Turbine design engineers have to ensure that film cooling can provide sufficient protection to turbine blades from the hot mainstream gas, while keeping the losses low. Film cooling hole design parameters include inclination angle (α), compound angle (β), hole inlet geometry, and hole exit geometry. The influence of these parameters on aerodynamic loss and net heat flux reduction is investigated, with loss being the primary focus. Low-speed flat plate experiments have been conducted at momentum flux ratios of IR = 0.16, 0.64, and 1.44. The film cooling aerodynamic mixing loss, generated by the mixing of mainstream and coolant, can be quantified using a three-dimensional analytical model that has been previously reported by the authors. The model suggests that for the same flow conditions, the aerodynamic mixing loss is the same for holes with different α and β but with the same angle between the mainstream and coolant flow directions (angle κ). This relationship is assessed through experiments by testing two sets of cylindrical holes with different α and β: one set with κ = 35 deg, and another set with κ = 60 deg. The data confirm the stated relationship between α, β, κ and the aerodynamic mixing loss. The results show that the designer should minimize κ to obtain the lowest loss, but maximize β to achieve the best heat transfer performance. A suggestion on improving the loss model is also given. Five different hole geometries (α = 35.0 deg, β = 0 deg) were also tested: cylindrical hole, trenched hole, fan-shaped hole, D-Fan, and SD-Fan. The D-Fan and the SD-Fan have similar hole exits to the fan-shaped hole but their hole inlets are laterally expanded. The external mixing loss and the loss generated inside the hole are compared. It was found that the D-Fan and the SD-Fan have the lowest loss. This is attributed to their laterally expanded hole inlets, which lead to significant reduction in the loss generated inside the holes. As a result, the loss of these geometries is ≈ 50% of the loss of the fan-shaped hole at IR = 0.64 and 1.44.

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