Exercises — Ablative cooling — charring, blowing

Kya: Hum definition choose karte hain. Answer: (b) ablative cooling. Kyun: "Passive" matlab material khud kaam karta hai — woh decompose hota hai aur apna gas khud blow karta hai. Regenerative cooling (Regenerative Cooling) cold propellant ko channels se pump karta hai; uske liye plumbing aur pressure chahiye — active hai. Isliye (a) nahi. Koi bhi pumped film cooling bhi active hai, isliye (c) bhi nahi.

Kya: Cold structure se bahar hot gas ki taraf path trace karo.

• Virgin material → (ii) abhi bhi thanda aur unreacted.
• Pyrolysis zone → (iii) resin decompose hoti hai, gas release hoti hai.
• Char layer → (i) porous carbon insulator.
• Boundary layer → (iv) gas cushion. Yeh order kyun: heat bahar se aati hai, isliye sabse hot, sabse zyada transformed zone (char, phir boundary layer) upar hai; thanda intact resin sabse neeche hai.

Kyunki bonds todna, vaporizing karna aur gas blow karna har ek woh energy consume karta hai jo otherwise structure tak pahunchti, aur nikalta gas physically hot gas ko surface ko touch karne se rokta hai. Heat structure ko garam karne ki jagah surface ko destroy karne mein spend hoti hai.

Kya: $Q^{\ast }={\dot{q}}_{net}/{\dot{m}}^{\prime \prime }$ use karo, isliye ${\dot{m}}^{\prime \prime }={\dot{q}}_{net}/Q^{\ast }$. Kyun: $Q^{\ast }$ hai "joules rejected per kg lost". Joules-per-second-per-area ko joules-per-kg se divide karne par kg-per-second-per-area milta hai — exactly ${\dot{m}}^{\prime \prime }$. Units force karte hain: $\frac{{\text{W/m}}^2}{\text{J/kg}}=\frac{{\text{J/s⋅m}}^2}{\text{J/kg}}={\text{kg/m}}^2\text{s}$. ✓ ${\dot{m}}^{\prime \prime }=\frac{4\times 10^6}{1.0\times 10^7}=0.40{\text{kg/m}}^2\text{s}$ Matlab: har second, surface ka har square metre 0.4 kg material us heat ko handle karne ke liye kho deta hai.

Kya (step 1): mass loss ko receding depth mein convert karo $\dot{s}={\dot{m}}^{\prime \prime }/{\rho }_v$ se. Kyun: mass per area per second ÷ (mass per volume) = volume per area per second = length per second — woh speed jis par surface andar ki taraf eat karta hai. $\dot{s}=\frac{0.40}{1600}=2.5\times 10^{-4}\text{m/s}=0.25\text{mm/s}$ Kya (step 2): time se multiply karo (steady flux ⇒ constant speed). $s=\dot{s}\cdot t=2.5\times 10^{-4}\times 80=0.020\text{m}=20\text{mm}$ Matlab: tumhe margin ke saath is burn se bachne ke liye 2 cm se mota liner chahiye.

Kya: $B^{\prime }=\frac{{\dot{m}}^{\prime \prime }{c}_p}{h_0}$ mein plug karo. Yeh grouping kyun: ${\dot{m}}^{\prime \prime }{c}_p$ ki units hain ${\text{W/m}}^2\text{K}$ (ek "blowing conductance"); $h_0$ se divide karne par (same units) ek pure number milta hai — hum kitna blow kar rahe hain un se compare karke jitna gas heat andar leak karta hai. $B^{\prime }=\frac{0.30\times 1200}{900}=\frac{360}{900}=0.40$ Matlab: $B^{\prime }=0.4$ moderate blowing hai — matter karne ke liye enough lekin blow-off se kaafi dur.

Kya: $\frac{h}{h_0}=\frac{\ln (1+B^{\prime })}{B^{\prime }}$ use karo. Yeh formula kyun, koi simpler cheez kyun nahi: blowing heat ko linearly subtract nahi karta — injected gas boundary layer ko mota karta hai (Boundary Layer Theory), aur wall injection ke saath 1-D energy balance solve karne par exactly logarithm milta hai. Wahi log ki wajah se cooling saturate hoti hai. $\frac{h}{h_0}=\frac{\ln (1.4)}{0.4}=\frac{0.33647}{0.4}=0.8412$ Matlab: heating un-blown value ke lagbhag 84% tak reduce ho jaati hai, matlab roughly $16\%$ ki cut. Modest — kyunki $B^{\prime }$ chota hai.

Kya: dono points par factor evaluate karo. $B^{\prime }=0.4:\frac{\ln 1.4}{0.4}=0.841B^{\prime }=4:\frac{\ln 5}{4}=\frac{1.6094}{4}=0.402$ $B^{\prime }=0.4$ se $B^{\prime }=4$ jaane par (blowing mein $10\times$ increase) $h/h_0$ sirf $0.84$ se $0.40$ tak drop karta hai — roughly half ho jaata hai, tenfold nahi. Kyun: figure mein curve $\ln (1+B^{\prime })/B^{\prime }$ hai, jo $B^{\prime }$ badhne par flatten hoti hai (red curve zero ki taraf dheere jhukti hai). Physically, jab boundary layer already moti ho, extra gas ke paas push karne ke liye kam hot gas bachti hai — diminishing returns. Tum kam extra protection ke liye zyada tezi se erode karte ho.

(a) $B^{\prime }\to 0$: chote $B^{\prime }$ ke liye $\ln (1+B^{\prime })\approx B^{\prime }-\frac{B^{\prime 2}}{2}$ use karo. Phir $\frac{\ln (1+B^{\prime })}{B^{\prime }}\approx \frac{B^{\prime }-B^{\prime 2}/2}{B^{\prime }}=1-\frac{B^{\prime }}{2}\to 1.$ Isliye $h\to h_0$: koi blowing nahi ⇒ koi reduction nahi. Sanity check pass. (b) $B^{\prime }\to \infty$: numerator $\ln (1+B^{\prime })$ $\ln B^{\prime }$ ki tarah grow karta hai (slow), denominator $B^{\prime }$ ki tarah (fast), isliye ratio $\to 0$. Heating band ho jaati hai — yeh blow-off limit hai. Matlab: formula dono ends par sahi behave karta hai, yahi wajah hai ki hum beech mein ise trust karte hain.

"Blocking term" kya hai? Yeh parent note se master formula ka teesra slot hai, $Q^{\ast }=\underset{\text{pyrolysis}}{\underbrace{Q_p}}+\underset{\text{sensible}}{\underbrace{c_p(T_w-T_i)}}+\underset{\text{blocking by blowing}}{\underbrace{\eta \frac{h_0}{{\dot{m}}^{\prime \prime }}(H_{gas}-H_w)}}.$ Blocking term $\eta \frac{h_0}{{\dot{m}}^{\prime \prime }}(H_{gas}-H_w)$ woh un-blown convective load $h_0(H_{gas}-H_w)$ hai jo blowing aane se rokti hai, mass rate ${\dot{m}}^{\prime \prime }$ se divide kiya jo blocking kiya — yaani joules blocked per kilogram lost — ek efficiency factor $\eta$ ke saath. Is problem mein woh poora group pehle se hmare liye evaluate ho gaya hai $5.5\times 10^6\text{J/kg}$ ke roop mein; hum ise given number maan lete hain. Hum teen ko kyun add karte hain: har term energy per kg measure karta hai jo material alag mechanism se handle karta hai (bonds todna, garam hona, gas blow karna). Woh overlap nahi karte, isliye simply add ho jaate hain.

• Sensible: $c_p(T_w-T_i)=1400\times (3000-300)=1400\times 2700=3.78\times 10^6\text{J/kg}$.
• Pyrolysis: $2.5\times 10^6\text{J/kg}$.
• Blocking: $5.5\times 10^6\text{J/kg}$. $Q^{\ast }=(2.5+3.78+5.5)\times 10^6=1.178\times 10^7\text{J/kg}.$ Dominant channel: blocking/blowing term ($5.5\times 10^6$) sabse bada hai — yeh confirm karta hai ki blowing ek main cooling channel hai, side effect nahi.

Kya (step 1): ${\dot{m}}^{\prime \prime }={\dot{q}}_{net}/Q^{\ast }$. ${\dot{m}}^{\prime \prime }=\frac{6\times 10^6}{1.178\times 10^7}=0.5093{\text{kg/m}}^2\text{s}$ Kya (step 2): burn mein total mass per area = ${\dot{m}}^{\prime \prime }\times t$. $m^{\prime \prime }=0.5093\times 120=61.1{\text{kg/m}}^2$ Kya (step 3): recession depth $s={\dot{m}}^{\prime \prime }t/{\rho }_v=m^{\prime \prime }/{\rho }_v$. $s=\frac{61.1}{1450}=0.04215\text{m}=42.2\text{mm}$ Matlab: surface lagbhag $42\text{mm}$ recede karti hai, isliye ek $30\text{mm}$ liner survive nahi karta — woh poora burn through ho jaata hai. Tumhe mota material ya zyada high-$Q^{\ast }$ ablator chahiye.

Kya: har ek ke liye, ${\dot{m}}^{\prime \prime }=\dot{q}/Q^{\ast }$, phir depth $s={\dot{m}}^{\prime \prime }t/{\rho }_v$, aur consumed mass $m^{\prime \prime }={\dot{m}}^{\prime \prime }t$.

Material A: ${\dot{m}}_A^{\prime \prime }=\frac{8\times 10^6}{1.2\times 10^7}=0.6667{\text{kg/m}}^2\text{s}$ $m_A^{\prime \prime }=0.6667\times 90=60.0{\text{kg/m}}^2,s_A=\frac{60.0}{1450}=0.04138\text{m}=41.4\text{mm}$

Material B: ${\dot{m}}_B^{\prime \prime }=\frac{8\times 10^6}{0.7\times 10^7}=1.1429{\text{kg/m}}^2\text{s}$ $m_B^{\prime \prime }=1.1429\times 90=102.9{\text{kg/m}}^2,s_B=\frac{102.9}{550}=0.1870\text{m}=187\text{mm}$

Depth winner: Material A sirf $41\text{mm}$ recede karta hai vs B ka $187\text{mm}$ — A thickness mein hugely jeetta hai (zyada $Q^{\ast }$ + zyada density). Consumed-mass winner: A $60{\text{kg/m}}^2$ khoता hai vs B ka $103{\text{kg/m}}^2$ — A bhi kam mass khoता hai, kyunki uska $Q^{\ast }$ bada hai (zyada joules per kg). Trade-off: A dono raw metrics mein jeetta hai, LEKIN B kaafi kam dense hai ($550$ vs $1450$). Agar mission mass-limited hai, toh ek mota B liner given standoff ke liye phir bhi per unit area kam weigh kar sakta hai, ya B sasta/shape karna aasaan ho sakta hai. Real selection $Q^{\ast }$, density, char stability (Re-entry Aerothermodynamics loads), aur manufacturability ko weigh karta hai — sirf depth ko nahi.

Kya / loop (Figure s02 ko clockwise follow karo):

1. Zyada heat aati hai → zyada pyrolysis → bada ${\dot{m}}^{\prime \prime }$.
2. Bada ${\dot{m}}^{\prime \prime }$ → bada blowing parameter $B^{\prime }={\dot{m}}^{\prime \prime }{c}_p/h_0$.
3. Bada $B^{\prime }$ → chota $h/h_0=\ln (1+B^{\prime })/B^{\prime }$ → kam heat aati hai (red arrow loop close karta hai). Isliye step 3 step 1 se ladta hai — ek negative feedback jo surface ko stabilise karta hai: ablator exactly usi waqt zyada blow karta hai jab zyada hit hota hai. Enthalpy $H$ kyun, temperature $T$ kyun nahi: re-entry temperatures par gas dissociate aur recombine karta hai (Convective Heat Transfer (Stanton number)). Heat transfer ke liye true driving potential enthalpy gap $H_{gas}-H_{wall}$ hai, kyunki recombination ki chemical energy bhi wall par deposit hoti hai — sirf $T$ use karne se woh chemical energy miss ho jaati hai aur isliye true heat load underestimate hota hai. Enthalpy thermal aur chemical energy ko ek number mein bundle karta hai, isliye ${\dot{q}}_0=h_0(H_{gas}-H_{wall})$ honest baseline hai. Matlab: ablator ek self-regulating system hai — koi sensors nahi, koi valves nahi, koi control loop nahi. Blowing ki physics automatically cap kart hai ki structure tak kitni heat pahunch sakti hai, aur yahi wajah hai ki passive ablation re-entry aur rocket nozzles ke liye itna robust hai.

• Heat load aur $Q^{\ast }$ se ${\dot{m}}^{\prime \prime }$? ::: ${\dot{m}}^{\prime \prime }={\dot{q}}_{net}/Q^{\ast }$.
• ${\dot{m}}^{\prime \prime }$ se recession speed? ::: $\dot{s}={\dot{m}}^{\prime \prime }/{\rho }_v$.
• Blowing parameter, aur $c_p,h_0$ ka matlab? ::: $B^{\prime }={\dot{m}}^{\prime \prime }{c}_p/h_0$; $c_p$ injected gas ki specific heat hai, $h_0$ un-blown transfer coefficient.
• Blowing reduction factor aur uski shape? ::: $\ln (1+B^{\prime })/B^{\prime }$; yeh saturate hoti hai (diminishing returns).
• $Q^{\ast }$ ke teen terms? ::: pyrolysis $Q_p$, sensible $c_p(T_w-T_i)$, blowing/blocking $\eta (h_0/{\dot{m}}^{\prime \prime })(H_{gas}-H_w)$.
• High $T$ par enthalpy $H$ kyun, temperature $T$ kyun nahi? ::: dissociation/recombination chemical energy add karta hai jo driving potential mein include honi chahiye.
• Self-limiting loop? ::: zyada heat → zyada ${\dot{m}}^{\prime \prime }$ → zyada $B^{\prime }$ → kam $h$ → kam heat.