Ablative cooling — charring, blowing: Worked examples

3.3.30 · D3 · Physics › Rocket Propulsion › Ablative cooling — charring, blowing

Shuru karne se pehle, yahan use hone wale saare symbols (sab parent se liye gaye hain) simple shabdon mein:

Definition Symbol glossary (koi bhi cheez is line se pehle use nahi hui)

$\overset{q}{˙}$ ::: heat flux — har second mein wall ke ek square metre se guzarne wale joules. Units $W/m^{2} = J/(m^{2} s)$ .
$Q^{*}$ ::: effective heat of ablation — jitna material tum jalate ho, uske har kilogram ke badle tum kitne joules bahar phenk sakte ho. Bada $Q^{*}$ = achha ablator. Units $J/kg$ .
$\overset{m}{˙}^{''}$ ::: mass loss rate per unit area — har second mein har square metre se nikalne wale kilograms. Units $kg/(m^{2} s)$ . (Do primes ka matlab sirf "per area" hai.)
$ρ_{v}$ ::: virgin (unburnt) material ki density. Units $kg/m^{3}$ .
$\overset{s}{˙}$ ::: recession rate — kitni tezi se (metres/second mein) surface andar ki taraf khaati hai.
$s$ ::: total recession depth — surface kitna andar kha gayi, metres mein.
$t$ ::: burn ya re-entry ka elapsed time, seconds mein.
$L$ ::: shield/liner thickness — available material ki poori depth, metres mein.
$h_{0}$ ::: convective heat-transfer coefficient bina blowing ke (dekho Convective Heat Transfer (Stanton number)).
$B^{'}$ ::: blowing parameter — injected mass aur jo flow carry kar sakta hai uska dimensionless ratio. $B^{'} = \overset{m}{˙}^{''} c_{p} / h_{0}$ .
$c_{p}$ ::: specific heat — 1 kg ko 1 K warm karne ke liye joules.
$Q_{p}$ ::: pyrolyze kiye gaye har kg mein absorb hone wali enthalpy (dekho Heat of Reaction and Pyrolysis). Positive $Q_{p}$ = reaction heat soakti hai (endothermic); negative matlab heat release hoti (exothermic).
$T_{i}$ ::: thande, unheated material ka initial temperature, kelvin (K) mein.
$T_{w}$ ::: wall (surface) temperature — exposed ablating face kitni hot hai, kelvin (K) mein.
$T_{g}$ ::: gas temperature — hot boundary-layer gas ka temperature, kelvin (K) mein.
$H_{g}$ ::: hot gas ki enthalpy per kg — gas ki total heat content (thermal + chemical), J/kg mein.
$H_{w}$ ::: wall par gas ki enthalpy per kg — bilkul surface par gas ki heat content, J/kg mein.

The scenario matrix

Is topic ke har problem ka koi na koi cell neeche hai. Do axes hain: kya pucha ja raha hai (rows mein) aur input space ka kaun sa edge probe ho raha hai (last column). Figure ko map ki tarah padho, table ko legend ki tarah; aage aane wala har example apne cell ki stamp ke saath aayega.

Figure — Ablative cooling — charring, blowing

#	Cell class	Degenerate / limiting edge jo probe karta hai
C1	Recession from a heat load (forward)	ordinary mid-range numbers
C2	Blowing factor, small $B^{'}$	limit $B^{'} \to 0$ ⇒ factor $\to 1$
C3	Blowing factor, large $B^{'}$	limit $B^{'} \to \infty$ ⇒ factor $\to 0$ (blow-off)
C4	Zero input	$\overset{q}{˙} = 0$ ya $\overset{m}{˙}^{''} = 0$ — kuch nahi hota
C5	Energy-budget split	teen independent J/kg sinks ka sum
C6	Inverse problem	survival diya hua hai, required $Q^{*}$ ya thickness solve karo
C7	Enthalpy vs temperature trap	high- $T$ dissociation — $H$ use karna zaroori, $T$ nahi
C8	Real-world word problem	re-entry shield, numbers khud chuno
C9	Exam twist	injection double karo — diminishing returns
C10	Sign-flip / exothermic edge	negative reaction enthalpy load mein add karta hai

C1 — Recession from a heat load (bread-and-butter case)

Worked example C1 · Forward recession

Ek silica-phenolic liner: $ρ_{v} = 1600 kg/m^{3}$ , $Q^{*} = 8.0 \times 1 0^{6} J/kg$ . Yeh $45 s$ ke burn ke liye $\overset{q}{˙}_{n e t} = 4 MW/m^{2}$ face karta hai. Surface kitni deep recede karti hai?

Forecast: compute karne se pehle guess karo — kya yeh 1 mm, 1 cm, ya 10 cm ke kareeb hoga?

Mass loss per area. $\overset{m}{˙}^{''} = \overset{q}{˙} / Q^{*}$ . Yeh step kyun? $Q^{*}$ hai "joules per kg thrown away", toh flux (joules per area·sec) ko isse divide karne par kg per area·sec milta hai. $\overset{m}{˙}^{''} = \frac{4 \times 1 0 ^{6}}{8.0 \times 1 0 ^{6}} = 0.5 kg/(m^{2} s)$
Mass ko depth mein convert karo. $\overset{s}{˙} = \overset{m}{˙}^{''} / ρ_{v}$ . Yeh step kyun? Density $ρ_{v}$ ke ek slab se per area mass loss karne ka matlab hai surface (mass/area)÷(mass/volume) = length per second ki dar se girti hai. $\overset{s}{˙} = \frac{0.5}{1600} = 3.125 \times 1 0^{- 4} m/s = 0.3125 mm/s$
Burn ke dauran total depth. $s = \overset{s}{˙} \times t = 3.125 \times 1 0^{- 4} \times 45 = 0.0141 m \approx 14.1 mm$ . Yeh step kyun? Recession rate $\overset{s}{˙}$ metres per second ki dar se khaata hai; surface burn ke dauran (near) constant rate par recede karti hai, toh total time $t$ se multiply karna rate ko total depth mein integrate karta hai.

Verify: Units chain: $\frac{W/m ^{2}}{J/kg} = \frac{J/(m ^{2} s)}{J/kg} = kg/(m^{2} s)$ ✔. Phir $\frac{kg/(m ^{2} s)}{kg/m ^{3}} = m/s$ ✔. ~1.4 cm loss ka matlab hai ek 2 cm liner margin ke saath survive karti hai — middle guess sahi nikla.

C2 — Blowing factor at small $B^{'}$ (limit $\to 1$ )

Worked example C2 · Weak blowing

Gentle injection $B^{'} = 0.1$ ke liye convective reduction factor $h / h_{0}$ compute karo, aur confirm karo ki yeh no-blowing value ke kareeb aata hai.

Forecast: almost koi gas inject nahi — kya factor 0, 0.5, ya 1 ke paas hona chahiye?

Model apply karo. $\frac{h}{h _{0}} = \frac{ln ( 1 + B ^{'} )}{B ^{'}}$ . Yeh step kyun? Yeh parent mein derived film/Couette blowing result hai — ek aur formula jo $B^{'}$ ko cooling se jodta ho, woh nahi. $\frac{h}{h _{0}} = \frac{l n ( 1.1 )}{0.1} = \frac{0.09531}{0.1} = 0.9531$
Interpret karo. Blowing ne heating mein sirf lagbhag $5%$ hi kaata — almost koi effect nahi.

Verify: Taylor limit — chote $x$ ke liye, $ln (1 + x) \approx x - x^{2} /2$ , toh $ln (1 + B^{'}) / B^{'} \approx 1 - B^{'} /2 = 1 - 0.05 = 0.95$ ✔. Jab $B^{'} \to 0$ toh factor $\to 1$ : no blowing, no reduction, bilkul $\overset{q}{˙}_{0} = h_{0} (\dots)$ se match karta hai.

C3 — Blowing factor at large $B^{'}$ (limit $\to 0$ , blow-off)

Worked example C3 · Strong blowing / blow-off

Ek tez pyrolysis $B^{'} = 10$ produce karti hai. Bare-wall heating ka kitna fraction bacha rehta hai?

Forecast: heavy gas injection — surviving fraction 0.5, 0.2, ya 0.05 ke kareeb hai?

Model apply karo. $\frac{h}{h _{0}} = \frac{ln ( 1 + B ^{'} )}{B ^{'}}$ . Yeh step kyun? Parent se wohi film/Couette blowing law — yahi ek relation hai jo blowing parameter $B^{'}$ ko heating reduction mein convert karta hai, har $B^{'}$ par valid hai, is bade wale par bhi. $\frac{h}{h _{0}} = \frac{l n ( 11 )}{10} = \frac{2.3979}{10} = 0.2398$
Interpret karo. Unblown flux ka sirf lagbhag $24%$ hi wall tak pahuncha — gas cushion ne zyada tar rok liya. Yeh "blow-off" regime ki onset hai.

Verify: Jab $B^{'} \to \infty$ , $ln (1 + B^{'})$ slow $ln B^{'}$ ki tarah grow karta hai jabki denominator fast $B^{'}$ ki tarah, toh ratio $\to 0$ . Hamara $0.24$ at $B^{'} = 10$ parent mein $B^{'} = 2$ par mile $0.55$ se kaafi neeche hai — monotone decreasing ✔.

C4 — Zero / degenerate input

Worked example C4 · "Kuch nahi hota" wale cases

Do degenerate scenarios jinpar ek careless formula blow up kar sakti hai: (a) $\overset{q}{˙}_{n e t} = 0$ (engine off, cold). (b) $B^{'} = 0$ (bilkul koi gas inject nahi).

Forecast: jab driver zero ho toh recession aur blowing factor dono ko kya hona chahiye?

(a) Zero heat flux. $\overset{m}{˙}^{''} = \overset{q}{˙} / Q^{*} = 0/ Q^{*} = 0$ , toh $\overset{s}{˙} = 0$ . Yeh step kyun? Koi heat nahi aarahi matlab koi bond nahi toota aur koi material nahi gaya — physically surface bas wahi baith jaati hai.
(b) Zero blowing. Naive tarike se $ln (1 + 0) /0 = 0/0$ undefined lagta hai — yeh ek trap hai. Yeh step kyun? Hume blindly plug in nahi karna, balki limit leni hai. $lim_{B^{'} \to 0} \frac{l n ( 1 + B ^{'} )}{B ^{'}} = 1$ Toh $h = h_{0}$ : koi gas inject nahi, wall ko full bare-wall convection feel hoti hai. Sensible hai, divide-by-zero disaster nahi.

Verify: $ln (1 + B^{'}) / B^{'}$ par L'Hôpital: top ka derivative $= 1/ (1 + B^{'}) \to 1$ , bottom ka derivative $= 1$ , ratio $\to 1$ ✔. Aur $\overset{m}{˙}^{''} = 0$ trivially $\overset{s}{˙} = 0$ deta hai ✔.

C5 — Energy budget split karna

Worked example C5 · Three-sink sum

Ek carbon-phenolic mein $Q_{p} = 2.5 \times 1 0^{6} J/kg$ , $c_{p} = 1400 J/(kg\cdotK)$ , surface $T_{w} = 3000 K$ par initial $T_{i} = 300 K$ se, aur ek blocking (blowing) term $5.5 \times 1 0^{6} J/kg$ ka hai. $Q^{*}$ nikalo.

Forecast: teen sinks mein se sabse bada kaun sa hoga?

Sensible heating sink. $c_{p} (T_{w} - T_{i})$ . Yeh step kyun? Ek kilogram pyrolyze hone se pehle use pehle $T_{i}$ se surface temperature $T_{w}$ tak heat hona padta hai — yeh warming energy maangti hai. $c_{p} Δ T = 1400 \times (3000 - 300) = 1400 \times 2700 = 3.78 \times 1 0^{6} J/kg$
Teen independent sinks ka sum lo. Yeh step kyun? Har channel — chemical decomposition ( $Q_{p}$ ), warming ( $c_{p} Δ T$ ), aur boundary-layer blocking — independently joules remove karta hai, toh ye add hote hain. $Q^{*} = 2.5 \times 1 0^{6} + 3.78 \times 1 0^{6} + 5.5 \times 1 0^{6} = 11.78 \times 1 0^{6} J/kg$

Verify: Teeno terms $J/kg$ carry karti hain (term 1 ke liye: $J/(kg\cdotK) \times K = J/kg$ ) ✔. Blocking term sabse bada hai ( $11.78$ mein se $5.5$ ), parent ki baat confirm hoti hai ki blowing ek main cooling channel hai, side effect nahi.

C6 — Inverse problem: survival ke liye design karo

Worked example C6 · Required

$Q^{*}$ ke liye backwards solve karo $L = 25 mm$ thickness aur $ρ_{v} = 1450 kg/m^{3}$ wali ek shield ko $t = 90 s$ re-entry ke dauran $\overset{q}{˙}_{n e t} = 6 MW/m^{2}$ survive karna hai — lekin apni thickness ka zyada se zyada aadha kho sakti hai. Material ko minimum $Q^{*}$ kitna chahiye?

Forecast: materials roughly $Q^{*} \sim 1 0^{6}$ se $1 0^{7}$ J/kg range mein hote hain — kya requirement us range ke andar padegi?

Allowed recession. $L = 25$ mm ka zyada se zyada aadha kho: $s_{ma x} = 12.5 mm = 0.0125 m$ . Yeh step kyun? Yeh design constraint ek length budget mein badal jaata hai.
Allowed recession rate. $\overset{s}{˙} = s_{ma x} / t = 0.0125/90 = 1.389 \times 1 0^{- 4} m/s$ . Yeh step kyun? Allowed loss ko burn ke upar evenly spread karo taaki maximum tolerable rate mile.
Allowed mass loss rate. $\overset{m}{˙}^{''} = ρ_{v} \overset{s}{˙} = 1450 \times 1.389 \times 1 0^{- 4} = 0.2014 kg/(m^{2} s)$ . Yeh step kyun? Length budget ko density ke zariye mass budget mein convert karo (C1 step 2 ka ulta).
Required heat of ablation. $Q^{*} = \overset{q}{˙} / \overset{m}{˙}^{''} = 6 \times 1 0^{6} /0.2014 = 2.98 \times 1 0^{7} J/kg$ . Yeh step kyun? Unknown $Q^{*}$ ke liye $\overset{m}{˙}^{''} = \overset{q}{˙} / Q^{*}$ rearrange karo.

Verify: $\approx 3.0 \times 1 0^{7} J/kg$ carbon-phenolic ke typical $\sim 1 \times 1 0^{7}$ se upar hai — toh yeh harsh case ek premium ablator ya moti shield maangta hai. Units: $\frac{W/m ^{2}}{kg/(m ^{2} s)} = J/kg$ ✔. Sanity: plug back karo — $Q^{*} = 3.0 \times 1 0^{7}$ par, $\overset{m}{˙}^{''} = 6 \times 1 0^{6} /3.0 \times 1 0^{7} = 0.2 kg/(m^{2} s)$ , $\overset{s}{˙} = 0.2/1450 = 1.38 \times 1 0^{- 4}$ m/s, 90 s mein $= 12.4$ mm ≈ half ✔.

C7 — Enthalpy vs temperature trap

Worked example C7 · High heat par

$T$ kyun underestimate karta hai Hot gas $T_{g} = 6000 K$ par, wall $T_{w} = 3000 K$ par, $h_{0} = 400 W/(m^{2} K)$ . Naive temperature-based flux versus enthalpy-based flux, jahan true enthalpy difference $H_{g} - H_{w} = 1.8 \times 1 0^{7} J/kg$ hai aur mass-transfer coefficient $h_{0} / c_{p}$ hai jahan $c_{p} = 2000 J/(kg\cdotK)$ .

Forecast: 6000 K par air partly dissociated hoti hai — kya enthalpy-based flux naive wale se choti hogi ya badi?

Naive temperature flux. $\overset{q}{˙}_{T} = h_{0} (T_{g} - T_{w}) = 400 \times (6000 - 3000) = 1.2 \times 1 0^{6} W/m^{2}$ . Yeh step kyun? Yeh tempting textbook form hai — hamara baseline.
Enthalpy-based flux. $\overset{q}{˙}_{H} = (h_{0} / c_{p}) (H_{g} - H_{w})$ use karo. Yeh step kyun? Jab gas dissociate aur recombine hoti hai, chemical enthalpy — sirf temperature nahi — heating drive karta hai; correct driving potential enthalpy difference $H_{g} - H_{w}$ hai. $\overset{q}{˙}_{H} = \frac{400}{2000} \times 1.8 \times 1 0^{7} = 0.2 \times 1.8 \times 1 0^{7} = 3.6 \times 1 0^{6} W/m^{2}$
Compare karo. $\overset{q}{˙}_{H} / \overset{q}{˙}_{T} = 3.6/1.2 = 3.0$ — enthalpy form teen guna load predict karta hai. Yeh step kyun? Hum ratio calculate karte hain taaki error ki size pata chale jo ek temperature-only estimate karega; agar designer $\overset{q}{˙}_{T}$ par trust kare toh shield teen guna under-build hogi, toh gap quantify karna is trap ka poora point hai.

Verify: Step 2 ke units: $\frac{W/(m ^{2} K)}{J/(kg\cdotK)} \times J/kg = \frac{W/(m ^{2} K) \cdot kg}{J} \times \frac{J}{kg}$ reduce hokar $W/m^{2}$ deta hai ✔. Enthalpy flux bada hai, bilkul parent ki warning wali baat: temperature akela heat load ko underestimate karta hai. Dekho Re-entry Aerothermodynamics.

C8 — Real-world word problem (re-entry shield)

Worked example C8 · Apollo-style shield sizing karna

Ek capsule heat shield peak $\overset{q}{˙}_{n e t} = 3 MW/m^{2}$ face karta hai jo $t = 400 s$ re-entry par average hai. Ablator carbon-phenolic hai: $ρ_{v} = 1450 kg/m^{3}$ , $Q^{*} = 1.2 \times 1 0^{7} J/kg$ . Engineers recession depth par $50%$ safety margin chahte hain. Unhe kaunsi shield thickness $L$ install karni chahiye?

Forecast: re-entry shields famous tor par moti hoti hain — guess karo: 2 cm, 6 cm, ya 20 cm?

Mass loss rate. $\overset{m}{˙}^{''} = \overset{q}{˙} / Q^{*} = 3 \times 1 0^{6} /1.2 \times 1 0^{7} = 0.25 kg/(m^{2} s)$ . Yeh step kyun? C1 waali hi forward chain — flux divided by heat of ablation.
Recession rate. $\overset{s}{˙} = \overset{m}{˙}^{''} / ρ_{v} = 0.25/1450 = 1.724 \times 1 0^{- 4} m/s$ . Yeh step kyun? Mass loss per area ko density se divide karo taaki kilograms lost ko metres of surface eaten mein badla ja sake, bilkul C1 step 2 ki tarah.
Total recession. $s = \overset{s}{˙} \times t = 1.724 \times 1 0^{- 4} \times 400 = 0.0690 m = 69.0 mm$ . Yeh step kyun? (Near-constant) rate ko poore re-entry time $t$ par integrate karo.
Margin add karo. $50%$ margin ke saath installed thickness $L = 1.5 \times s = 1.5 \times 69.0 mm = 103.4 mm \approx 103 mm$ (lagbhag $10.3 cm$ ) hai. Yeh step kyun? Design practice — heat load mein uncertainty ke against extra material rakho, toh bare-survival depth ko $1.5$ se multiply karo.

Verify: ~10 cm real Apollo-class shields se match karta hai (wo kaafi cm mote the) — "6–20 cm" bracket sahi tha, 2 cm nahi. C1 ke numbers se cross-check: same physics, longer burn ⇒ kaafi zyada deep loss ✔. Regenerative Cooling aur Radiative Cooling se contrast karo, jo wall kabhi consume nahi karte.

C9 — Exam twist: diminishing returns

Worked example C9 · Injection double karo

Ek exam poochta hai: " $B^{'} = 2$ par blowing $h / h_{0} = 0.549$ deta hai. Agar tum gas injection double karke $B^{'} = 4$ karo, toh kya heating phir se half ho jaati hai?" Compute karo aur comment karo.

Forecast: intuition kehta hai double gas → half heat. Trust karo ya nahi?

$B^{'} = 4$ par factor. $\frac{h}{h _{0}} = \frac{l n ( 5 )}{4} = \frac{1.6094}{4} = 0.4024$ Yeh step kyun? Doubled parameter par same $ln (1 + B^{'}) / B^{'}$ model directly evaluate karo.
Compare karo. $B^{'} = 2 \to 4$ jaane par factor sirf $0.549$ se $0.402$ gaya — lagbhag $0.147$ ki drop, halving nahi. Yeh step kyun? Trap yeh hai ki linearity expect karna; $lo g$ saturate ho jaata hai.

Verify: Ratio $0.402/0.549 = 0.73$ , yaani mass loss double karne ke bawajood humne pichli heating ka $73%$ rakh liya — yeh ek bura trade hai. Yeh parent ki teesri "common mistake" hai: zyada blowing linearly help nahi karta — tum sirf double-fast erode karte ho ek modest gain ke liye ✔. $B^{'} = 1$ compare karo: $ln 2/1 = 0.693$ — curve flatate rehta hai.

C10 — Sign-flip edge: ek exothermic reaction load mein add karta hai

Worked example C10 · Jab chemistry tumhare khilaf kaam kare

Har surface reaction heat soak nahi karta. Char incoming oxygen ke saath react kar sakta hai ( $C + O_{2}$ ), aur woh reaction heat release karta hai — ek exothermic contribution. Ek aisa case model karo jahan pyrolysis weakly endothermic hai, $Q_{p} = + 0.8 \times 1 0^{6} J/kg$ , lekin surface oxidation $Q_{o x} = - 1.5 \times 1 0^{6} J/kg$ wall mein wapas dump karti hai (minus sign ka matlab "heat added, removed nahi"). Sensible heating $c_{p} (T_{w} - T_{i})$ hai jahan $c_{p} = 1400 J/(kg\cdotK)$ , $T_{w} = 2600 K$ , $T_{i} = 300 K$ . Is budget ke liye blowing ignore karo. Kya effective heat of ablation phir bhi positive hai, aur ek negative $Q^{*}$ term physically kya matlab rakhta hai?

Forecast: heat-releasing reaction mix mein hone ke saath, kya $Q^{*}$ pure-endothermic case se bada aayega ya chota — aur kya yeh kabhi negative ho sakta hai?

Sensible heating sink (hamesha ek genuine sink). $c_{p} Δ T = 1400 \times (2600 - 300) = 1400 \times 2300 = 3.22 \times 1 0^{6} J/kg$ . Yeh step kyun? Material ko $T_{i}$ se $T_{w}$ tak warm hona hi padega, chahe chemistry kuch bhi kare — yeh term kabhi sign nahi badlegi.
Saare reaction enthalpies unke signs ke saath add karo. $Q^{*} = c_{p} Δ T + Q_{p} + Q_{o x}$ . Yeh step kyun? $Q^{*}$ joules-per-kg ka ek signed sum hai: endothermic terms positive hain (wo incoming heat remove karte hain), exothermic wale negative hain (wo extra heat inject karte hain), toh honest budget signs rakhta hai. $Q^{*} = 3.22 \times 1 0^{6} + 0.8 \times 1 0^{6} + (- 1.5 \times 1 0^{6}) = 2.52 \times 1 0^{6} J/kg$
Surviving sign interpret karo. $Q^{*} = + 2.52 \times 1 0^{6} J/kg$ phir bhi positive hai lekin ek purely endothermic ablator se kaafi chota — oxidation ne $1.5 \times 1 0^{6} J/kg$ protection chura li. Agar oxidation term dono sinks ka combined amount kabhi exceed kar le, $Q^{*}$ negative ho jaata — matlab reaction wall mein zyada heat pump karta hai jitna mass loss carry away karta hai — ek runaway jahan ablating faster cheezein aur buri kar deta hai. Yeh step kyun? $Q^{*}$ ka sign go/no-go flag hai: positive = protective, negative = self-heating catastrophe.

Verify: Units sab $J/kg$ ✔. Sanity: exothermic term hata do ( $Q_{o x} = 0$ ) aur tum $3.22 + 0.8 = 4.02 \times 1 0^{6} J/kg$ recover karte ho, jo hamare $2.52 \times 1 0^{6}$ se bada hai — confirming karta hai ki exothermic reaction ne figure of merit bilkul forecast ki tarah ghata diya. Isi liye Boundary Layer Theory mein char surface tak oxygen transport ek design worry hai, bonus nahi.

Upar ki figure pura blowing curve $ln (1 + B^{'}) / B^{'}$ plot karti hai jisme cell C2, parent ka $B^{'} = 2$ , C9 ka $B^{'} = 4$ , aur C3 mark hain. Left-to-right padho: $1$ se shuru hota hai (no blowing, C4b), dheere sag karta hai, phir $0$ ki taraf flat hota jaata hai — C9 aur blow-off C3 ke peeche "diminishing returns" ki geometric picture.

Recall

Recall Answers cover karo

C1 mein, pehle $\overset{q}{˙}$ ko $Q^{*}$ se kyun divide karte hain? ::: kyunki $Q^{*}$ joules-per-kg hai, toh flux÷ $Q^{*}$ mass-loss rate per area deta hai.
$lim_{B^{'} \to 0} ln (1 + B^{'}) / B^{'}$ kya hai aur yeh kyun matter karta hai? ::: yeh $1$ ke equal hai; yeh formula ko ek fake $0/0$ se bachata hai aur matlab hai "no blowing ⇒ full bare-wall heating".
$Q^{*}$ mein teen energy sinks ko sum kyun karna padta hai? ::: pyrolysis, sensible heating, aur blocking har ek independently joules remove karta hai.
C7 mein enthalpy flux ne temperature flux ko kyun beat kiya? ::: dissociation chemical energy store karta hai jise temperature akela ignore karta hai, toh $T$ load underestimate karta hai.
Kya $B^{'}$ ko 2 se 4 double karne par heating half ho jaati hai? ::: nahi — $lo g$ saturate ho jaata hai; yeh sirf $0.549$ se $0.402$ tak aata hai.
Ek negative $Q^{*}$ term (C10) ka kya matlab hai? ::: ek exothermic reaction wall mein heat add karta hai; agar yeh sinks se zyada ho jaaye, $Q^{*}$ negative ho jaata hai aur ablation self-heating runaway ban jaata hai.