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

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. A two-row Curtis stage with a single set of nozzles feeding both moving rows.

Find. Optimum blade speed, all velocity-triangle quantities, work per stage, total power and blade efficiency.

Approach. For a velocity-compounded wheel with $n$ rows the optimum blade speed is $U=\tfrac{V_{s1}\cos\alpha}{2n}$; build symmetric, frictionless triangles row by row, then sum the Euler work.

1. Optimum blade speed. With $n=2$, $U=\dfrac{V_{s1}\cos20^{\circ}}{2\cdot2}=\dfrac{1411\cdot0.9397}{4}=\boxed{331.5\ \text{m/s}}$.
2. Row-1 inlet triangle. Whirl $V_{w1}=1411\cos20^{\circ}=1325.9$ and flow $V_{f}=1411\sin20^{\circ}=482.6\ \text{m/s}$ give $V_{r1}=\sqrt{(V_{w1}-V_B)^{2}+V_f^{2}}=1105\ \text{m/s}$ at $\beta_1=25.9^{\circ}$.
3. Row-1 exit triangle. Symmetric frictionless blades keep $V_{r2}=V_{r1}=1105$ m/s at $\beta_2=\beta_1=25.9^{\circ}$ but reverse the relative whirl, so the absolute exit whirl is $V_B-V_{r2}\cos\beta_2=331.5-994.4=-662.9$ m/s (against the wheel) and $V_{s2}=\sqrt{662.9^{2}+482.6^{2}}=820.0\ \text{m/s}$ at $36.1^{\circ}$ to the wheel plane. Hence $w_1=V_B\,\Delta V_{w}=331.5(1325.9+662.9)=659.3\ \text{kJ/kg}$.
4. Fixed row and row-2 triangles. The symmetric frictionless fixed blades return the flow at the same speed and angle but in the direction of rotation, $V_{s3}=820.0$ m/s at $36.1^{\circ}$, so $V_{r3}=\sqrt{(662.9-331.5)^{2}+482.6^{2}}=585.5\ \text{m/s}$ at $\beta_3=55.5^{\circ}$, and $V_{r4}=585.5$ m/s at $\beta_4=55.5^{\circ}$. The absolute exit whirl is $331.5-331.5\approx0$, so $V_{s4}=482.6$ m/s axial — zero exit whirl, which is exactly the minimum-exit-kinetic-energy condition asked for in (a). Row-2 work $w_2=331.5(662.9-0)=219.8\ \text{kJ/kg}$, the classic 3 : 1 split of a two-row Curtis stage.
5. Total work and power. $w=w_1+w_2=879\ \text{kJ/kg}$; for 100 kg/s, $P=879\cdot100=87.9\ \text{MW}$.
6. Blade efficiency. $\eta_b=\dfrac{w}{\tfrac12 V_{s1}^{2}}=0.883=\cos^{2}20^{\circ}$ — the exact ideal result.

The blade efficiency collapsing to $\cos^{2}\alpha$ is the elegant self-check of the ideal velocity-compounded stage: for symmetric frictionless blading at the optimum speed the whole two-row wheel converts exactly $\cos^{2}20^{\circ}=88.3\%$ of the jet kinetic energy to work, independent of the individual row angles. The 3 : 1 work split shows why Curtis staging is used only where a large single pressure drop must be absorbed at low rotational speed: the second row does far less work than the first.