Ablative cooling — charring, blowing: Question bank

3.3.30 · D5 · Physics › Rocket Propulsion › Ablative cooling — charring, blowing

Is page ke symbols — test se pehle inhe samjho

Traps se pehle, har symbol ko earn karte hain. Yeh sab parent note se aate hain; yahan inhe seedha likha hai taaki neeche koi line surprise na kare.

Definition Gatekeeper symbols

$\overset{m}{˙}^{''}$ — mass-injection flux: pyrolysis gas ke kilograms jo ek square metre wall se har second bahar push hote hain. Double prime $^{''}$ ek bookkeeping mark hai jiska matlab hai "per unit area"; ek dot ka matlab hai "per unit time". Toh $\overset{m}{˙}^{''}$ padhte hain mass, per second, per square metre aur iske units hain $kg/m^{2} s$ .
$h_{0}$ — un-blown convective heat-transfer coefficient: hot gas wall mein heat kitni strongly dump karta — agar koi gas inject nahi ho rahi — $\overset{q}{˙}_{0} = h_{0} (T_{g a s} - T_{w a l l})$ se define hota hai. Bada $h_{0}$ = zyada fierce heating. Units: $W/m^{2} K$ . Subscript $0$ ek reminder hai: "blowing se pehle heating abhi neeche nahi aayi."
$h$ — actual (blown) convective heat-transfer coefficient: $h_{0}$ jaisi hi quantity lekin gas inject hote waqt measure ki gayi, $\overset{q}{˙} = h (T_{g a s} - T_{w a l l})$ se define hoti hai. Kyunki blowing hot gas ko door dhakelta hai, $h$ $h_{0}$ se chhoti nikalti hai; ratio $h / h_{0}$ (hamesha $\leq 1$ ) exactly wahi hai jo heating bacha hai.
$H$ — gas ki specific enthalpy: har kilogram mein store total energy, including dissociated chemical bonds mein chhipi energy, $J/kg$ mein. Hot free-stream gas ke liye $H_{g}$ likhte hain aur wall ke bilkul paas gas ke liye $H_{w}$ . $H$ isliye use karte hain na ki temperature, kyunki re-entry mein heat broken bonds mein chhip sakti hai bina $T$ badhaye.
$c_{p}$ — specific heat at constant pressure: ek kilogram substance ko ek kelvin warm karne ke liye joules chahiye, $J/kg\cdotK$ mein. Yeh temperature change aur energy ke beech ka bridge hai: heat $= c_{p} \times Δ T$ per kilogram.

Definition Do "heat per kilogram" symbols jinpe traps tikhe hain

$Q_{p}$ — pyrolysis enthalpy: ek kilogram virgin resin ko gas plus char mein decompose karne ke liye absorbed energy, $J/kg$ mein. Kyunki reaction endothermic hai (woh heat peeta hai), $Q_{p}$ ek genuine cooling channel hai — woh swallowed energy kabhi structure tak nahi pahunchti. Socho "joules per kilogram eaten, per kilogram broken apart."
$Q^{*}$ — effective heat of ablation (master figure of merit): har kilogram material lost ke liye reject ki gayi total heat, $Q^{*} \equiv \overset{q}{˙}_{n e t} / \overset{m}{˙}_{t o t a l}^{''}$ , $J/kg$ mein. Yeh teeno cooling channels ko bundle karta hai — pyrolysis $Q_{p}$ , sensible heating $c_{p} (T_{w} - T_{i})$ , aur blowing/blocking term. Bada $Q^{*}$ = handled joule per little mass sacrificed = ek achha ablator (carbon-phenolic $\sim 1 0^{7} J/kg$ tak pahunchta hai).

Ab do derived quantities jinpe neeche sab kuch tikha hai:

Formula Blocking factor

$ln (1 + B^{'}) / B^{'}$ — log kahan se aaya KYA: blowing ke saath, effective coefficient $h_{0}$ se $h$ tak girti hai jahan $\frac{h}{h _{0}} = \frac{l n ( 1 + B ^{'} )}{B ^{'}} .$ KYUN log aur straight line nahi: near-wall layer ko model karo gas ki tarah jo wall ke across flow kar rahi hai jabki ek chhoti wall-normal blowing velocity $v_{w}$ (outward speed, $m/s$ mein, jis par pyrolysis gas surface se bahar jaati hai) baahir dhakelta hai. Maano $y$ wall se seedha bahar maapi gayi distance hai ( $m$ mein), aur $α$ gas ki thermal diffusivity hai ( $m^{2} /s$ mein — heat diffusion se kitni tezi se failti hai). Wall-normal energy balance ek convection–diffusion equation hai jiska enthalpy profile ka solution exponential shape mein hai, $\propto e^{(v_{w} y) / α}$ , kyunki convection (heat bahar le jaana speed $v_{w}$ par) aur diffusion (use rate $α$ par failana) ek exponential standoff tak ladte hain. Jo heat actually wall tak pahunchti hai woh us exponential ke wall par slope ( $y = 0$ ) ke proportional hai, aur us exponential ko invert karke woh slope solve karna exactly wahi hai jo ek logarithm nikalta hai — $ln (1 + B^{'})$ . $B^{'}$ se divide karke yeh ek fractional reduction ban jata hai. Figure s01 dekho: curve $1$ se start hoti hai (no blowing, no help) aur bend ho kar $0$ ki taraf saturate karti hai — woh flattening hi poora reason hai ki "zyada blowing" diminishing returns deta hai.

Figure — Ablative cooling — charring, blowing

True or false — justify karo

Recall Answers cover karo — reveal karne se pehle apna reason bolo

Ek ablative material bina mass khoye survive karne ke liye design kiya jaata hai. ::: False — yeh jaan-boojhkar mass khoone ke liye design kiya jaata hai; har jaata hua gram energy le jaata hai. Survival ka matlab hai neeche ki structure thandi rehna, ablator ka intact rehna nahi. Ablative cooling kaam karne ke liye pumps aur coolant lines chahiye. ::: False — yeh ek passive method hai jisme koi moving parts nahi hain; compare karo Regenerative Cooling se, jo propellant ko wall channels se pump karta hai. Ek achhe ablator ki high effective heat of ablation $Q^{*}$ hoti hai. ::: True — $Q^{*}$ joules rejected per kilogram lost hai, toh high $Q^{*}$ ka matlab hai handled joule per little mass sacrificed — tum slowly erode hote ho. Charring melting jaisi ek phase change hai. ::: False — charring endothermic chemical decomposition (pyrolysis) hai: bonds toot jaate hain aur resin porous carbon plus gas ban jaata hai. Koi liquid phase zaroori nahi hai. Char layer ek damage hai jise minimize karna chahiye. ::: False — char star hai: ek low-conductivity insulator, ek radiator, aur ek reaction barrier. Tum chahte ho ki woh moti aur attached rahe. Blowing tabhi cool karti hai jab gas wall ko touch kar rahi hoti hai. ::: False — jaane ke baad, gas boundary layer ko thicken karti hai, wall ke paas hot gas ko cool gas se replace karti hai, jo wall temperature gradient ko soften karta hai aur incoming flux ko kaatta rehta hai. Pyrolysis ka endothermic hona hi use cooling ke liye useful banata hai. ::: True — endothermic ka matlab hai yeh proceed karne ke liye heat absorb karta hai; woh absorbed energy $Q_{p}$ incoming flux se churi jaati hai, toh structure tak kam pahunchti hai. Blowing rate double karne se cooling benefit roughly double ho jaati hai. ::: False — blocking factor $ln (1 + B^{'}) / B^{'}$ (jahan $B^{'} = \overset{m}{˙}^{''} c_{p} / h_{0}$ ) Figure s01 ki tarah saturate karta hai; ek point ke baad extra injection almost koi heat reduction nahi kharidta aur tum sirf tezi se erode hote ho.

Error dhundho

Recall Har statement mein ek flaw chhupa hai — use name karo

"Heat mainly radiation ke zariye wall tak aati hai, toh hum insulator use karte hain." ::: Dominant arrival mode convection hai hot boundary-layer gas se, $\overset{q}{˙}_{co n v} = h (T_{g a s} - T_{w a l l})$ ; ablators isse fight karte hain, haalaanki Radiative Cooling re-radiation part handle karta hai. "Accuracy ke liye re-entry temperatures par bhi $\overset{q}{˙} = h (T_{g} - T_{w})$ use karo." ::: High temperature par gas dissociate aur recombine hoti hai, isliye true driving potential specific-enthalpy difference $H_{g} - H_{w}$ (J/kg) hai; akela temperature load ko underestimate karta hai — yeh ek Re-entry Aerothermodynamics subtlety hai. "Blowing parameter $B^{'} = h_{0} / (\overset{m}{˙}^{''} c_{p})$ hai." ::: Inverted hai — yeh $B^{'} = \overset{m}{˙}^{''} c_{p} / h_{0}$ hai: outgoing-gas heat capacity over un-blown coefficient $h_{0}$ . Zyada injected mass flux $\overset{m}{˙}^{''}$ → bada $B^{'}$ . " $B^{'} \to 0$ hone par, blowing saari heat shut off kar deti hai." ::: Ulta hai — $B^{'} \to 0$ ka matlab hai koi injection nahi, toh $ln (1 + B^{'}) / B^{'} \to 1$ aur $h \to h_{0}$ (full heating). Bada $B^{'}$ heating shut off karta hai. " $Q^{*}$ sirf pyrolysis enthalpy $Q_{p}$ count karta hai." ::: Yeh sensible heating $c_{p} (T_{w} - T_{i})$ aur blowing/blocking term bhi include karta hai; typical carbon-phenolic mein blocking term dominate kar sakta hai. "Higher virgin density $ρ_{v}$ ka matlab hamesha front tezi se recede karta hai." ::: Ulta: $\overset{q}{˙}_{p y r o} = ρ_{v} \frac{d s}{d t} Q_{p}$ , toh fixed heat flux ke liye higher $ρ_{v}$ $d s / d t$ ko smaller banata hai — front slower recede karta hai. "Carbon-phenolic apni surface par melt hokar survive karta hai." ::: Yeh melt nahi karta; yeh char hota hai phir sublime/react karta hai. Melting is class of ablator ke liye wrong mental model hai.

Why questions

Recall "Kyun" ka jawab ek ya do sentences mein do

Hum un-blown flux ko temperatures $T$ ki jagah specific enthalpies $H$ (J/kg) mein kyun likhte hain? ::: Kyunki dissociation aur recombination energy per kilogram store karte hain bina temperature change kiye, toh enthalpy general driving potential hai; $T$ tabhi safe proxy hai jab gas chemically frozen ho. Char layer ki low thermal conductivity kyun zyada matter karti hai uske carbon hone se? ::: Low conductivity ise ek growing insulating blanket banati hai jo cold structure tak conduction throttle karti hai; dekho ki conduction gradient par kaise depend karta hai Boundary Layer Theory mein. $ln (1 + B^{'}) / B^{'}$ ek logarithm kyun hai aur linear kyun nahi? ::: Kyunki near-wall enthalpy profile blowing ke under exponential solve hoti hai, aur us exponential se wall slope read karna log mein invert hota hai — yahi factor ko saturate karta hai (Figure s01); zyada detail mein Convective Heat Transfer (Stanton number) mein. Blowing effective heat-transfer coefficient $h$ ko kyun reduce karta hai? ::: Injected gas $\overset{m}{˙}^{''}$ hot gas ko door dhakelta hai aur thermal boundary layer ko thicken karta hai, toh near-wall temperature gradient gentler hoti hai aur $h$ (jo us gradient ke saath scale karta hai) $h_{0}$ se neeche gir jaata hai. Ablative cooling ko "passive" kyun kehte hain agar itni chemistry ho rahi hai? ::: "Passive" ka matlab hai koi external machinery nahi — koi pumps, valves, ya coolant loops nahi; response khud incoming heat se self-driven hota hai. Bada endothermic $Q_{p}$ ablator ko kyun better banata hai? ::: Har kilogram pyrolyzed per zyada energy absorbed ka matlab hai front same heat load ke liye slower recede karta hai, toh liner zyada der tak chalti hai — yeh directly Heat of Reaction and Pyrolysis se jud ta hai. Effective heat of ablation $Q^{*}$ ek "figure of merit" kyun hai na ki fixed material constant? ::: Kyunki yeh pyrolysis, sensible heating, aur blowing/blocking term ko bundle karta hai, aur blocking part flow conditions ( $h_{0}$ , $\overset{m}{˙}^{''}$ , enthalpies) par depend karta hai — toh $Q^{*}$ environment ke saath shift karta hai, na ki sirf material ke saath.

Edge cases

Recall Limits, zeros, aur degenerate scenarios

Zero injection hone par blocking factor ka kya hoga, $\overset{m}{˙}^{''} = 0$ ? ::: Tab $B^{'} = 0$ aur $ln (1 + B^{'}) / B^{'} \to 1$ , toh $h = h_{0}$ : bina blowing ke koi blocking nahi, aur tumhe full un-blown convective flux milta hai. $B^{'} \to \infty$ hone par kya hoga (enormous injection)? ::: $ln (1 + B^{'}) / B^{'} \to 0$ , toh convective heating essentially shut off ho jaati hai — "blow-off" limit — lekin tum extremely fast erode ho rahe ho, toh yeh free win nahi hai. Agar char mechanically waqt se pehle spall off ho jaaye? ::: Tum premature insulating, radiating, reaction-blocking layer kho dete ho; virgin material suddenly full flux ke saamne expose ho jaata hai aur recession accelerate ho jaata hai — yeh classic failure mode hai. Agar $Q_{p} = 0$ ho (ek material jo bina energy absorb kiye decompose karta hai)? ::: Pyrolysis energy-sink lever gayab ho jaata hai, $\overset{q}{˙}_{p y r o} = 0$ , aur $Q^{*}$ apna pehla term kho deta hai; cooling phir poori tarah sensible heating aur blowing par lean karti hai, ablator ko bahut kamzor bana deti hai. Agar gas chemically frozen ho (koi dissociation nahi) — kya tab temperature use kar sakte hain? ::: Haan; jab koi dissociation/recombination nahi hoti, enthalpy aur temperature ek saath track karte hain, toh $\overset{q}{˙} = h (T_{g} - T_{w})$ acceptable hai — enthalpy form bas sab cases mein safe rehta hai. Burn ki bilkul shuruaat mein kya hoga, jab abhi koi char nahi bana? ::: Surface abhi bhi virgin hai, koi insulating blanket nahi aur abhi koi blowing nahi ( $\overset{m}{˙}^{''}$ ~ 0), toh material par initial heat flux sabse zyada hai; char aur gas cushion phir build up hote hain aur progressively load kaatte hain. Agar do ablators same net heat reject karte hain lekin ek bahut kam mass khoye, kaun better hai aur kyun? ::: Jo kam mass khoya — uska $Q^{*}$ zyada hai (zyada joules per kilogram), toh ek patla, halka liner same mission survive kar leta hai.