Film cooling — effectiveness, coverage fraction: Worked examples

Before we start, the two tools we keep re-using, restated in plain words so no symbol is unearned:

Formula The two working equations

Wall temperature is a blend of hot gas and coolant, weighted by effectiveness $η$ (a number from 0 to 1): $T_{a w} = (1 - η) T_{h o t} + η T_{coo l}$ Effectiveness decays exponentially with distance $x$ down the wall: $η (x) = exp (- \frac{h _{g} x}{m ˙ _{c} c _{p}})$ Coverage fraction is the share of chamber length $L$ where $η$ stays above a design floor $η_{min}$ : $f = \frac{m ˙ _{c} c _{p}}{h _{g} L} ln \frac{1}{η _{min}}$

Symbols, each in words: $T_{h o t}$ = hot-gas recovery temperature (from Adiabatic wall temperature & recovery factor); $T_{coo l}$ = injected coolant temperature; $T_{a w}$ = the adiabatic wall temperature the film produces; $h_{g}$ = gas-side Convective heat transfer coefficient in $W/m^{2} K$ ; $\overset{m}{˙}_{c}$ = coolant mass flow per unit wall width, $kg/(m\cdots)$ ; $c_{p}$ = coolant specific heat, $J/kgK$ ; $x$ = distance downstream from the injection slot, metres.

The scenario matrix

Every film-cooling problem is one of these cells. The examples below are tagged with the cell they cover.

Cell	What makes it special	Example
A. Normal blend	$0 < η < 1$ , find $T_{a w}$	Ex 1
B. Extreme $η = 1$	fresh film at the slot ( $x = 0$ )	Ex 2
C. Extreme $η = 0$	film fully spent, $x \to \infty$	Ex 2
D. Decay at a station	plug into the exponential	Ex 3
E. Degenerate $\overset{m}{˙}_{c} \to 0$	no coolant — limiting behaviour	Ex 4
F. Degenerate $\overset{m}{˙}_{c} \to \infty$	flooding the wall — limiting behaviour	Ex 4
G. Coverage / slot count	word problem: how many slots?	Ex 5
H. Inverse / hidden variable	exam twist: solve for $x$ or $\overset{m}{˙}_{c}$	Ex 6
I. Threshold sensitivity	change $η_{min}$ , see super-linear cost	Ex 7
J. Sign / sanity trap	$T_{coo l} > T_{h o t}$ ? $η > 1$ ? spot the impossible	Ex 8

The one picture the whole page lives inside

Before any numbers, look at the master figure. It shows all three ideas visually at once, and each worked example points back to a piece of it.

Figure — Film cooling — effectiveness, coverage fraction

Left panel (the blend): a vertical thermometer from $T_{coo l}$ (bottom) to $T_{h o t}$ (top). The magenta dot is $T_{a w}$ ; as $η$ slides from 0 to 1, that dot slides from the hot top down to the cool bottom. This is Ex 1 and Ex 2 — the wall temperature is just where the dot sits.
Middle panel (the decay): the magenta curve $η (x) = e^{- h_{g} x / (\overset{m}{˙}_{c} c_{p})}$ starting at 1 and sagging toward 0. The violet dashed line is the floor $η_{min}$ ; where the curve crosses it (orange line) is $x_{p r o t}$ ; the shaded orange area is the protected stretch. This is Ex 3–6.
Right panel (coverage cost): coverage $f$ plotted against the floor $η_{min}$ . Notice it plunges as you demand a stricter floor — the log curvature. This is Ex 7.

Worked examples

Worked example Example 1 — Cell A: normal blend, find the wall temperature

$T_{h o t} = 3200$ K, $T_{coo l} = 800$ K, and at some station $η = 0.35$ . Find $T_{a w}$ .

Forecast: guess — will the wall be closer to 800 or 3200? (Only 35% protected, so lean hot.)

Write the blend. $T_{a w} = (1 - η) T_{h o t} + η T_{coo l}$ . Why this step? $η$ is defined as "fraction of the way from hot toward cool", so it must appear as the weight on the cool stream — this is the magenta dot on the left panel of the figure.
Plug in. $T_{a w} = (0.65) (3200) + (0.35) (800) = 2080 + 280 = 2360$ K. Why this step? Weights $0.65$ and $0.35$ sum to 1 — a true weighted average, never outside $[800, 3200]$ . On the thermometer the dot sits 35% of the way down from the top.

Verify: $2360$ lies between $800$ and $3200$ ✓, and it is on the hot side of the midpoint ( $2000$ ), matching our forecast since $η < 0.5$ . Units: all Kelvin ✓.

Worked example Example 3 — Cell D: effectiveness at a downstream station

$h_{g} = 1200 W/m^{2} K$ , $\overset{m}{˙}_{c} = 0.06 kg/(m\cdots)$ , $c_{p} = 2000 J/kgK$ , station $x = 0.20$ m. Find $η$ , then $T_{a w}$ using $T_{h o t} = 3200$ K, $T_{coo l} = 800$ K.

Forecast: the exponent will be a small-ish number, so guess $η$ somewhere around a half?

Form the dimensionless group. $\frac{h _{g} x}{m ˙ _{c} c _{p}} = \frac{1200 \times 0.20}{0.06 \times 2000} = \frac{240}{120} = 2.0$ . Why this step? This exponent is the only thing $η$ depends on — big group ⇒ dead film, small group ⇒ fresh film. On the middle panel of the figure it is how far right along the sagging curve you have travelled.
Exponentiate. $η = e^{- 2.0} = 0.135$ . Why this step? The decay law says effectiveness falls off like $e^{- (group)}$ ; a group of 2 already drops us to ~14% — read the curve's height at that station.
Blend the temperature. $T_{a w} = (1 - 0.135) (3200) + (0.135) (800) = 2768 + 108 = 2876$ K. Why this step? Convert the abstract $η$ into the physical quantity an engineer cares about — the metal temperature (back to the left panel's dot, now near the top).

Verify: Group is dimensionless: $\frac{( W/m ^{2} K ) ( m )}{( kg/m\cdots ) ( J/kgK )} = \frac{W/m\cdotK}{W/m\cdotK} = 1$ ✓ (since $J/s = W$ ). $η = 0.135 \in [0, 1]$ ✓. $T_{a w} = 2876$ sits between 800 and 3200 ✓.

Worked example Example 4 — Cells E & F: degenerate coolant flow

Keep $h_{g} = 1200$ , $c_{p} = 2000$ , $x = 0.20$ m. Ask: what happens to $η$ as (E) $\overset{m}{˙}_{c} \to 0$ and (F) $\overset{m}{˙}_{c} \to \infty$ ?

Forecast: no coolant vs. drowning the wall — you already know the physics, now watch the formula agree.

Case E, $\overset{m}{˙}_{c} \to 0$ : exponent $\frac{h _{g} x}{m ˙ _{c} c _{p}} \to + \infty$ , so $η = e^{- \infty} \to 0$ . Why this step? With no cold "stuff" injected, hot gas instantly wins — the wall sees full flame. Formula gives $T_{a w} \to T_{h o t} = 3200$ K. On the middle panel the curve collapses to the floor immediately. Sane.
Case F, $\overset{m}{˙}_{c} \to \infty$ : exponent $\to 0$ , so $η = e^{0} = 1$ . Why this step? An infinite curtain of coolant never gets eaten — perfect protection at every $x$ . Formula gives $T_{a w} \to T_{coo l} = 800$ K. The middle-panel curve becomes a flat line at 1. Sane.
Middle check, $\overset{m}{˙}_{c} = 0.06$ : we already got $η = 0.135$ in Ex 3, which lies between the two limits 0 and 1. Why this step? Confirms the function is monotone: more coolant ⇒ higher $η$ , no reversals.

Verify: Both limits reproduce the endpoints from Ex 2 ( $T_{coo l}$ and $T_{h o t}$ ). The physical picture ("no towel" vs "infinite towel") matches the math exactly. ✓

Worked example Example 5 — Cell G: word problem — how many injection slots?

A chamber is $L = 0.90$ m long. One slot delivers $\overset{m}{˙}_{c} = 0.05 kg/(m\cdots)$ , $c_{p} = 2000$ , and the gas side has $h_{g} = 1500 W/m^{2} K$ . The design rule: nowhere may $η$ drop below $η_{min} = 0.25$ . How many slots do you need?

Forecast: guess the number of slots before computing — 2? 5? 10?

Protected length of one slot. Use the boxed definition of $x_{p r o t}$ from the top of the page: $x_{p r o t} = \frac{m ˙ _{c} c _{p}}{h _{g}} ln \frac{1}{η _{min}} = \frac{0.05 \times 2000}{1500} ln \frac{1}{0.25} = \frac{100}{1500} ln 4$ . Why this step? $x_{p r o t}$ is exactly the length over which one slot keeps $η \geq η_{min}$ ; beyond it the film has fallen below the floor (the orange line on the middle panel) and a fresh slot must take over.
Evaluate. $\frac{100}{1500} = 0.06667$ ; $ln 4 = 1.3863$ ; $x_{p r o t} = 0.06667 \times 1.3863 = 0.09242$ m. Why this step? Turn the symbolic length into metres so we can compare to $L$ ; this is the width of the shaded orange band.
Count slots. slots $= ⌈ L / x_{p r o t} ⌉ = ⌈ 0.90/0.09242 ⌉ = ⌈ 9.74 ⌉ = 10$ . Why this step? Each slot buys $x_{p r o t}$ of protected wall; we need to cover the whole $0.90$ m, so we round up. (The ceiling $⌈ y ⌉$ just means "round $y$ up to the next whole number" — e.g. $⌈ 9.74 ⌉ = 10$ — because you cannot install a fractional slot and cannot leave a bare strip.)

Verify: Coverage fraction of one slot $f = x_{p r o t} / L = 0.09242/0.90 = 0.1027$ (~10%). Ten slots give $10 \times 0.1027 = 1.027 \geq 1$ ✓ — just enough. Nine slots would give $0.924 < 1$ , leaving a hot gap, so 10 is right. ✓

Worked example Example 6 — Cell H: exam twist, solve for the hidden coolant flow

Design requirement: a single slot must keep $η \geq 0.30$ over the first $x = 0.15$ m, with $h_{g} = 1500$ , $c_{p} = 2000$ . What minimum $\overset{m}{˙}_{c}$ do you need?

Forecast: we're running the decay law backwards — do you expect $\overset{m}{˙}_{c}$ around 0.05 or 0.5?

Set effectiveness to the floor at the target station. $exp (- \frac{h _{g} x}{m ˙ _{c} c _{p}}) = 0.30$ . Why this step? The worst spot is the far edge $x = 0.15$ ; if $η = 0.30$ exactly there, everything upstream is safer. On the middle panel, we are forcing the curve to pass through the point $(0.15, 0.30)$ .
Take $ln$ of both sides. $- \frac{h _{g} x}{m ˙ _{c} c _{p}} = ln 0.30 = - 1.204$ , so $\frac{h _{g} x}{m ˙ _{c} c _{p}} = 1.204$ . Why this step? $ln$ is the exact inverse of $exp$ — it "unwraps" the exponent so we can isolate $\overset{m}{˙}_{c}$ .
Solve for $\overset{m}{˙}_{c}$ . $\overset{m}{˙}_{c} = \frac{h _{g} x}{c _{p} \times 1.204} = \frac{1500 \times 0.15}{2000 \times 1.204} = \frac{225}{2408} = 0.0934 kg/(m\cdots)$ . Why this step? Algebraic rearrangement isolates the unknown we were asked for.

Verify: Feed $\overset{m}{˙}_{c} = 0.0934$ back into the forward law: exponent $= \frac{1500 \times 0.15}{0.0934 \times 2000} = \frac{225}{186.8} = 1.204$ , and $e^{- 1.204} = 0.300$ ✓ — matches the requirement exactly.

Worked example Example 7 — Cell I: threshold sensitivity (super-linear cost)

With $h_{g} = 1500$ , $c_{p} = 2000$ , $\overset{m}{˙}_{c} = 0.05$ , $L = 0.50$ m, compare the coverage fraction for $η_{min} = 0.10$ versus a stricter $η_{min} = 0.50$ .

Forecast: demanding a higher minimum effectiveness — does coverage go up or down, and by how much?

Loose floor $η_{min} = 0.10$ . $f = \frac{m ˙ _{c} c _{p}}{h _{g} L} ln \frac{1}{0.10} = \frac{100}{750} ln 10 = 0.1333 \times 2.3026 = 0.3070$ (~31%). Why this step? A low bar is easy to clear, so the film "counts as protecting" for longer — the leftmost, tall part of the right panel curve.
Strict floor $η_{min} = 0.50$ . $f = \frac{100}{750} ln \frac{1}{0.50} = 0.1333 \times 0.6931 = 0.0924$ (~9%). Why this step? Insisting on half-strength protection everywhere shrinks the qualifying region dramatically — the right panel curve has plunged.
Compare. Ratio $= 0.3070/0.0924 = 3.32$ : raising the floor from 0.10 to 0.50 cut coverage by a factor >3. Why this step? Because coverage rides on $ln (1/ η_{min})$ , and $ln 10/ ln 2 = 3.32$ — the log makes strict thresholds punishingly expensive.

Verify: $\frac{100}{750} = 0.13 \overline{3}$ ✓; both $f$ values lie in $[0, 1]$ ✓; the ratio $ln 10/ ln 2 = 3.322$ matches the coverage ratio ✓. This is the "diminishing returns / super-linear cost" behaviour: pushing the floor up costs coolant far faster than linearly.

Worked example Example 8 — Cell J: the sign / sanity trap (two red flags + the

$T_{coo l} > T_{h o t}$ edge case) An intern reports: "I measured $T_{h o t} = 2800$ K, $T_{coo l} = 1000$ K, and computed $η = 1.4$ , giving a wall at $T_{a w} = 280$ K." Find both red flags, then explain the related edge case where $T_{coo l} > T_{h o t}$ .

Forecast: two red flags are hiding in the intern's numbers — find them before reading on.

Red flag #1 — $η$ is outside $[0, 1]$ . Use the inverted form derived above, $η = \frac{T _{h o t} - T _{a w}}{T _{h o t} - T _{coo l}}$ , which is a fraction of the way from hot to cool, so it is physically bounded to $[0, 1]$ . $η = 1.4 > 1$ means "the wall is cooler than the coolant" — impossible for a film that can only blend the two streams, never refrigerate below the coldest one. Why this step? Any $η > 1$ or $η < 0$ is an instant "stop, you made an error" signal. (This inverted formula is just the blend rearranged — see the definition callout near the top.)
Red flag #2 — $T_{a w}$ falls outside the $[T_{coo l}, T_{h o t}]$ bracket. Check the blend the intern's $η$ implies: $T_{a w} = (1 - 1.4) (2800) + (1.4) (1000) = (- 0.4) (2800) + 1400 = - 1120 + 1400 = 280$ K. But the bracket rule from Ex 2 says a real wall must satisfy $T_{coo l} \leq T_{a w} \leq T_{h o t}$ , i.e. $1000 \leq T_{a w} \leq 2800$ . The value $280$ K sits below $T_{coo l} = 1000$ K. Why this step? The negative weight $(1 - η) = - 0.4$ pushed the "average" below both inputs — a weighted average with a negative weight is not an average at all. On the left panel of the figure the dot would fall clean off the bottom of the thermometer. This is the second independent way the same error shows up: flag #1 is in $η$ , flag #2 is in $T_{a w}$ , and both point to the same bad data.
The related edge case — $T_{coo l} > T_{h o t}$ . Suppose instead the "coolant" is hotter than the core gas, say $T_{coo l} = 2000$ K while $T_{h o t} = 1500$ K. Then the denominator $T_{h o t} - T_{coo l} = 1500 - 2000 = - 500$ is negative, so the definition of $η$ flips sign meaning: "cooling" now heats the wall. The blend still works — e.g. $η = 0.35$ gives $T_{a w} = (0.65) (1500) + (0.35) (2000) = 975 + 700 = 1675$ K — but notice $1675$ K is now between $1500$ and $2000$ , i.e. the bracket has simply reordered to $[T_{h o t}, T_{coo l}]$ . The film raises the wall temperature; it is no longer coolant. Why this step? It exposes the hidden assumption behind every earlier example — $T_{coo l} < T_{h o t}$ . Break it and the physics inverts: your "film" becomes a heater, so always confirm the injected stream is genuinely colder than the core gas before calling $η$ an "effectiveness".

Verify: Flag #1: $η = 1.4 \in / [0, 1]$ ✓ impossible. Flag #2: $T_{a w} = 280$ K $< T_{coo l} = 1000$ K, outside $[1000, 2800]$ ✓ impossible. Edge case: with $T_{coo l} = 2000 > T_{h o t} = 1500$ and $η = 0.35$ , $T_{a w} = 1675$ K lies inside the reordered bracket $[1500, 2000]$ ✓ — a consistent (but non-cooling) blend. Any physical film requires $T_{coo l} < T_{h o t}$ and $η \in [0, 1]$ ; violate either and one of the two brackets breaks.

Active recall

Recall Which matrix cell does each trap belong to?

"Wall colder than the coolant" ::: Cell J — impossible, $η$ left $[0, 1]$ and $T_{a w}$ left its bracket.
"No coolant injected" ::: Cell E — $\overset{m}{˙}_{c} \to 0$ , $η \to 0$ , $T_{a w} \to T_{h o t}$ .
"How many slots for a full chamber?" ::: Cell G — round $L / x_{p r o t}$ up (ceiling).
"Solve for the coolant flow given a target $η$ " ::: Cell H — invert with $ln$ .
"Coolant hotter than the core gas" ::: Cell J edge case — denominator flips sign, film becomes a heater.

Recall Quick numeric checks

Ex1 $T_{a w}$ ? ::: 2360 K Ex3 $η$ ? ::: $e^{- 2} = 0.135$ Ex5 slot count? ::: 10 slots ( $x_{p r o t} = 0.0924$ m) Ex6 minimum $\overset{m}{˙}_{c}$ ? ::: 0.0934 kg/(m·s) Ex7 coverage ratio? ::: 3.32× (that is $ln 10/ ln 2$ ) Ex8 edge-case $T_{a w}$ ( $T_{coo l} = 2000$ , $T_{h o t} = 1500$ , $η = 0.35$ )? ::: 1675 K

Connections

Parent: Film cooling — effectiveness, coverage fraction (index 3.3.29) — formulas derived there, exercised here.
Convective heat transfer coefficient — supplies $h_{g}$ , the decay driver in Ex 3–6.
Stanton number — the dimensionless group $h_{g} x / (\overset{m}{˙}_{c} c_{p})$ is a Stanton-length parameter.
Adiabatic wall temperature & recovery factor — defines $T_{h o t}$ and $T_{a w}$ used in every blend.
Boundary layer & entrainment — the physics behind the exponential decay.
Combustion chamber thermal design — where slot-counting (Ex 5) feeds the real layout.