Question 3 of 8: Steam Turbine Blade Efficiency (10 marks)

Question 3 — Steam Turbine Blade Efficiency (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.

Given. A single symmetrical, frictionless impulse stage: steam leaves the nozzles at 300 m/s inclined 25° to the direction of blade motion, the blades move at 100 m/s, and 24 kg/s of steam passes the stage.

Find. The absolute exhaust velocity, the tangential force on the blades, the work per kilogram of steam, the inlet and exhaust kinetic energies, the blade (diagram) efficiency, and the stage power.

Approach. Resolve the inlet absolute velocity into whirl and axial components, subtract the blade speed to obtain the relative velocity, mirror it about the axial direction (symmetrical, frictionless blades), add the blade speed back to recover the absolute exhaust velocity, then apply the paper's force, work and efficiency relations from page 20.

1. Resolve the inlet absolute velocity. Measuring $\theta$ from the blade-motion direction as the attachment requires, $$V_{S1}\cos\theta = 300\cos 25^\circ = 271.9\ \text{m}\,\text{s}^{-1}, \qquad V_{S1}\sin\theta = 300\sin 25^\circ = 126.8\ \text{m}\,\text{s}^{-1}$$ The first is the whirl (tangential) component that does the work; the second is the axial component that carries the steam through the passage and is unchanged by frictionless symmetrical blades.
2. Build the relative inlet velocity. Subtracting the blade speed from the whirl component gives the relative whirl $271.9 - 100 = 171.9\ \text{m}\,\text{s}^{-1}$, so $$V_{R1} = \sqrt{171.9^{2} + 126.8^{2}} = 213.6\ \text{m}\,\text{s}^{-1}, \qquad \phi = \arctan\frac{126.8}{171.9} = 36.41^\circ$$ For shock-free entry the blade inlet angle must equal this $\phi$, and by the symmetry condition the outlet angle $\gamma$ takes the same value.
3. Part (a) — mirror the relative velocity and recover the exhaust. Frictionless symmetrical blades reverse the relative whirl without changing its magnitude, so the exit relative whirl is $-171.9\ \text{m}\,\text{s}^{-1}$ and the exit absolute whirl is $-171.9 + 100 = -71.89\ \text{m}\,\text{s}^{-1}$ (that is, the steam still has a small backward swirl). With the axial component unchanged, $$V_{S2} = \sqrt{71.89^{2} + 126.8^{2}}$$ $$\boxed{V_{S2} = 145.8\ \text{m}\,\text{s}^{-1} \ \text{at}\ \delta = 119.6^\circ}$$
4. Part (b) — impulse force from the change of whirl. The page-20 relation is $F = M(V_{S1}\cos\theta - V_{S2}\cos\delta)$, and because $\cos\delta$ is negative the two contributions add: $$\Delta V_w = 271.9 - (-71.89) = 343.8\ \text{m}\,\text{s}^{-1}, \qquad F = 24 \times 343.8$$ $$\boxed{F = 8251\ \text{N} = 8.251\ \text{kN}}$$ For symmetrical frictionless blades this reduces to the compact form $\Delta V_w = 2(V_{S1}\cos\theta - V_B) = 2(271.9 - 100)$, which reproduces 343.8 m/s exactly.
5. Part (c) — energy transferred to the blades. Multiplying the change of whirl by the blade speed, $$w = \left(V_{S1}\cos\theta - V_{S2}\cos\delta\right)V_B = 343.8 \times 100 = 34\,378\ \text{J}\,\text{kg}^{-1}$$ $$\boxed{w = 34.38\ \text{kJ}\,\text{kg}^{-1}}$$ The aid sheet's alternative energy form $w = \tfrac{1}{2}\left[(V_{S1}^{2} - V_{S2}^{2}) + (V_{R2}^{2} - V_{R1}^{2})\right]$ gives the identical value, the second bracket vanishing because the blades are frictionless.
6. Part (d) — inlet and exhaust kinetic energies. Using $E_{KE} = V^{2}/2$ per unit mass, $$\frac{V_{S1}^{2}}{2} = \frac{300^{2}}{2} = 45\,000\ \text{J}\,\text{kg}^{-1}, \qquad \frac{V_{S2}^{2}}{2} = \frac{145.8^{2}}{2} = 10\,622\ \text{J}\,\text{kg}^{-1}$$ $$\boxed{\text{inlet } 45.00\ \text{kJ}\,\text{kg}^{-1}, \qquad \text{exhaust } 10.62\ \text{kJ}\,\text{kg}^{-1}}$$ Their difference, $45.00 - 10.62 = 34.38\ \text{kJ}\,\text{kg}^{-1}$, equals the work of step 5 exactly — a complete energy audit, valid only because the blades are frictionless.
7. Part (e) — blade efficiency. Blade (diagram) efficiency compares the work extracted with the kinetic energy delivered by the nozzles: $$\eta_b = \frac{w}{V_{S1}^{2}/2} = \frac{34\,378}{45\,000}$$ $$\boxed{\eta_b = 0.7640 = 76.40\ \%}$$ The classical closed form for a symmetrical impulse stage, $\eta_b = 4\rho(\cos\theta - \rho)$ with the blade-speed ratio $\rho = V_B/V_{S1} = 0.3333$, returns the same 0.7640. It also shows the stage is running below its optimum: $\rho_{\text{opt}} = \tfrac{1}{2}\cos\theta = 0.4532$ would give $\eta_b = \cos^{2}\theta = 82.14\ \%$.
8. Part (f) — stage power. Multiplying the specific work by the mass flow, $$P = wM = 34\,378 \times 24$$ $$\boxed{P = 825\,083\ \text{W} = 825.1\ \text{kW}}$$ Independently, $P = FV_B = 8251 \times 100 = 825.1\ \text{kW}$, confirming the force and work calculations against each other.