Steam Turbine Impulse Blades: 22-Mec-A6 Fluid Machinery, May 2014, Q3

Question 3: Steam Turbine Impulse Blades (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. V_B = 100 m/s; V_S1 = 300 m/s; θ = 25°; M = 24 kg/s; symmetric frictionless blades (V_R2 = V_R1, γ = φ).

Find. (a) exhaust velocity V_S2, (b) blade force F, (c) work per kg, (d) inlet/exhaust KE, (e) blade efficiency, (f) power.

Combined inlet/exhaust velocity diagram at scale (1 cm = 20 m/s). Inlet apex above the blade-speed line, exhaust apex below; the exhaust whirl reverses direction (counter-swirl).

Approach. Resolve the inlet absolute velocity into whirl and flow components, build the inlet relative velocity, mirror it (symmetric, frictionless) for the exit, then apply the momentum (whirl-change) equation for force and work.

Inlet velocity triangle. Whirl and flow components of the absolute inlet velocity: $$V_{w1}=V_{S1}\cos\theta=300\cos25^\circ=271.9\ \text{m/s},\qquad V_{f1}=V_{S1}\sin\theta=126.8\ \text{m/s}.$$ The relative inlet velocity (subtract blade speed from the whirl component): $$V_{R1}=\sqrt{(V_{w1}-V_B)^2+V_{f1}^2}=\sqrt{171.9^2+126.8^2}=213.6\ \text{m/s},\quad \phi=36.4^\circ.$$
Exit velocity triangle. Symmetric, frictionless: $V_{R2}=V_{R1}=213.6$ m/s at $\gamma=\phi=36.4^\circ$. The exit relative whirl reverses, so the absolute exit whirl is $$V_{w2}=V_B-V_{R2}\cos\gamma=100-171.9=-71.9\ \text{m/s}\ \ (\text{counter-swirl}),$$ and with flow component $V_{f2}=126.8$ m/s the (a) absolute exhaust velocity is $$\boxed{V_{S2}=\sqrt{71.9^2+126.8^2}=145.8\ \text{m/s}.}$$
(b) Impulse force. The blade force equals the mass flow times the change in whirl velocity ($\Delta V_w=V_{w1}-V_{w2}=271.9+71.9=343.8$ m/s): $$F=M\,\Delta V_w=24(343.8)=8.25\ \text{kN}.$$
(c) Energy transferred. Work per unit mass is the force per unit mass times blade speed, $$w=V_B\,\Delta V_w=100(343.8)=34.4\ \text{kJ/kg}.$$
(d) Kinetic energies. $$\text{KE}_{in}=\frac{V_{S1}^2}{2}=\frac{300^2}{2}=45.0\ \text{kJ/kg},\qquad \text{KE}_{out}=\frac{V_{S2}^2}{2}=\frac{145.8^2}{2}=10.6\ \text{kJ/kg}.$$
(e) Blade (diagram) efficiency. Work divided by inlet kinetic energy: $$\eta_b=\frac{w}{\text{KE}_{in}}=\frac{34.4}{45.0}=76.4\%.$$
(f) Power developed. $P=wM=34.4\times24=825\ \text{kW}$ (equivalently $F\,V_B=8251\times100$).

Question 3 — results
Quantity	Value
(a) Absolute exhaust velocity V_S2	145.8 m/s
(b) Impulse force F	8.25 kN
(c) Energy to blades w	34.4 kJ/kg
(d) Inlet / exhaust KE	45.0 / 10.6 kJ/kg
(e) Blade efficiency	76.4 %
(f) Stage power	825 kW

Topic. Single-stage impulse turbine — velocity triangles and the whirl-momentum (Euler) work equation.

Key relations. $F=M(V_{S1}\cos\theta-V_{S2}\cos\delta)=M\,\Delta V_w$; $w=V_B\,\Delta V_w$; $P=wM$; blade efficiency $\eta_b=2w/V_{S1}^2$.

Why this works. In a pure impulse stage all the enthalpy drop occurs in the nozzle, so the moving blade merely deflects the jet at constant relative speed (here idealised as frictionless). The force on the blade is the rate of change of tangential momentum, and the power is that force moving at blade speed. Because the exit whirl reverses, the two whirl contributions add, which is why a single impulse row can extract a large fraction of the jet energy.

Common pitfall. Sign of the exit whirl: when $V_{R2}\cos\gamma > V_B$ the absolute exit whirl points backward, so $\Delta V_w$ is a sum, not a difference. Reading velocities off a diagram drawn to the wrong scale also drifts every downstream number.

Source. Dixon & Hall, Turbomachinery, Ch. 4 (axial turbines, impulse stage); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory, Ch. 7.

Question 3 of 8: Steam Turbine Impulse Blades

Question 3: Steam Turbine Impulse Blades (10 marks)