Question 5 of 8: Gas Turbine Blades (Power Turbine)

Question 5: Gas Turbine Blades (Power Turbine) (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. The power-turbine first-stage geometry and the peak-load operating data (halved to one gas turbine) are tabulated below.

Find. Mean blade speed, the stage velocity triangle, and the stage/turbine power by two independent routes, compared with the rated value.

Approach. The equal stator/rotor exit angles (60°) with 30° inlets are the signature of a 50% reaction stage; angles are from axial. Get $U$ at the mean radius, close the triangle for the axial velocity, then compute Euler work $w=U\,\Delta C_w$ (three like stages) and compare with $\dot m c_p\Delta T$.

1. (a) Mean blade velocity. $$U=\omega r_m=\frac{2\pi\cdot 3000}{60}\cdot\frac{1.275}{2}=\boxed{200.3\ \text{m/s}}$$
2. Close the triangle (50% reaction). With angles from axial, $U=C_a(\tan\alpha_1-\tan\beta_1)$, so $$C_a=\frac{U}{\tan60^\circ-\tan30^\circ}=\frac{200.3}{1.1547}=173.5\ \text{m/s}$$
3. (c) Velocities. $$C_1=\frac{C_a}{\cos60^\circ}=346.9\ \text{m/s}=W_2,\qquad W_1=\frac{C_a}{\cos30^\circ}=200.3\ \text{m/s}=C_2$$ (the mirror symmetry of 50% reaction). The whirl components follow: $C_{w1}=C_a\tan60^\circ=300.4$ m/s and $W_{w1}=C_{w1}-U=100.1$ m/s at inlet; at exit the rotor turns the relative flow to $\beta_2=60^\circ$ on the other side of the axial direction, $W_{w2}=-C_a\tan60^\circ=-300.4$ m/s, so $$C_{w2}=U+W_{w2}=200.3-300.4=-100.1\ \text{m/s}$$ i.e. the stage discharges with $\alpha_2=30^\circ$ of counter-swirl — which is exactly the $\alpha_0=30^\circ$ the next stator receives.
4. (d) Power from gas velocities. Because the exit whirl opposes the blade motion, the whirl change is a sum, not a difference: $$\Delta C_w=C_{w1}-C_{w2}=C_a(\tan\alpha_1+\tan\alpha_2)=173.5(1.7321+0.5774)=400.6\ \text{m/s}$$$$w=U\,\Delta C_w=200.3\cdot 400.6=\boxed{80.2\ \text{kJ/kg}}\quad(=2U^2)$$ The reference-sheet forms agree: $w=(C_1\sin\alpha_1+C_2\sin\alpha_2)U=(300.4+100.1)200.3$ and $w=\tfrac{1}{2}[(C_1^2-C_2^2)+(W_2^2-W_1^2)]$ both give 80.2 kJ/kg. For three like stages $w=240.7$ kJ/kg, so $$P_d=\dot m\,w=139\cdot 240.7\times10^{3}=\boxed{33.5\ \text{MW}}$$
5. (e) Power from temperature change. $$P_e=\dot m\,c_p\,\Delta T=139\cdot 1148\cdot(682-483)=\boxed{31.8\ \text{MW}}$$
6. (f) Comparison. The specified net output is 30.43 MW per gas turbine. The two independent estimates agree to about 5% of each other — (d) 33.5 MW from the velocity triangles, (e) 31.8 MW from the measured temperature drop — and both sit 4–10% above the specification. That ordering is the expected one: (d) and (e) are both gas-path (internal) powers, whereas the quoted 60 860 kW is the net station output, i.e. after bearing, coupling, gearbox, generator and auxiliary losses. (d) is the highest of the three because the blade angles are given only as approximate and a single mean-radius triangle assumes the same turning from root to tip, which slightly over-states the work; (e) is the most trustworthy figure because it rests on measured temperatures and the quoted gas $c_p$ rather than on nominal geometry.