Transpiration cooling: Worked examples — Physics deep dive

This page is the "no gap left" companion to Transpiration Cooling. The parent gave you the master result. Here we deliberately hunt down every kind of input the formula can face — normal numbers, zeros, infinities, and the sneaky edge cases — and grind each one to a number you can trust.

Recall The two formulas we lean on the whole way

$T_{w} = \frac{η h _{0} T _{g} + G c _{p} T _{c}}{η h _{0} + G c _{p}}, ϕ = \frac{G c _{p}}{η h _{0} + G c _{p}} = \frac{1}{1 + \frac{η h _{0}}{G c _{p}}} .$ Read $ϕ$ as "how far the wall has cooled, on a 0-to-1 scale, from the gas toward the coolant."

Before touching numbers, let us name the ingredients ONE more time in plain words so no symbol is unearned:

Definition The five knobs (in words)

$T_{g}$ — temperature of the hot gas hammering the wall (kelvin, K).
$T_{c}$ — temperature the coolant enters at, cold (K).
$T_{w}$ — the wall's steady temperature — the thing we usually solve for (K).
$h_{0}$ — the bare heat-transfer coefficient: how many watts cross each square metre of wall for every 1 K of gas-to-wall temperature gap, with no coolant blowing. Units $W m^{- 2} K^{- 1}$ . (See Convective Heat Transfer.)
$η$ — the blowing reduction factor: a pure number between 0 and 1 that says what fraction of that heat still gets through once coolant sweats out. $η = 1$ = no help, $η \to 0$ = perfect blanket.
$G$ — coolant mass flux: kilograms of coolant pushed through each square metre of wall per second ( $kg m^{- 2} s^{- 1}$ ).
$c_{p}$ — specific heat of the coolant: joules to warm 1 kg of it by 1 K.

The two products $η h_{0}$ and $G c_{p}$ both have units $W m^{- 2} K^{- 1}$ — they are conductances. That is why they can be added and compared in one fraction.

The scenario matrix

Every situation this topic can throw at you is one of the cells below. The examples that follow are labelled by cell so you can see the coverage is complete.

Cell	What makes it special	Covered by
A. Normal	all knobs finite and typical	Ex 1
B. Design inverse	solve for the coolant $G$ to hit a target $T_{w}$	Ex 2
C. Zero coolant ( $G = 0$ )	degenerate: no blowing at all	Ex 3
D. Perfect / infinite coolant ( $G \to \infty$ )	limiting behaviour, the floor	Ex 3
E. No blowing benefit ( $η = 1$ )	coolant is only a heat sink, no film	Ex 4
F. Non-linearity / diminishing returns	doubling $G$ ≠ halving $Δ T$	Ex 5
G. Real-world word problem	throat area → total coolant, thrust cost	Ex 6
H. Exam twist	$η$ depends on $G$ (variable blowing)	Ex 7
I. Consistency / sanity	recompute $T_{w}$ from $ϕ$ , cross-check units	Ex 8

We visualise the whole map first, so you can see where each case lives on the cooling curve.

Cell A — the everyday case

Worked example Example 1 — the plain wall temperature (Cell A)

$T_{g} = 3000 K$ , $T_{c} = 300 K$ , $c_{p} = 1200 J k g^{- 1} K^{- 1}$ , $h_{0} = 5000 W m^{- 2} K^{- 1}$ , $η = 0.4$ , $G = 2 kg m^{- 2} s^{- 1}$ .

Forecast: the two conductances will be close (2000 vs 2400), so $T_{w}$ should land between $T_{g}$ and $T_{c}$ , slightly nearer the cold side — guess ~1500 K.

Gas conductance $η h_{0} = 0.4 \times 5000 = 2000$ . Why this step? This is the effective heat delivery per kelvin after the coolant thins the boundary layer.
Coolant conductance $G c_{p} = 2 \times 1200 = 2400$ . Why this step? This is how much heat the coolant can soak up per kelvin, per square metre.
Weighted average: $T_{w} = \frac{2000 ( 3000 ) + 2400 ( 300 )}{2000 + 2400} = \frac{6 720 000}{4400} \approx 1527 K$ . Why this step? $T_{w}$ is exactly the conductance-weighted average of the two temperatures.

Verify: $ϕ = \frac{2400}{4400} = 0.545$ . Then $T_{g} - ϕ (T_{g} - T_{c}) = 3000 - 0.545 (2700) = 1527 K$ ✔. Between 300 and 3000, closer to cold — matches the forecast.

Cell B — designing for a target

Worked example Example 2 — how much coolant to hit a target (Cell B)

Same gas/coolant as Ex 1, $η = 0.4$ held constant. We demand $T_{w} \leq 1200 K$ . Find the smallest $G$ .

Forecast: 1200 K is colder than the 1527 K we got with $G = 2$ , so we must push more coolant — guess $G$ a bit above 3.

Target effectiveness: $ϕ = \frac{3000 - 1200}{3000 - 300} = \frac{1800}{2700} = 0.667$ . Why this step? When the target is a wall temperature, $ϕ$ is the cleanest handle — it turns a temperature into a dimensionless goal.
Invert $ϕ$ : from $ϕ = \frac{G c _{p}}{η h _{0} + G c _{p}}$ algebra gives $G c_{p} = \frac{ϕ}{1 - ϕ} η h_{0}$ . Why this step? We know $ϕ$ and want $G$ , so isolate the coolant conductance.
Numbers: $G c_{p} = \frac{0.667}{0.333} \times 2000 = 2 \times 2000 = 4000$ , so $G = \frac{4000}{1200} \approx 3.33 kg m^{- 2} s^{- 1}$ . Why this step? Divide the required conductance by $c_{p}$ to get mass flux.

Verify: put $G = 3.33$ back: $G c_{p} = 4000$ , $T_{w} = \frac{2000 ( 3000 ) + 4000 ( 300 )}{2000 + 4000} = \frac{7 200 000}{6000} = 1200 K$ ✔ exactly the target.

Cells C & D — the two degenerate ends

These are the "what happens at the extremes" cases. They are where students panic because a $0$ or an $\infty$ appears — but the formula stays perfectly well-behaved.

Worked example Example 3 — turn the coolant off, then to infinity (Cells C & D)

Same numbers as Ex 1.

Forecast: with $G = 0$ the wall must be as hot as the gas (nothing removes heat). With $G \to \infty$ the wall must chill down to the coolant temperature. So expect $T_{w} \to 3000$ and $T_{w} \to 300$ .

Zero coolant, $G = 0$ (Cell C): $G c_{p} = 0$ , so $T_{w} = \frac{2000 ( 3000 ) + 0}{2000 + 0} = 3000 K$ , and $ϕ = \frac{0}{2000 + 0} = 0$ . Why this step? With no coolant conductance, the weighted average has all its weight on the gas — the wall melts.
Infinite coolant, $G \to \infty$ (Cell D): divide top and bottom by $G c_{p}$ : $T_{w} = \frac{\frac{η h _{0}}{G c _{p}} T _{g} + T _{c}}{\frac{η h _{0}}{G c _{p}} + 1} \to \frac{0 + T _{c}}{0 + 1} = T_{c} = 300 K$ , and $ϕ \to 1$ . Why this step? As coolant conductance dominates, the weighted average collapses onto the cold side — that is the floor.
The floor is real: $T_{w}$ can never go below $T_{c} = 300 K$ no matter how much coolant you dump. Why this step? It sets a hard limit designers respect — you cannot cool below your coolant's inlet temperature.

Verify: at $G = 0$ , $ϕ = 0$ ✔; the $G \to \infty$ limit gives $ϕ = 1$ , $T_{w} = T_{c} = 300$ ✔. Both endpoints match the forecast and bracket every finite answer.

Cell E — heat sink only, no film

Worked example Example 4 — coolant that soaks heat but forms no blanket (

$η = 1$ , Cell E) Imagine the coolant absorbs heat inside the wall but does NOT thin the boundary layer, so blowing gives no benefit: $η = 1$ . Keep $T_{g} = 3000$ , $T_{c} = 300$ , $h_{0} = 5000$ , $c_{p} = 1200$ , $G = 2$ .

Forecast: losing the film means the gas delivers full heat ( $η h_{0} = 5000$ instead of 2000), so the wall runs hotter than Ex 1's 1527 K — guess ~1900 K.

Gas conductance $η h_{0} = 1 \times 5000 = 5000$ . Why this step? With $η = 1$ the full bare coefficient acts — this is the worst case for a given $h_{0}$ .
Coolant conductance $G c_{p} = 2 \times 1200 = 2400$ (unchanged). Why this step? The heat-sink capacity does not depend on $η$ .
Wall temperature $T_{w} = \frac{5000 ( 3000 ) + 2400 ( 300 )}{5000 + 2400} = \frac{15 720 000}{7400} \approx 2124 K$ . Why this step? Same weighted average with the new gas conductance.

Verify: $ϕ = \frac{2400}{7400} = 0.324$ — much less effective than Ex 1's 0.545, and $2124 > 1527$ ✔. This is exactly why the film effect matters: it drops $η$ and buys you hundreds of kelvin. Compare with Film Cooling, which relies on this blanket mechanism heavily.

Cell F — the diminishing-returns trap

Worked example Example 5 — double the coolant, does the wall halve? (Cell F)

Back to Ex 1 numbers, but now change $G$ from 2 to 4.

Forecast: linear intuition whispers "double coolant → half temperature," i.e. from 1527 down to ~760. Write that guess down — then watch it fail.

New coolant conductance $G c_{p} = 4 \times 1200 = 4800$ . Why this step? Only $G$ changed, so only this product updates.
New wall temperature $T_{w} = \frac{2000 ( 3000 ) + 4800 ( 300 )}{2000 + 4800} = \frac{7 440 000}{6800} \approx 1094 K$ . Why this step? Re-evaluate the same weighted average.
Measure the drop: $1527 - 1094 = 433 K$ , not the ~767 K a halving would demand. Why this step? The average is pinned above the floor $T_{c} = 300$ , so extra coolant buys ever-smaller gains.

Verify: halving 1527 would give 763.5 K; the true answer 1094 K is well above it ✔. The curve in figure s01 is concave, flattening toward $T_{c}$ — that shape is the whole warning.

Cell G — a full real-world word problem

Worked example Example 6 — total coolant for a nozzle throat, and its thrust cost (Cell G)

A throat has porous wall area $A = 0.05 m^{2}$ . Use Ex 1 conditions and require $T_{w} = 1200 K$ . From Ex 2 the needed mass flux is $G = 3.33 kg m^{- 2} s^{- 1}$ . Total propellant flow is $\overset{m}{˙}_{prop} = 20 kg s^{- 1}$ . (a) Find the total coolant flow $\overset{m}{˙}_{c}$ . (b) What fraction of propellant is spent cooling?

Forecast: a small patch times a few kg per m² per s → well under 1 kg/s, so maybe under 1% of a 20 kg/s engine. Guess ~0.8%.

Total coolant $\overset{m}{˙}_{c} = G A = 3.33 \times 0.05 = 0.1665 kg s^{- 1}$ . Why this step? Mass flux times area = mass flow; this is the definition $G = \overset{m}{˙}_{c} / A$ rearranged.
Coolant fraction $\frac{m ˙ _{c}}{m ˙ _{prop}} = \frac{0.1665}{20} = 0.0083 = 0.83%$ . Why this step? Coolant is propellant you divert from thrust, so its fraction is the real cost — it lowers effective Specific Impulse.
Interpret: under 1% of flow keeps the throat at a survivable 1200 K. Why this step? This is precisely why transpiration cooling wins: tiny mass penalty for a huge temperature drop.

Verify: $0.1665/20 = 0.008325 \approx 0.83%$ ✔, matching the forecast; and $\overset{m}{˙}_{c} = G A$ has units $(kg m^{- 2} s^{- 1}) (m^{2}) = kg s^{- 1}$ ✔.

Cell H — the exam twist: $η$ falls as $G$ rises

The parent warned that real blowing makes $η$ decrease with $G$ (thicker blanket). Here is that twist made concrete.

Worked example Example 7 — variable blowing,

$η = 0.4 - 0.05 G$ (Cell H) Ex 1 conditions, but now $η$ is not fixed: $η (G) = 0.4 - 0.05 G$ (valid for our range). Take $G = 2$ .

Forecast: at $G = 2$ , $η$ drops below 0.4, so the gas delivers less heat than in Ex 1 → the wall should be cooler than 1527 K. Guess ~1350 K.

Evaluate $η$ : $η = 0.4 - 0.05 (2) = 0.3$ . Why this step? Blowing at $G = 2$ has already thickened the blanket, weakening the gas coupling.
Gas conductance $η h_{0} = 0.3 \times 5000 = 1500$ . Why this step? Lower $η$ means lower effective delivery — this is the "double win" from the parent note.
Coolant conductance $G c_{p} = 2 \times 1200 = 2400$ (as before). Why this step? Heat-sink capacity is unchanged by $η$ .
Wall temperature $T_{w} = \frac{1500 ( 3000 ) + 2400 ( 300 )}{1500 + 2400} = \frac{5 220 000}{3900} \approx 1338 K$ . Why this step? Same weighted average with the improved gas conductance.

Verify: $1338 < 1527$ ✔ — variable blowing beats the constant- $η$ estimate, exactly as the parent predicted; $ϕ = \frac{2400}{3900} = 0.615 > 0.545$ ✔ (more effective). Forecast ~1350 was close.

Cell I — closing the loop on consistency

Worked example Example 8 — recover

$T_{w}$ two independent ways (Cell I) Using Ex 1's answer, prove $T_{w}$ from the boxed formula and from the $ϕ$ -definition agree, and check units.

Forecast: they must match — $ϕ$ was derived from the same balance — but doing it both ways catches arithmetic slips.

From the box: $T_{w} = 1527 K$ (Ex 1). Why this step? Our reference value.
From $ϕ$ : $ϕ = \frac{G c _{p}}{η h _{0} + G c _{p}} = \frac{2400}{4400} = 0.5 \overline{45}$ , then $T_{w} = T_{g} - ϕ (T_{g} - T_{c}) = 3000 - 0.5455 (2700) = 1527 K$ . Why this step? $ϕ$ is defined so $T_{g} - T_{w} = ϕ (T_{g} - T_{c})$ — an independent route.
Units audit: $η h_{0}$ and $G c_{p}$ are both $W m^{- 2} K^{- 1}$ ; in the fraction $\frac{( W m ^{- 2} K ^{- 1} ) ( K )}{W m ^{- 2} K ^{- 1}} = K$ . Why this step? A temperature formula must return kelvin, and it does.

Verify: both routes give 1527 K ✔; final units are K ✔. The formula is internally consistent — trust it across all cells above.

Recall Coverage self-check

Every cell A–I hit? ::: A(1) B(2) C&D(3) E(4) F(5) G(6) H(7) I(8) — all nine cells covered. Why can $T_{w}$ never fall below $T_{c}$ ? ::: Because $T_{w}$ is a conductance-weighted average of $T_{g}$ and $T_{c}$ ; the smallest a weighted average of two numbers can be is the smaller number itself. What single dimensionless group controls everything? ::: The conductance ratio $\frac{G c _{p}}{η h _{0}}$ , which sets $ϕ = \frac{1}{1 + η h _{0} / ( G c _{p} )}$ .

Related schemes worth contrasting: Regenerative Cooling, Ablative Cooling, and the boundary-layer machinery behind $h_{0}$ in Boundary Layer Theory.