3.1.16 · Physics › Compressible Flow & Aerodynamics
Jab ek supersonic flow convex corner ke around turn karti hai (wall flow se door jhukti hai),
to flow smoothly Mach waves ki ek infinite fan ke through expand hoti hai. Har wave itni weak hoti hai ki
process isentropic hoti hai (koi entropy nahi badhti, shock ke unlike). Flow speed up karti hai, Mach number
badhta hai , aur pressure / temperature / density girte hain .
WHY isentropic? Compression turn (concave corner) mein characteristics converge karti hain → wo ek shock mein pile up ho jaati hain. Expansion turn mein wo diverge karti hain → ek fan mein spread ho jaati hain, har Mach wave infinitesimally weak hoti hai, isliye Δ s → 0 .
Contrast: Oblique shock = sudden, lossy turn flow mein . Expansion fan = gradual, lossless turn flow se door .
Definition Mach wave & Mach angle
Mach wave supersonic flow mein sabse weakest possible disturbance hai. Ye Mach angle
μ = arcsin ( M 1 )
local flow direction ke saath banata hai.
sin μ = 1/ M kyun hai
Ek disturbance speed of sound a par travel karta hai; flow use speed V par downstream carry karti hai.
Time t mein sound signal radius a t ka circle spread karta hai, jabki uska source V t move karta hai.
Envelope (Mach line) mein sin μ = V t a t = V a = M 1 hota hai.
Jaise flow corner ke around accelerate karti hai, M badhta hai, isliye μ = arcsin ( 1/ M ) ghatta hai . Pehli Mach wave (at M 1 ) steep hoti hai, aakhri (at M 2 ) shallow hoti hai → waves ek fan mein spread ho jaati hain.
Wall ko ek infinitesimal angle d θ se ek Mach wave ke across turn karo. d θ ko Mach number
d M mein change se relate karo. Phir integrate karo taaki finite M change ke liye total turn mile.
Step 1 — Ek Mach wave ke across geometry.
Mach line ke saath velocity component unchanged rehta hai (sirf normal component badalta hai, bilkul ek
bahut weak oblique shock ki tarah). Speed V hai, d θ se turn kar rahi hai. Velocity
triangle se standard result ek Mach wave ke across:
d θ = M 2 − 1 V d V
Ye step kyun? Mach wave angle μ par hai. Infinitesimal turn se pehle/baad velocity resolve karo aur tan μ = 1/ M 2 − 1 use karo to exactly yahi milta hai. Factor
M 2 − 1 isliye aata hai kyunki sirf supersonic flow (M > 1 ) Mach waves ko support karta hai.
Step 2 — d V / V ko d M / M mein convert karo.
Kyunki V = M a hai, log lo aur differentiate karo:
ln V = ln M + ln a ⇒ V d V = M d M + a d a
Perfect gas ke liye, a = γ R T , aur isentropic / adiabatic energy relation deta hai
T T 0 = 1 + 2 γ − 1 M 2 .
a ∝ T ko T 0 constant rakh ke differentiate karne par milta hai
a d a = 1 + 2 γ − 1 M 2 − 2 γ − 1 M d M .
Combine karne par:
V d V = 1 + 2 γ − 1 M 2 1 M d M .
Ye step kyun? Hum sab kuch M ke terms mein chahte hain, jo natural variable hai, kyunki hum result ko M ke against tabulate karte hain.
Step 3 — Assemble karo aur integrate karo.
d θ = 1 + 2 γ − 1 M 2 M 2 − 1 M d M .
Prandtl–Meyer function ν ( M ) define karo M = 1 se integrate karke (jahan θ = 0 hai):
ν ( M 2 ) = ν ( M 1 ) + θ kyun kaam karta hai
ν sonic se turn hai. Fan sirf ν mein difference par depend karta hai, isliye θ ka turn
simply existing ν ( M 1 ) mein θ add kar deta hai. Kyunki process isentropic hai, p 0 aur T 0
kabhi nahi badlte — sirf stagnation values ke ratios badlte hain.
Worked example Example 1 — Air (
γ = 1.4 ) ko M 1 = 2.0 se θ = 1 0 ∘ se turn karo
Step 1: ν ( 2.0 ) = 26.3 8 ∘ (ν formula se).
Kyun? Ye woh angle hai jo M = 2 tak sonic se pahunchne mein already "used up" ho gaya hai.
Step 2: ν ( M 2 ) = 26.3 8 ∘ + 1 0 ∘ = 36.3 8 ∘ .
Kyun? Corner 1 0 ∘ turning add karta hai; expansion ν badhata hai.
Step 3: ν ( M 2 ) = 36.3 8 ∘ invert karo ⇒ M 2 ≈ 2.385 .
Kyun? ν M mein monotonic hai, isliye inversion unique hai.
Step 4 (properties): 1 + 0.2 M 2 ke saath (note 0.2 M 2 2 = 0.2 ( 2.385 ) 2 ≈ 1.138 ):
p 1 p 2 = ( 1 + 0.2 ( 5.69 ) 1 + 0.2 ( 4 ) ) 3.5 = ( 2.138 1.8 ) 3.5 ≈ 0.55 .
Kyun? Expansion ⇒ pressure apni original value ke lagbhag 55% tak girta hai . Mach angle 3 0 ∘ se 24. 8 ∘ tak shrink ho jaata hai.
Worked example Example 2 — Maximum possible turn (vacuum mein expand karo)
Jaise M → ∞ , ν ( M ) → ν m a x = 2 π ( γ − 1 γ + 1 − 1 ) .
γ = 1.4 ke liye: ν m a x = 9 0 ∘ ( 6 − 1 ) ≈ 130.4 5 ∘ .
Kyun? Ye asymptotic limit hai — M = 1 se start hone wali stream ko ≈ 130. 5 ∘ se zyada turn nahi kar sakte; us ke baad gas ko M = ∞ aur zero pressure (vacuum) chahiye hoga.
Worked example Example 3 — Flow already
M 1 = 1 par hai, θ = 2 0 ∘ turn karo
ν ( M 2 ) = 0 + 2 0 ∘ ⇒ M 2 ≈ 1.78 .
Kyun? Sonic se start karte hue, ν 1 = 0 hai, isliye M 2 directly ν = 2 0 ∘ se padhte hain.
Common mistake "Expansion fan shock ki tarah entropy increase karti hai."
Ye kyun sahi lagta hai: Dono shocks aur fans supersonic flow ko turn karte hain aur M , p , T bahut badlte hain.
Fix: Fan ek continuous set of infinitesimally weak Mach waves hai, har ek Δ s ∝ ( strength ) 3 → 0 . Shock ek single finite discontinuity hai. Isliye fan isentropic hai (p 0 constant); shock nahi hai.
Common mistake "Expansion flow ko slow karta hai kyunki pressure girta hai."
Ye kyun sahi lagta hai: Lower pressure matlab "less push" lagta hai.
Fix: Nozzle/expansion mein, lower static pressure matlab gas internal energy ko kinetic energy mein convert karti hai: M aur V badhte hain , T ghatta hai . Stagnation values fixed rehti hain.
Common mistake "Turn ke liye
ν ( M 2 ) = ν ( M 1 ) − θ use karo."
Ye kyun sahi lagta hai: Compression se confuse ho jaate hain.
Fix: Expansion (convex, flow door turn karti hai) ⇒ θ add karo: ν 2 = ν 1 + θ . Sirf (hypothetical reverse) compression subtract karega.
Common mistake "Mach angle fan ke through badhta hai."
Ye kyun sahi lagta hai: Cheezein "khul" jaati hain.
Fix: μ = arcsin ( 1/ M ) aur M badhta hai , isliye μ ghatta hai — waves downstream shallow hoti jaati hain.
Recall Feynman: ek 12-saal ke bache ko samjhao
Socho tum itni tez bhaag rahe ho ki hawa time par raasta nahi de sakti — tum "supersonic" ho.
Ab tumhare aage ki wall suddenly door jhuk jaati hai. Hawa ko failne ki jagah mil jaati hai, isliye ye speed up
karti hai aur thinner aur cooler ho jaati hai, jaise paani faster beh jaata hai jab pipe chaudi hoti hai. Ye aaram se
ek fan of super-thin invisible "ripples" ke through hota hai, isliye kuch bhi mess up ya waste nahi hota (koi heat
nahi jaati) — yahi "isentropic" ka matlab hai. Prandtl–Meyer number ν bas ek score hai ki kitna turn kiya hai — wall ko 1 0 ∘ bend karo aur apne score mein 10 add karo, phir apni nayi speed dekho.
Mnemonic Direction yaad rakho
"Expand = Away = Add" — flow door turn karti hai, ν mein θ add karte ho, M upar jaata hai, p neeche jaata hai, μ neeche jaata hai.
Aur "Fan, not bang" : Fan = isentropic; bang (shock) = entropy.
Prandtl–Meyer expansion fan supersonic flow ke saath kya karta hai? Use convex corner ke around turn karta hai, isentropically accelerate karta hai (M↑) jabki p, T, ρ girte hain.
Expansion fan isentropic kyun hai lekin shock nahi? Fan infinitesimally weak Mach waves ka continuum hai (Δs∝strength³→0); shock ek single finite discontinuity hai jismein finite entropy rise hoti hai.
Mach angle formula batao. μ = arcsin ( 1/ M ) .
Turn angle aur velocity ko link karne wala differential relation kya hai? Prandtl–Meyer function ν ( M ) likho. ν = γ − 1 γ + 1 arctan γ + 1 γ − 1 ( M 2 − 1 ) − arctan M 2 − 1 .
Angle θ ke turn par ν kaise apply karte ho? ν ( M 2 ) = ν ( M 1 ) + θ , phir M 2 ke liye invert karo.
γ = 1.4 ke liye M=1 se start karte hue maximum turn angle kya hai?ν m a x ≈ 130.4 5 ∘ (jaise M → ∞ ).
Fan ke through Mach angle badhta hai ya ghatta hai? Ghatta hai, kyunki M badhta hai aur μ = arcsin ( 1/ M ) .
Isentropic expansion fan ke across kya constant rehta hai? Stagnation pressure p 0 , stagnation temperature T 0 , entropy s .
M = 2 , γ = 1.4 par ν kya hai?Lagbhag 26.3 8 ∘ .
Oblique shock waves — compression counterpart; θ–β–M relation vs ν(M).
Mach waves and Mach cone — fan ka building block.
Isentropic flow relations — M 2 known hone par p, T, ρ get karne ke liye use hota hai.
Method of characteristics — fans characteristics ke saath simple-wave regions hain.
Nozzle design and overexpansion — underexpanded hone par nozzle lips par fans.
Entropy and the second law — kyun weak waves Δ s → 0 deti hain.
turns around convex corner
each infinitesimally weak
from sound circle vs source motion
contrast, lossy turn into flow
Mach angle mu equals arcsin 1 over M
d theta equals sqrt M2 minus 1 times dV over V
Prandtl-Meyer function nu of M