Aerodynamic heating — recovery temperature, heat flux

3.1.28 · Physics › Compressible Flow & Aerodynamics

1. Heat kahan se aati hai (first principles)

Derivation — energy equation se. Steady adiabatic flow mein jisme koi kaam nahi ho raha, total enthalpy conserve hoti hai:

$h + \frac{u ^{2}}{2} = h_{0} = const$

Yeh step kyun? Ek streamtube ke liye first law, jisme na heat add ho rahi hai aur na shaft work, kehta hai ki enthalpy aur kinetic energy conserve hoti hai. Perfect gas ke liye $h = c_{p} T$ , toh:

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

Ab $c_{p} = \frac{γ R}{γ - 1}$ aur sound speed $a^{2} = γ R T$ use karo, $M = u / a$ ke saath:

$\frac{u ^{2}}{2 c _{p} T} = \frac{u ^{2} ( γ - 1 )}{2 γ R T} = \frac{( γ - 1 )}{2} \frac{u ^{2}}{a ^{2}} = \frac{γ - 1}{2} M^{2}$

Yeh step kyun? Hum sab kuch Mach number ke terms mein chahte hain, jo compressible flow ka natural variable hai.

$\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2}$

2. Wall $T_{0}$ par KYUN NAHI hoti — recovery factor

Stagnation relation ke saath combine karne par adiabatic wall (recovery) temperature milti hai:

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

Yeh form kyun? $T_{0} - T_{e} = T_{e} \frac{γ - 1}{2} M_{e}^{2}$ se, $r$ se multiply karo aur $T_{e}$ add karo.

$r < 1$ KYUN: $P r < 1$ matlab heat momentum se tez diffuse hoti hai, toh kuch dissipated heat near-wall region se nikal jaati hai pehle ki woh poori tarah "recover" ho sake.

3. Wall par heat flux

Agar $T_{w} < T_{r}$ : heat wall mein jaati hai ( $\overset{q}{˙}_{w} > 0$ ) → wall garam hoti hai / cooling zaroori hai.
Agar $T_{w} = T_{r}$ : adiabatic wall, zero net flux (yahi $T_{r}$ ko define karta hai).
Agar $T_{w} > T_{r}$ : wall aslaan gas ko heat deti hai.

Figure — Aerodynamic heating — recovery temperature, heat flux

4. Worked examples

5. Common mistakes

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho tum car ki khidki se haath bahar nikal ke daud rahe ho. Dheemi speed: thandi hawa. Bahut tez: haath ki hatheli garam lagti hai kyunki tum hawa ko rok ke ragar rahe ho, aur ragad se heat banti hai. Ek rocket ITNA tez jaata hai ki hawa ko itna zyada dabata hai ki hawa laal-garam ho jaati hai. "Recovery temperature" bas yeh hai ki tumhari hatheli kitni garam hogi agar tum use kabhi thanda na karo. Agar tumhari hatheli us se thandi hai, toh heat usme aati rehti hai — wahi flow "heat flux" hai. Spaceships ke aage ek special shield lagaate hain taaki shield garam ho, andar ke logon ki jagah.

Connections

Stagnation properties & isentropic relations — $T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}$ ka source
Boundary layers & viscous dissipation — recovery factor ka origin
Prandtl number & thermal boundary layer — kyun $r$ , $P r$ par depend karta hai
Reynolds analogy & Stanton number — $\overset{q}{˙}_{w}$ ko skin friction se link karta hai
Hypersonic re-entry & thermal protection systems — engineering application
Normal & oblique shock heating — in formulas ko feed karne wala post-shock $T_{e}$

High-speed surface par kya heat mein convert hota hai, jisse temperature badhti hai?

Flow ki directed kinetic energy, jo wall par gas ke ruk jaane (no-slip + viscous dissipation) par dissipate hoti hai.

Stagnation temperature ratio formula?

T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}

c_{p} T + u^{2} /2 = c_{p} T_{0}

se derived.

Recovery factor

r

ki definition?

r = (T_{r} - T_{e}) / (T_{0} - T_{e})

— adiabatic wall par dynamic temperature rise ka kitna fraction recover hota hai.

Laminar vs turbulent air flow mein recovery factor?

Laminar

r = P r \approx 0.84

; turbulent

r = P r^{1/3} \approx 0.89

(air,

P r = 0.71

Recovery temperature formula?

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

T_{r}

T_{0}

se kam kyun hoti hai?

Kyunki

P r < 1

: heat momentum se tez conduct hoti hai, toh

r < 1

aur saari KE recover nahi hoti.

Convective heat-flux law aur uska driving potential?

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

; driver

T_{r} - T_{w}

hai,

T_{0} - T_{w}

T_{e} - T_{w}

NAHI.

Adiabatic wall kya define karta hai?

T_{w} = T_{r}

, jisse zero net heat flux

\overset{q}{˙}_{w} = 0

milta hai.

Wall ko cool karne ka

T_{r}

aur heat flux par effect?

T_{r}

unchanged rehta hai (flow se set hota hai); flux

h (T_{r} - T_{w})

T_{w}

ke girne se badh jaata hai.

Heat flux ka Stanton-number form?

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

, jisme Reynolds analogy se

S t \approx C_{f} /2