Question 3 of 8: Steam Turbine Blades — Velocity Diagram

Question 3: Steam Turbine Blades — Velocity Diagram (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. ṁ = 30 kg/s; α₁ = 20°; V_S1 = 450 m/s; V_B = 250 m/s; symmetrical blade β₂ = β₁; V_R2 = 0.95 V_R1.

Find. Exit absolute velocity V_S2 and angle δ; stage power; blade (diagram) efficiency.

Approach. Resolve the inlet absolute velocity into whirl and axial components, build the inlet relative velocity, mirror it (with the 0.95 friction factor) through the symmetrical blade, then recombine to get the exit absolute velocity; the change of whirl gives the tangential force, hence power and efficiency.

1. Inlet components. $V_{w1}=V_{S1}\cos20^\circ=422.9$ m/s, $V_{a1}=V_{S1}\sin20^\circ=153.9$ m/s.
2. Inlet relative velocity. $V_{R1}=\sqrt{(V_{w1}-V_B)^2+V_{a1}^2}=\sqrt{172.9^2+153.9^2}=231.5$ m/s at $\beta_1=\tan^{-1}\!\dfrac{153.9}{172.9}=41.7^\circ$.
3. Outlet relative velocity (friction + symmetry). $V_{R2}=0.95\,V_{R1}=219.9$ m/s at $\beta_2=\beta_1=41.7^\circ$. Its whirl component (opposing the blade) is $V_{R2}\cos\beta_2=164.2$ m/s and axial $V_{R2}\sin\beta_2=146.2$ m/s.
4. (a) Exit absolute velocity. $V_{w2}=V_B-V_{R2}\cos\beta_2=250-164.2=85.8$ m/s; $V_{a2}=146.2$ m/s. Then $V_{S2}=\sqrt{85.8^2+146.2^2}$, giving $\boxed{V_{S2}=170\ \text{m/s at }\delta=59.6^\circ}$ to the plane of rotation.
5. (b) Power from change of whirl. $\Delta V_w=V_{w1}-V_{w2}=422.9-85.8=337.1$ m/s; tangential force $F=\dot m\,\Delta V_w=30(337.1)=10.11$ kN; $P=F\,V_B=10.11\times10^{3}(250)$, so $\boxed{P=2.53\ \text{MW}}$.
6. (c) Blade efficiency. $\eta_b=\dfrac{V_B\,\Delta V_w}{V_{S1}^2/2}=\dfrac{250(337.1)}{450^2/2}$, so $\boxed{\eta_b=83.2\%}$.