Film cooling — effectiveness, coverage fraction: Exercises

3.3.29 · D4 · Physics › Rocket Propulsion › Film cooling — effectiveness, coverage fraction

Shuru karne se pehle, yeh sirf chaar relations hain jo tum kabhi bhi plug in karoge. Inhe ek baar padho taaki koi symbol anjaan na rahe.

Yeh symbols plain words mein kya matlab rakhte hain, taaki tumhe kabhi guess na karna pade:

Definition Symbol dictionary

$T_{h o t}$ ::: hot core gas ki recovery / adiabatic temperature — yeh woh temperature hai jo wall feel karegi bina film ke (the enemy).
$T_{coo l}$ ::: us cool gas ki temperature jo hum wall ke saath inject karte hain (the ally).
$T_{a w}$ ::: adiabatic wall temperature — woh temperature jo wall actually reach karti hai, assuming wall mein zero conduction.
$η$ ::: effectiveness. $1$ = perfect protection (wall $T_{coo l}$ par hai), $0$ = koi protection nahi (wall $T_{h o t}$ par hai).
$h_{g}$ ::: gas-side convective heat-transfer coefficient (W m $^{- 2}$ K $^{- 1}$ ) — kitni tezi se hot gas film ko heat karti hai.
$\overset{m}{˙}_{c}$ ::: coolant mass flow per unit wall width (kg m $^{- 1}$ s $^{- 1}$ ) — humne kitna thanda "stuff" spend kiya.
$c_{p}$ ::: coolant ki specific heat (J kg $^{- 1}$ K $^{- 1}$ ).
$x$ ::: injection slot se downstream distance (m).
$L$ ::: protect ki jaane wali total chamber length (m).
$η_{min}$ ::: sabse chhoti effectiveness jo design abhi bhi "protected" bolti hai.

Neeche ek figure hai jo har exercise ka map hai. Ise aise padho: cyan curve relation (C) hai — effectiveness $η$ slot par $1$ se shuru hoti hai ( $x = 0$ , top-left) aur daayein jaate-jaate $0$ ki taraf jhukti hai. Amber dashed horizontal line ek chosen threshold $η_{min}$ hai; amber dotted vertical line $x_{p r o t}$ mark karti hai, woh exact distance jahan curve us threshold ko cross karti hai. Uske baayein shaded region "protected" stretch hai jiska length relation (D) mein jaata hai. Yeh picture yaad rakho: L2 cyan curve par ek single point dhundta hai, L3 dhundta hai kahan curve amber line se milti hai, aur L4/L5 is curve ki copies stack karte hain.

Figure — Film cooling — effectiveness, coverage fraction

Level 1 — Recognition

Kya tum definitions padh ke sahi formula identify kar sakte ho?

L1.1 Ek wall $T_{a w} = T_{coo l}$ reach karti hai. $η$ kya hai? Physically iska matlab kya hai?

L1.2 Injection slot par ( $x = 0$ ), relation (C) se $η (0)$ evaluate karo. Kya value physical intuition se match karti hai?

L1.3 Chaar relations (A)–(D) mein se kaun sa use karoge is sawaal ka jawab dene ke liye: "Chamber ka kitna fraction $η = 0.3$ se upar protected hai?"

Recall Solutions — L1

L1.1 $T_{a w} = T_{coo l}$ ko relation (A) mein daalo: $η = \frac{T _{h o t} - T _{coo l}}{T _{h o t} - T _{coo l}} = 1.$ $η = 1$ matlab perfect protection — wall utni hi thandi hai jitni injected gas. Yeh best case hai, slot par hi (figure ke top-left corner mein).

L1.2 (C) mein $x = 0$ set karo: exponent ban jaata hai $- \frac{h _{g} \cdot 0}{m ˙ _{c} c _{p}} = 0$ , aur $e^{0} = 1$ . Toh $η (0) = 1$ . ✔ Match karta hai: bilkul slot par film pure coolant hai, abhi koi hot gas mix nahi hua, toh protection total hai — yahi woh jagah hai jahan cyan curve top axis ko touch karti hai.

L1.3 "Chamber ka fraction jo threshold se upar protected hai" exactly coverage question hai → relation (D) use karo $η_{min} = 0.3$ ke saath.

Level 2 — Application

Ek formula mein numbers daalo, answer nikalto, units clean rakho.

L2.1 $T_{h o t} = 3200$ K, $T_{coo l} = 800$ K, aur kisi station par $η = 0.5$ . $T_{a w}$ nikalo.

L2.2 $h_{g} = 1200$ W m $^{- 2}$ K $^{- 1}$ , $\overset{m}{˙}_{c} = 0.04$ kg m $^{- 1}$ s $^{- 1}$ , $c_{p} = 2500$ J kg $^{- 1}$ K $^{- 1}$ , aur $x = 0.20$ m. $η (x)$ nikalo.

L2.3 L2.2 ke numbers ke liye, kis distance $x$ par effectiveness $η = 0.5$ tak gir jaati hai?

Recall Solutions — L2

L2.1 Relation (B) use karo: $T_{a w} = (1 - 0.5) (3200) + (0.5) (800) = 1600 + 400 = 2000 K .$ Film ne wall ka $3200 - 2000 = 1200$ K kata.

L2.2 Pehle dimensionless exponent: $\frac{h _{g} x}{m ˙ _{c} c _{p}} = \frac{1200 \times 0.20}{0.04 \times 2500} = \frac{240}{100} = 2.4.$ Phir $η = e^{- 2.4} = 0.0907 \approx 0.091$ . Film yahan almost khatam ho gayi hai — yeh cyan curve ke daayein end par ek single point hai.

L2.3 (C) mein $η = 0.5$ set karo aur $x$ ke liye solve karo. Dono sides ka $ln$ lo: $- \frac{h _{g} x}{m ˙ _{c} c _{p}} = ln 0.5 \Rightarrow x = \frac{m ˙ _{c} c _{p}}{h _{g}} ln \frac{1}{0.5} .$ $x = \frac{0.04 \times 2500}{1200} \times ln 2 = \frac{100}{1200} \times 0.6931 = 0.0833 \times 0.6931 = 0.0578 m .$ Toh effectiveness lagbhag 5.8 cm mein half ho jaati hai — yeh film ka "half-life length" hai.

Level 3 — Analysis

Do relations combine karo, ya reason karo ki ek knob ko badalne se answer kaise change hota hai.

L3.1 (Coverage). $h_{g} = 1500$ W m $^{- 2}$ K $^{- 1}$ , $\overset{m}{˙}_{c} = 0.05$ kg m $^{- 1}$ s $^{- 1}$ , $c_{p} = 2000$ J kg $^{- 1}$ K $^{- 1}$ , $L = 0.60$ m, $η_{min} = 0.25$ . Coverage fraction $f$ nikalo.

L3.2 (Sensitivity). L3.1 ke liye, agar coolant flow $\overset{m}{˙}_{c}$ double karo toh protected length $x_{p r o t}$ kaise change hoti hai? Factor do aur ek line mein explain karo.

L3.3 (Threshold cost). L3.1 ke baseline ke liye, $η_{min} = 0.5$ aur $η_{min} = 0.1$ ke liye required protected length compare karo. Kaun sa threshold zyada lamba protected region maangta hai, aur kis ratio mein?

Recall Solutions — L3

L3.1 Relation (D) use karo. Yeh figure mein amber-line-crossing hai: hum chahte hain kahan cyan curve $η_{min} = 0.25$ se milti hai. Pehle prefactor $\frac{m ˙ _{c} c _{p}}{h _{g}} = \frac{0.05 \times 2000}{1500} = \frac{100}{1500} = 0.06667$ m. Phir: $x_{p r o t} = 0.06667 \times ln \frac{1}{0.25} = 0.06667 \times ln 4 = 0.06667 \times 1.3863 = 0.09242 m .$ $f = \frac{x _{p r o t}}{L} = \frac{0.09242}{0.60} = 0.1540 \approx 15.4%.$

L3.2 Relation (D) mein, $x_{p r o t} = \frac{m ˙ _{c} c _{p}}{h _{g}} ln (1/ η_{min})$ . Threshold term $ln (1/ η_{min})$ fixed hai, toh $x_{p r o t} \propto \overset{m}{˙}_{c}$ . $\overset{m}{˙}_{c}$ double karne se $x_{p r o t}$ double hoti hai (factor $2$ ). Zyada thanda budget → proportionally zyada lambi protection, fixed threshold ke liye.

L3.3 Prefactor common hai, toh ratio sirf logs se set hota hai: $\frac{x _{p r o t} ( η _{min} = 0.1 )}{x _{p r o t} ( η _{min} = 0.5 )} = \frac{l n ( 1/0.1 )}{l n ( 1/0.5 )} = \frac{l n 10}{l n 2} = \frac{2.3026}{0.6931} = 3.322.$ Ek zyada lenient threshold ( $η_{min} = 0.1$ ek kamzor film ko accept karti hai) lagbhag 3.3× zyada lambe region ki allow karti hai. Equivalently, ek zyada strong film maangna ( $η_{min} = 0.5$ ) protected region ko chota kar deta hai.

Level 4 — Synthesis

Teen ya zyada relations chain karo, ya design variable ke liye solve karo.

L4.1 (Coolant flow design karo). Tumhe poora chamber protect karna hai, $f = 1$ , $L = 0.40$ m par $η_{min} = 0.30$ ke saath, $h_{g} = 1600$ W m $^{- 2}$ K $^{- 1}$ aur $c_{p} = 1800$ J kg $^{- 1}$ K $^{- 1}$ ke saath. Single slot se kitna coolant flow $\overset{m}{˙}_{c}$ (per unit width) chahiye?

L4.2 (Coverage edge par wall temperature). L4.1 ka $\overset{m}{˙}_{c}$ use karte hue, aur $T_{h o t} = 3300$ K, $T_{coo l} = 700$ K ke saath, coverage edge $x = L$ par exactly $T_{a w}$ nikalo (jahan $η = η_{min} = 0.30$ hai).

L4.3 (Slots count karo). L3.1 ke parameters rakho ( $x_{p r o t} = 0.09242$ m per slot) lekin ab chamber $L = 0.60$ m hai. Agar har naaya slot $η = 1$ re-establish karta hai aur $η_{min}$ tak girne se pehle ek $x_{p r o t}$ length protect karta hai, toh poori length cover karne ke liye kitne slots chahiye? Round up karo.

Recall Solutions — L4

L4.1 "Ek slot se poora chamber" matlab $f = 1$ , yaani $x_{p r o t} = L$ . Relation (D) ko $\overset{m}{˙}_{c}$ ke liye rearrange karke $x_{p r o t} = L$ set karo: $L = \frac{m ˙ _{c} c _{p}}{h _{g}} ln \frac{1}{η _{min}} \Rightarrow \overset{m}{˙}_{c} = \frac{h _{g} L}{c _{p} l n ( 1/ η _{min} )} .$ $\overset{m}{˙}_{c} = \frac{1600 \times 0.40}{1800 \times l n ( 1/0.30 )} = \frac{640}{1800 \times 1.20397} = \frac{640}{2167.15} = 0.2953 kg m^{- 1} s^{- 1} .$ Yeh ek bada flow hai (~0.30 kg per metre of width per second) — ek single slot rarely poora chamber cover karta hai, isliye real engines mein several slots stack kiye jaate hain (dekho L4.3).

L4.2 Coverage edge par $η = 0.30$ , relation (B) use karo: $T_{a w} = (1 - 0.30) (3300) + (0.30) (700) = 0.70 \times 3300 + 0.30 \times 700 = 2310 + 210 = 2520 K .$ Sabse kamzor allowed film par bhi wall flame temperature se 780 K neeche hai.

L4.3 Har slot $x_{p r o t} = 0.09242$ m cover karta hai. Slots ki sankhya: $N = ⌈ \frac{L}{x _{p r o t}} ⌉ = ⌈ \frac{0.60}{0.09242} ⌉ = ⌈ 6.492 ⌉ = 7 slots .$ Tumhe 7 slots chahiye poore 0.60 m chamber ko $η \geq 0.25$ par cover karne ke liye.

Level 5 — Mastery

Ek relation prove karo, ya un limits par reason karo jahan algebra mislead kar sakta hai.

L5.1 (ODE se derive karo). Entrainment model $\frac{d ( Δ T )}{d x} = - \frac{h _{g}}{m ˙ _{c} c _{p}} Δ T$ se shuru karo jahan $Δ T \equiv T_{h o t} - T_{a w}$ aur $Δ T (0) = T_{h o t} - T_{coo l}$ , relation (C) derive karo: $η (x) = e^{- h_{g} x / (\overset{m}{˙}_{c} c_{p})}$ .

L5.2 (Two-slot effectiveness — limiting behaviour). Slot 1 $x = 0$ par inject karta hai; ek doosra identical slot $x = d$ par inject karta hai, local film ko wahan $η = 1$ par reset karta hai. Slots ke beech, combined effectiveness $η_{1} (x) = e^{- k x}$ hai jahan $k \equiv h_{g} / (\overset{m}{˙}_{c} c_{p})$ . Slot 2 se bilkul pehle, $η$ girkar $e^{- k d}$ ho jaati hai; bilkul baad, woh $1$ par jump karti hai. Dikhao ki slot 2 ko $d = x_{p r o t}$ par rakhna (jahan $η_{1}$ $η_{min}$ hit karti hai) $[0, 2 d]$ par har jagah effectiveness $\geq η_{min}$ rakhta hai, aur un do locations identify karo jahan $η = η_{min}$ exactly hai.

L5.3 (Zero-flow aur infinite-flow limits). Relation (C) se, fixed $x > 0$ ke liye $lim_{\overset{m}{˙}_{c} \to 0^{+}} η (x)$ aur $lim_{\overset{m}{˙}_{c} \to \infty} η (x)$ evaluate karo, aur dono ka physical meaning batao.

Recall Solutions — L5

L5.1 Pehle shorthand kisi bhi integration se pehle fix karo: $Δ T_{0} \equiv Δ T (0) = T_{h o t} - T_{coo l}$ , slot par $Δ T$ ki value (poora hot-to-cool gap, kyunki wahan film pure coolant hai). Ab ODE separable hai — humne separation isliye choose kiya kyunki $Δ T$ ka derivative $Δ T$ ke proportional hai, exponential decay ki pehchaan. Dono sides ko $Δ T$ se divide karo aur $d x$ se multiply karo: $\frac{d ( Δ T )}{Δ T} = - \frac{h _{g}}{m ˙ _{c} c _{p}} d x .$ Left ko $Δ T_{0}$ se $Δ T$ tak aur right ko $0$ se $x$ tak integrate karo: $ln \frac{Δ T}{Δ T _{0}} = - \frac{h _{g} x}{m ˙ _{c} c _{p}} \Rightarrow Δ T = Δ T_{0} e^{- h_{g} x / (\overset{m}{˙}_{c} c_{p})} .$ Ab $Δ T_{0} = T_{h o t} - T_{coo l}$ se divide karo. Kyunki $η = \frac{T _{h o t} - T _{a w}}{T _{h o t} - T _{coo l}} = \frac{Δ T}{Δ T _{0}}$ : $η (x) = e^{- h_{g} x / (\overset{m}{˙}_{c} c_{p})} . ■$

L5.2 Neeche figure argument ko sawtooth ki tarah dikhata hai: cyan curve $1$ se girती hai, aur $x = d$ par amber vertical line mark karti hai slot 2 ko jo $η$ ko $1$ par snap back karta hai, jiske baad identical fall repeat hoti hai. Is reasoning ko uske against follow karo. $[0, d]$ par: $η (x) = e^{- k x}$ ek decreasing function hai, toh is interval par uska minimum right endpoint $x = d$ par reach hota hai, $e^{- k d}$ ke barabar. $d = x_{p r o t}$ choose karna matlab $e^{- k d} = η_{min}$ (yahi $x_{p r o t}$ ki definition hai). Isliye $[0, d]$ par, $η \geq η_{min}$ , equality sirf right endpoint $x = d$ par reach hoti hai. $x = d$ par slot 2 $η$ ko $1$ par reset karta hai, aur $[d, 2 d]$ par wahi girती curve right endpoint $x = 2 d$ par minimum $η_{min}$ deti hai. Toh $η \geq η_{min}$ poore $[0, 2 d]$ par, equality exactly do endpoints $x = d$ aur $x = 2 d$ par hold karti hai. Isliye slots ek $x_{p r o t}$ apart space kiye jaate hain — har slot ki protection exactly wahan khatam hoti hai jahan agli shuru hoti hai.

L5.3 Exponent $- \frac{h _{g} x}{m ˙ _{c} c _{p}}$ hai, aur sirf $\overset{m}{˙}_{c}$ vary karta hai ( $x > 0$ fixed ke saath).

$\overset{m}{˙}_{c} \to 0^{+}$ : denominator shrink hota hai, toh exponent $- \frac{h _{g} x}{m ˙ _{c} c _{p}} \to - \infty$ , aur $η \to e^{- \infty} = 0$ . Koi coolant nahi → koi protection nahi: wall $T_{h o t}$ par baith jaati hai. Physically sensible.
$\overset{m}{˙}_{c} \to \infty$ : denominator blow up hota hai, toh exponent $\to 0^{-}$ , aur har finite $x$ ke liye $η \to e^{0} = 1$ . Infinite coolant → poori wall fully protected hai coolant temperature par. Yeh bhi sensible hai — aur yeh confirm karta hai ki formula dono extremes par correctly behave karta hai.

Connections

Adiabatic wall temperature & recovery factor — $T_{a w}$ aur $T_{h o t}$ supply karta hai jo in exercises mein use hoti hain.
Convective heat transfer coefficient — $h_{g}$ jo yahan har decay drive karta hai.
Stanton number — dimensionless group $h_{g} x / (\overset{m}{˙}_{c} c_{p})$ Stanton-number × length hai.
Boundary layer & entrainment — L5.1 mein decay ODE ke peeche mixing physics.
Regenerative cooling — real wall temperature set karta hai jab conduction allow hoti hai.
Combustion chamber thermal design — jahan slot-counting (L4.3, L5.2) actually use hoti hai.