Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂ - P₀₁: Visual walkthrough

3.1.12 · D2 · Physics › Compressible Flow & Aerodynamics › Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂

Step 0 — Hum dekh kya rahe hain?

KYA HAI. Ek box (ek control volume) banao jo shock ke dono taraf ho. Gas left se state 1 mein andar aata hai aur right se state 2 mein bahar jaata hai. Shock khud beech mein ek patli vertical line hai.

KYUN. Control volume bas space ka ek fixed region hota hai jise hum dekhne ka faisla karte hain. Jo kuch uske left face se cross karta hai, uska hisaab right face par dena padega. Yahi bookkeeping poora khel hai — koi forces nahi, koi jaadu nahi, bas "jo andar jaata hai woh bahar aana chahiye."

PICTURE. Neeche ki figure mein, blue hai fast upstream gas, orange hai slow aur squeezed downstream gas, aur red line hai shock.

Figure — Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂ - P₀₁

Har quantity ek subscript carry karta hai jo batata hai ki woh kis side par hai:

Step 1 — Mass pile up nahi ho sakti (continuity)

KYA HAI. Har second mein har face cross karne wale gas ke kilograms gino. Left face se andar equal hai right face se bahar: $\underbrace{\rho_1 u_1}_{\substack{\text{kg/s per m}^2\\ \text{entering}}}=\underbrace{\rho_2 u_2}_{\substack{\text{kg/s per m}^2\\ \text{leaving}}}\tag{1}$

Term by term: $ρ$ hai har cubic metre mein kitna gas hai, $u$ hai us gas ke kitne metres har second sweep karte hain. Inhe multiply karo aur milta hai kilograms per second sweeping past, face ke har square metre par. Woh number dono sides par identical hona chahiye — warna mass box ke andar hamesha ke liye build up ho jaata.

KYUN. Paper-thin shock ke andar kuch create ya destroy nahi hota. Yeh sabse tight constraint hai jo hume milta hai aur yeh $ρ$ ko $u$ se link karta hai: agar gas slow ho jaata hai ( $u_{2} < u_{1}$ ), to use zaroor denser hona padega ( $ρ_{2} > ρ_{1}$ ) product ko fixed rakhne ke liye.

PICTURE. Wahi "mass current" arrow, dono sides par mota, lekin right par gas slower hai (short arrow) aur denser hai (dots ek saath crowded).

Step 2 — Momentum sirf ek net push se badalta hai

KYA HAI. Box par net pressure force equal hai bahar flowing momentum ke change ke: $\underbrace{P_1+\rho_1 u_1^2}_{\text{"push" on left}}=\underbrace{P_2+\rho_2 u_2^2}_{\text{"push" on right}}\tag{2}$

Term by term: $P$ hai static pressure jo face par push kar raha hai. $ρ u^{2}$ hai momentum flux — momentum ( $ρ u$ , mass current) jo face ke across speed $u$ par carry hota hai. Inka sum impulse function kehlaata hai, aur yeh dono sides par same hota hai.

KYUN. Flow ke liye Newton's second law: (net force) = (momentum out) − (momentum in). Hamare frictionless, walls-do-no-work box par sirf do faces ke pressures ki forces hain. To poora balance equation (2) tak reduce ho jaata hai. Yahi force karta hai $P$ ko rise karne ke liye jab $u$ girta hai.

PICTURE. Do competing effects arrows ke roop mein drawn: pressure left-to-right push kar raha hai, momentum across carry ho raha hai. Sum ek balance mein see-saw ki tarah level rehta hai.

Step 3 — Energy carry hoti hai, lost nahi hoti (adiabatic)

KYA HAI. Koi heat andar nahi aata, koi shaft work nahi karta, to har kilogram per total energy conserved hai: $\underbrace{c_p T_1}_{\text{stored heat}}+\underbrace{\tfrac12 u_1^2}_{\text{motion}}=\underbrace{c_p T_2}_{\text{stored heat}}+\underbrace{\tfrac12 u_2^2}_{\text{motion}}\tag{3}$

Term by term: $c_{p} T$ hai enthalpy — woh thermal energy jo har kilogram carry karta hai. $\frac{1}{2} u^{2}$ hai kinetic energy jo har kilogram carry karta hai. Jaise flow slow hota hai ( $u$ girta hai), woh lost motion energy kahin nahi ja sakti sirf heat mein, to $T$ badhta hai. Energy bas kapde badal leti hai.

KYUN. Shock itna patla hai ki heat leak nahi kar sakta aur kisi piston par work nahi karta. To thermal + kinetic energy ka sum frozen hai. Yeh exactly woh statement hai ki stagnation temperature $T_{0}$ conserved hai (dekho Stagnation Properties T0 and P0).

PICTURE. Ek two-bar "energy budget": left par, lamba motion bar + short heat bar. Right par bars ki heights swap ho gayi hain, lekin total column height identical hai.

Step 4 — Sab kuch Mach number mein rewrite karo

KYA HAI. Hamare paas teen laws hain lekin yeh $P, ρ, u, T$ mix karte hain. Clean move hai Mach number mein translate karna, ideal gas ke baare mein do facts use karke: $a^{2} = γ R T, u = M a \Rightarrow ρ u^{2} = ρ γ R T M^{2} = γ P M^{2} .$

Woh last identity ne $P = ρR T$ use kiya $ρR T$ ko $P$ se swap karne ke liye. Ab $ρ u^{2} = γ P M^{2}$ ko momentum law (2) mein feed karo: $P_1\underbrace{(1+\gamma M_1^2)}_{\text{depends on }M_1}=P_2\underbrace{(1+\gamma M_2^2)}_{\text{depends on }M_2}\;\Rightarrow\;\boxed{\frac{P_2}{P_1}=\frac{1+\gamma M_1^2}{1+\gamma M_2^2}}\tag{4}$

KYUN yeh tool aur koi nahi? Mach number woh ek dimensionless dial hai jo measure karta hai "flow kitna supersonic hai." Kyunki ek shock ki poori personality is baat se set hoti hai ki flow sound se kitna aage hai, sab kuch $M$ mein express karne se answer sirf $M_{1}$ par depend karta hai — har doosra variable ( $P, T, ρ$ ) ek consequence ban jaata hai. Isliye humne Mach number choose kiya, maan lo absolute velocity ke bajaaye.

PICTURE. Ek conversion chart: left par messy $P, ρ, u, T$ box, ek arrow labelled " $u = M a, a^{2} = γ R T$ ", aur right par tidy $M$ -only box.

Energy (3) par wohi substitution karo. Kyunki $T_{0} = T (1 + \frac{γ - 1}{2} M^{2})$ conserved hai: $\frac{T_2}{T_1}=\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}\tag{5}$ Yahan $1 + \frac{γ - 1}{2} M^{2}$ woh factor hai jo static temperature $T$ ko uske stagnation value $T_{0}$ mein turn karta hai — yeh Mach language mein likha "kinetic energy tax" hai.

Aur mass (1), $ρ u = \frac{P M γ}{R T}$ use karke: $\frac{P_1 M_1}{\sqrt{T_1}}=\frac{P_2 M_2}{\sqrt{T_2}}\tag{6}$

Step 5 — Teeno ko ek mein fold karo

KYA HAI. Ab hamare paas teen Mach-flavoured equations hain (4), (5), (6). Pressure ratio (4) aur temperature ratio (5) ko mass equation (6) mein substitute karo, phir square roots hatane ke liye dono sides square karo. Saare $P$ 's aur $T$ 's cancel ho jaate hain aur hum ek single equation ke saath bacht jaate hain jo relate karta hai sirf $M_{1}$ aur $M_{2}$ ko.

KYUN. Teen equations hain, lekin hum ek clean relationship chahte hain. $P$ aur $T$ ko eliminate karna standard algebra move hai: combine karte raho jab tak sirf woh do quantities survive nahi kar letein jo tumhe care hai.

PICTURE. Teen streams (mass, momentum, energy) ek funnel mein flow karte hue aur ek equation ke roop mein drip karte hue.

Woh single equation ek quadratic in $M_{2}^{2}$ hai. Ek quadratic ke do roots hote hain:

Step 6 — Physics check karo: supersonic in, subsonic out

KYA HAI. (7) mein numbers plug karo aur dekho kya nikalta hai. Air ke liye ( $γ = 1.4$ ):

$M_{1}$	$M_{2}$ from (7)
$1.0$	$1.000$
$2.0$	$0.577$
$5.0$	$0.415$
$\to \infty$	$\to 0.378$

KYUN. Formula tabhi trustworthy hota hai jab woh sahi behave kare. $M_{1} > 1$ wali har entry $M_{2} < 1$ deti hai: supersonic hamesha subsonic ban jaata hai. Yeh by hand nahi daala gaya — yeh (7) se naturally nikalta hai, aur yeh second law se enforce hota hai (doosri taraf shock entropy decrease kar deta, jo forbidden hai).

PICTURE. Curve $M_{2} (M_{1})$ line $M_{2} = M_{1}$ ko exactly ek baar $M_{1} = 1$ par cross karta hai, phir uske neeche bend ho jaata hai aur $0.378$ ki taraf flatten ho jaata hai.

Step 7 — Edge cases (reader ko kabhi stranded mat chodo)

KYA HAI & KYUN & PICTURE, teeno limits ek figure par:

Mach par strong-shock cap ka ek famous partner hai: density bhi $\frac{γ + 1}{γ - 1} = 6$ par saturate hoti hai (dekho Rankine–Hugoniot Relations). Dono ceilings ek hi source se aate hain — mass conservation plus ek finite velocity drop.

Ek-picture summary

Yeh final figure poora safar compress karta hai: teen conservation laws (mass, momentum, energy) left se enter karte hain; ideal-gas + Mach translation unhe beech mein squeeze karta hai; ek single quadratic right par nikalta hai; uska physical root equation (7) hai; aur uska behaviour hai supersonic-in / subsonic-out curve. Us ek relation se, parent note mein har property ratio back-substitution se follow karta hai.

Recall Feynman: poora walkthrough plain words mein

Socho gas ek box mein sound se tez rush kar rahi hai. Teen unbreakable rules apply hote hain. Rule one: andar ke kilograms bahar ke kilograms ke equal hain — to agar gas andar slow ho jaati hai, to use denser hona padega. Rule two: sirf ek cheez gas ka momentum change kar sakti hai woh hai pressure jo dono ends par push kar raha hai — to jaise yeh slow hoti hai, pressure zaroor climb karta hai. Rule three: koi heat andar ya bahar nahi jaata — to gas jo motion energy lose karti hai woh seedha heat mein turn ho jaati hai, aur woh warm up hoti hai. In rules ko combine karna aasaan banane ke liye, hum inhe ek number use karke rewrite karte hain, Mach number, jo bas yeh kehta hai ki gas sound se kitna tez ja rahi hai. Jab hum teeno rewritten rules ko blend karte hain aur bachi hui cheezein cancel karte hain, hum ek single tidy equation ke saath bacht jaate hain: downstream Mach number sirf upstream Mach number par depend karta hai. Uske do answers hain — "kuch nahi hua" (boring) aur "ek real shock" (jo hum chahte hain). Real wala hamesha fast-than-sound flow ko slower-than-sound flow mein turn karta hai, kabhi ulta nahi, kyunki ulta karna entropy tidy kar deta, aur nature kabhi entropy free mein tidy nahi karta.

Recall Self-test

Hum teen laws ko Mach number ke terms mein kyun rewrite karte hain? ::: Kyunki ek shock ka poora behaviour is baat se set hota hai ki flow sound se kitna aage hai; $M$ use karne se answer sirf $M_{1}$ par depend karta hai aur $P, T, ρ$ mere consequences ban jaate hain. $M_{2}^{2}$ ke liye quadratic ke do roots hain — fake wala kaun sa hai aur kyun? ::: $M_{2} = M_{1}$ , "no shock" solution; yeh algebra satisfy karta hai lekin represent karta hai flow ko jo unchanged pass through karta hai. Jab $M_{1} \to \infty$ , air ke liye $M_{2}$ kya approach karta hai? ::: $M_{2} \to \frac{γ - 1}{2 γ} \approx 0.378$ — downstream Mach floor out ho jaata hai, kabhi zero nahi pahunchta. Normal shock subsonic flow mein kyun nahi ho sakta? ::: Iske liye $M_{1} < 1 \to M_{2} > 1$ chahiye, matlab $Δ s < 0$ , jo second law forbid karta hai.