Film cooling — effectiveness, coverage fraction: Exercises

Before we start, here are the only four relations you will ever plug into. Read them once so no symbol is a stranger.

Here is what those symbols mean, in plain words, so you never have to guess:

Definition Symbol dictionary

$T_{h o t}$ ::: the recovery / adiabatic temperature of the hot core gas — what the wall would feel with no film (the enemy).
$T_{coo l}$ ::: temperature of the cool gas we inject along the wall (the ally).
$T_{a w}$ ::: adiabatic wall temperature — the temperature the wall actually reaches, assuming zero conduction into the wall.
$η$ ::: effectiveness. $1$ = perfect protection (wall at $T_{coo l}$ ), $0$ = no protection (wall at $T_{h o t}$ ).
$h_{g}$ ::: gas-side convective heat-transfer coefficient (W m $^{- 2}$ K $^{- 1}$ ) — how fast hot gas heats the film.
$\overset{m}{˙}_{c}$ ::: coolant mass flow per unit wall width (kg m $^{- 1}$ s $^{- 1}$ ) — how much cold "stuff" we spent.
$c_{p}$ ::: specific heat of the coolant (J kg $^{- 1}$ K $^{- 1}$ ).
$x$ ::: distance downstream from the injection slot (m).
$L$ ::: total chamber length to be protected (m).
$η_{min}$ ::: the smallest effectiveness the design still calls "protected".

The single figure below is the map every exercise lives on. Read it as follows: the cyan curve is relation (C) — effectiveness $η$ starting at $1$ at the slot ( $x = 0$ , top-left) and sagging toward $0$ as you move right. The amber dashed horizontal line is a chosen threshold $η_{min}$ ; the amber dotted vertical line marks $x_{p r o t}$ , the exact distance where the curve crosses that threshold. The shaded region to its left is the "protected" stretch whose length feeds relation (D). Keep this picture in mind: L2 finds a single point on the cyan curve, L3 finds where the curve meets the amber line, and L4/L5 stack copies of this curve.

Figure — Film cooling — effectiveness, coverage fraction

Level 1 — Recognition

Can you read the definitions and identify the right formula?

L1.1 A wall reaches $T_{a w} = T_{coo l}$ . What is $η$ ? What does this physically mean?

L1.2 At the injection slot ( $x = 0$ ), evaluate $η (0)$ from relation (C). Does the value match physical intuition?

L1.3 Which of the four relations (A)–(D) would you use to answer: "What fraction of the chamber is protected above $η = 0.3$ ?"

Recall Solutions — L1

L1.1 Put $T_{a w} = T_{coo l}$ into relation (A): $η = \frac{T _{h o t} - T _{coo l}}{T _{h o t} - T _{coo l}} = 1.$ $η = 1$ is perfect protection — the wall is as cool as the injected gas. This is the best case, at the slot itself (the top-left corner of the figure above).

L1.2 Set $x = 0$ in (C): the exponent becomes $- \frac{h _{g} \cdot 0}{m ˙ _{c} c _{p}} = 0$ , and $e^{0} = 1$ . So $η (0) = 1$ . ✔ It matches: right at the slot the film is pure coolant, no hot gas has mixed in yet, so protection is total — this is where the cyan curve touches the top axis.

L1.3 "Fraction of the chamber protected above a threshold" is exactly the coverage question → use relation (D) with $η_{min} = 0.3$ .

Level 2 — Application

Plug numbers into one formula, get an answer, keep units clean.

L2.1 $T_{h o t} = 3200$ K, $T_{coo l} = 800$ K, and at some station $η = 0.5$ . Find $T_{a w}$ .

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 $^{- 1}$ , and $x = 0.20$ m. Find $η (x)$ .

L2.3 For the numbers in L2.2, at what distance $x$ has effectiveness fallen to $η = 0.5$ ?

Recall Solutions — L2

L2.1 Use relation (B): $T_{a w} = (1 - 0.5) (3200) + (0.5) (800) = 1600 + 400 = 2000 K .$ The film chopped $3200 - 2000 = 1200$ K off the wall.

L2.2 First the dimensionless exponent: $\frac{h _{g} x}{m ˙ _{c} c _{p}} = \frac{1200 \times 0.20}{0.04 \times 2500} = \frac{240}{100} = 2.4.$ Then $η = e^{- 2.4} = 0.0907 \approx 0.091$ . The film is almost spent here — this is a single point far down the right of the cyan curve.

L2.3 Set $η = 0.5$ in (C) and solve for $x$ . Take $ln$ of both sides: $- \frac{h _{g} x}{m ˙ _{c} c _{p}} = ln 0.5 \Rightarrow x = \frac{m ˙ _{c} c _{p}}{h _{g}} ln \frac{1}{0.5} .$ $x = \frac{0.04 \times 2500}{1200} \times ln 2 = \frac{100}{1200} \times 0.6931 = 0.0833 \times 0.6931 = 0.0578 m .$ So effectiveness halves in about 5.8 cm — this is the film's "half-life length."

Level 3 — Analysis

Combine two relations, or reason about how one knob changes the answer.

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, $η_{min} = 0.25$ . Find the coverage fraction $f$ .

L3.2 (Sensitivity). For L3.1, how does the protected length $x_{p r o t}$ change if you double the coolant flow $\overset{m}{˙}_{c}$ ? Give the factor and explain in one line.

L3.3 (Threshold cost). For L3.1's baseline, compare the protected length required for $η_{min} = 0.5$ versus $η_{min} = 0.1$ . Which threshold demands a longer protected region, and by what ratio?

Recall Solutions — L3

L3.1 Use relation (D). This is the amber-line-crossing in the figure above: we want where the cyan curve meets $η_{min} = 0.25$ . First the prefactor $\frac{m ˙ _{c} c _{p}}{h _{g}} = \frac{0.05 \times 2000}{1500} = \frac{100}{1500} = 0.06667$ m. Then: $x_{p r o t} = 0.06667 \times ln \frac{1}{0.25} = 0.06667 \times ln 4 = 0.06667 \times 1.3863 = 0.09242 m .$ $f = \frac{x _{p r o t}}{L} = \frac{0.09242}{0.60} = 0.1540 \approx 15.4%.$

L3.2 In relation (D), $x_{p r o t} = \frac{m ˙ _{c} c _{p}}{h _{g}} ln (1/ η_{min})$ . The threshold term $ln (1/ η_{min})$ is fixed, so $x_{p r o t} \propto \overset{m}{˙}_{c}$ . Doubling $\overset{m}{˙}_{c}$ doubles $x_{p r o t}$ (factor $2$ ). More cold budget → proportionally longer protection, for a fixed threshold.

L3.3 The prefactor is common, so the ratio is set purely by the logs: $\frac{x _{p r o t} ( η _{min} = 0.1 )}{x _{p r o t} ( η _{min} = 0.5 )} = \frac{l n ( 1/0.1 )}{l n ( 1/0.5 )} = \frac{l n 10}{l n 2} = \frac{2.3026}{0.6931} = 3.322.$ A more lenient threshold ( $η_{min} = 0.1$ accepts a weaker film) allows a region about 3.3× longer. Equivalently, demanding a stronger film ( $η_{min} = 0.5$ ) shrinks the protected region.

Level 4 — Synthesis

Chain three or more relations, or solve for a design variable.

L4.1 (Design the coolant flow). You must protect the whole chamber, $f = 1$ , over $L = 0.40$ m at $η_{min} = 0.30$ , with $h_{g} = 1600$ W m $^{- 2}$ K $^{- 1}$ and $c_{p} = 1800$ J kg $^{- 1}$ K $^{- 1}$ . What coolant flow $\overset{m}{˙}_{c}$ (per unit width) is required from a single slot?

L4.2 (Wall temperature at coverage edge). Using the $\overset{m}{˙}_{c}$ from L4.1, and $T_{h o t} = 3300$ K, $T_{coo l} = 700$ K, find $T_{a w}$ exactly at the coverage edge $x = L$ (where $η = η_{min} = 0.30$ ).

L4.3 (Count the slots). Keep L3.1's parameters ( $x_{p r o t} = 0.09242$ m per slot) but now the chamber is $L = 0.60$ m. If each fresh slot re-establishes $η = 1$ and protects one $x_{p r o t}$ length before dropping to $η_{min}$ , how many slots are needed to cover the full length? Round up.

Recall Solutions — L4

L4.1 "Whole chamber from one slot" means $f = 1$ , i.e. $x_{p r o t} = L$ . Set $x_{p r o t} = L$ in relation (D) rearranged for $\overset{m}{˙}_{c}$ : $L = \frac{m ˙ _{c} c _{p}}{h _{g}} ln \frac{1}{η _{min}} \Rightarrow \overset{m}{˙}_{c} = \frac{h _{g} L}{c _{p} l n ( 1/ η _{min} )} .$ $\overset{m}{˙}_{c} = \frac{1600 \times 0.40}{1800 \times l n ( 1/0.30 )} = \frac{640}{1800 \times 1.20397} = \frac{640}{2167.15} = 0.2953 kg m^{- 1} s^{- 1} .$ That is a big flow (~0.30 kg per metre of width per second) — a single slot rarely covers a whole chamber, which is exactly why real engines stack several slots (see L4.3).

L4.2 At the coverage edge $η = 0.30$ , use relation (B): $T_{a w} = (1 - 0.30) (3300) + (0.30) (700) = 0.70 \times 3300 + 0.30 \times 700 = 2310 + 210 = 2520 K .$ Even at the weakest allowed film the wall is 780 K below flame temperature.

L4.3 Each slot covers $x_{p r o t} = 0.09242$ m. Number of slots: $N = ⌈ \frac{L}{x _{p r o t}} ⌉ = ⌈ \frac{0.60}{0.09242} ⌉ = ⌈ 6.492 ⌉ = 7 slots .$ You need 7 slots to blanket the whole 0.60 m chamber at $η \geq 0.25$ .

Level 5 — Mastery

Prove a relation, or reason at the limits where the algebra could mislead.

L5.1 (Derive from the ODE). Starting from the entrainment model $\frac{d ( Δ T )}{d x} = - \frac{h _{g}}{m ˙ _{c} c _{p}} Δ T$ with $Δ T \equiv T_{h o t} - T_{a w}$ and $Δ T (0) = T_{h o t} - T_{coo l}$ , derive relation (C): $η (x) = e^{- h_{g} x / (\overset{m}{˙}_{c} c_{p})}$ .

L5.2 (Two-slot effectiveness — limiting behaviour). Slot 1 injects at $x = 0$ ; a second identical slot injects at $x = d$ , resetting the local film to $η = 1$ there. Between the slots, the combined effectiveness is $η_{1} (x) = e^{- k x}$ with $k \equiv h_{g} / (\overset{m}{˙}_{c} c_{p})$ . Just before slot 2, $η$ has dropped to $e^{- k d}$ ; just after, it jumps back to $1$ . Show that placing slot 2 at $d = x_{p r o t}$ (where $η_{1}$ hits $η_{min}$ ) keeps effectiveness $\geq η_{min}$ everywhere on $[0, 2 d]$ , and identify the two locations where $η = η_{min}$ exactly.

L5.3 (Zero-flow and infinite-flow limits). From relation (C), evaluate $lim_{\overset{m}{˙}_{c} \to 0^{+}} η (x)$ and $lim_{\overset{m}{˙}_{c} \to \infty} η (x)$ for a fixed $x > 0$ , and state the physical meaning of each.

Recall Solutions — L5

L5.1 First fix the shorthand before any integration: let $Δ T_{0} \equiv Δ T (0) = T_{h o t} - T_{coo l}$ , the value of $Δ T$ at the slot (the full hot-to-cool gap, since the film there is pure coolant). Now the ODE is separable — we chose separation because the derivative of $Δ T$ is proportional to $Δ T$ itself, the signature of exponential decay. Divide both sides by $Δ T$ and multiply by $d x$ : $\frac{d ( Δ T )}{Δ T} = - \frac{h _{g}}{m ˙ _{c} c _{p}} d x .$ Integrate the left from $Δ T_{0}$ to $Δ T$ and the right from $0$ to $x$ : $ln \frac{Δ T}{Δ T _{0}} = - \frac{h _{g} x}{m ˙ _{c} c _{p}} \Rightarrow Δ T = Δ T_{0} e^{- h_{g} x / (\overset{m}{˙}_{c} c_{p})} .$ Now divide by $Δ T_{0} = T_{h o t} - T_{coo l}$ . Since $η = \frac{T _{h o t} - T _{a w}}{T _{h o t} - T _{coo l}} = \frac{Δ T}{Δ T _{0}}$ : $η (x) = e^{- h_{g} x / (\overset{m}{˙}_{c} c_{p})} . ■$

L5.2 The figure below shows the argument as a sawtooth: the cyan curve falls from $1$ , and at $x = d$ the amber vertical line marks slot 2 snapping $η$ back up to $1$ , after which the identical fall repeats. Follow the reasoning against it. On $[0, d]$ : $η (x) = e^{- k x}$ is a decreasing function, so its minimum on this interval is reached at the right endpoint $x = d$ , equal to $e^{- k d}$ . Choosing $d = x_{p r o t}$ means $e^{- k d} = η_{min}$ (that is the definition of $x_{p r o t}$ ). Hence on $[0, d]$ , $η \geq η_{min}$ , with equality reached only at the right endpoint $x = d$ . At $x = d$ slot 2 resets $η$ to $1$ , and on $[d, 2 d]$ the same falling curve gives its minimum $η_{min}$ at the right endpoint $x = 2 d$ . So $η \geq η_{min}$ across all of $[0, 2 d]$ , with equality holding at exactly the two endpoints $x = d$ and $x = 2 d$ . This is why slots are spaced one $x_{p r o t}$ apart — each slot's protection ends precisely where the next begins.

L5.3 The exponent is $- \frac{h _{g} x}{m ˙ _{c} c _{p}}$ , and only $\overset{m}{˙}_{c}$ varies (with $x > 0$ fixed).

$\overset{m}{˙}_{c} \to 0^{+}$ : the denominator shrinks, so the exponent $- \frac{h _{g} x}{m ˙ _{c} c _{p}} \to - \infty$ , and $η \to e^{- \infty} = 0$ . No coolant → no protection: the wall sits at $T_{h o t}$ . Physically sensible.
$\overset{m}{˙}_{c} \to \infty$ : the denominator blows up, so the exponent $\to 0^{-}$ , and $η \to e^{0} = 1$ for every finite $x$ . Infinite coolant → the whole wall is fully protected at the coolant temperature. Also sensible — and it confirms the formula behaves correctly at both extremes.

Connections

Adiabatic wall temperature & recovery factor — supplies $T_{a w}$ and $T_{h o t}$ used throughout these exercises.
Convective heat transfer coefficient — the $h_{g}$ that drives every decay here.
Stanton number — the dimensionless group $h_{g} x / (\overset{m}{˙}_{c} c_{p})$ is Stanton-number × length.
Boundary layer & entrainment — the mixing physics behind the decay ODE in L5.1.
Regenerative cooling — sets the real wall temperature once conduction is allowed.
Combustion chamber thermal design — where slot-counting (L4.3, L5.2) actually gets used.