Nozzle thermodynamics — isentropic expansion from chamber to exit: Visual walkthrough

3.3.11 · D2 · Physics › Rocket Propulsion › Nozzle thermodynamics — isentropic expansion from chamber to

Yeh page parent result ko — the isentropic nozzle equations — bilkul zero se rebuild karta hai. Koi bhi formula tab tak nahi aata jab tak hum use dekh nahi lete. End tak aap samjhenge, pictures mein, ki chamber ka hot gas ek fast jet mein kyun badalta hai, aur tube pehle kyun squeeze hoti hai, phir flare.

Hum ek badi idea ki taraf build kar rahe hain: thermal chaos in → directed speed out, aur area woh knob hai jo is trade ko control karta hai.

Prerequisites jin par hum rely karenge, har ek tab introduce hoga jab zarurat ho: Ideal Gas Law, Compressible Flow Fundamentals, Mach Number and Sound Speed, Entropy and Reversibility.

Step 1 — "Gas in a chamber" actually kya hota hai

WHAT. Socho ek sealed hot box: combustion chamber. Andar countless molecules hain jo random directions mein fly kar rahi hain, walls se bang kar rahi hain. Woh banging pressure hai; banging ki average speed temperature hai. Abhi kuch bhi ek direction mein point nahi karta — motion pure chaos hai.

WHY. Kuch bhi accelerate karne se pehle, hum un do "stored" quantities ko naam denge jo hum spend karenge: pressure $P$ aur temperature $T$ . Hum unki chamber values ko subscript $0$ se bhi naam denge — woh state jahan gas (almost) rest par hai. Yeh stagnation conditions hain: $P_{0}$ , $T_{0}$ .

PICTURE. Red box dekho: arrows har direction mein point kar rahe hain. Woh randomness energy ka woh fuel tank hai jise hum redirect karne wale hain.

Figure — Nozzle thermodynamics — isentropic expansion from chamber to exit

Step 2 — Ek conservation law: energy gayab nahi ho sakti

WHAT. Jab gas chamber chhodti hai aur speed up hoti hai, hum uski energy track karte hain. Do buckets exist karte hain:

thermal energy per kilogram, likha $h$ (enthalpy kehte hain) — chaos,
kinetic energy per kilogram, $\frac{1}{2} V^{2}$ — directed motion, jahan $V$ speed hai.

Total ko stagnation enthalpy $h_{0}$ kehte hain — woh enthalpy jo gas ki hoti agar use fully rest par laya jaata, taaki uski saari energy thermal bucket mein baith jaaye. Ek ideal gas ke liye yeh total hai $h_{0} = c_{p} T_{0}$ (at-rest, chamber state ki thermal energy). Total kabhi nahi badlti:

$thermal (chaos) h + directed motion \frac{V ^{2}}{2} = total, fixed = c_{p} T_{0} h_{0}$

WHY yeh tool aur koi dusra kyun nahi? Hum pooch rahe hain "speed kahan se aati hai?" Energy conservation wahi ek law hai jo kehta hai ek bucket exactly utna hi bharta hai jitna doosra khaali hota hai. Yeh seedha sawal ka jawab deta hai: speed thermal energy se kharidi jaati hai.

WHY $h$ aur sirf $T$ kyun nahi? Ideal gas ke liye dono proportional hain: $h = c_{p} T$ , jahan $c_{p}$ specific heat at constant pressure hai — kitni energy ek kilogram gas ko ek degree raise karti hai jab woh freely expand kar sakti ho. $h = c_{p} T$ aur $h_{0} = c_{p} T_{0}$ substitute karke:

$thermal now c_{p} T + \frac{V ^{2}}{2} = thermal at rest c_{p} T_{0}$

PICTURE. See-saw: jaise kinetic bucket (red) bharta hai, thermal bucket exactly utna hi khaali hota hai. Total bar height constant rehti hai.

Step 3 — Speed measure karne ka smart tarika: Mach number

WHAT. Raw speed m/s mein awkward hai — "fast" gas par depend karta hai. Iske bajaye speed ko usi gas mein sound ki speed $a$ se compare karo. Unka ratio Mach number hai:

$M = \frac{V}{a}, a = γ R T$

$V$ = flow speed, $a$ = local sound speed,
$R$ = gas constant (gas kitna "springy" hai), $T$ = local temperature,
$γ = c_{p} / c_{v}$ = heat capacity ratio, gas ke liye ek fixed number (rocket exhaust ke liye ≈1.2). Yahan $c_{v}$ specific heat at constant volume hai — ek kilogram ko ek degree raise karne ki energy jab gas expand nahi kar sakti. Kyunki freely expanding gas pushing work bhi karti hai, hamesha $c_{p} > c_{v}$ , isliye $γ > 1$ .

WHY yeh tool? Sound woh fastest hai jitni ek pressure signal travel kar sakti hai. Flow $M < 1$ (subsonic, signals usse aage) hai ya $M > 1$ (supersonic, nahi kar sakti) — yeh nozzle ka poora behaviour flip kar deta hai — Step 6 dekho. Mach number woh switch hai. Mach Number and Sound Speed dekho.

PICTURE. Ek runner (flow) aur sound wave ke beech race. Mach 1 se neeche wave aage rehti hai; Mach 1 se upar runner wave ko peeche chhodta hai, ek cone banaate hue.

Ab Step 2 ka energy law $M$ use karke rewrite karo. $c_{p} = \frac{γ R}{γ - 1}$ substitute karo (jo $c_{p} - c_{v} = R$ aur $γ = c_{p} / c_{v}$ se milta hai) aur $V = M a$ with $a^{2} = γ R T$ :

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

Har term ko $\frac{γ R T _{0}}{γ - 1}$ se divide karo aur $T$ factor karo:

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

jo rearrange hota hai:

Step 4 — "No waste" rule: isentropic flow

WHAT. Hum journey ke baare mein do cheezein assume karte hain:

Adiabatic — walls se koi heat leak nahi hoti (gas milliseconds mein chali jaati hai; time hi nahi hai).
Reversible — koi friction nahi, koi shockwave flow ko scramble nahi kar rahi.

Ek saath iska matlab hai entropy $s$ constant rehta hai: $Δ s = 0$ . Yeh isentropic flow hai. Entropy and Reversibility dekho.

WHY. Constant entropy woh bridge hai jo pressure ko temperature se link karta hai. Iske bina, $T$ jaanna $P$ ke baare mein kuch nahi batata. Iske saath — aur provided gas calorically perfect ho, matlab $c_{p}$ aur $c_{v}$ (aur isliye $γ$ ) poori expansion mein constant hain — woh dono ek saath lock ho jaate hain:

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

left: temperature ratio (Step 3), right: pressure ratio ek special power tak raised,
exponent $\frac{γ - 1}{γ}$ woh hai jo "constant entropy" cost karta hai.

PICTURE. Do dials, $T$ aur $P$ , ek rigid rod se wired jo labelled hai " $s =$ const". Ek ko ghuma, doosra ek fixed curve ke along follow karna padega.

Invert karo (power $γ / (γ - 1)$ tak raise karo) aur Step 3 plug in karo:

Step 5 — Tube ki shape kyun badalni padti hai: the area rule

WHAT. Mass pile up nahi ho sakta: jo per second andar flow karta hai woh per second bahar flow karta hai. Density $ρ$ , speed $V$ , area $A$ ke saath:

$ρ V A = \overset{m}{˙} = constant (the co n t in u i t y law)$

Nozzle ke andar ek tiny step lo aur demand karo yeh constant rahe. Ab hum master shape equation khud build karte hain, teen facts se jo hmare paas pehle se hain.

WHY (derivation, step by step).

(a) Continuity "percent-change" form mein. Agar $ρ V A$ kabhi nahi badlta, toh uske teen factors ke fractional changes cancel hone chahiye. $ρ V A =$ const ka differential lete hue aur $ρ V A$ se divide karke:

$\frac{d ρ}{ρ} + \frac{d V}{V} + \frac{d A}{A} = 0$

Har term ek "percent change" hai: agar area 1% badhti hai aur density 1% girta hai, speed untouched rahti hai. Yeh sirf bookkeeping hai — product fixed hai, isliye percent changes zero sum karte hain.

(b) Momentum (Euler) smooth, frictionless flow ke liye. Gas ko aage push karna ek pressure drop cost karta hai. Steady flow ke liye nozzle ke along, $d P = - ρ V d V$ . Percent form mein rearrange karke:

$\frac{d ρ}{ρ} = \frac{d P}{ρ} \cdot \frac{1}{a ^{2}} \cdot \frac{a ^{2} ρ}{ρ} = - \frac{V ^{2}}{a ^{2}} \frac{d V}{V} = - M^{2} \frac{d V}{V}$

Yahan hum do cheezein use karte hain jo hmare paas pehle se hain: isentropic flow ke liye sound speed obey karta hai $a^{2} = d P / d ρ$ (ek pressure bump sound speed par travel karta hai), aur $M = V / a$ Step 3 se. Words mein: faster flow matlab thinner gas, aur thinning $M^{2}$ se amplify hoti hai — yahi exact reason hai ki supersonic gas itni violently thin hoti hai.

(c) (b) ko (a) mein substitute karo. $d ρ / ρ$ ko $- M^{2} d V / V$ se replace karo:

\;\Longrightarrow\; (1-M^2)\frac{dV}{V}=-\frac{dA}{A}$$ $(1-M^2)$ se divide karo aur sign flip karo: > [!formula] Area–velocity rule > $$\frac{dV}{V} = \frac{1}{\,M^2 - 1\,}\,\frac{dA}{A}$$ > $dV$ = tiny speed change, $dA$ = tiny area change, aur $M^2-1$ = woh sign-flipping gatekeeper jo humne abhi build kiya. Poori kahani us denominator mein chupi hai. **PICTURE (logic ka).** Tube ke andar do competing effects ladte hain: tube ko wider banana gas ko *slow* karta hai (zyada room), lekin gas ka thin hona ($-M^2\,dV/V$) use *speed up* karta hai. Mach 1 se neeche room-effect jeetta hai; Mach 1 se upar thinning-effect jeetta hai. Unka tie $M=1$ hai. **WHY it matters.** Denominator $M^2-1$ *sign flip* karta hai jab $M$ 1 cross karta hai, isliye area mein *wahi* change speed par opposite effects produce karta hai: | Regime | $M^2-1$ | Speed up karne ke liye ($dV>0$) hume chahiye... | Tube shape | |---|---|---|---| | Subsonic $M<1$ | negative | $dA<0$ | **converging** (squeeze) | | Supersonic $M>1$ | positive | $dA>0$ | **diverging** (flare) | | Sonic $M=1$ | zero | $dA=0$ | **throat** (narrowest) | **Formula se table padho.** Jab $M<1$, $M^2-1<0$; $dV>0$ paane ke liye right side positive hona chahiye, isliye $dA$ *negative* hona chahiye (squeeze). Jab $M>1$, $M^2-1>0$; ab $dA>0$ (flare) se $dV>0$ milta hai. Formula aur table ab exactly agree karte hain. **PICTURE.** Classic hourglass profile: ek waist tak squeeze karo, phir flare karo. Red waist ==throat== hai, woh ek spot jahan $M=1$ hota hai. ![[deepdives/dd-physics-3.3.11-d2-s05.png]] > [!intuition] Counter-intuitive bit > Mach 1 se neeche, squeeze karna gas ko speed up karta hai (garden hose ki tarah). Mach 1 se upar gas itni fast *thin* ho jaati hai ki tube ko widen karna woh cheez hai jo use accelerate karta hai. Throat woh unique handover point hai. --- ## Step 6 — Throat Mach 1 par force hota hai **WHAT.** Throat par $dA=0$. Area rule tab demand karta hai ya toh $dV=0$ ya $M^2-1=0$. Ek aise flow ke liye jo *continuously accelerate* karta rahe, sirf ek consistent choice hai $M=1$ exactly. Hum throat quantities ko ek star se mark karte hain: $T^*, P^*, \rho^*$. **WHY.** Yeh *choke point* hai: yeh maximum mass flow fix karta hai aur reference area $A^*$ set karta hai jiske against baki sab measure hota hai. Steps 3–4 mein $M=1$ set karo: > [!formula] Critical (throat) ratios > $$\frac{T^*}{T_0}=\frac{2}{\gamma+1},\quad > \frac{P^*}{P_0}=\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}},\quad > \frac{\rho^*}{\rho_0}=\left(\frac{2}{\gamma+1}\right)^{\frac{1}{\gamma-1}}$$ $\gamma=1.2$ ke liye: $T^*/T_0 = 0.909$, $P^*/P_0 = 0.564$. Waist par gas abhi bhi 91% utni hot hai lekin sirf 56% chamber pressure par hai. **PICTURE.** Ek gauge jo chamber (M=0) se throat (M=1) tak sweep karta hai: temperature bar mushkil se dip karta hai, pressure bar hard drop karta hai, aur red needle exactly Mach 1 par land karta hai waist par. ![[deepdives/dd-physics-3.3.11-d2-s06.png]] --- ## Step 7 — Geometry ko performance mein convert karna: area–Mach map **WHAT.** Kisi bhi point par continuity ko throat par continuity ke saath combine karo ($\rho V A = \rho^* V^* A^*$). Har isentropic ratio substitute karne par woh equation milti hai jo ek designer actually use karta hai: > [!formula] Area–Mach relation > $$\frac{A}{A^*}=\frac{1}{M}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$$ - $A/A^*$ = ==area ratio==, woh geometry jo aap metal mein carve karte ho, - $M$ = exit Mach number, woh performance jo aap chahte ho, - exponent $\frac{\gamma+1}{2(\gamma-1)}$: $\gamma=1.2$ ke liye yeh equals $\frac{2.2}{0.4}=5.5$ — ek bada power, isliye area throat se steeply grow karti hai. **WHY.** Yeh dono directions mein chalta hai: desired exit Mach *choose* karo → required flare ratio *pao*. Gaur karo ki $A/A^*$ ka **minimum 1 hai $M=1$ par** aur *dono* $M<1$ aur $M>1$ ke liye badhta hai — wahi area do baar appear karta hai, ek baar subsonic, ek baar supersonic. Aap konsa paate ho yeh depend karta hai ki aap throat se pehle ho ya baad mein. (Exactly 1 ki minimum value VERIFY mein machine-checked hai.) **PICTURE.** $A/A^*$ vs $M$ ka U-shaped curve, $(1,1)$ par red mein bottom karta hua. Left arm = converging (subsonic), right arm = diverging (supersonic). ![[deepdives/dd-physics-3.3.11-d2-s07.png]] --- ## Step 8 — Worked numbers, curve par dekhe gaye Parent ka Example 2 use karte hue: $T_0=3500\text{ K}$, $P_0=10\text{ MPa}$, $\gamma=1.2$, exit $M_e=3$. Pehle recurring group note karo $1+\frac{\gamma-1}{2}M_e^2 = 1+0.1\times9 = 1.9$. > [!example] $M_e=3$ par exit conditions > **Temperature:** $\displaystyle T_e=\frac{3500}{1.9}=1842\text{ K}$ > **Pressure:** $\displaystyle P_e=\frac{P_0}{(1.9)^{\gamma/(\gamma-1)}}=\frac{10}{(1.9)^{6}}\text{ MPa}=0.213\text{ MPa}=213\text{ kPa}$ > **Area ratio:** $\displaystyle \frac{A_e}{A^*}=\frac13\left[\tfrac{2}{2.2}(1.9)\right]^{5.5}=\frac13(1.727)^{5.5}=6.7$ **WHAT it means.** Mach 3 tak pahunchne ke liye, exit throat area ka **≈6.7×** hona chahiye. Kyunki $P_e=213\text{ kPa} > P_a=101\text{ kPa}$, jet abhi bhi lip par outward push kar rahi hai — ==overexpanded== yahan galat word hai; exhaust pressure ambient se *zyada* hai, isliye nozzle **under-expanded** hai: jet lip chhodne ke *baad* expand karti rehti hai. [[Nozzle Flow Regimes]] aur [[Specific Impulse Optimization]] dekho ki $P_e$ ko $P_a$ se match karna goal kyun hai. --- ## Ek-picture summary Upar sab kuch, ek single chamber-to-exit strip par: tube ek red throat tak squeeze hoti hai phir flare karti hai; neeche, teen tracks hain jo temperature girte, pressure crash hote, aur Mach number waist par 1 seedha climb karte dikhate hain. ![[deepdives/dd-physics-3.3.11-d2-s08.png]] > [!recall]- Feynman retelling — plain words mein bolo > Ek hot box se shuru karo jisme gas har direction mein fail rahi hai — woh chaos pressure aur temperature hai. Energy gayab nahi ho sakti, isliye agar main gas ko *ek direction* mein move karwata hun, uski randomness (temperature) exactly utni hi drop honi chahiye. Har second ek changing pipe se utni hi gas flow karwane ke liye, jab gas slow hai main pipe *squeeze* karke speed up karta hun; lekin jab woh narrowest waist par sound speed hit karta hai, rules flip ho jaate hain — ab main pipe *widen* karta hun use accelerate karte rehne ke liye. Woh waist hamesha Mach 1 hoti hai. Kyunki maine assume kiya no friction aur no heat leaks (constant entropy) aur ek gas jiska $\gamma$ kabhi drift nahi karta, temperature, pressure aur speed sab ek number se locked hain, Mach number: woh exit Mach choose karo jo main chahta hun, aur ek formula mujhe exactly batata hai exit throat se kitna wider hona chahiye. Gas thando, uska pressure girado, aur bahar nikalti hai ek supersonic jet — woh jet thrust hai. > [!recall]- Quick self-test > Nozzle mein exactly Mach number 1 kahan hai? ::: Throat par, minimum-area point par. > Gas speed up hone par temperature kyun girta hai? ::: Energy conservation — thermal enthalpy kinetic energy mein convert hoti hai, isliye $T$ girta hai jaise $V^2$ badhta hai. > Mach 1 se neeche, gas accelerate karne ke liye tube squeeze karo ya flare karo? ::: Squeeze (converging): $dA<0$ se $dV>0$ milta hai jab $M<1$. > Mach 1 se upar, squeeze ya flare? ::: Flare (diverging): $dA>0$ se $dV>0$ milta hai jab $M>1$. > "Isentropic" assumption se kya milta hai? ::: Constant entropy (constant $\gamma$ ke saath) $P$, $T$, $\rho$ ko lock karta hai, ek Mach number ko teeno fix karne deta hai. > [!mnemonic] Flip yaad rakho > **"Slow squeeze, fast flare, sonic at the waist."** > Mach 1 se neeche speed up karne ke liye *squeeze* karo; Mach 1 se upar speed up karne ke liye *flare* karo; handover — Mach 1 — narrowest waist par hota hai, throat par. --- **Continue:** [[Thrust Equation Derivation]] yahan mili exit velocity use karta hai · [[Combustion Chamber Thermodynamics]] $P_0,T_0$ supply karta hai · [[Nozzle Flow Regimes]] over/underexpansion explain karta hai.