Euler's Equation and Velocity Triangles

Euler's turbomachinery equation, velocity triangles, flow and loading coefficients, and the degree of reaction

Farnborough, 1926

In July 1926 A. A. Griffith of the Royal Aircraft Establishment published "An Aerodynamic Theory of Turbine Design". The turbines and compressors of the day were miserably inefficient, and Griffith showed why: their blades were "flying stalled". He proposed giving them proper aerofoil shapes — treating each blade as a small wing — which would dramatically improve their performance (A. A. Griffith). Lesson 4-1 showed what was at stake: below about 70 % efficiency a gas turbine cannot even drive its own compressor. Griffith's paper said that an axial machine with well-designed blades could clear that bar comfortably. Every axial compressor and turbine since is a stack of Griffith's little wings. This lesson gives the one equation that tells you how much work they do.

A stage is a row of rotating blades — the rotor — and a row of fixed ones — the stator (called vanes or nozzle guide vanes in a turbine). Only the rotor exchanges work with the gas. The stator turns the flow back so that the next rotor receives it at the right angle, and in a compressor it converts some of the flow's kinetic energy into pressure.

The work a rotor does depends only on how much it changes the gas's swirl — the velocity component

C_\theta

in the direction of rotation.

w = U_2 C_{\theta 2} - U_1 C_{\theta 1} \quad\xrightarrow{\ U_1 = U_2 = U\ }\quad w = U\,(C_{\theta 2} - C_{\theta 1})

Euler's turbomachinery equation: the work done on each kilogram of gas by a rotor with blade speed U. For an axial stage the radius barely changes, so U is the same at inlet and exit.

Derivation

Euler's equation from angular momentum

Take a control volume around the rotor. The torque on the gas equals the rate at which its angular momentum about the axis changes — the angular form of Newton's second law. Gas enters at radius

r_1

with swirl

C_{\theta 1}

and leaves at

r_2

with

C_{\theta 2}

\tau = \dot m\,(r_2 C_{\theta 2} - r_1 C_{\theta 1}).

The rotor spins at

\Omega

, so the power it delivers is

\tau\Omega

, and with

U = \Omega r

the work per kilogram is

w = \frac{\tau\Omega}{\dot m} = U_2 C_{\theta 2} - U_1 C_{\theta 1}.

Nothing else about the blades appears: their shape matters only through the swirl they produce. For an adiabatic rotor this work is the rise in stagnation enthalpy,

c_p\,\Delta T_t = w

💡Velocity triangles

The blades see the gas in their own rotating frame. Its velocity relative to the blade,

\vec W

, and its absolute velocity,

\vec C

, differ by the blade speed:

\vec C = \vec W + \vec U.

Drawn at the rotor inlet and exit, these vectors form the velocity triangles. Angles are measured from the axial direction:

\alpha

for the absolute flow,

\beta

for the relative flow. With the axial velocity

C_x

usually about constant through a stage, the triangles fix everything: the swirl, the work, and how much the blades must turn the relative flow.

📐A compressor stage

Mean blade speed

U = 350

m/s, axial velocity

C_x = 175

m/s, flow entering at

\alpha_1 = 20°

, rotor turning the relative flow to

\beta_2 = 45°

(Lab 1):

Inlet swirl $C_{\theta 1} = 63.7$ m/s; the blade sees the flow at $\beta_1 = 58.6°$ , $W_1 = 335.6$ m/s.
Exit swirl $C_{\theta 2} = 175$ m/s.
Work $w = 350 \times (175 - 63.7) =$ 38.96 kJ/kg: a stagnation temperature rise of 38.8 K.

A 40:1 compressor needs about 740 kJ/kg (Lesson 6-2). One stage does a twentieth of that.

💡Three dimensionless numbers

Flow coefficient $\phi = C_x/U$ : how fast the gas moves through compared with the blades.
Stage loading $\psi = w/U^2 = \Delta C_\theta/U$ : how hard the stage works. Compressors run about 0.3–0.45, turbines 1–2.5.
Degree of reaction $R$ : the share of the stage's static enthalpy rise that happens in the rotor. A 50 % stage shares it equally between rotor and stator and has symmetric triangles.

For the stage above:

\phi = 0.5

\psi = 0.318

R = 0.659

R = \frac{h_2 - h_1}{h_3 - h_1} = 1 - \frac{C_{\theta 1} + C_{\theta 2}}{2U} = \frac{\phi}{2}\left(\tan\beta_1 + \tan\beta_2\right)

Degree of reaction of a repeating stage with constant axial velocity (C3 = C1).

Derivation

The degree of reaction

In the rotor's frame no work is done, and at fixed radius the rothalpy

h + \tfrac12 W^2 - \tfrac12 U^2

is conserved. With

U

unchanged, the rotor's static enthalpy rise is

h_2 - h_1 = \tfrac12\left(W_1^2 - W_2^2\right), \qquad W^2 = C_x^2 + (C_\theta - U)^2 .

For a repeating stage the gas leaves the stator as it entered the rotor, so the stage's static rise equals its stagnation rise,

h_3 - h_1 = U(C_{\theta 2} - C_{\theta 1})

. Dividing, the

C_x^2

terms cancel and

R = \frac{(C_{\theta 1} - U)^2 - (C_{\theta 2} - U)^2}{2U(C_{\theta 2} - C_{\theta 1})} = 1 - \frac{C_{\theta 1} + C_{\theta 2}}{2U}.

Writing the swirls as

C_\theta = U - C_x\tan\beta

gives the relative-angle form. Lab 3 lets SymPy do the algebra.

📐The same work, less pressure

Feed the stage above with real air (CoolProp, stage efficiency 0.90; Lab 2). The first stage, from 288 K, gives a pressure ratio of 1.492. Repeat it ten times: the last stage gives only 1.207, because the air arrives at it hotter — and compressing hotter air takes more work per unit of pressure ratio. Ten stages give 14.9 overall and an exit temperature of 667 K. That is why compressors need ever more stages for each extra point of pressure ratio.

For since the fabric of the universe is most perfect, and is the work of a most wise Creator, nothing whatsoever takes place in the universe in which some relation of maximum and minimum does not appear.

– Leonhard Euler, on the calculus of variations (1744) (Wikiquote)

🕰Griffith's other theory

The same A. A. Griffith is the father of fracture mechanics: his theory of how cracks make brittle materials fail far below their theoretical strength was later extended to almost all materials (A. A. Griffith). It returns in Lesson 9-2, in the burst of a turbine disc.

💡Where you will use this

Lesson 6-2 uses the triangles to find how hard a compressor stage can be loaded, 6-3 applies them to turbines. The same equation governs centrifugal compressors (where

U_2 \ne U_1

does much of the work), pumps, fans and wind turbines.

📚Further reading

S. Larry Dixon and Cesare A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed., Butterworth-Heinemann, 2014), on axial compressors and turbines.
Nicholas A. Cumpsty, Compressor Aerodynamics (Longman, 1989; Krieger, 2004).
H. I. H. Saravanamuttoo et al., Gas Turbine Theory (7th ed., Pearson, 2017), on axial-flow compressors.

↓But there is a problem

One compressor stage gave 39 kJ/kg. Why not turn the flow harder and make each stage do twice as much? Because a compressor's blades work against a rising pressure, and if they are asked for too much the flow breaks away. The F-14A Tomcat's engines did that in combat — at the worst possible moments.