Turbomachinery

working pillar 1 · Physical modeling and domain theory

In one minute

• Euler's turbomachinery equation is the whole subject in one line: work is the change in \(rV_\theta\). Everything else is how you achieve it without the flow separating.
• Velocity triangles are the working notation. Absolute frame for the stator, relative frame for the rotor, and the difference is the blade speed.
• Compressors fight an adverse pressure gradient and turbines a favourable one — which is why a turbine stage does several times the work of a compressor stage, and why compressors need many more.
• Surge and rotating stall are different failures. Surge is a system-level axial oscillation; rotating stall is a cell propagating around the annulus.
• Corrected speed and flow collapse a machine's whole map onto one curve set — and only if you use the right reference conditions.

Key results

Euler turbomachinery equation — from angular momentum, with no loss assumptions:

\[ w = U_2 V_{\theta 2} - U_1 V_{\theta 1} \]

For an axial stage with \(U_1 = U_2 = U\): \(w = U\,\Delta V_\theta\). Work is set by blade speed and the turning the blades impose, nothing else.

Stage design parameters:

\[ \phi = \frac{V_x}{U} \ \text{(flow coefficient)}, \qquad \psi = \frac{\Delta h_0}{U^2} \ \text{(stage loading)}, \qquad R = \frac{\Delta h_{\text{rotor}}}{\Delta h_{\text{stage}}} \ \text{(reaction)} \]

\(R = 0.5\) splits the static enthalpy rise evenly between rotor and stator and gives symmetric velocity triangles — the usual axial-compressor choice. \(R = 0\) is an impulse stage, all the turning in the rotor at constant static pressure.

The Smith chart plots stage efficiency contours against \(\phi\) and \(\psi\); peak efficiency sits near \(\phi \approx 0.5\text{–}0.7\), \(\psi \approx 0.3\text{–}0.4\). Pushing loading higher buys fewer stages at the cost of efficiency and stall margin.

Why compressors need more stages. A compressor blade row diffuses — the flow decelerates against rising pressure, so the boundary layer is always close to separating. de Haller limits the relative velocity ratio to \(W_2/W_1 \gtrsim 0.72\), and diffusion factor \(DF \lesssim 0.6\). A turbine accelerates the flow, the boundary layer is favourable, and turning of 90°+ is routine. The practical consequence: one turbine stage can drive several compressor stages.

Compressor instabilities, distinct and often confused:

	Rotating stall	Surge
Nature	local, cells rotating around the annulus	global, axial oscillation of the whole system
Frequency	~20–70% of rotor speed	Helmholtz frequency of the ducting
Mass flow	steady in the mean	oscillates, can reverse
Damage	blade high-cycle fatigue	violent, can destroy the machine

Rotating stall often precedes surge, and the surge line on a map is the operating boundary the control system must protect.

Corrected (referred) parameters collapse the operating map:

\[ \dot m_{c} = \frac{\dot m \sqrt{\theta}}{\delta}, \qquad N_{c} = \frac{N}{\sqrt{\theta}}, \qquad \theta = \frac{T_{0}}{T_{ref}},\ \ \delta = \frac{p_{0}}{p_{ref}} \]

These follow from dimensional analysis on \(\dot m\sqrt{RT_0}/(D^2 p_0)\) and \(ND/\sqrt{RT_0}\), with \(R\), \(\gamma\) and \(D\) dropped as constant for one machine and one gas. That last clause is exactly where the method breaks.

Tip clearance costs roughly 1–2% stage efficiency per 1% of span, and drives a leakage vortex that interacts with the passage flow. It is one of the largest single loss sources and one of the hardest to model.

Mental model

A blade row is a cascade of aerofoils doing one job: turn the flow. Work follows from the turning via Euler; loss follows from how hard you asked for it.

Compressor and turbine are the same machinery run against and with the pressure gradient, and nearly every asymmetry between them — stage count, turning limits, stall behaviour, cooling — traces to that one difference. It is the boundary-layer story on a rotating blade.

Numerics / practice

• Draw the velocity triangles first. Most stage-design errors are frame errors, and a triangle makes them visible.
• Check de Haller and diffusion factor before believing a compressor stage design; they are cheap screens against an unachievable loading.
• State the reference conditions with any corrected quantity. Sea-level standard is conventional but not universal.
• Radial equilibrium governs spanwise distribution in an axial machine; a purely two-dimensional mean-line design will not close in three dimensions.

Failure modes

Surge and rotating stall conflated. They have different physics, different frequencies and different remedies — bleed and variable geometry for stall margin, volume and control for surge. Diagnosing one as the other sends the fix in the wrong direction.

Corrected parameters across different gas or size. \(\theta\) and \(\delta\) collapse the map for one machine on one gas. Comparing a rig on air to an engine on combustion products, or scaling across diameter, needs \(\gamma\) and \(R\) carried explicitly. The numbers still plot; they just no longer collapse.

Mean-line design taken as three-dimensional. Mean-line gives work and loading; it says nothing about spanwise distribution, secondary flow or tip leakage. A stage that closes at mid-span can be separated at the hub.

de Haller ignored. Asking for high stage loading is arithmetically easy and aerodynamically impossible. A design that violates \(W_2/W_1 > 0.72\) will separate, and the mean-line calculation will not say so.

Tip clearance omitted from a CFD model. Clearance flow is a first-order loss and changes the whole passage structure. A sealed-tip computation overpredicts efficiency by points and misses the leakage vortex entirely.

Off-design ignored. Machines spend most of their life away from the design point. A stage optimised at one condition can have unacceptable stall margin at part speed, where the front stages stall and the rear stages choke.

Total and static conditions mixed. Efficiency defined total-to-total, total-to-static and static-to-static differ materially, especially for the last turbine stage where exit kinetic energy is or is not counted. State which.

Worked example

Why a turbine drives many compressor stages:

import numpy as np

U = 350.0                      # blade speed, m/s
cp = 1005.0

# Compressor: turning limited by diffusion. Take a realistic dVtheta.
dV_comp = 90.0                 # m/s of turning a compressor stage can manage
dV_turb = 500.0                # a turbine stage accelerates, so far more turning

w_comp = U * dV_comp
w_turb = U * dV_turb
print(f"compressor stage work : {w_comp/1e3:7.1f} kJ/kg   (dT0 = {w_comp/cp:5.1f} K)")
print(f"turbine stage work    : {w_turb/1e3:7.1f} kJ/kg   (dT0 = {w_turb/cp:5.1f} K)")
print(f"stages driven per turbine stage: {w_turb/w_comp:.1f}")

print()
print(f"{'stage loading psi':>18}{'de Haller W2/W1':>18}{'feasible?':>12}")
for psi in (0.25, 0.35, 0.45, 0.60):
    # crude 50%-reaction estimate of the relative velocity ratio
    phi = 0.55
    W1 = np.hypot(phi, 0.5 + psi/2); W2 = np.hypot(phi, 0.5 - psi/2)
    print(f"{psi:18.2f}{W2/W1:18.3f}{'yes' if W2/W1 > 0.72 else 'NO':>12}")

compressor stage work :    31.5 kJ/kg   (dT0 =  31.3 K)
turbine stage work    :   175.0 kJ/kg   (dT0 = 174.1 K)
stages driven per turbine stage: 5.6

 stage loading psi   de Haller W2/W1   feasible?
              0.25             0.800         yes
              0.35             0.734         yes
              0.45             0.676          NO
              0.60             0.603          NO

One turbine stage drives roughly six compressor stages — which is why a high-pressure spool has one or two turbine stages and eight to ten compressor stages.

The de Haller column shows the ceiling. At this flow coefficient the limit falls between \(\psi = 0.35\) and \(0.45\) — crossing \(W_2/W_1 = 0.72\) at about \(\psi = 0.37\). Ask for more loading and the stage cannot be built, however attractive the reduced stage count looks on paper.

That limit is strongly dependent on flow coefficient, which is the part worth carrying away. Repeating the calculation across \(\phi\) gives a maximum loading of 0.30 at \(\phi = 0.45\), 0.37 at 0.55, 0.46 at 0.65 and 0.57 at 0.75 — so loading and flow coefficient cannot be chosen independently. That coupling is exactly why the Smith chart is a two-dimensional map rather than a pair of limits, and why this screen is a screen rather than a design rule.

Connections

• Propulsion — the cycle these components realise.
• Aerodynamics — cascade aerofoils and loading limits.
• Fluids › Stability and transition — bypass transition is the norm on blades.
• Thermal › Heat transfer — turbine blade cooling.

Sources

• Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery.
• Cumpsty, Compressor Aerodynamics.
• Archive: Mechanical/ME422, Mechanical/ME404, Aerospace/AERSP410; see the course archive.