/s.
Step 1 — Γ compute karo. Γ = 1.22 ( 2.22 2 ) 0.44 2.22 = 1.1045 × ( 0.9009 ) 5.045 .
Yeh step kyun? Γ saari γ -dependence pack karta hai; ek baar compute karo, baar baar reuse karo. ( 0.9009 ) 5.045 ≈ 0.5905 , isliye Γ ≈ 0.6522 .
Step 2 — "hot & light" term compute karo. R u T 0 / M = 8.314 × 3500/0.022 = 1.3227 × 1 0 6 = 1150.1 m/s .
Yeh step kyun? Yeh square-root term unit mass per raw energy content hai; Γ sirf isko scale karta hai.
Step 3 — Divide karo. c ∗ = 1150.1/0.6522 ≈ 1763 m/s .
Yeh step kyun? c ∗ = Γ 1 R u T 0 / M — hamare master formula ka right face.
Verify: Parent ke H 2 / O 2 value se ratio: 1763/2143 = 0.82 . Sirf bada change M ka tha 16→22, matlab 16/22 = 0.853 ; chhota T 0 drop (3500/3600 = 0.986 ) aur Γ shift baaki explain karta hai. Units: J/kg = m 2 / s 2 = m/s. ✓
c ∗ padho
Ek firing mein p 0 = 6.5 MPa (chamber pressure), throat diameter d t = 50 mm (toh throat area A t follow karta hai), m ˙ = 18 kg/s (mass flow) milta hai. c ∗ nikalo.
Forecast: bada throat, high pressure — chamber healthy lagti hai. Guess ~700 m/s (yeh measured face hai; agar combustion kharaab ho toh yeh theoretical se bahut neeche ho sakta hai).
Step 1 — Throat area. A t = 4 π d t 2 = 4 π ( 0.050 ) 2 = 1.9635 × 1 0 − 3 m 2 .
Yeh step kyun? Definition ko area chahiye, diameter nahi; round throat deta hai 4 π d 2 .
Step 2 — Left face apply karo. c ∗ = m ˙ p 0 A t = 18 6.5 × 1 0 6 × 1.9635 × 1 0 − 3 .
Yeh step kyun? Koi nozzle data nahi chahiye — isliye c ∗ ko is tarah define kiya gaya tha (dekho Choked Flow and the Throat : throat wahan hai jahan flow choke hoti hai aur m ˙ lock hota hai).
Step 3 — Arithmetic. Numerator = 12763 N ; c ∗ = 12763/18 ≈ 709 m/s .
Yeh step kyun? Bas final divide.
Verify: Units: kg/s Pa ⋅ m 2 = kg/s N = kg/s kg⋅m/s 2 = m/s. ✓ (Yeh low value ek diagnostic flag karega — bhaari products ya kharaab mixing.)
Worked example Propellant feed size karo
Tumhe c ∗ = 1750 m/s chahiye p 0 = 5 MPa (chamber pressure) par, 30 mm diameter ke throat se. Chamber kitna mass flow m ˙ demand karta hai?
Forecast: c ∗ = p 0 A t / m ˙ rearrange karo toh m ˙ = p 0 A t / c ∗ milta hai. Zyaada c ∗ (accha propellant) → same push ke liye kam mass flow. Kuch kg/s guess karo.
Step 1 — Throat area. A t = 4 π ( 0.030 ) 2 = 7.069 × 1 0 − 4 m 2 .
Yeh step kyun? Same geometry rule Example 2 jaisi.
Step 2 — Definition invert karo. m ˙ = c ∗ p 0 A t = 1750 5 × 1 0 6 × 7.069 × 1 0 − 4 .
Yeh step kyun? c ∗ ek ratio hai; iske teen variables mein se kisi ke liye bhi solve karna pure algebra hai.
Step 3 — Arithmetic. Numerator = 3534.3 N ; m ˙ = 3534.3/1750 ≈ 2.02 kg/s .
Verify: Plug back karo: c ∗ = 5 × 1 0 6 × 7.069 × 1 0 − 4 /2.02 = 1750 m/s. ✓ Units: m/s Pa⋅m 2 = m/s N = kg/s. ✓
c ∗ se T 0 infer karo
Ek monopropellant measured c ∗ = 1200 m/s deta hai, M = 0.020 kg/mol (molecular weight), γ = 1.25 (specific heat ratio) ke saath. Chamber temperature T 0 estimate karo (assume karo perfect combustion, toh measured = theoretical).
Forecast: c ∗ ∝ T 0 isliye T 0 ∝ ( c ∗ ) 2 — ~1200 m/s value ko square karne se kuch hazaar K hint milta hai.
Step 1 — Γ . Γ = 1.25 ( 2.25 2 ) 0.5 2.25 = 1.1180 × ( 0.8889 ) 4.5 = 1.1180 × 0.6288 = 0.7030 .
Yeh step kyun? T 0 isolate karne se pehle Γ chahiye; yeh sirf γ par depend karta hai.
Step 2 — T 0 ke liye rearrange karo. c ∗ = Γ 1 R u T 0 / M se, dono sides square karo:
T 0 = R u ( c ∗ Γ ) 2 M .
Yeh step kyun? Square-root hi ek obstacle hai; square karna isse undo karta hai aur T 0 ke liye linearly solve karne deta hai.
Step 3 — Plug in karo. c ∗ Γ = 1200 × 0.7030 = 843.6 ; squared = 7.117 × 1 0 5 ; times M / R u : T 0 = 8.314 7.117 × 1 0 5 × 0.020 ≈ 1712 K .
Verify: Forward-check karo T 0 = 1712 se: 8.314 × 1712/0.020 = 7.117 × 1 0 5 = 843.6 ; Γ = 0.7030 se divide karo → 1200 m/s. ✓ (Dekho Adiabatic Flame Temperature — is tarah ek stand estimate thermochemistry ko cross-check karta hai.)
Worked example Chamber ko grade do
Theory predict karta hai c t h ∗ = 1763 m/s (Example 1 ka engine). Stand measure karta hai c m e a s ∗ = 1640 m/s . η c ∗ nikalo aur interpret karo.
Forecast: measured hamesha ≤ theoretical hota hai (losses sirf subtract karte hain). Guess ~93 %.
Step 1 — Ratio. η c ∗ = c t h ∗ c m e a s ∗ = 1763 1640 .
Yeh step kyun? Efficiency defined hai measured-over-ideal ke roop mein; yeh nozzle ko poori tarah strip out kar deta hai kyunki dono chamber quantities hain.
Step 2 — Compute karo. η c ∗ = 0.9302 ≈ 93.0 % .
Yeh step kyun? Direct division.
Verify: 0.9302 × 1763 = 1640 m/s ✓. Interpretation: 7 % deficit matlab combustion ne sirf ~0.9 3 2 = 86 % ideal energy release ki (kyunki c ∗ ∝ T 0 , T 0 mein efficiency η 2 hai) — ek real diagnostic jo nozzle mask nahi kar sakta.
Yahan function ki geometry matter karti hai, isliye hum ek picture dekhte hain.
Figure 1 — Γ versus γ . Horizontal axis specific heat ratio γ hai (1.05 se 1.67 tak); vertical axis Vandenkerckhove function Γ hai. Blue curve Γ ( γ ) hai, jo roughly 0.62 se 0.73 tak gently rise karti hai. Curve par teen labelled dots hain — red γ = 1.10 , Γ = 0.6255 par; yellow γ = 1.40 , Γ = 0.6847 par; green γ = 1.67 , Γ = 0.7262 par. Ek faint white horizontal band real-gas range Γ ≈ 0.63 –0.73 mark karta hai, aur curve poore plot mein uske andar rehti hai — yeh visual proof hai ki γ , c ∗ ∝ 1/Γ ko sirf weakly affect karta hai.
Intuition Upar ki figure kaise padhen
Blue curve ko left se right follow karo: jaise γ badhta hai, Γ climb karta hai lekin kabhi narrow white band nahi chhodta. Kyunki c ∗ ∝ 1/Γ , woh tight band matlab hai ki γ c ∗ par sabse kamzor lever hai — coloured dots woh teen cases hain jo hum ab compute karte hain.
γ → 1 aur γ bada hone par Γ ka kya hota hai?
Γ evaluate karo γ = 1.10 , 1.40 , 1.67 par aur trend describe karo.
Forecast: figure dikhata hai Γ gently γ ke saath badhta hai. Real gases narrow band Γ ≈ 0.63 –0.73 mein hote hain.
Step 1 — γ = 1.10 (red dot). Γ = 1.10 ( 2.10 2 ) 0.20 2.10 = 1.0488 × ( 0.95238 ) 10.5 = 1.0488 × 0.5964 = 0.6255 .
Yeh step kyun? Exponent 2 ( γ − 1 ) γ + 1 , γ → 1 par blow up karta hai (denominator → 0 ), isliye hume carefully evaluate karna hai na ki "apne dimaag mein limit lena."
Step 2 — γ = 1.40 (yellow dot, air). Γ = 1.40 ( 2.40 2 ) 0.80 2.40 = 1.1832 × ( 0.83333 ) 3 = 1.1832 × 0.5787 = 0.6847 .
Yeh step kyun? γ = 1.40 diatomic air ke liye everyday value hai; isse check karna curve ko ek aisi value par anchor karta hai jo zyaadatar readers pehle se jaante hain, aur confirm karta hai ki Γ ne γ = 1.10 result se climb kiya hai.
Step 3 — γ = 1.67 (green dot, monatomic, e.g. He). Γ = 1.67 ( 2.67 2 ) 1.34 2.67 = 1.2923 × ( 0.74906 ) 1.9925 = 1.2923 × 0.5619 = 0.7262 .
Yeh teen kyun? Yeh har real combustion gas ko bracket karte hain. Trend: Γ γ ke saath badhta hai , lekin slowly — poora span 0.63→0.73 hai.
Verify: Kyunki c ∗ ∝ 1/Γ , ek bada γ (bada Γ ) same T 0 / M ke liye chhota c ∗ deta hai. Parent note ne kaha tha "γ weakly enter karta hai" — figure aur yeh teen numbers prove karte hain ki band tight hai. ✓ (Dekho Vandenkerckhove Function Γ aur Isentropic Flow Relations .)
Worked example Edge cases kya kehte hain?
m ˙ → 0 , A t → 0 , aur T 0 → 0 discuss karo dono forms se.
Forecast: zero se divide karne par ya toh blow up hona chahiye ya physically-meaningful "kuch nahi" milna chahiye.
Step 1 — m ˙ → 0 fixed p 0 , A t par. c ∗ = p 0 A t / m ˙ → + ∞ .
Yeh step kyun? Physically real combustion mein impossible — yeh kehta hai "mass ki ek whisper se infinite quality," ek warning ki tum p 0 nahi sambhaal sakte agar flow starve ho. Ek real chamber ka p 0 collapse ho jaata hai agar m ˙ → 0 , isliye ratio finite rehta hai. Lesson: teen variables independent nahi hain .
Step 2 — A t → 0 . Fixed m ˙ ke liye, choking require karta hai ki pressure bina limit ke climb kare; c ∗ = p 0 A t / m ˙ finite rehta hai kyunki p 0 ∝ 1/ A t . Formula throat size mein scale-invariant hai.
Yeh step kyun? Confirm karta hai ki c ∗ ek material property hai, geometry property nahi — hole chhhota karne se propellant quality nahi badalti.
Step 3 — T 0 → 0 . Theoretical form: c ∗ = Γ 1 R u T 0 / M → 0 .
Yeh step kyun? Thanda gas mein push karne ke liye koi thermal energy nahi — c ∗ sahi se vanish hota hai. Sanity confirm hua.
Verify: Teeno limits consistent hain dono faces ke beech agar tum physical coupling p 0 ( T 0 , m ˙ ) yaad rakhte ho. Koi contradiction nahi. ✓
Worked example Kya hume fuel-rich run karna chahiye?
Stoichiometric par ek H 2 / O 2 engine T 0 = 3600 K , M = 18 g/mol (pure H 2 O ) par burn karta hai. Fuel-rich run karne se T 0 girkar 3300 K ho jaata hai lekin M ghatakar 14 g/mol (excess H 2 ) ho jaata hai. Dono ke liye γ = 1.22 lo (toh Γ = 0.6522 Example 1 ke γ se). Kaun sa zyaada c ∗ deta hai?
Forecast: parent note claim karta hai fuel-rich jeetta hai kyunki 1/ M T 0 loss ko beat karta hai. Numbers se confirm karte hain.
Step 1 — Stoichiometric c ∗ . 8.314 × 3600/0.018 = 1.6628 × 1 0 6 = 1289.5 ; c ∗ = 1289.5/0.6522 = 1977 m/s .
Yeh step kyun? Baseline case, formula ka right face.
Step 2 — Fuel-rich c ∗ . 8.314 × 3300/0.014 = 1.9597 × 1 0 6 = 1399.9 ; c ∗ = 1399.9/0.6522 = 2147 m/s .
Yeh step kyun? Same formula, naye T 0 , M .
Step 3 — Compare karo. Fuel-rich zyaada hai: 2147 > 1977 , ek + 8.6 % gain.
Yeh step kyun? Ratio test: c s t o i c h ∗ c r i c h ∗ = 3600/18 3300/14 = 200 235.7 = 1.179 = 1.086 — exactly + 8.6 % , aur Γ cancel ho jaata hai.
Verify: 1.179 = 1.0857 ; 1977 × 1.0857 = 2146.5 ≈ 2147 ✓. Conclusion: fuel-rich run karo — molecular-weight lever temperature penalty ko beat karta hai, bilkul jaisa Propellant Selection and Molecular Weight predict karta hai.
Definition Specific impulse
I s p (is example ke liye zaroori)
I s p (seconds) rocket ka fuel-economy score hai: propellant ka ek unit-weight kitne seconds tak ek unit of thrust produce kar sakta hai. Yeh effective exhaust velocity c e (exhaust jet ki real average speed) se I s p g 0 = c e se link hota hai, jahan g 0 = 9.81 m/s 2 standard gravity hai. Aur c e khud split hota hai c e = c ∗ C F mein — chamber quality times nozzle amplification.
Worked example Chamber grade se delivered performance tak
Ek chamber ka c ∗ = 1763 m/s hai (Example 1). Nozzle C F = 1.65 (thrust coefficient) deliver karta hai. g 0 = 9.81 m/s 2 lo. Effective exhaust velocity c e aur specific impulse I s p nikalo.
Forecast: parent kehta hai c e = c ∗ C F aur c e ≈ 1.5 –1.9 × c ∗ . C F = 1.65 ke saath expect karo c e ≈ 2900 m/s aur I s p ≈ 300 s.
Step 1 — Effective exhaust velocity. c e = c ∗ C F = 1763 × 1.65 = 2909 m/s .
Yeh step kyun? Clean split: chamber (c ∗ ) times nozzle (Thrust Coefficient CF ). Yahi reason hai c ∗ invent kiya gaya tha.
Step 2 — Specific impulse. I s p = c e / g 0 = 2909/9.81 = 296.5 s .
Yeh step kyun? I s p g 0 = c e definition hai (dekho Specific Impulse Isp ); g 0 se divide karna velocity ko seconds mein convert karta hai.
Verify: 296.5 × 9.81 = 2909 ✓; aur c e / c ∗ = 2909/1763 = 1.65 = C F ✓. Units: m/s ÷ (m/s²) = s ✓. Teeno links consistent hain.
Recall Self-test: har ek ko ek matrix cell se match karo
Kaun sa cell m ˙ ke liye invert karne ko kehta hai? ::: Cell C (Example 3).
Kaun sa cell dikhata hai ki c ∗ → 0 physically sense karta hai? ::: Cell G, T 0 → 0 limit (Example 7).
Kaun sa cell numerically prove karta hai ki fuel-rich jeetta hai? ::: Cell H (Example 8).
Kaun sa cell c ∗ ko I s p se link karta hai? ::: Cell I (Example 9).
Kaun se cell mein Γ answer se cancel ho jaata hai? ::: Cell H — ek hi γ par do c ∗ ka ratio .