Hypersonic flow — Mach 5+, high temperature effects: Question bank

3.1.27 · D5 · Physics › Compressible Flow & Aerodynamics › Hypersonic flow — Mach 5+, high temperature effects

Shuru karne se pehle, ek shared vocabulary reminder, taaki neeche koi symbol unexplained na rahe:

True or false — justify karo

Recall T/F — "Mach 5 exact boundary hai jahan flow hypersonic hoti hai."

False — ye ek rule of thumb hai. Asli boundary ye hai ki kaun si physics ko ab ignore nahi kar sakte (thin shock layers, real-gas chemistry); ek slender cold body M6 par 'supersonic' behave kar sakta hai, ek blunt hot body M4 par hi 'hypersonic' ho sakta hai. :::

T/F — "Jaise Mach number bina kisi limit ke badhta hai, real air mein stagnation temperature

T_{0}

bhi bina kisi limit ke badhti hai."

False — calorically-perfect formula

T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}

unbounded growth predict karta hai, lekin real air vibration, dissociation aur ionization mein energy dump karta hai, isliye actual temperature bahut dheere badhti hai aur ideal prediction se kaafi neeche level off ho jaati hai.

T/F — "Ek hypersonic shock ke peeche, energy lost hoti hai kyunki molecular bonds todna energy kharach karta hai."

False — total energy (enthalpy) phir bhi conserved rehti hai; bond-breaking sirf thermal energy ko chemical potential energy mein relocate karta hai, isliye temperature girta hai chahe kuch bhi lost nahi hota.

T/F — "Active degrees of freedom

f

ki badi sankhya

γ

ko bada banati hai."

False — kyunki

γ = 1 + 2/ f

, energy chupaane ke zyada drawers hone ka matlab hai ==bada

f

aur isliye chhota

γ

==, jo high temperature par

1.4

1.1

–

1.3

tak gir jaata hai.

T/F — "Calorically-perfect air (

γ = 1.4

) ke liye, ek bahut strong normal shock ke peeche density Mach number zyada hone par kitni bhi badh sakti hai."

False — ratio

\frac{γ + 1}{γ - 1} = 6

par cap ho jaata hai jab

M \to \infty

; sirf ==

γ

ko lower karne wale real-gas effects== hi ise

15

–

20

tak push karte hain. Dekho Rankine–Hugoniot Relations.

T/F — "Newtonian pressure coefficient

C_{p} = 2 sin^{2} θ

Mach number par strongly depend karta hai."

False — ye sirf geometry par depend karta hai; Mach number barely enter karta hai, aur yahi hypersonic Mach-independence principle ka poora point hai.

T/F — "Ek sharper (chhote radius wala) nose stagnation heat flux ko reduce karta hai."

False — heat flux

q_{w} \propto 1/ R_{n}

ki tarah scale karta hai, isliye ==chhota nose radius matlab zyada heating==; yahi exact reason hai ki re-entry vehicles blunt hote hain. Dekho Boundary Layers & Aerodynamic Heating.

T/F — "Ek hypersonic body ke shadow (leeward) region mein, Newtonian theory negative pressure (suction) predict karta hai."

False — ye wahan

C_{p} = 0

predict karta hai, kyunki koi bhi particles surface tak nahi pahunchte; model mein sirf koi wall se nahi takra raha, koi pulling force nahi hai.

Spot the error

Spot the error: "

M = 10

par,

γ = 1.4

, isliye

T_{0} / T = 21

aur re-entry air really mein lagbhag

4600

K tak pahunchti hai."

Arithmetic sahi hai lekin physics galat hai: un temperatures par ==

γ

1.4

nahi rehta== aur real-gas energy absorption ka matlab hai actual temperature kaafi kam hai; calorically-perfect number ise badly overpredicts karta hai.

Spot the error: "Kyunki shock layer thin hai, hypersonic flow basically incompressible hai."

Layer thin isliye hai kyunki shock ke across density enormously jump kar gayi — ye most compressible regime possible hai, incompressible ka bilkul ulta.

Spot the error: "Newton ka impact model purana aur disproven hai, isliye hypersonic pressure ke liye kabhi use nahi karna chahiye."

Ye low speed par galat hai lekin hypersonics ke liye accurate hai, precisely isliye kyunki thin shock layer flow ko particles ki tarah behave karne deta hai jo wall se takraate hain aur slide ho jaate hain.

Spot the error: "Hum energy equation

c_{p} T + \frac{1}{2} V^{2} = c_{p} T_{0}

c_{p} T

se divide karte hain

M

paane ke liye — lekin hum ye skip karke directly

V

use kar sakte hain, same baat hai."

Kar sakte ho, lekin divide karne ka point ye hai ki

V^{2} / (2 c_{p} T)

ko ==dimensionless

\frac{γ - 1}{2} M^{2}

== mein convert kiya jaaye

a^{2} = γ R T

use karke; yahi woh

M^{2}

scaling expose karta hai jo regime ko name deta hai.

Spot the error: "Ionization radio blackout isliye cause karta hai kyunki hot plasma physically antenna ko melt karta hai."

Koi melting implied nahi hai — plasma mein free electrons radio waves ko reflect aur absorb karte hain, jo communication cut karta hai chahe antenna survive kare ya na kare. Dekho Real Gas Thermodynamics & Dissociation.

Spot the error: "Kyunki lift hypersonic speeds par Mach number par barely depend karti hai, aerodynamic heating bhi Mach-independent hona chahiye."

Sirf pressure/force coefficients Mach-independent hote hain; heating speed ke saath steeply badhti hai kyunki heat flux kinetic energy dump ko track karta hai, jo

V

ke saath badhta rehta hai.

Why questions

Why is the hypersonic regime defined by physics rather than a clean Mach threshold?

Kyunki vibration aur dissociation jaisi effects temperature ke saath gradually turn on hoti hain, aur shock ke peeche temperature

M^{2}

ke saath scale karta hai; koi sharp switch nahi hai, sirf

M \approx 5

ke around ek band hai jahan unhe ignore karna possible nahi rehta.

Temperature ko (pressure ko nahi, maslan) real-gas effects ke trigger ke roop mein kyun single out kiya jaata hai?

Kyunki degrees of freedom temperature se unlock hote hain: vibration

800

K ke aaspaas, O

_{2}

dissociation

2500

K ke aaspaas, ionization

9000

K ke aaspaas — ye thresholds thermal hain, aur

T_{0}

hi hai jo hypersonic flow mein blow up karta hai.

Ek blunt nose wall ko ek sharp nose se cooler kyun rakhta hai, chahe wo zyada flow rokta ho?

Blunt body ek detached bow shock surface se door push karta hai, flow ki zyaatar heat shock layer / air mein dump karta hai instead of ek tip par concentrate karne ke.

Same air ka free stream mein

γ \approx 1.4

aur shock ke peeche

γ \approx 1.2

kyun ho sakta hai?

Shock gas ko itna heat karta hai ki naye internal modes (vibration, dissociation) activate ho jaate hain,

f

badhta hai aur

γ

ghatta hai; upstream cold air mein sirf translation aur rotation active hote hain.

γ

kam karne se shock layer thinner kyun hoti hai?

Chhota

γ

limiting density ratio

\frac{γ + 1}{γ - 1}

ko badhata hai, isliye mass conservation same flow ko aur bhi thinner sheet mein force karta hai shock aur body ke beech. Dekho Normal and Oblique Shock Waves.

Stagnation-temperature derivation mein sirf internal energy nahi, enthalpy conserved quantity kyun hoti hai?

Steady adiabatic flow mein bina shaft work ke, energy balance mein flow work shamil hota hai, aur enthalpy

h = e + p / ρ

internal energy ko us flow work ke saath bundle karta hai, isliye

h + \frac{1}{2} V^{2}

natural constant hai. Dekho Stagnation Properties & Isentropic Relations.

Edge cases

Edge case: Ek surface exactly flow ke parallel hai,

θ = 0

. Newtonian theory kya deta hai?

C_{p} = 2 sin^{2} 0 = 0

— ek grazing surface koi impact pressure feel nahi karta kyunki parcels us par zero normal momentum lose karte hain.

Edge case: Ek surface directly flow mein face kar rahi hai,

θ = 9 0^{\circ}

C_{p}

kya hai?

C_{p} = 2 sin^{2} 9 0^{\circ} = 2

, maximum Newtonian value, kyunki har parcel apni poori normal velocity head-on lose karta hai.

Edge case: Calorically-perfect gas ke liye

M \to \infty

ki limit —

T_{0} / T

aur

ρ_{2} / ρ_{1}

ka kya hota hai?

T_{0} / T

bina bound ke badhta hai (

\frac{γ - 1}{2} M^{2}

term dominate karta hai), lekin

ρ_{2} / ρ_{1}

\frac{γ + 1}{γ - 1}

par saturate ho jaata hai== (

γ = 1.4

ke liye 6) — temperature unbounded hai jabki compression capped hai.

Edge case: Exactly

M = 0

(still air) par

T_{0} / T

kya hai?

T_{0} / T = 1

, matlab ==

T_{0} = T

== — koi motion nahi hai jo heat mein convert ho, isliye stagnation aur static temperatures coincide karte hain.

Edge case: Bahut low temperature limit (bahut cold gas, sirf translation + rotation active) mein, ek diatomic molecule ke liye maximum

γ

kya hai?

f = 5

ke saath,

γ = 1 + 2/5 = 1.4

— ye ceiling hai jab vibration frozen out ho; tum diatomic air ko is se upar nahi le ja sakte.

Edge case: Agar ionization fully reverse ho jaaye (electrons recombine kar lein) jaise gas downstream cool hoti hai, kya released energy 'nayi' hai?

Nahi — ye wahi energy hai jo chemical/ionization pool se wapas di jaati hai, temperature restore karta hai jaise drawers band hote hain; total enthalpy poore time conserved rahi.