Visual walkthrough — Film cooling — effectiveness, coverage fraction

3.3.29 · D2 · Physics › Rocket Propulsion › Film cooling — effectiveness, coverage fraction

Shuru karne se pehle poora safar ek map ki tarah dekho:

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Step 1 — Scene draw karo: ek hot wall aur ek patli cool film

KYA HAI. Rocket combustion chamber ki andar wali wall ko ek flat floor ki tarah socho. Uske kaafi upar hot core gas behta hai — jalte hue propellant ka flame temperature wala gas. Bilkul floor ke saath, ek slot se inject hoke, ek patli ribbon mein cool gas (yaani film) behta hai. Gas left → right jaata hai; wall ke saath ki distance ko $x$ kehte hain.

KYU. Koi bhi algebra karne se pehle hum yeh agree kar lete hain ki har symbol kaunsi physical cheez ko naam deta hai. Is picture mein teen temperatures hain, aur baaki sab unhi se banta hai:

• $T_{hot}$ — woh temperature jo wall ko bina film ke raw hot gas se milti. (Precisely, yeh core stream ka recovery temperature hai.)
• $T_{cool}$ — fresh film gas ka temperature bilkul wahan jahan woh slot se nikalta hai.
• $T_{aw}$ — woh temperature jo wall ko actually milti hai: adiabatic wall temperature, matlab "wall koi heat conduct karke nahi le jaati, toh woh bas us temperature par float karti hai jo uske paas wala gas impose karta hai."

PICTURE.

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Step 2 — Ek important quantity naam do: coolness gap $\Delta T$

KYA HAI. Teen temperatures track karne ki jagah hum ek single number track karte hain: wall hot gas se kitni zyada thandi hai. Isse coolness gap bolte hain:

$\underset{\text{coolness gap}}{\underbrace{\Delta T}}\equiv \underset{\text{hot gas temp}}{\underbrace{T_{hot}}}-\underset{\text{wall temp now}}{\underbrace{T_{aw}}}$

• $\Delta T$ bada → wall aag se bahut zyada thandi hai → film kaam kar rahi hai.
• $\Delta T=0$ → wall aag jitni hi hot hai → film khatam.

KYU. Film ka poora point wall aur aag ke beech ka difference hai. Difference ek number hota hai, do ki jagah ek cheez track karna aasaan hai. Aur — sabse zaroori baat — mixing ki physics is difference par depend karti hai, jaise hum aage dekhenge.

PICTURE. Neeche wali figure mein vertical green arrow hi $\Delta T$ hai. Dekho kaise woh $x$ badhne par chhotata jaata hai.

Slot par ($x=0$) film bilkul fresh coolant hai, toh wall $T_{cool}$ par hoti hai aur gap sabse bada hota hai jitna kabhi hoga:

$\Delta T_0\equiv T_{hot}-T_{cool}(\text{gap at }x=0\text{ par}).$

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Step 3 — Gap kyun chhotata hai: hot gas film mein entrain hota hai

KYA HAI. Jaise film downstream behti hai, film aur hot gas ke beech ki turbulent boundary unchi-neech hoti hai — yeh hot gas ko cool ribbon mein khainch leti hai (entrain karti hai). (Yeh entrainment process hai.) Har hot gas ka parcel jo mix hota hai woh film ko warm karta hai, jo wall ko warm karta hai, jo gap $\Delta T$ ko chhotata hai.

KYU. Humein ek rate rule chahiye: gap kitni tezi se chhotata hai? Do competing effects hain:

• Driving force gap khud hai: jitna bada temperature difference, utni tezi se heat film mein jaati hai. Yeh bilkul convective heat transfer ka idea hai — heat flux $\propto h_g\Delta T$, jahan $h_g$ gas-side heat transfer coefficient hai (units ${\text{W/m}}^2\text{K}$: watts of heat per square metre per degree of gap).
• Resistance film ki heat capacity hai: ek heavy, high-specific-heat film entrained heat ko zyada temperature rise ke bina sokhti hai. Yeh capacity per unit wall width ${\dot{m}}_c{c}_p$ hai, jahan ${\dot{m}}_c$ coolant mass flow per unit width ($\text{kg/(m⋅s)}$) hai aur $c_p$ uski specific heat ($\text{J/kgK}$).

PICTURE. Red arrows hot gas ko blue film mein khenchte hue dikhate hain; film phaiki (warm) hoti jaati hai jaise zyada arrows girte hain.

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Step 4 — Rate rule ko equation mein likho

KYA HAI. Step 3 ki picture ko symbols mein badlo. Ek chhotesi distance $dx$ par, gap $d(\Delta T)$ se change hoti hai:

$\underset{\begin{array}{c}\text{gap kitni tezi se}\\ \text{per metre change hota hai}\end{array}}{\underbrace{\frac{d(\Delta T)}{dx}}}=-\underset{\begin{array}{c}\text{heating strength}\\ ÷\text{ film capacity}\end{array}}{\underbrace{\frac{h_g}{{\dot{m}}_c{c}_p}}}\underset{\begin{array}{c}\text{abhi ka}\\ \text{gap}\end{array}}{\underbrace{\Delta T}}$

Har piece kyun.

• Minus sign ($-$): gap sirf chhotata hi hai — hot gas kabhi film ko thanda nahi kar sakta. Sign check pass.
• Right side par $\Delta T$: driving force hi gap hai. Bada gap → tezi se chhotao; chhota gap → dheere chhotao. Jab $\Delta T=0$ toh rate $0$ hai aur wahan hi rehta hai — film un-mix nahi ho sakti. Sahi hai.
• $h_g/({\dot{m}}_c{c}_p)$: fixed "decay strength." Zyada $h_g$ (fierce hot gas) → tezi se decay. Zyada ${\dot{m}}_c{c}_p$ (mota, thanda film) → dheere decay. Dono intuition se match karte hain.

Ek differential equation kyun, sirf algebra kyun nahi? Kyunki shrink rate current gap par depend karta hai, jo khud bhi badal raha hai — chicken-and-egg loop. Algebra fixed numbers ke liye kaam karta hai; differential equation precisely usi cheez ke liye bana hua tool hai jab "rate abhi kitni cheez hai uspar depend karta hai." Woh signature ("rate $\propto$ amount") exponential decay ki pehchaan hai.

PICTURE. Green curve ka slope har point par $-\frac{h_g}{{\dot{m}}_c{c}_p}$ times uski height ke barabar hai.

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Step 5 — ODE solve karo: exponential saamne aata hai

KYA HAI. Variables separate karo (gap-wali cheez left par, distance-wali cheez right par) aur slot se ($x=0$, gap $\Delta T_0$) kisi bhi station $x$ tak (gap $\Delta T$) integrate karo:

${\int }_{\Delta T_0}^{\Delta T}\frac{d(\Delta T)}{\Delta T}=-\frac{h_g}{{\dot{m}}_c{c}_p}{\int }_0^{x}dx$

Left integral hai $\ln (\Delta T)-\ln (\Delta T_0)=\ln \frac{\Delta T}{\Delta T_0}$; right hai $-\frac{h_{g}x}{{\dot{m}}_c{c}_p}$. Toh

$\ln \frac{\Delta T}{\Delta T_0}=-\frac{h_{g}x}{{\dot{m}}_c{c}_p}\Rightarrow \boxed{\Delta T=\Delta T_0{e}^{-\frac{h_{g}x}{{\dot{m}}_c{c}_p}}}$

KYU. "$\frac{d(\text{stuff})}{\text{stuff}}$" integrate karne par hamesha $\ln$ milta hai; $\ln$ undo karne par $e$ milta hai. Isliye exponential yahan unavoidable hai — yeh akela function hai jiska rate of change khud uske proportional hai.

Exponent padhna. Group $\frac{h_{g}x}{{\dot{m}}_c{c}_p}$ ek pure number hai (ek Stanton number-times-length parameter). Jab yeh $1$ hota hai, gap $1/e\approx 37\%$ tak gir jaata hai; jab $4.5$ hota hai, toh lagbhag $1\%$ tak.

PICTURE. Ek textbook exponential decay — shuru mein steep, phir lambi tail.

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Step 6 — Effectiveness $\eta$ mein rescale karo

KYA HAI. Parent note ki effectiveness bas original gap ka kya fraction abhi bhi bacha hai hai:

$\eta \equiv \frac{T_{hot}-T_{aw}}{T_{hot}-T_{cool}}=\frac{\Delta T}{\Delta T_0}.$

Step 5 ke boxed result ko $\Delta T_0$ se divide karo aur $\Delta T_0$ cancel ho jaata hai:

$\boxed{\eta (x)=\exp \left(-\frac{h_{g}x}{{\dot{m}}_c{c}_p}\right)}$

Yeh rescaling kyun. $\Delta T_0$ engine se engine badlata hai, lekin $\eta$ dimensionless aur universal hai: yeh hamesha $1$ (poori protection) se shuru hota hai $x=0$ par aur hamesha $0$ (koi protection nahi) ki taraf decay karta hai. Har curve ek hi shape par collapse ho jaata hai.

Edge cases — sab check karo:

• $x=0$: exponent $=0$, $\eta =e^0=1$. Fresh film, wall = coolant. ✓
• $x\to \infty$: exponent $\to +\infty$, $\eta \to 0$. Film poori khatam, wall = hot gas. ✓
• ${\dot{m}}_c\to 0$ (koi coolant nahi): kisi bhi $x>0$ ke liye exponent $\to \infty$, toh $\eta \to 0$ har jagah — film nahi matlab koi protection nahi, turant. ✓
• ${\dot{m}}_c\to \infty$ (infinite coolant): exponent $\to 0$, $\eta \to 1$ har jagah — infinitely thick film kabhi warm nahi hoti. ✓
• $h_g\to 0$ (hot gas heat transfer nahi kar sakta): exponent $\to 0$, $\eta \to 1$ — film ko warm karne ke liye kuch nahi. ✓

Har limiting scenario sensibly behave karta hai, toh formula trustworthy hai.

PICTURE. Same curve jaise Step 5 mein lekin vertical axis $\Delta T$ se $\eta$ relabel hui, $1\to 0$ jaati hai, aur design threshold ${\eta }_{min}$ ek horizontal line ki tarah khinchi hai.

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Step 7 — Protection kahan khatam hoti hai? Logarithm

KYA HAI. Engineers ek minimum acceptable effectiveness ${\eta }_{min}$ choose karte hain (maan lo $0.2$). Film wall ko sirf station $x_{prot}$ tak "protect" karti hai jahan $\eta$ pehli baar ${\eta }_{min}$ tak gir jaata hai. $\eta (x_{prot})={\eta }_{min}$ set karo aur solve karo:

$\exp \left(-\frac{h_g{x}_{prot}}{{\dot{m}}_c{c}_p}\right)={\eta }_{min}\Rightarrow \frac{h_g{x}_{prot}}{{\dot{m}}_c{c}_p}=\ln \frac{1}{{\eta }_{min}}\Rightarrow x_{prot}=\frac{{\dot{m}}_c{c}_p}{h_g}\ln \frac{1}{{\eta }_{min}}.$

Coverage fraction $f$ woh protected length hai chamber length $L$ ke fraction ke roop mein:

$\boxed{f=\frac{x_{prot}}{L}=\frac{{\dot{m}}_c{c}_p}{h_{g}L}\ln \frac{1}{{\eta }_{min}}}$

Logarithm kyun. Exponential undo karne ke liye logarithm lete hain — literally $\ln$ yahi kaam karta hai. Step 6 mein exponential decay ka matlab hai log tab hi aata hai jab hum poochhte hain "kaunse $x$ par $\eta$ kisi given value tak pahunchta hai?" Physically: kyunki protection exponentially khatam hoti hai, zyada ${\eta }_{min}$ maangna coolant rapidly diminishing returns ke saath leta hai — $\ln (1/{\eta }_{min})$ sirf dheere dheere badhta hai jaise ${\eta }_{min}\to 1$.

Degenerate check: agar ${\eta }_{min}=1$ toh $\ln (1/1)=\ln 1=0$, toh $x_{prot}=0$ — perfect protection maangna matlab film sirf bilkul slot par hi qualify karti hai. Sensible hai.

PICTURE. Threshold line decay curve ko $x_{prot}$ par kaatti hai; shaded strip $[0,x_{prot}]$ covered wall hai.

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Step 8 — Numbers daalo (parent ka Example 3, dekha hua)

KYA HAI. $h_g=1500{\text{W/m}}^2\text{K}$, ${\dot{m}}_c=0.05\text{kg/(m⋅s)}$, $c_p=2000\text{J/kgK}$, $L=0.50$ m, ${\eta }_{min}=0.2$ lo.

$\frac{{\dot{m}}_c{c}_p}{h_g}=\frac{0.05\times 2000}{1500}=\frac{100}{1500}=0.0667\text{m},\ln \frac{1}{0.2}=\ln 5=1.609$ $x_{prot}=0.0667\times 1.609=0.1073\text{m},f=\frac{0.1073}{0.50}=0.215\approx 21\%.$

Yeh kyun matter karta hai. Ek slot sirf ~21% wall ko $\eta \ge 0.2$ par cover karta hai. Poori chamber protect karne ke liye tum lagbhag $1/f\approx 5$ slots distribute karte ho — engineers literally aisi hi injection slots count karte hain.

PICTURE. Poori chamber ka bar, pehle slot ka covered strip (~21%) shaded, aur baaki hissa bharne ke liye chaar extra slots.

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Ek-picture summary

Recall Feynman retelling — poora walkthrough seedhe shabdon mein

Tum jalne wali hot wall ke saath cool gas spray karte ho (Step 1). Sirf yeh matter karta hai ki wall aag se kitni thandi hai — use gap kaho (Step 2). Lekin aag teri cool ribbon mein ghuste rehti hai aur use warm karti rehti hai, aur woh tab aur tezi se ghuste hai jab gap bada ho (Step 3). "Change jo bacha hai uspe proportional hai" exponential fade ka exact recipe hai (Steps 4–5). Gap ko rescale karo taki woh hamesha $1$ se shuru ho aur tumhe milta hai effectiveness $\eta =e^{-(\text{stuff})}$ — slot par full, downstream khatam (Step 6). Yeh dhundne ke liye ki film "give up" kab karti hai, poochhte ho ki $\eta$ kab tumhare minimum tak girta hai; exponential undo karne par ek logarithm aata hai, aur chamber length se divide karne par coverage fraction milti hai (Step 7). Numbers dikhate hain ki ek slot sirf wall ka lagbhag paanchwa hissa cover karta hai, toh use protect karne ke liye kabhi kabhi paanch slots chahiye (Step 8). Zyada coolant ya fresh slot = zyada covered wall.

Recall Khud se rebuild karo

Rate rule mein minus sign kyun hai yeh explain karo ::: Gap sirf chhotata hi hai — hot gas kabhi film ko thanda nahi karta — toh $d(\Delta T)/dx<0$. Kaunsa mathematical tool "rate proportional to amount" ko solution mein badalta hai, aur kyun? ::: $d(\Delta T)/\Delta T$ integrate karne par $\ln$ milta hai, aur usse undo karne par $e^{-\dots }$ milta hai — exponential akela function hai jo apne khud ke decay rate ke barabar hai. Coverage fraction mein logarithm kyun hota hai? ::: Kyunki $\eta$ exponentially decay karta hai; $x$ ke liye $\eta ={\eta }_{min}$ solve karne ke liye $\exp$ ka inverse chahiye, jo $\ln$ hai. $x=0$ aur $x\to \infty$ par $\eta$ kya hai? ::: $1$ (fresh film, wall = coolant) aur $0$ (spent film, wall = hot gas).

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Connections

• Parent: Film cooling effectiveness & coverage
• Convective heat transfer coefficient — $h_g$ deta hai, decay driver.
• Boundary layer & entrainment — woh mixing jo gap chhotata hai.
• Adiabatic wall temperature & recovery factor — $T_{aw}$ aur $T_{hot}$ define karta hai.
• Stanton number — exponent mein dimensionless group.
• Regenerative cooling — gas-side film physics ka wall-side complement.
• Combustion chamber thermal design — jahan slot-counting hoti hai.