Curtis (Velocity-Compounded) Impulse Turbine: 22-Mec-A6 Fluid Machinery, December 2013, Q4

Question 4: Curtis (Velocity-Compounded) 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. Nozzle exit (stage-1 inlet absolute) $V_{S1}=1411$ m/s at θ=20°; two moving rows; symmetric, frictionless blades; steam flow 100 kg/s.

Find. Optimum blade speed, the full set of stage velocities and angles, the work split between stages, total power and blade efficiency.

Stage 1 whirl change 1325.9 to -662.9 m/s across the first moving row.

Stage 2 removes the residual whirl (662.9 to ~0); exit is nearly axial, confirming optimum U.

Approach. For an $n$-row velocity-compounded stage the optimum blade speed is $U=V_{S1}\cos\theta/(2n)$. Build each combined velocity diagram (symmetric blades keep the relative speed constant); Euler work per row is $U\,\Delta V_w$; blade efficiency is total work over inlet kinetic energy.

(a) Optimum blade velocity. With $n=2$ moving rows: $$U=\frac{V_{S1}\cos\theta}{2n}=\frac{1411\cos20^\circ}{4}=\boxed{331.5\ \text{m/s}}$$
Stage-1 inlet triangle. Whirl $V_{w1}=V_{S1}\cos20^\circ=1325.9$, flow $V_{f}=V_{S1}\sin20^\circ=482.6$ m/s. Relative velocity $$V_{R1}=\sqrt{(1325.9-331.5)^2+482.6^2}=1105.3\ \text{m/s},\quad \beta_1=\tan^{-1}\frac{482.6}{994.4}=25.9^\circ$$
Stage-1 exit / stage-2 inlet. Symmetric, frictionless: $V_{R2}=V_{R1}=1105.3$ m/s, so the absolute exit whirl is $U-994.4=-662.9$ m/s and $V_{S2}=\sqrt{662.9^2+482.6^2}=820\ \text{m/s}$. The symmetric fixed (guide) blade reverses this whirl to $+662.9$ m/s entering stage 2.
Stage-2 exit. Repeating the triangle with $V_w=662.9$ gives a relative velocity $585.5$ m/s and an absolute exit whirl of $\approx 0$ — the discharge is essentially axial ($V\approx 482.6$ m/s), confirming minimum exit kinetic energy at the chosen $U$.
(d) Work per stage. Euler $w=U\,\Delta V_w$: $$w_1=331.5(1325.9+662.9)=\boxed{659.3\ \text{kJ/kg}},\quad w_2=331.5(662.9-0)=\boxed{219.7\ \text{kJ/kg}}$$ (the classic 3 : 1 split), total $w=879.0$ kJ/kg.
(e) Total power. $$P=\dot m\,w=100\cdot 879.0\times10^{3}=\boxed{87.9\ \text{MW}}$$
(f) Blade efficiency. $$\eta_b=\frac{w}{V_{S1}^2/2}=\frac{879.0\times10^{3}}{1411^2/2}=\boxed{88.3\%}=\cos^2\theta$$

Question 4 results
Quantity	Value
(a) Optimum blade velocity	331.5 m/s
Stage-1 relative velocity / β₁	1105.3 m/s / 25.9°
Stage-1 absolute exit velocity	820 m/s
(d) Work: stage 1 / stage 2	659.3 / 219.7 kJ/kg
(e) Total power (100 kg/s)	87.9 MW
(f) Blade efficiency	88.3% (= cos²20°)

Topic: Velocity-compounding (Curtis stage)

Key relations: Optimum $U=V_{S1}\cos\theta/(2n)$; per-row Euler work $w=U\,\Delta V_w$; symmetric frictionless blade $V_{R,out}=V_{R,in}$; $\eta_{b,max}=\cos^2\theta$.

Why this works: A Curtis stage drops a very high steam velocity in two moving rows separated by a fixed guide row that only redirects (does no work). Splitting the deceleration lets a modest blade speed absorb an enormous kinetic energy at low rotor stress. At the optimum blade speed the discharge whirl vanishes, so no kinetic energy is wasted and the work divides 3 : 1 between the rows.

Common pitfall: Using the single-row optimum $U=V\cos\theta/2$ (instead of $/2n$) doubles the blade speed and throws away the exit-KE benefit; also, forgetting that the guide row conserves speed but reverses whirl.

Source: Dixon & Hall §4 (axial steam turbines); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory (impulse staging).

Question 4 of 8: Curtis (Velocity-Compounded) Impulse Turbine

Question 4: Curtis (Velocity-Compounded) Impulse Turbine (10 marks)