Question 3 of 8: Steam Turbine Blades (Impulse Stage)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations — 16-Mec-A6 Fluid Machinery, May 2019. Closed-book, three hours. Section A is calculative (5 questions) and Section B is descriptive (3 questions); candidates answer four questions from Section A and two from Section B (six questions, 60 marks, 10 marks each). All eight questions are solved as a study resource.
Reference texts. Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory (6th ed.); Fox & McDonald, Introduction to Fluid Mechanics (10th ed.); F. M. White, Fluid Mechanics (8th ed.); Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.).
Paper. This is the 16-Mec-A6, May 2019 Fluid Machinery examination.
Given. Single impulse stage: $\dot m=30\;\text{kg/s}$, nozzle angle $\theta=20^\circ$ (from the plane of rotation), $C_1=V_{s1}=450\;\text{m/s}$, blade speed $U=V_B=250\;\text{m/s}$, symmetric moving blade ($\beta_2=\beta_1$), blade friction $k=V_{R2}/V_{R1}=0.95$.
Given data
$\dot m$
$\theta$
$C_1$
$U$
$k$
blade
30 kg/s
20°
450 m/s
250 m/s
0.95
symmetric
Find. The exit absolute velocity $C_2$ and its direction $\delta$, the stage power, and the blade efficiency.
Approach. Build the inlet triangle from $C_1$, $\theta$ and $U$; carry the relative velocity through the symmetric blade with the 0.95 friction factor; close the outlet triangle to read $C_2$ and $\delta$; then power $=\dot m\,U\,\Delta C_w$ and blade efficiency $=U\,\Delta C_w/\tfrac12 C_1^2$.
Combined inlet/outlet velocity diagram (angles from the plane of rotation). The relative velocity is turned through the symmetric bucket and shortened by friction ($W_2=0.95\,W_1$); the whirl swings from $C_{w1}$ forward to a much smaller $C_{w2}$, and that large change $\Delta C_w$ produces the work.
Inlet triangle. Resolve the jet: whirl $C_{w1}=C_1\cos\theta=450\cos20^\circ=422.9\;\text{m/s}$, axial (flow) $C_{a1}=C_1\sin\theta=153.9\;\text{m/s}$. The inlet relative whirl is $C_{w1}-U=172.9\;\text{m/s}$, so
$$W_1=\sqrt{(C_{w1}-U)^2+C_{a1}^2}=231.4\;\text{m/s},\qquad \beta_1=\arctan\frac{C_{a1}}{C_{w1}-U}=41.7^\circ.$$
Carry the relative velocity through the blade. The symmetric bucket turns the flow so the exit relative angle equals the inlet ($\beta_2=41.7^\circ$) while friction shortens it:
$$W_2=k\,W_1=0.95\times231.4=219.8\;\text{m/s}.$$
Its components are $W_{w2}=W_2\cos\beta_2=164.2\;\text{m/s}$ (now pointing against blade motion) and $C_{a2}=W_2\sin\beta_2=146.2\;\text{m/s}$.
Outlet triangle — exit absolute velocity. The absolute exit whirl is $C_{w2}=U-W_{w2}=250-164.2=85.8\;\text{m/s}$ (still in the direction of motion), so
$$C_2=\sqrt{C_{w2}^2+C_{a2}^2}=\boxed{169.5\;\text{m/s}},\qquad \delta=\arctan\frac{C_{a2}}{C_{w2}}=\boxed{59.6^\circ}$$ measured from the plane of rotation.
Change of whirl and power. The whirl change across the stage is $\Delta C_w=C_{w1}-C_{w2}=422.9-85.8=337.1\;\text{m/s}$. The tangential force is $\dot m\,\Delta C_w$, so
$$\dot W=\dot m\,U\,\Delta C_w=30\times250\times337.1=\boxed{2528\;\text{kW}}.$$
Blade (diagram) efficiency. Referred to the kinetic energy of the entering jet,
$$\eta_b=\frac{U\,\Delta C_w}{\tfrac12 C_1^2}=\frac{250\times337.1}{\tfrac12(450)^2}=\boxed{0.832\;(83.2\%)}$$
