3.3.29 · D3Film cooling — effectiveness, coverage fraction
Parent note Film cooling ne tumhe do formulas di thi. Yeh page unhe har us situation ke against stress-test karta hai jo ek real chamber (ya exam) tumhare saamne phenk sakta hai: normal case, do extremes (η = 0 η = 1
Shuru karne se pehle, woh do tools jo hum baar baar use karte rahenge, simple words mein restate kiye gaye hain taaki koi bhi symbol unexplained na rahe:
Definition Protected lengthx p r o t η min x η ( x ) ≥ η min η ( x p r o t ) = η min ln x p r o t = h g m ˙ c c p ln η min 1 x p r o t f = x p r o t / L
Definition Usi effectiveness ko padhne ke do tarikeBlend formula ko invert karke η η T a w = ( 1 − η ) T h o t + η T coo l = T h o t − η ( T h o t − T coo l ) T h o t ( T h o t − T coo l ) η = T h o t − T coo l T h o t − T a w η wall ko hot gas se kitna pull down kiya gaya hai, poore hot-to-cool gap ke fraction ke roop mein . Yeh bilkul wahi equation hai jaise blend — bas rearrange ki gayi hai — isliye ek ke baare mein jo bhi sach hai woh doosre ke baare mein bhi sach hai. Hum is inverted form ko Ex 8 mein use karte hain.
Har film-cooling problem inhi cells mein se ek hai. Neeche ke examples us cell ke saath tagged hain jo woh cover karte hain.
Cell
Kya khaas baat hai
Example
A. Normal blend 0 < η < 1 T a w Ex 1
B. Extreme η = 1 slot par fresh film (x = 0
Ex 2
C. Extreme η = 0 film fully spent, x → ∞
Ex 2
D. Ek station par decay exponential mein plug in karo
Ex 3
E. Degenerate m ˙ c → 0 koi coolant nahi — limiting behaviour
Ex 4
F. Degenerate m ˙ c → ∞ wall ko flood karna — limiting behaviour
Ex 4
G. Coverage / slot count word problem: kitne slots chahiye?
Ex 5
H. Inverse / hidden variable exam twist: x m ˙ c
Ex 6
I. Threshold sensitivity η min Ex 7
J. Sign / sanity trap T coo l > T h o t η > 1 Ex 8
Koi bhi numbers se pehle, master figure dekho. Yeh teeno ideas ek saath visually dikhata hai, aur har worked example uske ek piece ki taraf point karta hai.
Left panel (the blend): T coo l T h o t T a w η Yeh Ex 1 aur Ex 2 hai — wall temperature sirf yeh hai ki dot kahan baitha hai. Middle panel (the decay): magenta curve η ( x ) = e − h g x / ( m ˙ c c p ) η min x p r o t Yeh Ex 3–6 hai. Right panel (coverage cost): coverage f η min Yeh Ex 7 hai.
Worked example Example 1 — Cell A: normal blend, wall temperature nikaloT h o t = 3200 T coo l = 800 η = 0.35 T a w
Forecast: guess karo — kya wall 800 ke kareeb hogi ya 3200 ke? (Sirf 35% protected, isliye hot ki taraf lean karo.)
Blend likho. T a w = ( 1 − η ) T h o t + η T coo l Yeh step kyun? η left panel par magenta dot hai.Plug in karo. T a w = ( 0.65 ) ( 3200 ) + ( 0.35 ) ( 800 ) = 2080 + 280 = 2360 Yeh step kyun? Weights 0.65 0.35 [ 800 , 3200 ]
Verify: 2360 800 3200 2000 η < 0.5
Worked example Example 2 — Cells B & C: ek saath do extremesSame T h o t = 3200 T coo l = 800 T a w η = 1 η = 0
Forecast: sabse thandi aur sabse garam walls jo film de sakti hai.
Case B, η = 1 T a w = ( 0 ) ( 3200 ) + ( 1 ) ( 800 ) = 800 Yeh step kyun? η = 1 perfect film hai — wall sirf coolant "dekhti" hai. T coo l Case C, η = 0 T a w = ( 1 ) ( 3200 ) + ( 0 ) ( 800 ) = 3200 Yeh step kyun? η = 0
Verify: Yeh exactly interval [ T coo l , T h o t ] Kisi bhi η ∈ [ 0 , 1 ]
Worked example Example 3 — Cell D: ek downstream station par effectivenessh g = 1200 W/m 2 K m ˙ c = 0.06 kg/(m⋅s) c p = 2000 J/kgK x = 0.20 η T h o t = 3200 T coo l = 800 T a w
Forecast: exponent ek chhota-sa number hoga, isliye guess karo η
Dimensionless group banao. m ˙ c c p h g x = 0.06 × 2000 1200 × 0.20 = 120 240 = 2.0 Yeh step kyun? Yeh exponent hi woh ek cheez hai jis par η middle panel par yeh hai ki tum sagging curve ke kitna right tak travel kar chuke ho.Exponentiate karo. η = e − 2.0 = 0.135 Yeh step kyun? Decay law kehta hai effectiveness e − ( group ) Temperature blend karo. T a w = ( 1 − 0.135 ) ( 3200 ) + ( 0.135 ) ( 800 ) = 2768 + 108 = 2876 Yeh step kyun? Abstract η
Verify: Group dimensionless hai: ( kg/m⋅s ) ( J/kgK ) ( W/m 2 K ) ( m ) = W/m⋅K W/m⋅K = 1 J/s = W η = 0.135 ∈ [ 0 , 1 ] T a w = 2876
Worked example Example 4 — Cells E & F: degenerate coolant flowh g = 1200 c p = 2000 x = 0.20 m ˙ c → 0 m ˙ c → ∞ η
Forecast: koi coolant nahi vs. wall ko dabaana — tum pehle se physics jaante ho, ab formula ko agree hote dekho.
Case E, m ˙ c → 0 exponent m ˙ c c p h g x → + ∞ η = e − ∞ → 0 Yeh step kyun? Koi cold "stuff" inject nahi kiya, hot gas turant jeet jaati hai — wall full flame feel karti hai. Formula deta hai T a w → T h o t = 3200 Case F, m ˙ c → ∞ exponent → 0 η = e 0 = 1 Yeh step kyun? Coolant ka infinite parda kabhi nahi khaata — har x T a w → T coo l = 800 Middle check, m ˙ c = 0.06 hume Ex 3 mein pehle se η = 0.135 Yeh step kyun? Confirm karta hai ki function monotone hai: zyada coolant ⇒ zyada η
Verify: Dono limits Ex 2 ke endpoints reproduce karte hain (T coo l T h o t
Worked example Example 5 — Cell G: word problem — kitne injection slots chahiye?Ek chamber L = 0.90 m ˙ c = 0.05 kg/(m⋅s) c p = 2000 h g = 1500 W/m 2 K η η min = 0.25
Forecast: compute karne se pehle slots ki number guess karo — 2? 5? 10?
Ek slot ki protected length. Page ke top se x p r o t x p r o t = h g m ˙ c c p ln η min 1 = 1500 0.05 × 2000 ln 0.25 1 = 1500 100 ln 4 Yeh step kyun? x p r o t η ≥ η min Evaluate karo. 1500 100 = 0.06667 ln 4 = 1.3863 x p r o t = 0.06667 × 1.3863 = 0.09242 Yeh step kyun? Symbolic length ko metres mein turn karo taaki L Slots count karo. slots = ⌈ L / x p r o t ⌉ = ⌈ 0.90/0.09242 ⌉ = ⌈ 9.74 ⌉ = 10 Yeh step kyun? Har slot x p r o t 0.90 round up karte hain. (Ceiling ⌈ y ⌉ y ⌈ 9.74 ⌉ = 10
Verify: Ek slot ka coverage fraction f = x p r o t / L = 0.09242/0.90 = 0.1027 10 × 0.1027 = 1.027 ≥ 1 0.924 < 1
Worked example Example 6 — Cell H: exam twist, hidden coolant flow ke liye solve karoDesign requirement: ek single slot ko x = 0.15 η ≥ 0.30 h g = 1500 c p = 2000 m ˙ c
Forecast: hum decay law ko ulta chala rahe hain — kya tumhe m ˙ c
Target station par effectiveness ko floor set karo. exp ( − m ˙ c c p h g x ) = 0.30 Yeh step kyun? Sabse bura spot far edge x = 0.15 η = 0.30 ( 0.15 , 0.30 ) Dono sides ka ln − m ˙ c c p h g x = ln 0.30 = − 1.204 m ˙ c c p h g x = 1.204 Yeh step kyun? ln exp m ˙ c m ˙ c m ˙ c = c p × 1.204 h g x = 2000 × 1.204 1500 × 0.15 = 2408 225 = 0.0934 kg/(m⋅s) Yeh step kyun? Algebraic rearrangement us unknown ko isolate karti hai jo hum se maanga gaya tha.
Verify: m ˙ c = 0.0934 = 0.0934 × 2000 1500 × 0.15 = 186.8 225 = 1.204 e − 1.204 = 0.300
Worked example Example 7 — Cell I: threshold sensitivity (super-linear cost)h g = 1500 c p = 2000 m ˙ c = 0.05 L = 0.50 η min = 0.10 η min = 0.50
Forecast: zyada minimum effectiveness demand karna — kya coverage upar ya neeche jaati hai, aur kitna?
Loose floor η min = 0.10 f = h g L m ˙ c c p ln 0.10 1 = 750 100 ln 10 = 0.1333 × 2.3026 = 0.3070 Yeh step kyun? Low bar clear karna aasaan hai, isliye film "protect karna count karta hai" zyada der tak — right panel curve ka sabse baayaan, tall part.Strict floor η min = 0.50 f = 750 100 ln 0.50 1 = 0.1333 × 0.6931 = 0.0924 Yeh step kyun? Har jagah half-strength protection par insist karna qualifying region ko dramatically shrink kar deta hai — right panel curve plunge ho gayi hai.Compare karo. Ratio = 0.3070/0.0924 = 3.32 Yeh step kyun? Kyunki coverage ln ( 1/ η min ) ln 10/ ln 2 = 3.32
Verify: 750 100 = 0.13 3 f [ 0 , 1 ] ln 10/ ln 2 = 3.322
Worked example Example 8 — Cell J: sign / sanity trap (do red flags +T coo l > T h o t T h o t = 2800 T coo l = 1000 η = 1.4 T a w = 280 Dono red flags nikalo, phir related edge case explain karo jahan T coo l > T h o t
Forecast: intern ke numbers mein do red flags chhupe hain — aage padhne se pehle unhe dhundho.
Red flag #1 — η [ 0 , 1 ] Upar derive kiya hua inverted form use karo, η = T h o t − T coo l T h o t − T a w fraction of the way hai, isliye physically [ 0 , 1 ] η = 1.4 > 1 blend kar sakti hai, kabhi bhi sabse thande se neeche refrigerate nahi kar sakti.
Yeh step kyun? Koi bhi η > 1 η < 0 Red flag #2 — T a w [ T coo l , T h o t ] Intern ke η T a w = ( 1 − 1.4 ) ( 2800 ) + ( 1.4 ) ( 1000 ) = ( − 0.4 ) ( 2800 ) + 1400 = − 1120 + 1400 = 280 T coo l ≤ T a w ≤ T h o t 1000 ≤ T a w ≤ 2800 280 T coo l = 1000 neeche hai.
Yeh step kyun? Negative weight ( 1 − η ) = − 0.4 doosra independent tarika hai: flag #1 η T a w Related edge case — T coo l > T h o t Suppose karo "coolant" core gas se garam hai, jaise T coo l = 2000 T h o t = 1500 T h o t − T coo l = 1500 − 2000 = − 500 negative hai, isliye η heat karti hai. Blend abhi bhi kaam karta hai — e.g. η = 0.35 T a w = ( 0.65 ) ( 1500 ) + ( 0.35 ) ( 2000 ) = 975 + 700 = 1675 1675 1500 2000 beech hai, yaani bracket sirf reorder ho gayi hai [ T h o t , T coo l ] Yeh step kyun? Yeh har pehle example ke peeche chhupa hua assumption expose karta hai — T coo l < T h o t η
Verify: Flag #1: η = 1.4 ∈ / [ 0 , 1 ] T a w = 280 < T coo l = 1000 [ 1000 , 2800 ] T coo l = 2000 > T h o t = 1500 η = 0.35 T a w = 1675 reordered bracket [ 1500 , 2000 ] T coo l < T h o t aur η ∈ [ 0 , 1 ]
Recall Har trap kis matrix cell se belong karta hai?
"Wall coolant se thandi" ::: Cell J — impossible, η [ 0 , 1 ] T a w
"Koi coolant inject nahi kiya" ::: Cell E — m ˙ c → 0 η → 0 T a w → T h o t
"Ek full chamber ke liye kitne slots?" ::: Cell G — L / x p r o t upar round karo (ceiling).
"Target η ln
"Coolant core gas se garam hai" ::: Cell J edge case — denominator sign flip karta hai, film heater ban jaati hai.
Recall Quick numeric checksEx1 T a w η e − 2 = 0.135 x p r o t = 0.0924 m ˙ c ln 10/ ln 2 T a w T coo l = 2000 T h o t = 1500 η = 0.35
"BRACKET pehle." Kisi bhi T a w T coo l T h o t η [ 0 , 1 ]
Parent: Film cooling — effectiveness, coverage fraction (index 3.3.29) — formulas wahan derive hue, yahan exercise kiye gaye.
Convective heat transfer coefficient — h g Stanton number — dimensionless group h g x / ( m ˙ c c p ) Adiabatic wall temperature & recovery factor — T h o t T a w Boundary layer & entrainment — exponential decay ke peeche physics.Combustion chamber thermal design — jahan slot-counting (Ex 5) real layout mein feed karta hai. 