Question 3 of 8: Curtis Type Impulse Turbine

Question 3: Curtis Type Impulse Turbine (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 two-row velocity-compounded (Curtis) impulse wheel:

Find. The blade speed that minimises the exit kinetic energy, the full set of absolute and relative velocities and blade angles for both rows drawn to scale, the work of each stage, the total power and the blade efficiency.

Approach. The optimum blade speed for an n-row velocity-compounded wheel follows from maximising w = ΣV_B ΔV_w; with that speed fixed, the two combined velocity diagrams close by themselves because zero friction plus symmetrical blades means each row returns the relative velocity unchanged in magnitude and reversed in swirl. Work per row is then V_BΔV_w and the blade efficiency is the total work divided by the nozzle kinetic energy.

1. Part (a) — optimum blade velocity. Writing the total tangential momentum change of n symmetrical frictionless moving rows in terms of the blade speed and differentiating gives the standard result that the optimum blade-speed ratio is V_B/V_S1 = cosθ/2n. With two rows, $$V_B=\frac{V_{S1}\cos\theta}{2n}=\frac{1411\cos 20^{\circ}}{4}=\frac{1325.9}{4} =\boxed{331.5\ \text{m/s}}$$ Compare a single-row (De Laval) wheel, where the optimum is V_S1cosθ/2 = 663 m/s. Halving the blade speed is exactly why Curtis compounding exists: 1411 m/s of steam can be absorbed at a rim speed a conventional disc can survive.
2. Part (b), (c) — resolve the nozzle jet and close the stage-1 inlet triangle. The axial component is set by the nozzle angle and never changes thereafter, because a frictionless symmetrical blade only reverses the swirl: $$V_a=V_{S1}\sin\theta=1411\sin 20^{\circ}=482.6\ \text{m/s},\qquad V_{w1}=V_{S1}\cos\theta=1325.9\ \text{m/s}$$ Subtracting the blade speed from the whirl gives the relative velocity entering row 1: $$V_{R1}=\sqrt{(V_{w1}-V_B)^2+V_a^2}=\sqrt{994.4^{2}+482.6^{2}}=1105\ \text{m/s}$$ at a blade angle, measured from the plane of rotation, of $$\phi_1=\tan^{-1}\!\left(\frac{482.6}{994.4}\right)=25.9^{\circ}$$ Because the row is symmetrical and frictionless, that same angle and the same 1105 m/s describe the flow leaving it.
3. Part (c) — steam leaving row 1 and entering row 2. Adding the blade velocity back to the reversed relative whirl gives the absolute velocity leaving row 1: $$V_{w2}=-(V_{w1}-V_B)+V_B=-994.4+331.5=-662.9\ \text{m/s}$$ $$V_{S2}=\sqrt{662.9^{2}+482.6^{2}}=820.0\ \text{m/s},\qquad \delta=\tan^{-1}\!\left(\frac{482.6}{662.9}\right)=36.1^{\circ}$$ The negative whirl says the steam is now moving against the direction of blade motion, which is precisely what the fixed row is there to correct. Being symmetrical and frictionless, the fixed blades return the steam at the same 820.0 m/s and the same 36.1°, but with the swirl restored to the direction of rotation.
4. Part (c) — close the stage-2 diagram. Repeating the construction with V_S3 = 820.0 m/s and whirl +662.9 m/s: $$V_{R3}=\sqrt{(662.9-331.5)^{2}+482.6^{2}}=\sqrt{331.5^{2}+482.6^{2}}=585.5\ \text{m/s}$$ $$\phi_2=\tan^{-1}\!\left(\frac{482.6}{331.5}\right)=55.5^{\circ}$$ and the absolute velocity leaving the second moving row has whirl $$V_{w4}=-331.5+331.5=0$$ so the steam leaves purely axially at 482.6 m/s. That is the signature of correct Curtis design at the optimum blade speed: all the useful swirl has been taken out, and the residual is the unavoidable axial component the nozzle angle imposed at the start.

5. Part (d) — work done by each row. Euler's turbine equation for an axial machine gives the specific work as the blade speed times the change in absolute whirl: $$w=V_B\,\Delta V_w$$ For the first row, the whirl swings from +1325.9 to −662.9 m/s, a change of 1988.8 m/s: $$w_1=331.5\times 1988.8=659\,260\ \text{J/kg}=\boxed{659.3\ \text{kJ/kg}}$$ For the second row, the swing is from +662.9 to 0: $$w_2=331.5\times 662.9=219\,750\ \text{J/kg}=\boxed{219.8\ \text{kJ/kg}}$$ Total w = 879.0 kJ/kg. The ratio is exactly 3 : 1, which is the classical Curtis result for two rows and a useful arithmetic check: an n-row wheel at its optimum speed splits the work in the odd-number ratio (2n − 1) : (2n − 3) : … : 1.
6. Part (e) — total power. Page 21 gives P = wM, so at 100 kg/s of steam $$P=879.0\times 100=87\,900\ \text{kW}=\boxed{87.9\ \text{MW}}$$
7. Part (f) — blade (diagram) efficiency. Blade efficiency compares the work extracted with the kinetic energy the nozzles delivered: $$\eta_b=\frac{w}{V_{S1}^{2}/2}=\frac{879\,010}{1411^{2}/2}=\frac{879\,010}{995\,460} =\boxed{0.883\ \ (88.3\ \text{per cent})}$$ The frictionless theory says this must equal cos²θ at the optimum blade speed, and cos²20° = 0.8830 — agreement to four figures, which confirms every velocity in the table above. The missing 11.7 per cent is the residual axial kinetic energy, 482.6²/2 = 116.4 kJ/kg, and 879.0 + 116.4 = 995.5 kJ/kg recovers the nozzle kinetic energy exactly.