Question 7 of 8: Turbine Blade Flow

Question 7: Turbine Blade Flow (10 marks)

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.

Part I — Pressure and velocity through four stages

In a pressure-compounded impulse turbine the total pressure drop is divided among several stages, each stage consisting of a fixed nozzle row followed by a moving row. The pressure therefore falls as a descending staircase: it drops in each fixed (nozzle) row, where the steam is accelerated, and stays constant across each moving row, where the impulse blades only turn the flow. The absolute velocity does the opposite — it rises sharply in each nozzle row (pressure energy converted to kinetic energy) and falls in each moving row as the steam gives up its kinetic energy to the blades. Over four stages this produces a repeated saw-tooth in velocity riding on a stepwise-descending pressure line.

Part II — Blade length and twist

(a) Why the blade lengthens and twists. As the steam expands through the turbine its pressure falls and its specific volume increases by orders of magnitude. To pass the same mass flow at a roughly constant axial velocity, the flow annulus area — and therefore the blade height — must grow steadily toward the low-pressure exhaust end. Once a blade is long, the blade speed $U=\omega r$ varies substantially from root to tip, because r changes appreciably over the blade height while ω is fixed. If the blade angle were constant, the incidence would be wrong everywhere except one radius, causing shock losses and uneven work. The blade is therefore twisted (a free-vortex design), its inlet and outlet angles changing continuously from root to tip so the relative flow enters each section smoothly and the work per unit mass stays reasonably uniform along the span.

(b) Root, mid, tip velocity diagrams. The Page 15 diagram (an impulse stage) is taken as the root section. With a free-vortex design the axial velocity is the same at every radius, the whirl varies as $C_w r=\text{const}$ (so it falls toward the tip), and the blade speed $U=\omega r$ rises. Illustrative proportions (radius ratio root : mid : tip = 1 : 1.5 : 2):

• (i) Root: large inlet whirl, small U; symmetric impulse triangle, blade inlet and exit angles both about 34°.
• (ii) Mid-height: whirl down to 1/1.5 of the root value, U up by 1.5; the relative inlet flow is now almost axial (β₁ ≈ 90°) while the exit angle closes to about 26°.
• (iii) Tip: whirl halved, U doubled; the relative inlet flow approaches from the other side of axial (β₁ ≈ 138°) and the exit angle closes further to about 21°, so the relative velocity rises across the tip section (reaction increases toward the tip).

The product U×(change of whirl) is the same at all three radii, so every section does equal work; the blade must therefore twist from an impulse-type section at the root to a reaction-type section at the tip.