Prandtl-Meyer expansion waves — isentropic, supersonic turning: Question bank

3.1.16 · D5 · Physics › Compressible Flow & Aerodynamics › Prandtl-Meyer expansion waves — isentropic, supersonic turni

Shuru karne se pehle, do symbols jo hum neeche har jagah use karte hain, plain words mein stated hain taaki kuch assume na ho:

$M$ = Mach number = flow speed $V$ divided by local speed of sound $a$ ; $M > 1$ ka matlab hai "supersonic" (apni khud ki awaaz se tez).
$ν (M)$ = Prandtl–Meyer function, wo angle (degrees ya radians mein) jitna stream pehle se turn away kar chuka hai sonic ( $M = 1$ ) se $M$ tak accelerate karne mein. Ise ek running "turn score" samjho.
$μ = arcsin (1/ M)$ = Mach angle, sabse kamzor wave ka flow ke relative tilt.

True or false — justify

Supersonic flow ko ek convex (door-bend hone wale) corner ke around turn karne se uska Mach number badhta hai.

True. Wall ka raasta dena gas ko phailne deta hai; wo internal energy ko speed mein convert karta hai, isliye

M

badhta hai jabki

p, T, ρ

girte hain.

Ek expansion fan flow ki entropy badhata hai.

False. Har Mach wave infinitesimally weak hoti hai aur uski entropy jump (strength)

^{3} \to 0

ki tarah scale hoti hai, isliye total

Δ s = 0

— fan isentropic hai, oblique shock ke unlike.

Kyunki expansion fan ke across static pressure girta hai, stagnation pressure

p_{0}

bhi girta hai.

False. Isentropic + adiabatic ka matlab hai

p_{0}

aur

T_{0}

conserved rehte hain; sirf static values aur stagnation se unke ratios change hote hain.

Mach angle

μ

fan ke through downstream jaane par bada hota jaata hai.

False.

M

badhta hai, aur

μ = arcsin (1/ M)

shrink hota hai kyunki

1/ M

shrink hota hai — waves shallower ho jaati hain, isliye fan flow direction mein khulta hai.

Expansion turn ke liye tum likhte ho

ν (M_{2}) = ν (M_{1}) - θ

False. Expansion turn add karta hai:

ν (M_{2}) = ν (M_{1}) + θ

. Subtract karna (hypothetical isentropic) compression direction hoga.

Ek Prandtl–Meyer fan flow ko kisi bhi angle se turn kar sakta hai agar corner ko bas itna sharp bana do.

False. Turn

ν_{m a x} - ν (M_{1})

pe cap hai; sonic se bhi total sirf

ν_{m a x} \approx 130.4 5^{\circ}

hai

γ = 1.4

ke liye, jo sirf

M \to \infty

(vacuum) par reach hota hai.

Fan mein individual Mach waves sab ek doosre ke parallel hoti hain.

False. Har wave local

M

ke liye

μ = arcsin (1/ M)

par hoti hai; kyunki

M

fan ke across change hota hai, har wave ka alag angle hota hai — wohi spread exactly reason hai ki ye ek fan hai.

Expansion fan ko reverse karna (same geometry ko backwards chalana) tumhe ek valid isentropic compression deta hai.

Half-true. Mathematically

ν_{2} = ν_{1} - θ

ek isentropic compression fan describe karta hai, lekin reality mein ek concave corner characteristics ko ek shock mein converge karta hai, isliye smooth reverse sirf ek idealised limit hai.

Ek Mach wave ke saath-saath velocity component unchanged rehta hai jab flow ise cross karta hai.

True. Sirf wave ke normal component change hoti hai (jaise ek bahut weak oblique shock); ye fact poore

d θ = M^{2} - 1 d V / V

derivation ka seed hai.

Ek subsonic flow (

M < 1

) bhi corner ke around Prandtl–Meyer expansion fan bana sakta hai.

False. Mach waves aur factor

M^{2} - 1

ke liye

M > 1

chahiye; subsonic flow smoothly aur continuously adjust hoti hai bina kisi wave structure ke.

Spot the error

"Fan gas ko cool karta hai, isliye total (stagnation) temperature

T_{0}

zaroor girni chahiye."

Static temperature

T

girta hai jab energy kinetic ban jaati hai, lekin

T_{0}

(jo internal aur kinetic energy dono count karta hai) adiabatic flow mein conserved hota hai. Static ko stagnation se confuse karna hi error hai.

"Kam pressure matlab gas par kam push, isliye flow fan ke through slow ho jaati hai."

Ulta hai. Favourable pressure gradient (peeche high, aage low) gas ko accelerate karta hai; lower static pressure speeding up ka result hai, brake nahi.

"Kyunki fan aur shock dono supersonic flow ko turn karte hain, hum ek strong expansion ko rarefaction shock treat kar sakte hain."

Aisa koi cheez exist nahi karti jaise stable rarefaction (expansion) shock — second law ise forbid karta hai (

Δ s < 0

). Expansions zaroor ek smooth fan mein spread honi chahiye.

"Pehli Mach wave shallow hai aur aakhri steep hai, kyunki flow slow ho rahi hai."

Dono facts ulte hain. Flow speed up karti hai, isliye pehli wave (lower

M_{1}

par) steep hai aur aakhri (higher

M_{2}

) shallow hai; fan downstream mein khulta hai.

M_{2}

nikalne ke liye,

ν (M_{1})

se isentropic pressure ratio subtract karo."

Nonsense units — tum ek angle mein pressure ratio add nahi kar sakte. Tum deflection angle

θ

ν (M_{1})

mein add karte ho,

ν

ko invert karke

M_{2}

nikalte ho, aur phir pressure ke liye isentropic relations use karte ho.

"Ek centred expansion fan ki finite thickness hoti hai kyunki usmein infinitely many waves hoti hain."

Sharp corner par centred ek fan geometrically vertex par zero-thickness ka hota hai — infinitely many waves sab ek point se emanate hoti hain; wo sirf corner se door jaane par spread apart hoti hain.

"Method of characteristics expansion fans handle nahi kar sakta kyunki characteristics collide hoti hain."

Ek expansion mein characteristics diverge hoti hain, jo exactly reason hai ki Method of characteristics cleanly kaam karta hai; collision (convergence) compression/shock case hai.

Why questions

Factor

M^{2} - 1

kyun appear hota hai

d θ = M^{2} - 1 d V / V

mein?

Ye Mach wave ke across velocity triangle mein

tan μ = 1/ M^{2} - 1

se aata hai; ye sirf

M > 1

ke liye real hai, encoding karta hai ki supersonic flow wave ke exist hone ke liye zaroori hai.

Hum

d θ

ko specifically

M = 1

se kyon integrate karte hain

ν

define karne ke liye?

M = 1

natural zero hai — sabse slow speed jo Mach waves support karti hai — isliye

ν (1) = 0

ek clean reference deta hai, kisi bhi real turn ko difference

ν (M_{2}) - ν (M_{1})

ki tarah handle karne deta hai.

ν (M)

monotonically increasing kyun hai, aur usse kya fark padta hai?

Integrand

\frac{M ^{2} - 1}{M ( 1 + \frac{γ - 1}{2} M ^{2} )}

sabhi

M > 1

ke liye positive hai, isliye

ν

sirf badhta hai; monotonicity guarantee karta hai ki har

ν

exactly ek

M

se map karta hai, inversion

ν \to M

ko unique banata hai.

Simple rule

ν_{2} = ν_{1} + θ

corner geometry ki details ko kyun ignore kar sakta hai?

Kyunki

ν

sonic-se-turn measure karta hai aur isentropic state sirf

M

par depend karta hai; flow apna path "bhool" jaata hai aur sirf total turned angle ki parwah karta hai, isliye tum bas

θ

add karo.

Maximum turn angle

γ

par kyun depend karta hai?

ν_{m a x} = \frac{π}{2} (\frac{γ + 1}{γ - 1} - 1)

M \to \infty

limit se aata hai;

γ

(heat-capacity ratio) set karta hai ki pressure vacuum hit karne se pehle gas kitna accelerate kar sakti hai.

Expansion convex corner se kyun tied hai aur shock concave se?

Convex wall ko door bend karta hai, characteristics ko diverge karne deta hai (fan); concave flow mein bend karta hai, unhe converge hone aur ek shock mein pile up hone par majboor karta hai.

Pressure ratio mein exponent sirf

γ

ki jagah

γ / (γ - 1)

kyun use hota hai?

Wo exponent isentropic relation

p / p_{0} = (T / T_{0})^{γ / (γ - 1)}

se aata hai; ye pressure ko temperature se ek constant-entropy path ke saath tie karta hai, jo exactly hai kya ek fan hai.

Edge cases

Exactly

M = 1

par

ν

aur

μ

kya hain?

ν (1) = 0^{\circ}

(abhi tak koi turning nahi) aur

μ = arcsin (1) = 9 0^{\circ}

— "Mach wave" flow ke normal hai, yaani ek sound front jise flow mushkil se outrun kar sakti hai.

M \to \infty

par

μ

ka kya hota hai?

μ = arcsin (1/ M) \to arcsin (0) = 0^{\circ}

; waves flow ke parallel ho jaati hain, isliye ek infinitely fast stream koi forward-tilting disturbance carry nahi karti.

Ek flow exactly

M_{1} = 1

se shuru hoti hai aur

θ = 2 0^{\circ}

turn leti hai —

M_{2}

kaise nikaalte ho?

Kyunki

ν (1) = 0

hai, simply

ν (M_{2}) = 2 0^{\circ}

;

M_{2}

ko directly

ν

ke inverse se read karo (approximately

M_{2} \approx 1.78

γ = 1.4

ke liye).

Agar tum

ν_{m a x}

se aage turn karne ki koshish karo toh flow ko physically kya limit karta hai?

Gas ko

M = \infty

, zero static pressure aur zero temperature — vacuum chahiye hoga; stream simply separate ho jaati hai aur extra angle fill nahi kar sakti, vacuum region chhod kar.

Kya ek flow ke liye fan possible hai jo sirf barely supersonic hai, maano

M_{1} = 1.01

Haan, lekin available turn almost full

ν_{m a x}

hai (kyunki

ν (1.01) \approx 0

), aur pehli Mach wave almost normal hai (

μ

near

9 0^{\circ}

) — fan apne start mein bahut "steep" hai.

Zero strength ki limit mein ek single Mach wave ki entropy change kya hai?

Leading order par exactly zero;

Δ s

(wave strength)

^{3}

ki tarah scale karta hai, isliye ek infinitesimal wave koi entropy contribute nahi karti — isliye infinitely many ka integral bhi

Δ s = 0

deta hai.

Agar opposite walls se do expansion fans milte hain, toh kya flow suddenly shock ho jaati hai?

Nahi — do expansions cross karte hue isentropic rehti hain; Method of characteristics intersection ko smoothly resolve karta hai. Compressive characteristics ka milna hi shock mein coalesce hota hai.

Over-expanded nozzle in fans se kaise relate karta hai?

Under-expanded jet exit hoti hai aur lower ambient pressure match karne ke liye expansion fans ke through door turn karti hai; over-expanded one instead oblique shocks ke through compress hoti hai — same corner, opposite pressure mismatch.

Recall Traps ki ek-line summary

Expand = Away = Add: $M ↑$ , $p, T ↓$ , $μ ↓$ , $p_{0}, T_{0}$ fixed, $Δ s = 0$ . Sab kuch ek $ν$ ka difference hai, turn cap $ν_{m a x}$ hai, aur sirf convex corners fan banate hain jabki concave corners bang karte hain.