Steam Turbine Blade Efficiency: 25-Nav-B5 Marine Control Systems, December 2016, Q5

Question 5: Steam Turbine Blade Efficiency (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. Symmetric frictionless impulse stage with $V_B=100\ \text{m/s}$, $V_{s1}=300\ \text{m/s}$, $\theta=25^\circ$, $M=24\ \text{kg/s}$, $V_{R2}=V_{R1}$, $\gamma=\beta_1$.

Find. Exhaust velocity, tangential (impulse) force, blade work per kg, inlet and exhaust kinetic energies, blade efficiency, and stage power.

Fig. 5.1 — Combined velocity diagram on the common blade-speed base VB. Inlet: Vs1 at 25°, relative VR1. Outlet (symmetric, frictionless): VR2=VR1 reversed, giving a backward-swirling exhaust Vs2.

Approach. Resolve the inlet triangle to get the relative velocity and blade inlet angle; mirror it for the symmetric frictionless outlet; the change of whirl gives the force, work, efficiency and power.

Inlet triangle. Whirl $V_{w1}=V_{s1}\cos\theta=300\cos25^\circ=271.9\ \text{m/s}$; flow $V_{a1}=V_{s1}\sin25^\circ=126.8\ \text{m/s}$. Relative velocity $$V_{R1}=\sqrt{(V_{w1}-V_B)^2+V_{a1}^2}=\sqrt{171.9^2+126.8^2}=213.6\ \text{m/s},\quad \beta_1=\tan^{-1}\tfrac{126.8}{171.9}=36.4^\circ$$
Symmetric frictionless outlet. $V_{R2}=V_{R1}=213.6\ \text{m/s}$ at $\beta_2=\beta_1$, so the absolute exit whirl is $V_{w2}=V_B-V_{R2}\cos\beta_1=100-171.9=-71.9\ \text{m/s}$ (backward), and the change of whirl is $\Delta V_w=V_{w1}-V_{w2}=271.9-(-71.9)=343.8\ \text{m/s}$.
Absolute exhaust velocity (a). Axial $V_{a2}=V_{R2}\sin\beta_1=126.8\ \text{m/s}$: $$V_{s2}=\sqrt{V_{w2}^2+V_{a2}^2}=\sqrt{71.9^2+126.8^2}=\boxed{145.8\ \text{m/s}}$$
Impulse force (b). Tangential force $$F=M\,\Delta V_w=24\times343.8=\boxed{8.25\ \text{kN}}$$
Energy to blades (c). $$w=V_B\,\Delta V_w=100\times343.8=34.38\times10^3\ \text{J/kg}=\boxed{34.38\ \text{kJ/kg}}$$
Kinetic energies (d). $$\text{KE}_{in}=\tfrac{V_{s1}^2}{2}=\tfrac{300^2}{2}=\boxed{45.0\ \text{kJ/kg}},\qquad \text{KE}_{out}=\tfrac{V_{s2}^2}{2}=\tfrac{145.8^2}{2}=\boxed{10.63\ \text{kJ/kg}}$$
Blade efficiency (e). $$\eta_b=\frac{w}{\text{KE}_{in}}=\frac{34.38}{45.0}=\boxed{0.764\ (76.4\%)}$$
Stage power (f). $$P=M\,w=24\times34.38=\boxed{825\ \text{kW}}$$

Question 5 results
Quantity	Value
Absolute exhaust velocity $V_{s2}$	145.8 m/s
Impulse force $F$	8.25 kN
Energy to blades	34.38 kJ/kg
Inlet / exhaust KE	45.0 / 10.63 kJ/kg
Blade efficiency	0.764 (76.4%)
Stage power	825 kW

Topic: Single-stage impulse (de Laval) turbine velocity diagram and blade (diagram) efficiency.

Key relations: $\Delta V_w=V_{w1}-V_{w2}$; $F=M\Delta V_w$; $w=V_B\Delta V_w$; $\eta_b=w/(\tfrac12 V_{s1}^2)$; symmetric frictionless blade $V_{R2}=V_{R1},\ \beta_2=\beta_1$.

Why this works: In an impulse stage all the pressure drop is in the nozzle, so the blade only turns the relative flow at constant relative speed (when frictionless). The driving force is the rate of change of tangential momentum, and the work is that force times blade speed. Blade efficiency compares the work extracted with the kinetic energy supplied by the nozzle.

Common pitfall: Dropping the sign of the exit whirl. Here $V_{w2}$ is negative (the steam leaves swirling backward), so $\Delta V_w$ is a sum of magnitudes $271.9+71.9$, not a difference — forgetting this roughly halves the force.

Source: Çengel & Boles / standard steam-turbine texts; Dixon & Hall, ch. 4 (axial turbine stages).

Question 5 of 8: Steam Turbine Blade Efficiency

Question 5: Steam Turbine Blade Efficiency (10 marks)