Chamber-to-exit relation - all quantities as f(M_e, γ)

3.3.12 · Physics › Rocket Propulsion

Physical Setup

Ek rocket combustion chamber ek converging-diverging nozzle ko feed karta hai:

Chamber (subscript 0): Nearly stagnant flow, $T_{0}, P_{0}, ρ_{0}$
Exit (subscript e): Supersonic expansion, $T_{e}, P_{e}, ρ_{e}, V_{e}, M_{e}$

Inke beech mein: isentropic expansion (reversible, adiabatic—koi heat loss nahi, koi shock nahi).

Relations ko First Principles se Derive Karna

Step 1: Temperature Ratio $T_{e} / T_{0}$

Energy conservation se shuru karo. Nozzle se flow karne wali ideal gas ke liye, stagnation (total) enthalpy barabar hoti hai static enthalpy aur kinetic energy ke sum ke:

$h_{0} = h_{e} + \frac{V _{e}^{2}}{2}$

Ideal gas ke liye, $h = c_{p} T$ , toh:

$c_{p} T_{0} = c_{p} T_{e} + \frac{V _{e}^{2}}{2}$

$c_{p} T_{e}$ se divide karo:

$\frac{T _{0}}{T _{e}} = 1 + \frac{V _{e}^{2}}{2 c _{p} T _{e}}$

Yeh step kyun? Hum $V_{e}$ ko Mach number ke form mein convert karna chahte hain. Speed of sound $a_{e} = γ R T_{e}$ , aur Mach number $M_{e} = V_{e} / a_{e}$ , toh:

$V_{e}^{2} = M_{e}^{2} a_{e}^{2} = M_{e}^{2} γ R T_{e}$

Aur, $c_{p} = γ R / (γ - 1)$ , isliye:

$\frac{V _{e}^{2}}{2 c _{p} T _{e}} = \frac{M _{e}^{2} γ R T _{e}}{2 \cdot \frac{γ R}{γ - 1} \cdot T _{e}} = \frac{M _{e}^{2} ( γ - 1 )}{2}$

Wapas substitute karo:

Physical meaning: Jaise $M_{e}$ badhta hai, static temperature $T_{e}$ girta hai kyunki kinetic energy, thermal energy ki keemat par badhti hai.

Step 2: Pressure Ratio $P_{e} / P_{0}$

Ideal gas ki isentropic flow ke liye, hamare paas isentropic relation hai:

$\frac{P}{P _{0}} = (\frac{T}{T _{0}})^{γ / (γ - 1)}$

Yeh form kyun? Thermodynamics se, $P V^{γ} = const$ aur ideal gas law $P V = n R T$ . Inhe combine karne par $P / T^{γ / (γ - 1)} = const$ milta hai.

Abhi nikala hua temperature ratio substitute karo:

$\frac{P _{e}}{P _{0}} = (\frac{T _{e}}{T _{0}})^{γ / (γ - 1)} = (\frac{1}{1 + \frac{γ - 1}{2} M _{e}^{2}})^{γ / (γ - 1)}$

Physical meaning: Pressure $M_{e}$ ke saath dramatically girta hai. Ek rocket nozzle high chamber pressure ko high exit velocity mein convert karta hai.

Step 3: Density Ratio $ρ_{e} / ρ_{0}$

Ideal gas law ko ratio form mein use karo:

$\frac{P _{e}}{P _{0}} = \frac{ρ _{e}}{ρ _{0}} \cdot \frac{T _{e}}{T _{0}}$

Rearrange karo:

$\frac{ρ _{e}}{ρ _{0}} = \frac{P _{e} / P _{0}}{T _{e} / T _{0}}$

Apne formulas substitute karo:

$\frac{ρ _{e}}{ρ _{0}} = \frac{( 1 + \frac{γ - 1}{2} M _{e}^{2} ) ^{- γ / (γ - 1)}}{( 1 + \frac{γ - 1}{2} M _{e}^{2} ) ^{- 1}}$

Exponents simplify karo:

$\frac{ρ _{e}}{ρ _{0}} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{- γ / (γ - 1) + 1} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{- 1/ (γ - 1)}$

Step 4: Exit Velocity $V_{e}$

Mach number ki definition se:

$M_{e} = \frac{V _{e}}{a _{e}} = \frac{V _{e}}{γ R T _{e}}$

$V_{e}$ ke liye solve karo:

$V_{e} = M_{e} γ R T_{e}$

$T_{e}$ ko chamber conditions ke terms mein express karo:

$T_{e} = \frac{T _{0}}{1 + \frac{γ - 1}{2} M _{e}^{2}}$

Isliye:

$V_{e} = M_{e} γ R T_{0} \cdot \frac{1}{1 + \frac{γ - 1}{2} M _{e}^{2}}$

$a_{0} = γ R T_{0}$ define karo (chamber conditions par speed of sound):

Step 5: Area Ratio $A_{e} / A^{*}$

Nozzle ki geometry bhi $M_{e}$ aur $γ$ se fix hoti hai. Mass conservation se shuru karo ( $\overset{m}{˙} = ρ A V = const$ ). Exit ko throat se compare karo (jahan $M = 1$ hota hai, $*$ se denote kiya):

$\frac{A _{e}}{A ^{*}} = \frac{ρ ^{*} V ^{*}}{ρ _{e} V _{e}}$

Yeh step kyun? Dono cross-sections se same mass flow hoti hai, toh chhota $ρ V$ bada area maangta hai. Har factor ko $M_{e}$ aur $γ$ ke terms mein apne ratios use karke express karo (throat ke liye $M = 1$ par evaluate karke), jo algebra ke baad classic area–Mach relation deta hai:

Physical meaning: Supersonic exit ke liye ( $M_{e} > 1$ ), diverging section ko chauda hona padta hai; bada $M_{e}$ bada area ratio maangta hai. Yeh flow relations ko actual nozzle shape se jodta hai.

Summary Table

Rocket nozzle mein isentropic flow ke liye, saari exit quantities sirf $M_{e}$ aur $γ$ par depend karti hain:

Quantity	Relation
Temperature	$\frac{T _{e}}{T _{0}} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{- 1}$
Pressure	$\frac{P _{e}}{P _{0}} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{- γ / (γ - 1)}$
Density	$\frac{ρ _{e}}{ρ _{0}} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{- 1/ (γ - 1)}$
Velocity	$V_{e} = M_{e} \frac{γ R T _{0}}{1 + \frac{γ - 1}{2} M _{e}^{2}}$
Area ratio	$\frac{A _{e}}{A ^{*}} = \frac{1}{M _{e}} [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M_{e}^{2})]^{\frac{γ + 1}{2 ( γ - 1 )}}$

Worked example Worked Example 1: H₂/O₂ Rocket

Diya hua: Chamber $T_{0} = 3500 K$ , $P_{0} = 70 bar$ , $γ = 1.2$ , exit $M_{e} = 3.0$ , $R = 520 J/(kg\cdotK)$

Nikaalein: $T_{e}, P_{e}, V_{e}$

Solution:

Temperature: $T_{e} = \frac{T _{0}}{1 + \frac{γ - 1}{2} M _{e}^{2}} = \frac{3500}{1 + \frac{0.2}{2} ( 9 )} = \frac{3500}{1 + 0.9} = \frac{3500}{1.9} = 1842 K$ Yeh step kyun? Factor $(1 + 0.1 \cdot 9) = 1.9$ kinetic energy ke thermal energy churaane ko account karta hai.
Pressure: $\frac{P _{e}}{P _{0}} = (1.9)^{- 1.2/0.2} = (1.9)^{- 6} = 0.0213$ $P_{e} = 70 \times 0.0213 = 1.49 bar$ Yeh step kyun? Exponent $- γ / (γ - 1) = - 6$ steep pressure drop cause karta hai—yahi thrust create karta hai!
Velocity: $V_{e} = 3.0 \times \frac{1.2 \times 520 \times 3500}{1.9} = 3.0 \times 1.149 \times 1 0^{6} = 3.0 \times 1072 = 3216 m/s$

Yeh step kyun? Note karo $\frac{1.2 \times 520 \times 3500}{1.9} = \frac{2.184 \times 1 0 ^{6}}{1.9} \approx 1.149 \times 1 0^{6}$ , jiska square root $\approx 1072 m/s$ hai. High $M_{e}$ aur high $T_{0}$ dono high exhaust velocity mein contribute karte hain—rocket performance ki key.

Worked example Worked Example 2:

$γ$ ke Sensitivity ka Analysis Setup: $M_{e} = 3.0$ fix rakho, $γ = 1.2$ (H₂ products) vs. $γ = 1.4$ (air) compare karo

Property	$γ = 1.2$	$γ = 1.4$
$T_{e} / T_{0}$	$1/1.9 = 0.526$	$1/2.8 = 0.357$
exponent $- γ / (γ - 1)$	$- 6$	$- 3.5$
$P_{e} / P_{0}$	$(1.9)^{- 6} = 0.0213$	$(2.8)^{- 3.5} = 0.0272$

Insight: Kam $γ$ (halke molecules) same $M_{e}$ ke liye kam temperature drop deta hai—isliye H₂ upper stages mein preferred hai (higher specific impulse). Note karo pressure-ratio exponent $- γ / (γ - 1)$ hai: $γ = 1.4$ ke liye yeh $- 3.5$ hai, nahi ki $- 35$ .

Recall Ek 12-Saal Ke Bachche Ko Explain Karo

Socho tumhare paas ek balloon hai jisme hot, high-pressure air hai (chamber). Tum usse chhodte ho, aur air opening se bahar nikalta hai (nozzle). Jaise air speed up hoti hai, teen cheezein hoti hain:

Woh thandi ho jaati hai (jaise spray can thanda ho jaata hai jab use karte ho—energy heat ki jagah speed mein jaati hai)
Pressure girta hai (kyunki air phail jaati hai)
Woh patli ho jaati hai (kam dense—molecules ke beech zyada space hoti hai)

Cool part yeh hai: agar tum mujhe batao kitni fast air exit par ja rahi hai (Mach number $M_{e}$ ), main exactly calculate kar sakta hoon kitni thandi, kitne kam pressure mein, aur kitni patli ho gayi—aur even kitna chauda nozzle wahan hona chahiye. Yeh ek recipe ki tarah hai: Mach number + gas type ( $γ$ ) → saari exit properties. Rocket scientists isi se nozzles design karte hain jo fuel se maximum thrust nikaal sake.

Connections

Isentropic Flow Relations — thermodynamic foundation
Converging-Diverging Nozzle — jahan yeh relations apply hoti hain
Area Ratio and Mach Number — $M_{e}$ se tied geometric design
Thrust Equation — $P_{e}, V_{e}$ in relations se use karta hai
Specific Impulse — $V_{e}$ par dependent performance metric
Characteristic Velocity c-star — chamber conditions $T_{0}, P_{0}$ use karta hai

#flashcards/physics

Chamber-to-exit temperature ratio kis par depend karta hai? :: Sirf exit Mach number $M_{e}$ aur specific heat ratio $γ$ par: $T_{e} / T_{0} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{- 1}$

Temperature kyun girta hai jab Mach number badhta hai?

Energy conservation: kinetic energy badhti hai thermal energy ki keemat par, isliye static temperature ghata hi chahiye

Pressure ratio exponent

γ

ke terms mein kya hai?

- γ / (γ - 1)

, toh

P_{e} / P_{0} = (1 + \frac{γ - 1}{2} M_{e}^{2})^{- γ / (γ - 1)}

Exit velocity chamber temperature par kaise depend karti hai?

V_{e} = M_{e} γ R T_{0} / (1 + \frac{γ - 1}{2} M_{e}^{2})

— zyada

T_{0}

se zyada

V_{e}

milti hai (square root dependence)

Nozzle ke liye area–Mach relation kya hai?

A_{e} / A^{*} = \frac{1}{M _{e}} [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M_{e}^{2})]^{(γ + 1) / (2 (γ - 1))}

— exit Mach number ko nozzle geometry se jodata hai

Kam

γ

rocket performance ko kyun benefit karta hai?

Kam

γ

matlab same Mach number ke liye kam temperature drop, jo higher specific impulse aur better performance mein translate hota hai

P_{0}

use karte time common mistake kya hai?

Stagnation pressure

P_{0}

(total pressure agar flow isentropically rok di jaaye) ko chamber mein static pressure se confuse karna; yeh approximately equal sirf tab hote hain jab velocity near zero ho

Agar

M_{e} = 3

aur

γ = 1.2

ho, toh exit par chamber temperature ka kitna fraction bachta hai?

T_{e} / T_{0} = 1/ (1 + 0.1 \times 9) = 1/1.9 \approx 0.526

ya approximately 53%

Saari exit conditions nikaalny ke liye aapko kaun si teen chamber properties chahiye?

Chamber temperature

T_{0}

, chamber pressure

P_{0}

, aur gas specific heat ratio

γ

(plus exit Mach number

M_{e}

)