Question 5 of 8: Steam-Turbine Stage Velocity Diagram
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 07-Mec-A6-1 Fluid Machinery, December 2014 — closed book, three hours, 60 marks. Section A is calculative (Q1–Q5) and Section B is descriptive (Q6–Q8); the rubric asks for four questions of Section A plus two of Section B (six questions, each of equal value, 10 marks). All eight questions are solved in full as a study resource. General constants supplied with the paper: g = 9.81 m/s², patm = 100 kPa, pvapour = 2.34 kPa (20 °C), ρwater = 1000 kg/m³, ρair = 1.21 kg/m³ (15 °C), cp,air = 1.005, cv,air = 0.718 kJ/kg·K.
Reference texts. S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); R. A. Sabersky, A. J. Acosta, E. G. Hauptmann & E. M. Gates, Fluid Flow: A First Course in Fluid Mechanics (4th ed.) — source of the pump-selection charts Figs 15.11/15.12; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory; R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics.
Check: Questions 3 rely on chart reads from the examination attachments (Figs 15.11 & 15.12). Values read from those charts (φe, σc, η) are quoted to the precision the graphs allow and are flagged where used; a candidate would read the same values off the supplied plots. The induction-motor pole count is not stated in Q3, so a standard 4-pole (1800 rev/min synchronous) machine is assumed — the governing method is unaffected.
Find. VS2 and its angle δ, the stage power, and the blade (diagram) efficiency.
Fig. Q5 - Combined inlet/outlet velocity diagram. Top triangle A-B-C (inlet), bottom A-B-D (outlet). Symmetric blade (γ=φ) with V_R2=0.95 V_R1.
Approach. Build the inlet triangle to get the relative velocity and the (symmetric) blade angle, scale the relative velocity down by 0.95 for the outlet triangle, resolve the outlet absolute velocity, then get power from the change in whirl velocity and efficiency from the ratio of blade work to inlet kinetic energy.
Inlet triangle — relative velocity and blade angle. Resolving VS1 into tangential and axial components, $V_{S1}\cos\theta=422.9$ and $V_{S1}\sin\theta=153.9\ \text{m/s}$. The relative velocity has tangential component $V_{S1}\cos\theta-V_B=172.9$ and the same axial component, so$$V_{R1}=\sqrt{172.9^2+153.9^2}=232\ \text{m/s},\qquad \tan\Phi=\frac{153.9}{172.9}\Rightarrow \Phi=41.7^\circ.$$
Outlet triangle. Friction reduces the relative velocity to $V_{R2}=0.95V_{R1}=220\ \text{m/s}$, and by symmetry it leaves at $\gamma=\Phi=41.7^\circ$ (opposite sense). Its tangential component is $V_{R2}\cos\gamma=164.2\ \text{m/s}$ (against blade motion) and axial $V_{R2}\sin\gamma=146.1\ \text{m/s}$. Adding the blade velocity gives the absolute exit velocity components $V_{S2,x}=V_B-164.2=85.8$ and $V_{S2,y}=146.1\ \text{m/s}$, so$$V_{S2}=\sqrt{85.8^2+146.1^2}=169\ \text{m/s},\qquad \tan\delta=\frac{146.1}{85.8}\Rightarrow \delta=59.6^\circ.$$$$\boxed{V_{S2}=169\ \text{m/s},\;\delta=59.6^\circ}$$
Power from change in whirl. The driving force is the rate of change of tangential (whirl) momentum. The forward whirl falls from $V_{S1}\cos\theta=422.9$ to $V_{S2}\cos\delta=85.8\ \text{m/s}$, so $\Delta V_w=337.1\ \text{m/s}$ and$$P=\dot m\,V_B\,\Delta V_w=30(250)(337.1)=2.53\times10^6\ \text{W}.$$$$\boxed{P=2528\ \text{kW}}$$The energy form $P=\tfrac{\dot m}{2}[(V_{S1}^2-V_{S2}^2)+(V_{R2}^2-V_{R1}^2)]$ gives the same 2528 kW.
Blade (diagram) efficiency. Referring the blade work to the kinetic energy of the incoming jet,$$\eta_b=\frac{V_B\,\Delta V_w}{\tfrac12 V_{S1}^2}=\frac{250(337.1)}{\tfrac12(450)^2}=0.832.$$$$\boxed{\eta_b=83.2\%}$$
