Question 2 of 8: Compressor Blade Angles

Question 2 — Compressor Blade Angles (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.

Find. Blade speeds at tip and root, the first-stage exit temperature, the specific work and the whirl components $C_{Y1}$ and $C_{Y2}$ at both radii, the inlet and outlet relative-flow (blade) angles from the scale diagrams, and a check of those angles against the photograph of the real blades.

Approach. Get the blade speeds from the rotational speed and the two diameters; take the stage temperature rise as one seventh of the whole LP rise computed in Question 1; convert that rise to specific work and invert Euler's equation to find the exit whirl at each radius; then close the velocity triangles with the axial velocity held constant.

1. Blade velocity at tip and root (part a). The angular velocity of the LP spool is $$\omega = \frac{2\pi N}{60} = \frac{2\pi \times 6500}{60} = 680.68\ \text{rad/s}$$ and the peripheral speed at any radius is $U = \omega r$, so $$\boxed{U_{\text{tip}} = 680.68 \times \tfrac{1.12}{2} = 381.2\ \text{m/s}\qquad U_{\text{hub}} = 680.68 \times \tfrac{0.38}{2} = 129.3\ \text{m/s}}$$ The tip runs at nearly three times the root speed, which is the whole reason a first-stage blade has to be twisted.
2. Temperature at the exit of the first stage (part b). Question 1 established that the whole LP spool raises the air by $\Delta T_{LP} = 144.60\ \text{K}$. Sharing that equally between the seven stages, $$\Delta T_{\text{stage}} = \frac{144.60}{7} = 20.66\ \text{K}$$ $$\boxed{T_{\text{exit,1}} = 353.15 + 20.66 = 373.8\ \text{K} = 100.7\ {}^{\circ}\text{C}}$$ A rise of about 21 K in one stage is a heavy but realistic loading for a transonic first stage of this era.
3. Specific work of the stage (part c). The stage is adiabatic, so all the work appears as stagnation enthalpy rise, and with the axial velocity unchanged through the rotor the static rise is the whole of it: $$w = c_p\,\Delta T_{\text{stage}} = 1005 \times 20.66 = 20\,761\ \text{J/kg} = 20.76\ \text{kJ/kg}$$
4. Whirl components at tip and root (part c continued). The paper states there is no prewhirl, so the absolute velocity at inlet is purely axial and $C_{Y1} = 0$ at every radius. Euler's compressor equation, in the page-21 form $w = U(C_{Y2}-C_{Y1})$, then gives the exit whirl directly. The work is the same at both radii — the stage is designed for constant work along the span — so the whirl varies inversely with blade speed: $$C_{Y2,\text{tip}} = \frac{20\,761}{381.2} = 54.5\ \text{m/s},\qquad C_{Y2,\text{hub}} = \frac{20\,761}{129.3} = 160.5\ \text{m/s}$$ $$\boxed{C_{Y1} = 0 \text{ at both radii};\quad C_{Y2,\text{tip}} = 54.5\ \text{m/s},\quad C_{Y2,\text{hub}} = 160.5\ \text{m/s}}$$ The product $C_{Y2}\,r$ is the same at both radii (30.5 and 30.5 m²/s), which is the free-vortex condition — a useful check that the constant-work assumption has been applied consistently.
5. Velocity diagrams and blade angles at the tip (part d). The axial component is held at $C_X = 184\ \text{m/s}$ because no exit annulus is given. All angles are measured from the axial direction, following the page-11 attachment. At inlet the relative velocity closes the triangle between the axial absolute velocity and the blade speed: $$\begin{aligned}\beta_1 &= \arctan\!\left(\frac{U}{C_X}\right) = \arctan\!\left(\frac{381.2}{184}\right) = 64.2^{\circ} \\ W_1 &= \sqrt{184^2 + 381.2^2} = 423.3\ \text{m/s}\end{aligned}$$ At exit the relative whirl is what is left of the blade speed after the air has been given its own whirl, $W_{Y2} = U - C_{Y2} = 381.2 - 54.5 = 326.7\ \text{m/s}$, so $$\beta_2 = \arctan\!\left(\frac{326.7}{184}\right) = 60.6^{\circ},\qquad W_2 = 375.0\ \text{m/s},\qquad C_2 = 191.9\ \text{m/s}$$ The de Haller number is $W_2/W_1 = 375.0/423.3 = 0.886$, comfortably above the 0.72 diffusion limit, so the tip section is lightly loaded.
Combined velocity diagram for the first-stage rotor at the tip. Inlet triangle above the blade-speed line, outlet below. The turning of the relative flow, from 64.2° to 60.6° off the axial direction, is only 3.6°: at this blade speed a very small deflection buys the whole stage work.
6. The same construction at the root. Nothing changes except the blade speed, but its effect is dramatic: $$\beta_1 = \arctan\!\left(\frac{129.3}{184}\right) = 35.1^{\circ},\qquad W_1 = 224.9\ \text{m/s}$$ and at exit $W_{Y2} = U - C_{Y2} = 129.3 - 160.5 = -31.2\ \text{m/s}$, a negative relative whirl, so $$\beta_2 = \arctan\!\left(\frac{-31.2}{184}\right) = -9.6^{\circ}$$ that is, 9.6° on the other side of the axial direction. The relative flow leaves the root section pointing slightly into the direction of blade motion. Collecting the diagram values, $W_2 = 186.6\ \text{m/s}$, $C_2 = 244.2\ \text{m/s}$, $\alpha_2 = 41.1^{\circ}$, and the de Haller number is $186.6/224.9 = 0.830$, still acceptable.
Combined velocity diagram at the hub. The inlet stagger is far smaller than at the tip (35.1° against 64.2°), and the exit relative velocity crosses to the other side of the axial line because the required whirl exceeds the local blade speed.
7. Comparison with the photograph of the real blades (part e). The page-12 photograph of the Olympus 593 first-stage LP blades shows exactly the feature the calculation predicts: a pronounced twist from root to tip, with the leading edges near the tip lying almost along the plane of rotation — that is, staggered some 60–65° from the axial direction — while the root sections are much closer to axial, around 35°. The calculated inlet angles of 64.2° at the tip and 35.1° at the root are therefore reasonably correct, and the direction of the twist (large stagger outboard, small stagger inboard) matches the photograph without ambiguity. The calculated blade camber is small, and this too is visible: the blades are thin and only gently curved, consistent with a few degrees of turning. The one feature the photograph does not support is the negative exit angle at the root, which is discussed in the note below.