Aerodynamic heating — recovery temperature, heat flux: Question bank

3.1.28 · D5 · Physics › Compressible Flow & Aerodynamics › Aerodynamic heating — recovery temperature, heat flux

Spot the error

"Heat flux ke liye driving temperature difference

T_{0} - T_{w}

hai kyunki

T_{0}

flow mein sabse hot temperature hai."

Sahi driver

T_{r} - T_{w}

hai.

T_{0}

heating ko over-predict karta hai, aur yeh sahi ho bhi nahi sakta kyunki yeh adiabatic wall (

T_{w} = T_{r} < T_{0}

) par nonzero flux deta — insulated surface ki definition ke viruddh.

"Kyunki air wall par rest mein aa jaati hai, isliye

T_{r} = T_{e} (1 + \frac{γ - 1}{2} M_{e}^{2})

Recovery factor

r

missing hai. Sahi form hai

T_{r} = T_{e} (1 + r \frac{γ - 1}{2} M_{e}^{2})

;

r

chhodna silently

r = 1

(full recovery) assume karta hai, jo disguise mein

T_{0}

wali galti hai.

r = \frac{T _{r} - T _{0}}{T _{e} - T _{0}}

Definition inverted hai. Hona chahiye

r = \frac{T _{r} - T _{e}}{T _{0} - T _{e}}

— dynamic rise (

T_{0} - T_{e}

) ka wo fraction jo edge temperature ke upar actually recover hota hai.

"Laminar flow ke liye

r = P r^{1/3}

use karo aur turbulent ke liye

r = P r

Dono swap hain. Laminar square root hai,

r = P r

; turbulent cube root hai,

r = P r^{1/3}

. Mnemonic: "Lam = root, Turb = cube."

"Kyunki

\overset{q}{˙}_{w} = h (T_{r} - T_{w})

hai, agar hum wall ko adiabatic banate hain to coefficient

h

zero ho jaana chahiye."

Nahi —

h

ek flow property hai aur finite rehta hai. Flux isliye vanish hoti hai kyunki temperature difference

T_{r} - T_{w}

zero ho jaata hai jab

T_{w}

rise karke

T_{r}

tak pahunch jaata hai, na ki

h

badalne ki wajah se.

"Recovery temperature aur stagnation temperature ek hi cheez hain kyunki dono flow ko rokne se aate hain."

Yeh recovery factor se alag hote hain:

T_{0}

ideal adiabatic deceleration assume karta hai bina sideways heat loss ke, jabki

T_{r}

woh hai jo ek real boundary layer wall tak deliver karta hai,

P r < 1

hone par

T_{r} < T_{0}

milta hai.

"Newton's law of cooling

\overset{q}{˙}_{w} = h (T_{r} - T_{w})

kehta hai wall hamesha cool hoti hai."

Yeh tabhi cool hoti hai jab

T_{w} > T_{r}

ho. Jab

T_{w} < T_{r}

ho to flux positive hoti hai — heat wall mein flow hoti hai aur yeh heat up hoti hai; "cooling" sirf law ka naam hai, guaranteed direction nahi.

Why questions

T_{r}

kyun

T_{e}

aur

T_{0}

ke beech hota hai, kisi ek extreme par nahi?

Boundary layer ke andar viscous dissipation heat add karta hai (

T_{0}

ki taraf push karta hai) jabki conduction heat ko sideways le jaati hai (

T_{e}

ki taraf push karti hai); wall balance point par settle hoti hai, isliye

T_{e} < T_{r} < T_{0}

(jab

P r < 1

Hum heat flux ke reference ke liye

T_{e}

T_{0}

ki jagah

T_{r}

kyun use karte hain?

T_{r}

woh temperature hai jo ek uncooled wall actually reach karegi, isliye yeh woh sahi thermal "potential" hai jis taraf gas wall ko drive karta hai; isse use karne par adiabatic case (

T_{w} = T_{r}

) correctly zero flux deta hai.

P r < 1

r < 1

kyun follow karta hai?

P r = μ c_{p} / k

momentum diffusion ko heat diffusion se compare karta hai;

P r < 1

matlab heat faster diffuse hoti hai, isliye dissipated energy near-wall region se escape ho jaati hai isse pehle ki sab wall heating ke roop mein recover ho sake — 100% se kam recover hota hai. Dekho Prandtl number & thermal boundary layer.

Thermal-protection design ke liye laminar se turbulent flow mein transition kyun itna critical hai?

Turbulence recovery factor ko (thoda) aur heat-transfer coefficient

h

ko (often 3–5×) badhata hai, isliye transition hone ki location local heat load ko multiply kar sakti hai, jo dictate karti hai ki shield kahan sabse moti honi chahiye. Dekho Hypersonic re-entry & thermal protection systems.

Engineering mein Stanton-number form

\overset{q}{˙}_{w} = ρ_{e} u_{e} c_{p} S t (T_{r} - T_{w})

kyun use ki jaati hai?

Yeh

h

ko ek dimensionless number mein package karta hai jo correlations par collapse hota hai (e.g. Reynolds analogy

S t \approx C_{f} /2

), jisse heat ko directly skin friction se predict kiya ja sake. Dekho Reynolds analogy & Stanton number.

Stagnation relation

T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}

purely Mach number ke terms mein kyun nikalti hai?

Energy conservation deti hai

T_{0} = T (1 + \frac{u ^{2}}{2 c _{p} T})

;

c_{p} = \frac{γ R}{γ - 1}

aur

a^{2} = γ R T

substitute karne par

u^{2} / a^{2}

turn ho jaata hai

M^{2}

mein, jo natural compressible-flow variable hai. Dekho Stagnation properties & isentropic relations.

Wall ko cool karne se remove ki jaane waali heat kyun badh sakti hai, jo problem ko aur bura banata lagta hai?

Kyunki

T_{r}

flow se fixed hai,

T_{w}

ghatane se difference

T_{r} - T_{w}

bada ho jaata hai, aur

\overset{q}{˙}_{w} = h (T_{r} - T_{w})

grow karta hai — thandi walls zyaada bada heat flux draw karti hain, halanki structure ko safe rakhti hain.

Edge cases

M_{e} \to 0

(bahut slow flow) ki limit mein

T_{r}

kya hoga?

Dynamic term

r \frac{γ - 1}{2} M_{e}^{2} \to 0

ho jaata hai, isliye

T_{r} \to T_{e}

— koi appreciable heating nahi, wall sirf local static temperature par baith jaati hai.

Agar wall ko deliberately

T_{r}

se upar rakha jaaye to heat flux ka kya hoga?

Tab

T_{r} - T_{w} < 0

hoga, isliye

\overset{q}{˙}_{w} < 0

: flux reverse ho jaata hai aur wall gas ko heat deti hai (e.g. ek hot surface cooler recovered flow mein radiating/convecting karta hai).

Agar kisi hypothetical gas mein

P r = 1

exactly ho, to

r

kya hoga?

Dono formulas dete hain

r = 1 = 1^{1/3} = 1

, isliye

T_{r} = T_{0}

— momentum aur heat equally diffuse hote hain, kuch escape nahi hota, aur full recovery achieve hoti hai.

P r > 1

waali gas ke liye (kuch oils/dense gases), kya

T_{r}

T_{0}

se exceed kar sakta hai?

Haan — tab

r > 1

hoga, aur

T_{r} = T_{e} (1 + r \frac{γ - 1}{2} M_{e}^{2})

T_{0}

se upar ja sakta hai; "recovery" overshoot kar jaata hai kyunki heat momentum se slower diffuse hoti hai, dissipated energy ko wall ke paas trap karti hai.

Exactly adiabatic operating point

T_{w} = T_{r}

par, kya boundary layer abhi bhi energy dissipate kar rahi hai?

Haan — viscous dissipation jaari rehta hai; yeh exactly sideways conduction se balance hota hai taki wall mein net flux zero rahe. Zero flux ka matlab zero physics nahi hai, bas ek steady balance hai.

Agar

γ \to 1

ho jaaye (bahut saare internal energy modes waali gas), to heating term ka kya hoga?

Factor

\frac{γ - 1}{2} \to 0

ho jaata hai, isliye

T_{0}

aur

T_{r}

dono

T_{e}

ke paas aa jaate hain — aisi gas kinetic energy ko internal modes mein soak kar leti hai bina kisi significant temperature rise ke.

Recall Jaane se pehle ek-line self-test

Cover karo aur jawab do: Wall heat flux ko kaun si temperature drive karti hai, aur zyaada hot $T_{0}$ kyun nahi? $T_{r} - T_{w}$ drive karta hai, kyunki $T_{r}$ (na ki $T_{0}$ ) woh hai jo ek uncooled wall actually reach karta hai, isliye yeh adiabatic wall par correctly zero flux deta hai. ::: Sahi hai agar tumne $T_{r}$ naam liya aur adiabatic-consistency reason diya.

Question bank — Aerodynamic heating — recovery temperature, heat flux

True or false — justify

Spot the error

Why questions

Edge cases