Area-Mach number relation A - A - = f(M) — isentropic flow: Question bank

3.1.6 · D5 · Physics › Compressible Flow & Aerodynamics › Area-Mach number relation A - A - = f(M) — isentropic flow

Intuition Is bank ko kaise use karein

Neeche har item ek misconception ya boundary case ko target karta hai Area–Mach relation ka. Answer cover karo, apni reasoning poori sentences mein zor se bolo, phir reveal karo. Agar sirf "true" ya "false" bol sakte ho bina kyun ke, to abhi idea tumhara nahi hua.

Notation reminder, taaki yahan kuch bhi unexplained na lage:

$A$ = duct ka local cross-sectional area (pipe us jagah kitni chaudi hai).
$A^{*}$ = sonic reference area, woh chaurai jo flow ko exactly Mach 1 par baithne ke liye chahiye hogi — same mass flow aur reservoir state ke liye — ek fixed yardstick, parent note mein define kiya gaya hai.
$M$ = Mach number = flow speed $V$ divided by local speed of sound $a$ .
$γ$ = ratio of specific heats ( $1.4$ air ke liye).
$T$ = local (static) temperature; $p$ = local static pressure; $ρ$ = local density — us station par chalti gas dwara measured.
Stagnation (reservoir) quantities $T_{0}, p_{0}, ρ_{0}$ = woh temperature, pressure aur density jo gas ki hoti agar usse smoothly (isentropically) rest par la diya jaye. Socho ek bada shant tank jo nozzle ko feed kar raha hai: uski still gas $T_{0}, p_{0}, ρ_{0}$ par hai. Jaise gas duct mein speed up hoti hai, uska static $T, p, ρ$ in reservoir values se neeche aa jaata hai — jitni zyada speed, utna zyada drop. Ye "0-subscript" numbers isentropic flow mein constant rehte hain, isliye ye handy references banate hain.

Intuition Curve ka guided walk — har visual feature ko algebra se jodo

Neeche diye figure ko feature by feature dekho, koi bhi "U-shape" ya "double-root" claim baad mein trust karne se pehle:

Red dot (valley floor). Yeh $M = 1$ par baitha hai, height $A / A^{*} = 1$ . $M = 1$ ko $f$ mein plug karo: bracket ban jaata hai $\frac{2}{γ + 1} (1 + \frac{γ - 1}{2}) = \frac{2}{γ + 1} \cdot \frac{γ + 1}{2} = 1$ , aur $1/ M$ front factor $1$ hai — to $f (1) = 1$ exactly. Yahi akela jagah hai jahan curve height 1 ko touch karta hai.
Left arm (lavender, left jaate jaate upar uthta hai). $M \to 0$ ki taraf $1/ M$ factor explode karta hai jabki bracket finite rehta hai, to curve upar shoot karta hai — left dashed annotation is asymptote ko mark karta hai.
Right arm (right jaate jaate upar uthta hai). Bade $M$ ki taraf bracket ( $M^{2}$ ki positive power) explode karta hai, curve ko phir se upar kheenchta hai — right annotation is second asymptote ko mark karta hai.
Green dashed horizontal line $A / A^{*} = 2$ par. Yeh curve ko do baar cross karti hai — ek baar left (subsonic) arm par, ek baar right (supersonic) arm par. Woh do butter-yellow squares ka visual proof hai ki ek area ratio do Mach numbers se map hoti hai.

Figure — Area-Mach number relation A - A - = f(M) — isentropic flow

Intuition

$f (M)$ kahan se aata hai — self-contained recap Taaki ye page apne aap mein complete lage, yahan derivation hai algebra ke saath, chhupaaya nahi:

Continuity kehta hai mass ikatta nahi ho sakti: $\overset{m}{˙} = ρ A V = const$ streamtube ke har jagah.
$A^{*}$ ki definition: us ek jagah par wahi constant evaluate karo jahan $M = 1$ (density $ρ^{*}$ , speed $a^{*}$ ), jo deta hai $ρ A V = ρ^{*} A^{*} a^{*}$ , isliye $\frac{A}{A ^{*}} = \frac{ρ ^{*}}{ρ} \cdot \frac{a ^{*}}{V}$ .
Isentropic stagnation relations (Isentropic Stagnation Relations) har factor ko pure functions of $M$ mein convert karti hain. Dono factors ko reservoir state $ρ_{0}, T_{0}$ ke through route karo (jo sonic point aur local point dono ke liye common hai, isliye cancel ho jaata hai):
- Density factor. $\frac{ρ _{0}}{ρ} = (1 + \frac{γ - 1}{2} M^{2})^{\frac{1}{γ - 1}}$ ke saath aur $M = 1$ par same se $\frac{ρ _{0}}{ρ ^{*}} = (1 + \frac{γ - 1}{2})^{\frac{1}{γ - 1}} = (\frac{γ + 1}{2})^{\frac{1}{γ - 1}}$ milta hai, divide karne par $\frac{ρ ^{*}}{ρ} = [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M^{2})]^{\frac{1}{γ - 1}} .$
- Speed factor. $V = M a$ aur $a = γ R T$ likho, to $\frac{a ^{*}}{V} = \frac{1}{M} \frac{T ^{*}}{T}$ ; $\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2}$ use karke (aur uska $M = 1$ value $\frac{γ + 1}{2}$ ) wahi cancellation deta hai $\frac{a ^{*}}{V} = \frac{1}{M} [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M^{2})]^{\frac{1}{2}} .$
- Dono ko multiply karo. Identical bracket dono mein appear karta hai, to unke exponents simply add ho jaate hain: $\frac{1}{γ - 1} + \frac{1}{2} = \frac{2 + ( γ - 1 )}{2 ( γ - 1 )} = \frac{γ + 1}{2 ( γ - 1 )}$ . Yahi woh jagah hai jahan mysterious exponent $\frac{γ + 1}{2 ( γ - 1 )}$ aur bracket form aata hai — yeh ek "density" power $\frac{1}{γ - 1}$ aur "square-root of temperature" power $\frac{1}{2}$ ka sum hai. Bacha hua $\frac{1}{M}$ speed factor se bahar baithta hai, aur tum exactly boxed $f (M)$ par land karte ho.
Alag sign-flip fact $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ (from momentum + continuity) wahi hai jo valley bottom ko exactly $M = 1$ par land karata hai: $M < 1$ ke neeche factor $(M^{2} - 1)$ negative hai to speed up karne ke liye area shrink hona chahiye; uske upar, positive hai to area grow hona chahiye; $M = 1$ par yeh zero hai, to area stationary hai — ek minimum.

Intuition Minimum

$M = 1$ par kyun baithta hai — derivative step by step kiya gaya Hum $f$ ko differentiate karte hain aur dekhte hain kahan slope zero hai. Clean trick hai $ln f$ ko differentiate karna (logarithm products aur powers ko sums mein turn kar deta hai, to messy bracket easy ho jaata hai), phir $\frac{f ^{'}}{f} = \frac{d}{d M} ln f$ use karo.

Log lo. $ln f = - ln M + \frac{γ + 1}{2 ( γ - 1 )} ln [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M^{2})] .$ Front wala $\frac{1}{M}$ ban gaya $- ln M$ ; bracket ki power ek coefficient ban gayi bahar.
Term by term differentiate karo. Pehla term deta hai $- \frac{1}{M}$ . Doosre ke liye, sirf inner $(1 + \frac{γ - 1}{2} M^{2})$ hi $M$ par depend karta hai; uska derivative hai $(γ - 1) M$ , aur inner expression se divide karne par (chain rule for a log) milta hai $\frac{f ^{'}}{f} = - \frac{1}{M} + \frac{γ + 1}{2 ( γ - 1 )} \cdot \frac{( γ - 1 ) M}{1 + \frac{γ - 1}{2} M ^{2}} .$
$(γ - 1)$ cancel karo aur common denominator par combine karo. $(γ - 1)$ cancel ho jaata hai, chhod ke $\frac{( γ + 1 ) M}{2 ( 1 + \frac{γ - 1}{2} M ^{2} )}$ . Dono terms ko denominator $M (1 + \frac{γ - 1}{2} M^{2})$ par rakho; numerator ban jaata hai $- (1 + \frac{γ - 1}{2} M^{2}) + \frac{γ + 1}{2} M^{2} = - 1 + M^{2} (\frac{γ + 1}{2} - \frac{γ - 1}{2}) = M^{2} - 1$ . Bracket mein $γ$ -terms exactly $1$ par cancel ho jaate hain — yahi wajah hai ki poori cheez $M^{2} - 1$ par collapse ho jaati hai. Isliye $\frac{df}{d M} = \frac{f ( M )}{M} \cdot \frac{M ^{2} - 1}{1 + \frac{γ - 1}{2} M ^{2}} .$
Sign padho. Kyunki $f (M) > 0$ , $M > 0$ , aur denominator hamesha positive hai, $df / d M$ ka sign $M^{2} - 1$ ka sign hai: $M < 1$ ke liye negative (curve floor ki taraf girta hai), $M > 1$ ke liye positive (curve door uthti hai), zero sirf $M = 1$ par. Woh single sign change hi U-shape ka floor hai $M = 1$ par jo tumne red dot ke roop mein dekha.

True or false — justify karo

A / A^{*}

ki har value exactly ek Mach number se correspond karti hai.

False. Kyunki

f (M)

pehle girta hai phir uthta hai (uska derivative

M = 1

par sign change karta hai),

A / A^{*} > 1

par horizontal line U-shaped curve ko do baar cut karti hai — ek subsonic, ek supersonic root; sirf

A / A^{*} = 1

(valley floor at

M = 1

) unique hai.

A^{*}

ek isentropic streamtube mein station se station tak change ho sakta hai.

False. Isentropic flow mein mass flow aur stagnation state conserved hote hain, isliye

A^{*}

ek constant hai — exactly isliye yeh yardstick ka kaam karta hai.

Choked converging–diverging nozzle ke throat par,

A = A^{*}

aur

M = 1

hota hai.

True. Jab nozzle choked hota hai to actual minimum (throat) area hi sonic area hoti hai, to geometric minimum reference se coincide karta hai aur wahan

M = 1

hota hai.

Agar purely subsonic flow kahin bhi

M = 1

tak nahi pahunchi, to

A^{*}

exist nahi karta.

False.

A^{*}

ek reference hai (woh area jo flow ko

M = 1

hit karne ke liye chahiye hoti); yeh mathematically ek constant ke roop mein exist karta hai chahe duct mein koi physical point sonic na ho.

A / A^{*}

kisi real duct mein kisi jagah

1

se kam ho sakta hai.

False. Curve ki minimum value

f (1) = 1

hai; baaki har Mach

A / A^{*} > 1

deta hai, isliye ratio kabhi one se neeche nahi jaata.

Jaise

M \to 0

, required area ratio

A / A^{*}

approach karta hai

1

ko.

False.

M \to 0

par

f (M)

mein

1/ M

factor blow up karta hai, isliye

A / A^{*} \to \infty

: lagbhag-zero speed par fixed mass flow carry karne ke liye tumhe enormous area chahiye hogi.

Ek given

A / A^{*} > 1

ke liye, subsonic root aur supersonic root same static pressure share karte hain.

False. Woh sirf same area ratio share karte hain.

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

ke through, zyada

M

matlab bahut kam

p / p_{0}

, isliye supersonic root bahut kam static pressure par hota hai (aur temperature bhi).

Area–Mach relation momentum conservation use karke derive ki gayi thi.

False. Yeh continuity (

ρ A V =

const), $A^{*}$ ki definition

M = 1

par, aur isentropic stagnation relations se aata hai — momentum woh hai jo separate area–velocity relation deta hai.

Error pakdo

"Kyunki

A / A^{*} = 2

hai, exit Mach number

2

hai."

Area ratio Mach number nahi hota.

f (M) = 2

mein plug karne par do roots milte hain (≈0.31 subsonic, ≈2.2 supersonic); tumhe equation solve karna hoga,

M

seedha padhna nahi.

"Kisi bhi flow ko speed up karne ke liye, area shrink karo — isliye nozzles narrow hoti hain."

Yeh sirf subsonic flow ke liye sach hai.

\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}

se, jab

M > 1

hota hai to factor

(M^{2} - 1)

positive ho jaata hai, isliye area aur speed saath badhte hain — accelerate karne ke liye tumhe area diverge (wide) karna hoga. Dekho Area-Velocity Relation dA-A = (M^2-1) dV-V.

"Back-pressure ko choking se neeche karne par throat se zyada mass pump hoti hai."

Ek baar choked (

M = 1

throat par),

\overset{m}{˙}

fix ho jaata hai reservoir

p_{0}, T_{0}

aur

A^{*}

se. Throat downstream changes nahi "sun" sakta kyunki pressure signals sirf speed of sound par travel karte hain; dekho Choked Flow & Maximum Mass Flow.

"Formula

A / A^{*} = 1

deta hai

M = 1

par, isliye

A^{*}

woh area hai jo actually sabse chota hota hai."

Sabse chota actual area

A^{*}

ke equal hota hai sirf tabhi jab flow wahan sonic ho. Purely subsonic duct mein minimum area par bhi

M < 1

hota hai, isliye wahan

f (M) > 1

aur uska area

A^{*}

se zyada hota hai.

"Nozzle ke diverging part mein accelerate ho rahi air energy kho rahi hai kyunki woh phail rahi hai."

Supersonic flow mein density aur temperature itni tezi se girte hain ki area badhne par speed badhti hai — total (stagnation) enthalpy conserved rehta hai; flow thermal energy ko kinetic energy mein trade karta hai, lose nahi karta.

"Kyunki

f (M)

ek proper function lagta hai, main ise algebraically invert kar sakta hoon

M

ke liye."

f (M)

ka koi closed-form inverse nahi hai aur yeh non-monotonic hai; tumhe numerically iterate karna hoga ya tables use karni hongi, aur physics se branch (sub- ya supersonic) choose karni hogi.

Why questions

A / A^{*}

dono

M \to 0

aur

M \to \infty

par infinity kyun jaata hai?

M \to 0

par flow lagbhag still hai (

V \to 0

), isliye

ρ A V

=const ke liye huge

A

chahiye (

1/ M

factor blow up karta hai);

M \to \infty

par density zero ki taraf collapse karti hai (

M^{2}

ki power ka bracket blow up karta hai), isliye phir se same mass pass karne ke liye

A

enormously bada hona chahiye. Yahi figure s01 mein U ke do rising arms hain.

A / A^{*}

curve ka minimum

M = 1

par kyun hai, kisi aur jagah kyun nahi?

Kyunki

df / d M \propto (M^{2} - 1)

(the

γ

-terms cancel hokar exactly

M^{2} - 1

chhod gaye), jo sirf

M = 1

par zero hai aur wahan negative se positive ho jaata hai — akela turning point (figure s01 mein red dot). Physically yahi woh jagah hai jahan area–velocity sign flip karta hai, isliye area exactly sonic point par stationary (minimum) hota hai.

Hum har station ko sonic state se compare kyun karte hain, reservoir se kyun nahi?

Sonic point woh unique state hai jo poore streamtube ke liye do conserved quantities ke saath fix hota hai — mass flow aur stagnation state — isliye uska area

A^{*}

ek single constant hai. Reservoir par

V = 0

hai, jo normalise karne ke liye koi throughput per unit area nahi deta; sonic state woh natural "one flow unit" hai jo

\overset{m}{˙} = ρ^{*} A^{*} a^{*}

ko continuity ko

M

akele ka clean function banata hai.

Geometry pehle se

A / A^{*}

fix kar deti hai, phir boundary/back-pressure branch kyun decide karta hai?

Geometry akele ek symmetric double answer deta hai (figure s01 mein U-curve ke saath green line ke do intersections); back-pressure

p_{b}

— figure s02 mein sketch ki gayi downstream push — pressure gap

p_{0} - p_{b}

set karta hai jo decide karta hai ki flow subsonic rahi ya sonic throat se guzri aur supersonic ho gayi. Sirf ends ki physics tie todhti hai.

Area–Mach relation isentropic flow assume kyun karta hai?

Constant

A^{*}

unchanging stagnation state par depend karta hai; friction (viscosity) aur shocks entropy badhate hain aur stagnation pressure girate hain, jo downstream

A^{*}

ko badhata hai, isliye single-yardstick derivation ab hold nahi karti. Real nozzle mein, boundary-layer growth aur heat transfer effective

A^{*}

ko drift karate hain, isliye designers discharge-coefficient corrections apply karte hain. Dekho Normal Shock Waves.

A / A^{*}

M

jaanne par baaki har property kyun unlock hoti hai?

Isentropic Stagnation Relations ke through,

M

akele

T / T_{0}

p / p_{0}

, aur

ρ / ρ_{0}

fix karta hai fixed reservoir values ke relative; area ratio woh ek hard-to-get piece hai, aur ek baar

M

mil jaaye to sab kuch thermodynamic follow karta hai.

Edge cases

Exactly

M = 1

par,

A / A^{*} = 1

kitne Mach numbers deta hai?

Exactly ek — yahi woh single point hai jahan subsonic aur supersonic branches valley floor par milte hain (

f

ka akela turning point), isliye double-root ambiguity khatam ho jaati hai.

Agar diverging section ke andar normal shock form ho jaaye to

A^{*}

ka kya hoga?

Flow shock ke across ab isentropic nahi rehti: entropy badhti hai, stagnation pressure girta hai, aur downstream

A^{*}

upstream wale se bada ho jaata hai — yardstick jump kar jaata hai. Dekho Normal Shock Waves.

Kya converging-only nozzle supersonic exit flow produce kar sakta hai?

Nahi. Ek converging duct subsonic flow ko zyada se zyada apne narrowest exit par

M = 1

tak accelerate kar sakta hai; supersonic jaane ke liye baad mein diverging section chahiye, yani ek Converging-Diverging (de Laval) Nozzle.

Agar do nozzles ka same

A_{e} / A^{*}

ho lekin absolute sizes alag hoon, to kya woh same Mach par exit karte hain?

Haan same

γ

ke liye —

f (M)

sirf ratio par depend karta hai, absolute area par nahi, isliye exit Mach identical hoga (mass flow size ke saath scale karta hai, lekin Mach nahi).

Exit area ratio

A_{e} / A^{*} \to 1^{+}

upar se as hone par exit Mach number kya hoga?

Dono roots

M = 1

ki taraf converge karte hain; subsonic root 1 ki taraf uthta hai aur supersonic root 1 ki taraf girta hai, exactly choked throat condition par milte hain — valley floor.

Jis duct mein flow throughout subsonic rehti hai, wahan

A^{*}

kabhi physically occupied hota hai?

Nahi — duct mein koi bhi point actually

M = 1

tak nahi pahunchta, isliye

A^{*}

purely mathematical reference hai jo wahan present har real area se chota hai.

Kya upar ke isentropic edge cases real nozzle mein kabhi exactly hold karte hain?

Kabhi perfectly nahi — viscosity (walls par friction), boundary-layer growth, aur koi bhi heat exchange sab entropy add karte hain, isliye real flows sirf isentropic

f (M)

ko approximate karte hain; relation ek idealised backbone hai jise practice mein empirical factors se correct kiya jaata hai.

Recall Ek-line self-test

Geometry akele nozzle station par Mach number kyun nahi bata sakti? ::: Kyunki $A / A^{*} = f (M)$ U-shaped hai aur 1 se upar har area ratio ko subsonic aur supersonic Mach dono se map karta hai; sirf back-pressure/boundary conditions real branch choose karte hain.