22-Mec-A6 Fluid Machinery · May 2013
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
The distinction is where the pressure drop — and hence the fluid acceleration — occurs. In a pure impulse stage the entire pressure drop takes place in the fixed nozzles; the high-speed jet then crosses the moving blades at essentially constant pressure. Because the relative speed is unchanged in magnitude (only its direction is reversed), the moving passages have a symmetrical, constant-area shape. The force on the blades is the reaction to changing the direction of the jet's momentum — the blades are pushed by the impulse of the deflected stream. The Pelton wheel and the Curtis/velocity-compounded steam stage are examples.
In a reaction stage the pressure drops in both the fixed and the moving rows. The moving passages are shaped like nozzles, so the relative velocity increases across them; the blades feel both an impulse force from turning the flow and an additional reaction force from accelerating the fluid relative to themselves (the same principle that drives a garden sprinkler or a rocket). In the common 50% reaction design the enthalpy drop is split equally between fixed and moving rows, making the two velocity triangles mirror images. Reaction stages have lower blade loading per stage and higher efficiency, but need more stages and must handle a pressure difference across the rotor (hence tip leakage and axial thrust). In velocity-diagram terms, the impulse triangle has VR2 = VR1 (bar friction) and is symmetric about the axial line, whereas the 50%-reaction triangle is symmetric about the vertical, with the relative velocity growing from inlet to outlet.
For a Pelton wheel the work done per unit mass is proportional to the product of the blade speed and the change of whirl, w = U(V − U)(1 + k cosθ), where θ is the bucket deflection angle and k the friction factor. At zero blade speed the buckets do not move, so no work is done however large the force — efficiency is zero. At the other extreme, when the blade speed equals the jet speed, the jet cannot catch the buckets, no water strikes them, the flow through the wheel and the force both vanish, and efficiency is again zero. Between these limits the efficiency rises and falls as a parabola. Differentiating w with respect to U and setting the result to zero gives the maximum at U = V/2 — the blade should recede at half the jet speed. Physically, at this point the water leaves the bucket with almost no absolute velocity, so nearly all the jet's kinetic energy has been transferred to the wheel. In practice friction and windage move the true optimum to about U/V = 0.46–0.47 and cap the peak efficiency near 88–90%, but the shape and the half-speed rule are as shown.
