Question 5 of 8: Compressor First Stage

Question 5: Compressor First Stage (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. First (N1) rotor stage of an industrial gas-turbine compressor:

The page-17 attachment defines both angles from the axial direction, and both velocity triangles share the blade speed at the mean diameter. The axial velocity is taken constant through the rotor, which is the standard assumption for a first stage of constant annulus area and the only one that makes part (b) solvable.

Find. Mean blade speed and axial velocity, the four velocities on the two triangles, the specific work and stage power, the enthalpy and temperature rise, and the isentropic pressure ratio of the stage.

Approach. Mean diameter and continuity give U and C_x; the two given angles then close the inlet and outlet triangles; page 21's compressor work w = U(C_Y2 − C_Y1) gives the work, which for an adiabatic rotor is the stagnation enthalpy rise, hence the temperature rise; the isentropic relation converts that into a pressure ratio.

1. Part (a) — mean blade speed. The mean diameter of the annulus is the arithmetic mean of hub and tip: $$D_m=\frac{0.480+1.120}{2}=0.800\ \text{m}$$ and the blade speed there is $$U=\frac{\pi D_mN}{60}=\frac{\pi(0.800)(6800)}{60}=\boxed{284.8\ \text{m/s}}$$
2. Part (a) — axial velocity from continuity. The annulus flow area is the difference of two circles: $$A=\frac{\pi}{4}\left(D_t^{2}-D_h^{2}\right)=\frac{\pi}{4}\left(1.120^{2}-0.480^{2}\right) =\frac{\pi}{4}(1.0240)=0.8042\ \text{m}^2$$ and continuity M = ρAC_x1 (page 20) gives $$C_{x1}=\frac{M}{\rho A}=\frac{136}{1.21\times 0.8042}=\boxed{139.8\ \text{m/s}}$$ The page-19 density of 1.21 kg/m³ is consistent with p/RT = 100 000/(287 × 288) = 1.210 kg/m³, so the two data sources agree to three figures.
3. Part (b) — close the rotor inlet triangle. The guide vanes leave the air at 30° to the axial, so the absolute velocity and its swirl component are $$C_1=\frac{C_{x1}}{\cos\alpha_1}=\frac{139.8}{\cos 30^{\circ}}=\boxed{161.4\ \text{m/s}},\qquad C_{Y1}=C_{x1}\tan\alpha_1=139.8\tan 30^{\circ}=80.7\ \text{m/s}$$ Subtracting that swirl from the blade speed gives the relative velocity seen by the rotor: $$W_{Y1}=U-C_{Y1}=284.8-80.7=204.2\ \text{m/s},\qquad W_1=\sqrt{139.8^{2}+204.2^{2}}=\boxed{247.4\ \text{m/s}}$$ at a relative angle of β₁ = tan⁻¹(204.2/139.8) = 55.6°.
4. Part (b) — close the rotor outlet triangle. Holding the axial velocity constant and applying the given relative exit angle of 40°, $$W_2=\frac{C_{x1}}{\cos\beta_2}=\frac{139.8}{\cos 40^{\circ}}=\boxed{182.4\ \text{m/s}},\qquad W_{Y2}=C_{x1}\tan\beta_2=139.8\tan 40^{\circ}=117.3\ \text{m/s}$$ Adding the blade speed back gives the absolute state leaving the rotor: $$C_{Y2}=U-W_{Y2}=284.8-117.3=167.6\ \text{m/s},\qquad C_2=\sqrt{139.8^{2}+167.6^{2}}=\boxed{218.2\ \text{m/s}}$$ leaving at α₂ = 50.2° to the axial. Note the two things that matter physically: the relative velocity has fallen from 247.4 to 182.4 m/s, which is the diffusion that raises the pressure, while the absolute velocity has risen from 161.4 to 218.2 m/s, which is the swirl the stator must later remove.
5. Part (c) — specific work and stage power. Page 21 gives the compressor work as $$w=U(C_{Y2}-C_{Y1})=284.8(167.6-80.7)=284.8\times 86.9=24\,750\ \text{J/kg} =\boxed{24.75\ \text{kJ/kg}}$$ so the power the shaft must supply to this one stage is $$P=wM=24.75\times 136=\boxed{3366\ \text{kW}}$$ The energy form of the same equation on page 21, w = [(C₂² − C₁²) + (W₁² − W₂²)]/2, returns the same 24.75 kJ/kg, which checks all four velocities at once.
6. Part (d) — enthalpy and temperature rise. The rotor is adiabatic and does work on the air, so the whole of that work appears as a rise in stagnation enthalpy: $$\Delta h_0=w=\boxed{24.75\ \text{kJ/kg}}$$ and with c_p = 1.005 kJ/kg°C from page 19, $$\Delta T=\frac{\Delta h_0}{c_p}=\frac{24.75}{1.005}=\boxed{24.6\ \text{K}}$$ so the stage leaves at T₂ = 288 + 24.6 = 312.6 K. About 25 K per stage is exactly the loading a subsonic axial stage is designed for — push much past 30 K and the diffusion becomes too severe for the boundary layer.
7. Part (e) — isentropic stage pressure ratio. The ratio of specific heats follows from the paper's own constants, k = c_p/c_v = 1.005/0.718 = 1.400, and page 21's isentropic relation inverted gives $$\frac{p_2}{p_1}=\left(\frac{T_2}{T_1}\right)^{k/(k-1)} =\left(\frac{312.6}{288.0}\right)^{3.50}=(1.0855)^{3.50}=\boxed{1.333}$$ A first-stage ratio of 1.33 against the machine's overall gas-generator pressure ratio of 14.1 (page 16) implies roughly nine equivalent stages spread across the two spools, which is consistent with the cross-section drawn in the attachment.