Turbine Cooling and Materials

Cooling effectiveness, film cooling, thermal barrier coatings, superalloys from equiaxed to single crystal, creep and Larson–Miller, and the cooling-air penalty

Hotter than the metal

The gas flowing over the first blades of a modern high-pressure turbine can be at 1300–1800 K; the melting point of the blade material is about 1300–1400 K (turbine blade). A bare blade would soften and fail in minutes. The blades survive for years. Air from the compressor runs through passages inside them and leaves through rows of tiny holes to lay a cool film over their surface. A ceramic coating a fraction of a millimetre thick shields the metal. And the metal itself, a nickel superalloy, is cast as a single crystal, with no grain boundaries along which it could creep. The Jumo 004's air-cooled steel blades of 1944 were the first step. This lesson follows the rest — and counts what the cooling air costs.

\varepsilon = \frac{T_g - T_m}{T_g - T_c}

Overall cooling effectiveness: how far the cooling pulls the blade metal temperature Tm from the gas temperature Tg toward the coolant temperature Tc (0 = uncooled, 1 = metal at coolant temperature).

💡Three ways to cool, one way to insulate

Convection: coolant flows through internal passages, often with ribs and pins to stir it.
Impingement: jets of coolant blast the inside of the leading edge, the hottest spot.
Film cooling: coolant leaves through holes in the surface and forms a protective layer — more effective than convection or impingement alone.
Thermal barrier coatings (TBCs): a thin ceramic layer with low conductivity, worth on the order of 100 K off the metal temperature.

Typical overall effectiveness is 0.5–0.7. Every percentage point costs cooling air.

📐How much cooling is enough?

Coolant from the compressor exit at 900 K and a metal limit of 1300 K (Lab 1):

Gas at 1900 K with $\varepsilon = 0.6$ : metal at 1300 K — just at the limit.
The effectiveness needed rises from 0.50 at 1700 K gas to 0.64 at 2000 K.
A TBC worth 120 K lowers the need: 0.545 instead of 0.600 at 1900 K.

💡From equiaxed to single crystal

Modern blades are nickel-based superalloys with chromium, cobalt and rhenium. At turbine temperatures metals creep — they stretch slowly under constant load — and creep is fastest along the boundaries between the crystal grains. Hence three generations:

Equiaxed: ordinary castings with grains in all directions.
Directionally solidified: grains grown along the blade, so no boundaries run across the main load.
Single crystal: one grain, no boundaries at all, typically with the crystal's ⟨001⟩ direction along the blade.

The breakthrough was as much the casting process as the alloys.

\mathrm{LMP} = \frac{T\,(C + \log_{10} t_r)}{1000}, \qquad C \approx 20

The Larson–Miller parameter: at a given stress, every combination of temperature T (K) and rupture time t_r (hours) with the same LMP is equivalent.

Derivation

Why the Larson–Miller parameter works

Creep is thermally activated: its rate follows an Arrhenius law,

\dot\varepsilon \propto e^{-Q/(\mathcal R T)}

, so the time to rupture at a fixed stress goes as

t_r = A\,e^{Q/(\mathcal R T)}

. Taking logarithms,

\log_{10} t_r = \log_{10} A + \frac{Q}{2.303\,\mathcal R\,T} \quad\Longrightarrow\quad T\left(\log_{10} t_r - \log_{10} A\right) = \frac{Q}{2.303\,\mathcal R} = \text{const}.

With

C = -\log_{10}A \approx 20

for many alloys,

T(C + \log_{10} t_r)

is constant at a given stress. Short, hot laboratory tests can then predict long, cooler service lives — and show how steeply life falls with temperature.

📐What a few kelvin are worth

With illustrative Larson–Miller values at the same stress (Lab 2):

At 1250 K a single-crystal blade lasts 10,000 h, a directionally solidified one about 1600 h, an equiaxed one about 100 h.
For the single crystal near 1280 K, every 16 K hotter halves the life.
At the same 10,000-hour life, the single crystal can run 104 K hotter than the equiaxed alloy.

💡The cooling-air penalty

Cooling air is compressed — it has cost compressor work — but it skips the burner and does little work in the turbine; mixing it back into the gas also loses stagnation pressure. Raising

T_{t4}

raises the ideal cycle's work and efficiency, but needs more cooling air. In a deliberately simple model (Lab 3,

\pi_c = 30

, metal limit 1300 K):

The coolant fraction grows from 2 % at 1400 K to 19 % at 2200 K.
Thermal efficiency peaks near 1950 K at 39.4 %, against 47 % for the same cycle uncooled.
Work per kilogram keeps rising — so designers still push $T_{t4}$ for a smaller, lighter core.

If you don't have a consensus that it's nonsense, you don't have a breakthrough.

– Burt Rutan, Time, 29 November 2004 (Wikiquote)

💡Ceramic matrix composites

Ceramic matrix composites — ceramic fibres in a ceramic matrix — are about a third the density of nickel alloys and tolerate higher temperatures, so they need far less cooling air. The GE9X uses them in two combustor liners, two nozzles and a turbine shroud; they are not yet used for the most highly stressed first-stage turbine blades.

💡Where you will use this

Lesson 7-3 applies film cooling to combustor liners. Lesson 8-2 shows why hot days and climbs are when turbines are hottest. Lesson 9-2 treats the disc that carries these blades.

📚Further reading

Roger C. Reed, The Superalloys: Fundamentals and Applications (Cambridge University Press, 2006).
Je-Chin Han, Sandip Dutta and Srinath Ekkad, Gas Turbine Heat Transfer and Cooling Technology (2nd ed., CRC Press, 2012).
Turbine blade on Wikipedia, on cooling methods and materials.

↓But there is a problem

Upstream of the turbine sits the combustor: a flame that must burn kerosene completely in air rushing through at tens of metres per second, at 40 bar, without ever going out. Pour more fuel into it, and something odd happens: beyond a point the flame stops getting hotter. The next unit is about why.