Question 2 of 8: Compressor Design

Question 2: Compressor Design (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. A multi-stage axial compressor for a gas turbine:

Find. The inlet hub, tip and mean blade diameters; the mean blade speed; the first-stage velocity diagram with all absolute and relative velocities; the stage work; and the number of stages needed for the overall pressure ratio.

Approach. Continuity at the intake fixes the annulus area and hence the two diameters; the mean diameter and the shaft speed give the blade speed; fifty per cent reaction then closes the velocity diagram with no further data, and Euler's equation converts the whirl change into stage work. Dividing the whole-machine temperature rise by the stage rise gives the stage count.

1. Part (a) — annulus area from continuity. The axial velocity carries the whole mass flow through the annulus, so $$A=\frac{\dot m}{\rho_1 C_x}=\frac{50}{1.19\times 160}=0.2626\ \text{m}^2$$ The page-13 density at 20 °C, 1.19 kg/m³, agrees with the ideal-gas value $p_1/RT_1=100000/(287\times 293.15)=1.189$ kg/m³ to one part in a thousand, so either may be used.
2. Tip and hub diameters. With the annulus written in terms of the tip diameter and the fixed hub/tip ratio, $$A=\frac{\pi}{4}\left(D_t^2-D_h^2\right)=\frac{\pi}{4}D_t^2\left(1-0.6^2\right)=0.5027\,D_t^2$$ Solving, $D_t=\sqrt{0.2626/0.5027}$, and the hub follows from the ratio: $$\boxed{D_t=0.723\ \text{m},\qquad D_h=0.434\ \text{m}}$$
3. Mean blade diameter. The mean (pitch-line) diameter is the arithmetic mean of the two: $$D_m=\tfrac{1}{2}\left(D_t+D_h\right)=\tfrac{1}{2}\left(0.723+0.434\right)$$ $$\boxed{D_m=0.578\ \text{m}}$$
4. Part (b) — mean blade velocity. The pitch line travels once round per revolution: $$U=\frac{\pi D_m N}{60}=\frac{\pi\times 0.5782\times 8000}{60}$$ $$\boxed{U=242.2\ \text{m/s}}$$ The tip speed is $\pi D_t N/60=302.8$ m/s, comfortably within the mechanical limit for a steel or titanium rotor.
5. Part (c) — rotor inlet velocities. The stator upstream discharges at 30° to the axial direction with the axial component unchanged, so the absolute inlet velocity and its whirl component are $$C_1=\frac{C_x}{\cos\alpha_1}=\frac{160}{\cos 30^\circ}=184.8\ \text{m/s},\qquad C_{y1}=C_x\tan\alpha_1=160\tan 30^\circ=92.4\ \text{m/s}$$ The Mach number at rotor inlet is C₁ divided by the local sonic velocity $\sqrt{kRT_1}=343$ m/s, that is 0.54, so the stage is safely subsonic.
6. Closing the diagram with fifty per cent reaction. A reaction of one-half means the static enthalpy rise is shared equally between rotor and stator, which forces the two triangles to be mirror images: $\beta_1=\alpha_2$ and $\beta_2=\alpha_1$. The mirror condition fixes the rotor exit whirl: $$C_{y2}=U-C_{y1}=242.2-92.4=149.8\ \text{m/s}$$ and the remaining velocities follow from the constant axial component, $$C_2=W_1=\sqrt{C_x^2+C_{y2}^2}=219.2\ \text{m/s},\qquad W_2=C_1=184.8\ \text{m/s}$$ with flow angles $\beta_1=\alpha_2=43.1^\circ$ and $\beta_2=\alpha_1=30^\circ$ from the axial. Figure 2.1 is this diagram drawn on the common blade-speed base, which is exactly the construction the recommended scale of 10 mm = 20 m/s produces on paper.
7. Part (d) — stage work from Euler's equation. The work is the blade speed times the change of whirl, the relation printed on page 15: $$w=U\left(C_{y2}-C_{y1}\right)=242.2\times\left(149.8-92.4\right)=13.92\times 10^{3}\ \text{J/kg}$$ $$\boxed{w=13.9\ \text{kJ/kg per stage}}$$ which corresponds to a stagnation temperature rise of $\Delta T_0=w/c_p=13.85$ K per stage — a normal, conservative subsonic stage loading.
8. Part (e) — overall temperature rise. For the whole machine at the specified pressure ratio of 5.0, the isentropic stagnation temperature rise is $$\Delta T_{0,\,overall}=T_1\left[\left(\frac{p_{out}}{p_1}\right)^{(k-1)/k}-1\right] =293.15\left(5^{0.2857}-1\right)=171.1\ \text{K}$$ so the total work requirement is $c_p\,\Delta T_{0}=1.005\times 171.1=172.0$ kJ/kg.
9. Number of stages. Every stage has the same velocity diagram and therefore the same work, so $$n=\frac{172.0}{13.92}=12.4$$ $$\boxed{n=13\ \text{stages}}$$ Thirteen stages give a small margin over the requirement, which the designer would absorb by trimming the stage loading slightly rather than by leaving one stage doing 40 % of its neighbours' work.