Intuition The big pictureWhen a fast vehicle slams into air, the air right at its surface gets brought to rest (no-slip).
All that kinetic energy of the flow turns into heat . So the surface "feels" a temperature much
hotter than the free-stream air — this is the recovery temperature . The wall then either heats
up or, if you cool it, sheds heat by convective heat flux .
WHY it matters: at Mach 6 the air "stagnation" temperature is ~1600 K — enough to melt aluminium.
Re-entry capsules, missiles and SR-71-class jets all live or die by managing this.
Definition Stagnation (total) temperatureIf you decelerate a moving gas adiabatically to rest, all its directed kinetic energy converts
to internal energy, raising temperature to the stagnation temperature T 0 T_0 T 0
Derivation — from the energy equation. Steady adiabatic flow with no work conserves total enthalpy:
h + u 2 2 = h 0 = const h + \frac{u^2}{2} = h_0 = \text{const} h + 2 u 2 = h 0 = const
Why this step? The first law for a streamtube with no heat added & no shaft work says enthalpy
plus kinetic energy is conserved. For a perfect gas h = c p T h = c_p T h = c p T
c p T + u 2 2 = c p T 0 ⇒ T 0 = T ( 1 + u 2 2 c p T ) c_p T + \frac{u^2}{2} = c_p T_0 \;\Rightarrow\; \boxed{T_0 = T\left(1 + \frac{u^2}{2 c_p T}\right)} c p T + 2 u 2 = c p T 0 ⇒ T 0 = T ( 1 + 2 c p T u 2 )
Now use c p = γ R γ − 1 c_p = \frac{\gamma R}{\gamma-1} c p = γ − 1 γ R a 2 = γ R T a^2 = \gamma R T a 2 = γ R T M = u / a M = u/a M = u / a
u 2 2 c p T = u 2 ( γ − 1 ) 2 γ R T = ( γ − 1 ) 2 u 2 a 2 = γ − 1 2 M 2 \frac{u^2}{2 c_p T} = \frac{u^2 (\gamma-1)}{2\gamma R T} = \frac{(\gamma-1)}{2}\frac{u^2}{a^2} = \frac{\gamma-1}{2}M^2 2 c p T u 2 = 2 γ R T u 2 ( γ − 1 ) = 2 ( γ − 1 ) a 2 u 2 = 2 γ − 1 M 2
Why this step? We want everything in terms of Mach number, the natural variable of compressible flow.
T 0 T = 1 + γ − 1 2 M 2 \boxed{\dfrac{T_0}{T} = 1 + \dfrac{\gamma-1}{2}M^2} T T 0 = 1 + 2 γ − 1 M 2
Intuition Two competing effects in the boundary layerInside the thin boundary layer the gas is slowed (→ heating by viscous dissipation ), but heat
also conducts away sideways. These don't balance perfectly. So the wall (adiabatic, no heat
drawn out) settles at a temperature between T T T T 0 T_0 T 0 recovery temperature T r T_r T r
Definition Recovery factorr r r r ≡ T r − T e T 0 − T e r \equiv \frac{T_r - T_e}{T_0 - T_e} r ≡ T 0 − T e T r − T e T e T_e T e r r r what fraction of the kinetic energy is "recovered" as wall heating .
Combining with the stagnation relation gives the adiabatic wall (recovery) temperature :
T r = T e ( 1 + r γ − 1 2 M e 2 ) \boxed{T_r = T_e\left(1 + r\,\frac{\gamma-1}{2}M_e^2\right)} T r = T e ( 1 + r 2 γ − 1 M e 2 )
Why this form? From T 0 − T e = T e γ − 1 2 M e 2 T_0 - T_e = T_e\frac{\gamma-1}{2}M_e^2 T 0 − T e = T e 2 γ − 1 M e 2 r r r T e T_e T e
WHY r < 1 r<1 r < 1 P r < 1 Pr<1 P r < 1 faster than momentum, so some dissipated heat escapes the
near-wall region before it can be fully "recovered."
Intuition Heat flows from hot to cold — but which "hot"?The driving temperature difference is not T 0 − T w a l l T_0 - T_{wall} T 0 − T w a l l not T e − T w a l l T_e - T_{wall} T e − T w a l l T r − T w a l l T_r - T_{wall} T r − T w a l l T r T_r T r uncooled wall would reach — the true thermal
potential the gas pushes the wall toward.
Definition Convective heat flux (Newton's law of cooling)q ˙ w = h ( T r − T w ) \boxed{\dot q_w = h\,(T_r - T_w)} q ˙ w = h ( T r − T w ) h h h [ W m − 2 K − 1 ] [\mathrm{W\,m^{-2}K^{-1}}] [ W m − 2 K − 1 ] T w T_w T w
If T w < T r T_w < T_r T w < T r into the wall (q ˙ w > 0 \dot q_w>0 q ˙ w > 0
If T w = T r T_w = T_r T w = T r adiabatic wall, zero net flux (this defines T r T_r T r
If T w > T r T_w > T_r T w > T r
Worked example (a) Recovery temperature of a Mach 3 jet at altitudeGiven M e = 3 M_e=3 M e = 3 T e = 223 T_e=223 T e = 223 γ = 1.4 \gamma=1.4 γ = 1.4 r = P r 1 / 3 = 0.71 1 / 3 = 0.892 r=Pr^{1/3}=0.71^{1/3}=0.892 r = P r 1/3 = 0.7 1 1/3 = 0.892
Step 1 — stagnation factor. γ − 1 2 M 2 = 0.2 × 9 = 1.8 \frac{\gamma-1}{2}M^2 = 0.2\times 9 = 1.8 2 γ − 1 M 2 = 0.2 × 9 = 1.8 Why? This is the full kinetic-energy heating fraction.
Step 2 — apply recovery. T r = 223 ( 1 + 0.892 × 1.8 ) = 223 ( 1 + 1.606 ) = 581 K T_r = 223(1 + 0.892\times1.8) = 223(1+1.606) = 581\ \text{K} T r = 223 ( 1 + 0.892 × 1.8 ) = 223 ( 1 + 1.606 ) = 581 K Why use r r r The wall recovers only ~89 % of the dynamic temperature rise.
For comparison T 0 = 223 ( 1 + 1.8 ) = 624 T_0 = 223(1+1.8)=624 T 0 = 223 ( 1 + 1.8 ) = 624 T r T_r T r T 0 T_0 T 0
Worked example (b) Heat flux into a cooled wallT r = 581 T_r=581 T r = 581 T w = 350 T_w=350 T w = 350 h = 900 W m − 2 K − 1 h=900\ \mathrm{W\,m^{-2}K^{-1}} h = 900 W m − 2 K − 1
Step 1. q ˙ w = h ( T r − T w ) = 900 × ( 581 − 350 ) = 900 × 231 \dot q_w = h(T_r-T_w) = 900\times(581-350) = 900\times231 q ˙ w = h ( T r − T w ) = 900 × ( 581 − 350 ) = 900 × 231 Why T r − T w T_r-T_w T r − T w The adiabatic-wall temperature is the true thermal driving potential.
Step 2. q ˙ w = 2.08 × 10 5 W / m 2 = 208 k W / m 2 \dot q_w = 2.08\times10^5\ \mathrm{W/m^2} = 208\ \mathrm{kW/m^2} q ˙ w = 2.08 × 1 0 5 W/ m 2 = 208 kW/ m 2
Worked example (c) Laminar vs turbulent recovery,M = 6 M=6 M = 6 T e = 250 T_e=250 T e = 250 γ − 1 2 M 2 = 0.2 × 36 = 7.2 \frac{\gamma-1}{2}M^2 = 0.2\times36 = 7.2 2 γ − 1 M 2 = 0.2 × 36 = 7.2 r = 0.71 = 0.843 r=\sqrt{0.71}=0.843 r = 0.71 = 0.843 T r = 250 ( 1 + 6.07 ) = 1768 T_r=250(1+6.07)=1768 T r = 250 ( 1 + 6.07 ) = 1768 r = 0.892 r=0.892 r = 0.892 T r = 250 ( 1 + 6.42 ) = 1856 T_r=250(1+6.42)=1856 T r = 250 ( 1 + 6.42 ) = 1856 Why turbulent hotter? Higher r r r h h h Why this step matters: transition location dominates TPS design.
Common mistake "The wall reaches the stagnation temperatureT 0 T_0 T 0 Why it feels right: the gas is brought to rest at the wall (no-slip), so surely T = T 0 T=T_0 T = T 0 The fix: no-slip stops bulk motion, but conduction leaks dissipated heat sideways
(P r < 1 Pr<1 P r < 1 T r = T e ( 1 + r γ − 1 2 M 2 ) T_r = T_e(1+r\frac{\gamma-1}{2}M^2) T r = T e ( 1 + r 2 γ − 1 M 2 ) r < 1 r<1 r < 1 below T 0 T_0 T 0
Common mistake "Heat flux driver isT 0 − T w T_0 - T_w T 0 − T w Why it feels right: T 0 T_0 T 0 The fix: the correct potential is T r − T w T_r - T_w T r − T w T 0 T_0 T 0 T w = T r T_w=T_r T w = T r T r T_r T r
Common mistake "Cooling the wall lowers the recovery temperature."Why it feels right: cooling should reduce all temperatures.
The fix: T r T_r T r M e , T e , r M_e, T_e, r M e , T e , r not T w T_w T w T w T_w T w increases the flux h ( T r − T w ) h(T_r-T_w) h ( T r − T w ) T r T_r T r
Recall Feynman: explain to a 12-year-oldImagine running with your hand out the car window. Slow speed: cool breeze. Really fast: your palm
feels warm because you're squashing and rubbing the air to a stop, and rubbing makes heat. A rocket
goes SO fast it squashes the air so hard the air gets red-hot. The "recovery temperature" is just
how hot your palm would get if you never cooled it. If your palm is colder than that, heat keeps
flowing into it — that flow is the "heat flux." Spaceships put a special shield in front so the
shield gets hot instead of the people inside.
Mnemonic Remember the chain"STAR-FLUX": ST agnation → A pply R ecovery factor → drive FLUX by T r − T w T_r-T_w T r − T w "Lam = root, Turb = cube" (r = P r r=\sqrt{Pr} r = P r r = P r 1 / 3 r=Pr^{1/3} r = P r 1/3
Stagnation properties & isentropic relations — source of T 0 / T = 1 + γ − 1 2 M 2 T_0/T = 1+\frac{\gamma-1}{2}M^2 T 0 / T = 1 + 2 γ − 1 M 2 Boundary layers & viscous dissipation — origin of the recovery factorPrandtl number & thermal boundary layer — why r r r P r Pr P r Reynolds analogy & Stanton number — links q ˙ w \dot q_w q ˙ w Hypersonic re-entry & thermal protection systems — engineering applicationNormal & oblique shock heating — post-shock T e T_e T e
What converts to heat at a high-speed surface, raising its temperature? The directed kinetic energy of the flow, dissipated as the gas is brought to rest at the wall (no-slip + viscous dissipation).
Stagnation temperature ratio formula? T 0 / T = 1 + γ − 1 2 M 2 T_0/T = 1 + \frac{\gamma-1}{2}M^2 T 0 / T = 1 + 2 γ − 1 M 2 , derived from
c p T + u 2 / 2 = c p T 0 c_pT + u^2/2 = c_pT_0 c p T + u 2 /2 = c p T 0 .
Definition of recovery factor r r r r = ( T r − T e ) / ( T 0 − T e ) r = (T_r - T_e)/(T_0 - T_e) r = ( T r − T e ) / ( T 0 − T e ) — fraction of dynamic temperature rise recovered at an adiabatic wall.
Recovery factor for laminar vs turbulent air flow? Laminar
r = P r ≈ 0.84 r=\sqrt{Pr}\approx0.84 r = P r ≈ 0.84 ; turbulent
r = P r 1 / 3 ≈ 0.89 r=Pr^{1/3}\approx0.89 r = P r 1/3 ≈ 0.89 (air,
P r = 0.71 Pr=0.71 P r = 0.71 ).
Recovery temperature formula? T r = T e ( 1 + r γ − 1 2 M e 2 ) T_r = T_e\left(1 + r\frac{\gamma-1}{2}M_e^2\right) T r = T e ( 1 + r 2 γ − 1 M e 2 ) .
Why is T r T_r T r T 0 T_0 T 0 Because
P r < 1 Pr<1 P r < 1 : heat conducts away faster than momentum, so
r < 1 r<1 r < 1 and not all KE is recovered.
Convective heat-flux law and its driving potential? q ˙ w = h ( T r − T w ) \dot q_w = h(T_r - T_w) q ˙ w = h ( T r − T w ) ; the driver is
T r − T w T_r - T_w T r − T w , NOT
T 0 − T w T_0-T_w T 0 − T w or
T e − T w T_e-T_w T e − T w .
What defines an adiabatic wall? T w = T r T_w = T_r T w = T r , giving zero net heat flux
q ˙ w = 0 \dot q_w=0 q ˙ w = 0 .
Effect of cooling the wall on T r T_r T r T r T_r T r unchanged (set by flow); flux
h ( T r − T w ) h(T_r-T_w) h ( T r − T w ) increases as
T w T_w T w drops.
Stanton-number form of heat flux? q ˙ w = ρ e u e c p S t ( T r − T w ) \dot q_w = \rho_e u_e c_p\,St\,(T_r-T_w) q ˙ w = ρ e u e c p S t ( T r − T w ) , with
S t ≈ C f / 2 St\approx C_f/2 S t ≈ C f /2 by Reynolds analogy.
Pr < 1 so heat diffuses faster
Air brought to rest at wall
Energy equation: h + u2/2 const
T0/T = 1 + half gamma-1 M2
Intuition Hinglish mein samjhoSocho ek jet ya rocket bahut tezi se hawa ko cheer raha hai. Surface ke bilkul paas hawa "no-slip"
ki wajah se ruk jaati hai, aur uski saari kinetic energy heat ban jaati hai. Isi se surface garam hota
hai. Agar hawa ko poori tarah rok do (adiabatically), toh jo temperature milta hai use stagnation
temperature T 0 = T ( 1 + γ − 1 2 M 2 ) T_0 = T(1+\frac{\gamma-1}{2}M^2) T 0 = T ( 1 + 2 γ − 1 M 2 )
Lekin asli wall thoda kam garam hota hai, kyunki boundary layer ke andar heat side me conduct hoke nikal
bhi jaati hai (Prandtl number P r < 1 Pr<1 P r < 1 recovery
temperature T r = T e ( 1 + r γ − 1 2 M e 2 ) T_r = T_e(1 + r\frac{\gamma-1}{2}M_e^2) T r = T e ( 1 + r 2 γ − 1 M e 2 ) r r r r = P r r=\sqrt{Pr} r = P r r = P r 1 / 3 r=Pr^{1/3} r = P r 1/3 r < 1 r<1 r < 1 T r T_r T r T 0 T_0 T 0
Ab heat flux: garmi hamesha hot se cold jaati hai, lekin "hot" ka matlab yahan T r T_r T r T 0 T_0 T 0 q ˙ w = h ( T r − T w ) \dot q_w = h(T_r - T_w) q ˙ w = h ( T r − T w ) T w T_w T w badh jaata hai (gap bada ho gaya). Aur dhyan do — cooling se T r T_r T r T r T_r T r
Yeh sab matter kyun karta hai? Re-entry capsule, missile, ya fast jet — sabka design isi heating par
tika hai. Heat shield isiliye lagate hain taaki shield garam ho, andar ke log nahi. Exam me bhi yeh
high-yield hai: T 0 T_0 T 0
