Choked flow — condition M = 1 at throat, maximum mass flow: Visual walkthrough

3.1.8 · D2 · Physics › Compressible Flow & Aerodynamics › Choked flow — condition M = 1 at throat, maximum mass flow

Step 1 — "Mass flow" ek hole se hoti kya hai?

KYA. Socho gas pipe ke ek slice se guzar rahi hai jiska cross-section area $A$ hai. Ek second mein kitne kilogram us slice ko cross karte hain?

KYUN. Choking ke baare mein sab kuch is ek number ke baare mein baat hai — kilograms per second, likha jaata hai $\overset{m}{˙}$ (dot matlab "per second"). Yeh peak kahan hai yeh jaanne se pehle, hume jaanna padega yeh hai kya.

PICTURE. Figure mein shaded slug of gas dekho. Ek second mein gas apni speed $V$ ke barabar distance aage jaati hai (ek second mein kitne metres chali). Toh jo slug us slice ko cross karti hai woh $V$ length aur $A$ face area ka cylinder hai: uska volume $A \times V$ hai. Density $ρ$ (har cubic metre mein kitne kilograms) se multiply karo aur kilograms per second milte hain.

Figure — Choked flow — condition M = 1 at throat, maximum mass flow

Teen cheezein ( $ρ$ , $A$ , $V$ ) $\overset{m}{˙}$ ko set karne mein lagin hain. Jab hum gas ko faster squeeze karte hain, $V$ badhta hai — lekin $ρ$ ghat jaata hai (fast gas patli ho jaati hai). Wahi tug-of-war poori kahaani hai. Hume ise describe karne ke liye ek single dial chahiye.

Step 2 — Woh single dial: Mach number $M$

KYA. Mach number $M$ flow speed $V$ ko local speed of sound $a$ se divide karna hai: $M = \frac{V}{a}, a = γ R T .$

KYUN. $ρ$ aur $V$ ko alag alag track karne ki jagah, hum dono ko ek variable $M$ ke terms mein likhenge. Tab $\overset{m}{˙}$ ek single dial ka function ban jaata hai, aur "maximum dhundho" ek one-variable calculus problem ban jaata hai. $M$ natural dial hai kyunki speed of sound $a$ exactly woh speed hai jis par pressure news travel karti hai — aur choking un news ke baare mein hai jo upstream nahi pahunch sakti.

PICTURE. Neeche wala dial: $M < 1$ subsonic (news upstream jaati hai, throat exit ko "sun" sakta hai), $M = 1$ sonic (news throat par freeze ho jaati hai), $M > 1$ supersonic (purely converging nozzle mein impossible). Har symbol earn kiya:

$γ$ = ratio of specific heats (air ke liye, $1.4$ ) — gas kitni springy hai.
$R$ = gas constant (air ke liye, $287 J kg^{- 1} K^{- 1}$ ).
$T$ = local temperature.

Step 3 — Gas ke speed up hone par temperature kaise girti hai

KYA. Steady flowing gas ke liye energy conservation se, flowing temperature $T$ aur reservoir ("stagnation") temperature $T_{0}$ aapas mein linked hain: $\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2} .$

KYUN. Ek bade tank mein parkhi hui gas ki saari energy heat ke roop mein hoti hai (temperature $T_{0}$ ). Jab woh flow karti hai, kuch heat motion ban jaati hai. Faster flow (bada $M$ ) ⇒ thandi gas (chhota $T$ ). Hume yeh chahiye kyunki $ρ$ aur $V$ dono $T$ par dependent hain, aur hum inhe $M$ ke terms mein chahte hain.

PICTURE. Curve dikhata hai $T / T_{0}$ $1$ se neeche slide karta hai ( $M = 0$ , still gas) jab $M$ badhta hai. Term by term: $\frac{T _{0}}{T} = 1 + = 0.2 for air \frac{γ - 1}{2} how fast, squared M^{2}$ " $+$ " keh raha hai: koi bhi motion $T_{0} / T$ mein add karta hai, toh $T$ sirf gir sakti hai. Air ke liye $M = 1$ par, $T_{0} / T = 1.2$ , matlab $T = 0.833 T_{0}$ — hamari critical temperature.

Step 4 — Gas ke speed up hone par density kaise girti hai

KYA. Isentropic (no-heat-loss, no-friction) flow use karte hue, density temperature ke baad ek power law follow karti hai: $\frac{ρ _{0}}{ρ} = (1 + \frac{γ - 1}{2} M^{2})^{\frac{1}{γ - 1}} .$

KYUN. Step 1 mein humne dekha $\overset{m}{˙} = ρ A V$ : hume jaanna padega ki jab $M$ badhta hai toh density $ρ$ kaise patli hoti hai. Isentropic flow $ρ$ ko $T$ se $ρ \propto T^{1/ (γ - 1)}$ ke zariye tie karta hai, toh wahi bracket dobara aata hai, bas ek naye power par raise hota hai.

PICTURE. Dono curves compare karo: temperature (Step 3) aur density (yeh step) dono neeche slide karte hain, lekin density zyada tezi se girti hai (air ke liye bada exponent: $1/ (γ - 1) = 2.5$ ). Yahi "fast gas thins out" effect hai jo $V$ se lada karega.

$ρ = ρ_{0} (1 + \frac{γ - 1}{2} M^{2})^{- \frac{1}{γ - 1} minus \Rightarrow ρ shrinks}$

Step 5 — $\overset{m}{˙}$ ko sirf $M$ ke function ke roop mein assemble karo

KYA. Steps 3–4 ko $\overset{m}{˙} = ρ A V$ mein substitute karo, $V = M a = M γ R T$ aur $ρ_{0} = p_{0} / (R T_{0})$ use karte hue. Sab kuch yahan collapse hota hai: $\overset{m}{˙} = fixed constants \frac{A p _{0}}{T _{0}} \frac{γ}{R} \cdot f (M) — the only part that moves \frac{M}{( 1 + \frac{γ - 1}{2} M ^{2} ) ^{\frac{γ + 1}{2 ( γ - 1 )}}}$

KYUN. Ek fixed reservoir ( $p_{0}, T_{0}$ ) aur ek fixed slice area $A$ ke liye, leading clump kabhi nahi badlta. Saara drama $f (M)$ mein hai. Peak mass flow dhundne ke liye hume bas $f (M)$ ka peak dhundna hai.

PICTURE. Numerator $M$ curve ko upar kheenchta hai (faster flow). Denominator (density-and-temperature bracket) ise neeche kheenchta hai (thinning gas). Chhote $M$ par, upar wala jeetta hai aur $\overset{m}{˙}$ badhta hai. Bade $M$ par, neeche wala jeetta hai aur $\overset{m}{˙}$ girta hai. Kahin beech mein ek hilltop hai.

Step 6 — Hilltop dhundho: slope ko zero karo

KYA. Peak wahan hai jahan curve momentarily flat ho — uska slope $\frac{d m ˙}{d M} = 0$ . Kyunki $f$ powers ka product/quotient hai, hum pehle logarithm lete hain (products ko sums mein badal deta hai, differentiate karna easy ho jaata hai), phir derivative ko zero set karte hain: $\frac{1}{M} - \frac{γ + 1}{2 ( γ - 1 )} \cdot \frac{( γ - 1 ) M}{1 + \frac{γ - 1}{2} M ^{2}} = 0.$

Derivative kyun use karein? Derivative woh tool hai jo sawaal ka jawaab deta hai "yeh curve kahan flat hai?" — exactly woh sawaal "maximum kahan hai?" Koi doosra tool hilltop directly nahi deta. Pehle $ln$ lena ek convenience hai: $ln f = ln M - \frac{γ + 1}{2 ( γ - 1 )} ln (1 + \frac{γ - 1}{2} M^{2})$ line-by-line differentiate hota hai.

PICTURE. Hill par tangent line exactly top par horizontal hoti hai. Equation saaf karo: $\frac{1}{M} = \frac{( γ + 1 ) M /2}{1 + \frac{γ - 1}{2} M ^{2}} \Rightarrow 1 + \frac{γ - 1}{2} M^{2} = \frac{γ + 1}{2} M^{2} \Rightarrow M^{2} = 1.$

Dono taraf se $\frac{γ - 1}{2} M^{2}$ term subtract karo: $1 = \frac{( γ + 1 ) - ( γ - 1 )}{2} M^{2} = \frac{2}{2} M^{2} = M^{2}$ . Toh $M = 1$ — $γ$ bilkul cancel ho jaata hai.

Step 7 — Edge cases: hilltop se hat ke kya hota hai

KYA. Teen regimes, aur ek degenerate limit.

KYUN. Contract: reader ko koi bhi aisa scenario kabhi nahi milna chahiye jo humne dikhaya na ho. Toh hum curve ka har hissa walk karte hain.

PICTURE.

$M < 1$ (peak ke left, subsonic): back pressure $p_{b}$ abhi bhi flow control karta hai. $p_{b}$ ghataao ⇒ hill par chadho ⇒ zyada $\overset{m}{˙}$ . Throat exit ko "sun" sakta hai.
$M = 1$ (the peak): throat pressure air ke liye $p^{*} = 0.528 p_{0}$ par lock ho jaata hai (Isentropic Flow Relations se $M = 1$ par). $p_{b}$ aur ghataane se kuch nahi hota — tum already top par ho.
$M > 1$ (peak ke right): purely converging duct mein reach nahi ho sakta. Supersonic jaane ke liye diverging section chahiye — ek Converging–Diverging (de Laval) Nozzle. Throat ke baad, zyada strong back pressure ek normal shock force kar sakta hai.
Degenerate $M = 0$ : still gas, $\overset{m}{˙} = 0$ — curve ka left end, kuch bhi flow nahi karta.

Ek-picture summary

Yeh single figure poori derivation stack karta hai: do competing effects ( $M$ ka up-pull, density ka down-pull), unka product $f (M)$ , aur $M = 1$ par vertical line jo peak nail karti hai — saath mein locked throat pressure $p^{*} = 0.528 p_{0}$ marked hai.

Recall Feynman retelling — walkthrough simple shabdon mein

Humne pucha: ek second mein kitne kilograms pipe ke ek slice se guzarte hain? Woh density times area times speed hai. Jab hum gas ko faster push karte hain, uski speed badhti hai — acha, zyada flow — lekin gas bhi patli hoti jaati hai — bura, kam flow. Is fight ko settle karne ke liye humne speed ko "Mach" mein measure kiya, flow speed ko speed of sound se compare kiya, kyunki sound exactly woh hai jis se pressure news travel karti hai. Humne patlaapan aur teziyon dono ko Mach ke terms mein likha, unhe multiply kiya, aur ek curve mili jo zero se shuru hoti hai, upar jaati hai, aur zero par wapas aati hai. Aisi curve ka ek top hota hai. Humne top dhundha yeh pooch ke "curve kahan flat hai?" — woh derivative zero set karna hai — aur Mach = 1 nikla, aur har gas wahi jawaab deta hai. Us top par throat pressure air ke liye tank pressure ka lagbhag $0.528$ par freeze ho jaata hai, aur throat deaf ho jaata hai chahe downstream kuch bhi ho. Woh frozen, maximum flow wohi hai jise hum choking kehte hain.

Recall Quick self-test

$\overset{m}{˙}$ ko $ρ$ aur $V$ alag alag ki jagah $M$ ke terms mein kyun likhte hain? ::: Taki dono competing effects (speed up, density down) ek variable mein aa jayein aur maximum ek one-variable calculus problem ban jaaye. Hilltop kaun sa tool dhundta hai, aur woh kaun sa sawaal answer karta hai? ::: Derivative zero set karna — yeh answer karta hai "curve kahan flat hai?", wahi jagah maximum hota hai. Kya choke Mach number gas par depend karti hai? ::: Nahi — $γ$ cancel ho jaata hai; $M = 1$ har perfect gas ke liye. Ek converging nozzle $M = 1$ se past kyun nahi ja sakta? ::: Area–Mach relation ko sonic se aage accelerate karne ke liye increasing area chahiye; converging duct sirf throat ki taraf shrink karta hai.