Hypersonic flow — Mach 5+, high temperature effects: Exercises

3.1.27 · D4 · Physics › Compressible Flow & Aerodynamics › Hypersonic flow — Mach 5+, high temperature effects

Symbols jo hum reuse karte hain (saare parent mein defined hain):

$M = V / a$ — Mach number, flow kitni baar sound se faster move karta hai.
$a = γ R T$ — temperature $T$ par gas mein speed of sound.
$γ$ — ratio of specific heats, $γ = c_{p} / c_{v} = 1 + 2/ f$ , jahan $f$ active "energy storage drawers" (degrees of freedom) count karta hai.
$T_{0}$ — stagnation temperature, gas ko rest mein laane par jo temperature milti.
$C_{p} = \frac{p - p _{\infty}}{\frac{1}{2} ρ _{\infty} V _{\infty}^{2}}$ — pressure coefficient, pressure rise ko stream ke "dynamic push" ke units mein measure kiya gaya.

Level 1 — Recognition

L1.1 Kya ye hypersonic hai?

Ek aircraft $V = 1700 m/s$ se $T = 220 K$ wali air mein $γ = 1.4$ aur $R = 287 J/(kg\cdotK)$ ke saath fly kar raha hai. $M$ compute karo. Kya flow hypersonic hai?

Recall Solution

WHAT: speed of sound nikalo, phir divide karo. WHY: "hypersonic" $M ≳ 5$ se define hota hai, isliye humein $M = V / a$ chahiye.

$a = γ R T = 1.4 \times 287 \times 220 = 88, 396 \approx 297.3 m/s$ $M = \frac{V}{a} = \frac{1700}{297.3} \approx 5.72$

Kyunki $M \approx 5.72 \geq 5$ , flow hypersonic hai. Real-gas heating ab matter karne lagti hai.

L1.2 Stagnation temperature, cold-gas estimate

Usi flight ke liye ( $M = 5.72$ , $T = 220 K$ , $γ = 1.4$ ), calorically-perfect formula $\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2}$ use karke $T_{0}$ estimate karo.

Recall Solution

WHAT: parent note ki boxed stagnation-temperature ratio mein numbers plug karo. WHY: ye stream ki kinetic energy ko ek temperature mein convert karta hai jo wall feel karegi agar gas adiabatically ruk jaye.

$\frac{T _{0}}{T} = 1 + \frac{1.4 - 1}{2} (5.72)^{2} = 1 + 0.2 \times 32.72 = 1 + 6.544 = 7.544$ $T_{0} = 7.544 \times 220 \approx 1660 K$

Ye already ~800 K vibration threshold se upar hai aur O $_{2}$ dissociation ke paas — perfect-gas number sirf ek upper estimate hai.

Level 2 — Application

L2.1 Inclined face par Newtonian pressure

Ek flat surface hypersonic stream se $θ = 2 0^{\circ}$ par inclined hai. Newtonian impact theory, $C_{p} = 2 sin^{2} θ$ , use karke iska pressure coefficient nikalo.

Figure — Hypersonic flow — Mach 5+, high temperature effects

Recall Solution

WHAT: $θ = 2 0^{\circ}$ ko Newtonian formula mein substitute karo. WHY this tool: parent note ne dikhaya ki ek thin shock layer mein, particles wall ke normal direction mein apna momentum simply kho dete hain; sirf flow se inclination $θ$ matter karta hai, isliye $C_{p} = 2 sin^{2} θ$ ko $M$ ya $γ$ ki zaroorat nahi. WHAT IT LOOKS LIKE: figure mein, plate ke perpendicular sirf velocity component (laal arrow) impact par destroy hoti hai; parallel part slide karta rehta hai.

$C_{p} = 2 sin^{2} (2 0^{\circ}) = 2 (0.3420)^{2} = 2 \times 0.11698 = 0.2340$

L2.2 Density ratio ceiling

Normal shock ke across jab $M \to \infty$ , density ratio $\frac{ρ _{2}}{ρ _{1}} = \frac{γ + 1}{γ - 1}$ ceiling tak pahunchi hai. Cold air ( $γ = 1.4$ ) aur dissociated air ( $γ = 1.15$ ) ke liye ye ceiling evaluate karo.

Recall Solution

WHAT: har $γ$ ko limiting ratio mein plug karo. WHY: parent note explain karta hai ki ye Rankine–Hugoniot Relations se aata hai jahan $u_{2} / u_{1} \to (γ - 1) / (γ + 1)$ aur mass conservation $ρ u = const$ hoti hai.

Cold air: $\frac{ρ _{2}}{ρ _{1}} = \frac{1.4 + 1}{1.4 - 1} = \frac{2.4}{0.4} = 6.0$ Dissociated air: $\frac{ρ _{2}}{ρ _{1}} = \frac{1.15 + 1}{1.15 - 1} = \frac{2.15}{0.15} \approx 14.33$

Real-gas $γ$ girta hai, ratio double se bhi zyada ho jata hai, aur shock layer body ke against aur bhi thin squeeze ho jati hai. Dekho Normal and Oblique Shock Waves.

Level 3 — Analysis

L3.1 Real gas temperature kyun lower karta hai

$M = 12$ , $T = 250 K$ par, calorically-perfect stagnation temperature ( $γ = 1.4$ ) ko ek real-gas estimate se compare karo jo effective $γ_{eff} = 1.20$ use karta hai (energy vibration aur dissociation mein drain ho rahi hai). $T_{0} / T$ kis factor se girta hai?

Recall Solution

WHAT: $T_{0} / T$ do baar compute karo, ek baar har $γ$ ke liye, aur ratio lo. WHY: parent ka steel-man warn karta tha ki vibration aur bond-breaking energy soakh lete hain, isliye lower $γ$ same kinetic energy ke liye smaller temperature rise model karta hai.

Perfect gas: $\frac{T _{0}}{T}_{1.4} = 1 + \frac{0.4}{2} (12)^{2} = 1 + 0.2 \times 144 = 1 + 28.8 = 29.8$ Effective: $\frac{T _{0}}{T}_{1.20} = 1 + \frac{0.20}{2} (12)^{2} = 1 + 0.10 \times 144 = 1 + 14.4 = 15.4$ Drop factor: $\frac{15.4}{29.8} \approx 0.517$

Naive tables temperature rise ko roughly do guna overpredict karte hain. Yahi exact reason hai ki thermal-protection design Real Gas Thermodynamics & Dissociation use karni chahiye, na ki calorically-perfect Stagnation Properties & Isentropic Relations.

L3.2 Degrees of freedom bookkeeping

Moderate temperature par air mein $γ = 1.40$ hota hai. Jab vibration fully activate hoti hai, $γ$ girke $1.29$ ho jata hai. $γ = 1 + 2/ f$ use karke, har case mein effective degrees of freedom $f$ nikalo aur confirm karo ki vibration ne expected amount add kiya.

Recall Solution

WHAT: $γ = 1 + 2/ f$ ko invert karke $f = \frac{2}{γ - 1}$ pao. WHY: $f$ active "storage drawers" count karta hai. Rearrange karne se pata chalta hai kitne open hain.

Cold air: $f = \frac{2}{1.40 - 1} = \frac{2}{0.40} = 5$ (3 translational + 2 rotational — bina vibration ke ek diatomic molecule.)

Vibrating air: $f = \frac{2}{1.29 - 1} = \frac{2}{0.29} \approx 6.90$

$f$ $5$ se badhkar approximately $7$ ho gaya: ek vibrational mode two half- $R$ pieces contribute karta hai (kinetic + oscillator ka potential), yaani $Δ f = 2$ , jo parent note ki table entry se match karta hai ( $γ \to 1.29$ , $f \to 7$ ).

Level 4 — Synthesis

L4.1 Angle of attack par flat plate ki lift

Ek flat plate Mach 8 par, angle of attack $α = 1 5^{\circ}$ se fly kar rahi hai. Newtonian theory use karke: (a) windward $C_{p}$ nikalo; (b) leeward $C_{p}$ nikalo; (c) normal-force coefficient $C_{N} \approx (C_{p, wind} - C_{p, lee}) cos α$ nikalo; (d) lift coefficient $C_{L} \approx C_{N} cos α$ nikalo. Mach number ke role par comment karo.

Recall Solution

WHAT: har face par $C_{p} = 2 sin^{2} θ$ apply karo, phir lift/normal directions par project karo. WHY the geometry (figure): windward face stream se $θ = α = 1 5^{\circ}$ par inclined hai; leeward face shadow mein hai. Pressure ko body-normal par aur phir vertical par project karne se normal force aur lift milti hai.

(a) Windward: $C_{p, wind} = 2 sin^{2} (1 5^{\circ}) = 2 (0.2588)^{2} = 2 \times 0.066987 = 0.13397$ (b) Leeward (shadow): $C_{p, lee} = 0$ . (c) Normal force: $C_{N} = (0.13397 - 0) cos (1 5^{\circ}) = 0.13397 \times 0.96593 = 0.12941$ (d) Lift: $C_{L} = C_{N} cos (1 5^{\circ}) = 0.12941 \times 0.96593 = 0.12500$

Comment: kahin bhi $M = 8$ appear nahi kiya. Yahi Mach-independence principle hai: high hypersonic Mach numbers par aerodynamic coefficients $M$ par depend karna band kar dete hain. Geometry rule karta hai.

L4.2 Stagnation heating aur nose radius

Stagnation heat flux $q_{w} \propto \frac{1}{R _{n}}$ scale karta hai, jahan $R_{n}$ nose radius hai. Capsule A ka $R_{n} = 1.0 m$ hai; ek sharp probe B ka $R_{n} = 0.02 m$ hai. $q_{w, B} / q_{w, A}$ ka ratio kya hai? Interpret karo.

Recall Solution

WHAT: scaling law ka ratio banao. WHY: parent note argue karta hai ki blunt bodies re-entry mein isliye survive karti hain kyunki heating bade nose radius ke saath girti hai. Dekho Boundary Layers & Aerodynamic Heating.

$\frac{q _{w, B}}{q _{w, A}} = \frac{R _{n, A}}{R _{n, B}} = \frac{1.0}{0.02} = 50 \approx 7.07$

Sharp probe blunt capsule se approximately 7 guna zyada stagnation heat flux suffer karta hai — yahi vivid quantitative reason hai ki capsules round hoti hain aur Shuttle ka belly blunt tha.

Level 5 — Mastery

L5.1 Measured density ratio se effective $γ$ reverse-engineer karna

Ek hypersonic experiment ek limiting normal-shock density ratio $\frac{ρ _{2}}{ρ _{1}} = 12.0$ measure karta hai (6 ki $γ = 1.4$ ceiling se kaafi zyada). Effective $γ_{eff}$ deduce karo jo real, partly dissociated gas describe karta hai, aur effective degrees of freedom $f$ nikalo.

Recall Solution

WHAT: $\frac{ρ _{2}}{ρ _{1}} = \frac{γ + 1}{γ - 1}$ ko invert karke $γ$ solve karo. WHY: density ratio $γ$ ka ek direct, measurable fingerprint hai; kyunki real-gas chemistry $γ$ lower karti hai, ek high ratio humein batata hai ki chemistry kitni far progress hui hai.

Maano $r = 12.0$ . Toh $r (γ - 1) = γ + 1 \Rightarrow γ (r - 1) = r + 1 \Rightarrow γ = \frac{r + 1}{r - 1} .$ $γ_{eff} = \frac{12 + 1}{12 - 1} = \frac{13}{11} \approx 1.1818$ Degrees of freedom: $f = \frac{2}{γ _{eff} - 1} = \frac{2}{0.1818} \approx 11.0$

Gas aise behave karta hai jaise uske ~11 active degrees of freedom hain — cold diatomic air ke 5 se kaafi zyada — jo vibration fully excited plus dissociation ke chemical storage kholne se consistent hai.

L5.2 Full re-entry temperature chain

Ek capsule $V = 7000 m/s$ se $T = 250 K$ , $γ = 1.4$ , $R = 287 J/(kg\cdotK)$ wali air mein enter karta hai. (a) $M$ nikalo. (b) Calorically-perfect $T_{0}$ nikalo. (c) Real gas energy dissociation mein dump karta hai isliye actual peak temperature sirf ambient se perfect-gas rise ka $40%$ hai. Real peak temperature estimate karo aur batao ki wahan kaunse real-gas effects active hain.

Recall Solution

WHAT & WHY: tools chain karo — speed of sound se Mach, boxed ratio se stagnation temperature, phir real-gas energy absorption ke liye downward correct karo.

(a) Speed of sound aur Mach: $a = 1.4 \times 287 \times 250 = 100, 450 \approx 316.9 m/s$ $M = \frac{7000}{316.9} \approx 22.09$

(b) Perfect-gas stagnation temperature: $\frac{T _{0}}{T} = 1 + 0.2 \times (22.09)^{2} = 1 + 0.2 \times 488.0 = 1 + 97.6 = 98.6$ $T_{0} = 98.6 \times 250 \approx 24, 650 K$

(c) Ambient se upar rise $= T_{0} - T = 24, 650 - 250 = 24, 400 K$ . Real rise $= 0.40 \times 24, 400 = 9760 K$ , isliye $T_{real} = 250 + 9760 = 10, 010 K .$ ~10,000 K par, air fully dissociated aur ionizing hai — plasma form hoti hai, jisse re-entry communications blackout hoti hai. Perfect-gas 24,650 K physically impossible hai; energy instead broken bonds aur free electrons mein lock ho jaati hai. Dekho Real Gas Thermodynamics & Dissociation.

Recall Self-test recap

Kaun sa formula Mach-independent hai? ::: Newtonian pressure coefficient $C_{p} = 2 sin^{2} θ$ — sirf geometry. High measured density ratio low $γ$ kyun imply karta hai? ::: Kyunki $ρ_{2} / ρ_{1} = (γ + 1) / (γ - 1)$ badhta hai jab $γ$ ghatta hai; real-gas chemistry $γ$ ko lower karti hai. Kaun si conserved quantity fix rehti hai jab temperature perfect-gas prediction se neeche girta hai? ::: Total enthalpy (energy) — ye sirf internal/chemical modes mein redistribute ho jaati hai.