Question 5 of 8: Gas Turbine Blades (Power Turbine)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: 07-Mec-A6 Fluid Machinery — National Examinations, May 2015. Closed book, three hours. Section A (calculative, Q1–Q5) and Section B (descriptive, Q6–Q8); candidates answer four of A and two of B, but all eight are solved here as a study resource. Each question 10 marks.
Reference texts: S.L. Dixon & C.A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed.; R.K. Turton, Principles of Turbomachinery, 2nd ed.; H. Cohen, G.F.C. Rogers & H.I.H. Saravanamuttoo, Gas Turbine Theory, 6th ed.; R.W. Fox, A.T. McDonald & P.J. Pritchard, Introduction to Fluid Mechanics, 8th ed.
Q1 (with Parts I and II interchanged), Q2 (Vanderkloof), Q4 (Curtis) and Q5 (Acacia/Port Rex) carry the same data; only Q3 (multi-jet Pelton) is unique. Q1 Part II additionally instructs the candidate to select a turbine efficiency, which is applied here.
Question 5: Gas Turbine Blades (Power Turbine) (10 marks)
Given. A 50 %-reaction free-turbine first stage on a twin-shaft industrial gas turbine.
Given data (per single gas turbine)
Stator angles α0/α1
30° / 60°
Rotor angles β1/β2
30° / 60°
Tip / root diameter
1500 mm / 1050 mm
Speed
3000 rev/min
Exhaust cp
1.148 kJ/kg°C
Exhaust flow / power (half of combined)
139 kg/s / 30.43 MW
Find. Mean blade speed, the velocity triangle, gas velocities, power by two methods and the discrepancy.
Figure 4. Combined inlet and exit velocity diagram for the first stage of the power turbine. The stage is 50 % reaction, so the rows are mirror images (α1 = β2 = 60°, β1 = α2 = 30°); the rotor turns the gas through the axial direction, so the exit whirl opposes U and the whirl change is the sum of the two whirl components.
Approach. Take the mean-diameter blade speed, exploit the symmetry of a 50 %-reaction stage to fix the axial velocity from $U=C_a(\tan\alpha_1-\tan\beta_1)$, then compute the Euler work and compare with the enthalpy-drop power.
Mean blade speed. $D_m=\tfrac{1.5+1.05}{2}=1.275\ \text{m}$, $U=\dfrac{\pi D_m N}{60}=\dfrac{\pi\cdot1.275\cdot3000}{60}=200.3\ \text{m/s}$.
Axial velocity (50 % reaction). All angles are measured from the axial direction. The rows are mirror images, so $U=C_a(\tan\alpha_1-\tan\beta_1)$ gives $C_a=\dfrac{U}{\tan60^{\circ}-\tan30^{\circ}}=173.4\ \text{m/s}$. Checking the degree of reaction, $\Lambda=\dfrac{C_a(\tan\beta_2-\tan\beta_1)}{2U}=0.50$ as expected.
Gas velocities (part c). $C_1=C_a/\cos60^{\circ}=346.9$ and $W_1=C_a/\cos30^{\circ}=200.3\ \text{m/s}$ at inlet; by the mirror symmetry $W_2=C_1=346.9$ and $C_2=W_1=200.3\ \text{m/s}$ at exit. All are comfortably subsonic (the sound speed in the exhaust gas at 682 °C is about 620 m/s).
Whirl change and stage work. The rotor turns the gas through the axial direction, so the exit whirl opposes U: $C_{w1}=C_a\tan60^{\circ}=300.4$ m/s and $C_{w2}=-C_a\tan30^{\circ}=-100.1$ m/s. The reference-sheet form $w=(C_1\sin\alpha_1+C_2\sin\alpha_2)U$ — note the plus sign — therefore gives $\Delta C_w=C_a(\tan60^{\circ}+\tan30^{\circ})=400.6\ \text{m/s}$ and $w_{stage}=U\,\Delta C_w=\boxed{80.2\ \text{kJ/kg}}$; for the three like stages $w=240.7\ \text{kJ/kg}$.
Power from gas velocities. With $\dot m=278/2=139\ \text{kg/s}$, $P_d=139\cdot240.7=\boxed{33.5\ \text{MW}}$.
Power from temperature drop. $P_e=\dot m c_p\Delta T=139\cdot1.148\cdot(682-483)=31.8\ \text{MW}$.
Comparison (part f). The specified per-turbine output is $60\,860/2=30.4\ \text{MW}$. The three figures fall in the order $\boxed{P_d=33.5>P_e=31.8>P_{spec}=30.4\ \text{MW}}$ — the ideal triangles exceed the measured enthalpy drop by 5.3 %, and the enthalpy drop exceeds the net output by a further 4.4 %.
Check: the three-stage count is given, not assumed — the question instructs the candidate to assume “the gas flow conditions are the same for the second and third stages of the power turbine”, and the free turbine is the N3 rotor of Attachment page 13. Two independent cross-checks confirm the triangles: (i) the ideal stage work corresponds to $\Delta T=w/c_p=69.9$ °C per stage, 209.6 °C over three stages against the measured 199 °C — the right size and on the correct (ideal-high) side; (ii) continuity over the annulus $A=\tfrac{\pi}{4}(1.5^{2}-1.05^{2})=0.901\ \text{m}^2$ with $C_a=173.4$ m/s and $\dot m=139$ kg/s gives $\rho=0.889\ \text{kg/m}^3$, i.e. about 244 kPa at 682 °C, a sensible free-turbine inlet pressure behind a gas generator of pressure ratio 14.1.
The point of parts (d)–(f) is that the three figures are a ladder of increasing realism, not a contradiction. The velocity-diagram power is the largest because it is an ideal mean-line calculation: it uses nominal blade angles at one radius, treats the annulus as loss-free, and ignores tip leakage, blade-row friction, disc windage and the radial (free-vortex) variation of the real blading. The temperature-drop power is what the gas actually gave up, so it sits 5.3 % below. The specified net output is lower again, by a further 4.4 %, because it is measured after mechanical and generator losses and after auxiliary loads. Agreement within about 10 % across all three is a good result for a mean-line estimate made from approximate blade dimensions, and the correct comment for part (f) is that the differences are losses and idealisations rather than an error in the diagram.
