Interactive Lecture
Understanding Energy Transfer in Turbines and Compressors
by Dr Arnaldo Delli Carri
Turbomachinery refers to devices that transfer energy between a rotating shaft and a fluid. This includes turbines (extract energy from fluid), compressors and pumps (add energy to fluid), and fans. The key to understanding these machines is analysing the velocity triangles at blade inlet and outlet.
Energy transfer in turbomachinery is governed by the change in angular momentum of the fluid as it passes through the blade row. Velocity triangles let you visualise and quantify this momentum exchange, directly linking blade geometry to performance.
At any point in a turbomachine, three velocity vectors form a triangle:
Velocity Components:
Vector Relationship:
\[\vec{V} = \vec{U} + \vec{W}\]
Each velocity is decomposed into:
The fundamental equation for energy transfer in turbomachinery:
\[w = U_2 V_{\theta 2} - U_1 V_{\theta 1}\]
where:
Power:
\[\dot{W} = \dot{m} \cdot w = \dot{m}(U_2 V_{\theta 2} - U_1 V_{\theta 1})\]
where \(\dot{m}\) is the mass flow rate (kg/s)
The blade geometry is defined by angles measured from the axial direction:
Absolute flow angle: \(\alpha\) (angle of \(\vec{V}\) from axial)
\[\tan \alpha = \frac{V_\theta}{V_z}\]
Relative flow angle: \(\beta\) (angle of \(\vec{W}\) from axial)
\[\tan \beta = \frac{W_\theta}{W_z}\]
For axial machines: \(V_z \approx W_z = V_a\) (constant axial velocity)
The degree of reaction indicates how much pressure change occurs in the rotor vs the stator:
\[R = \frac{\text{Pressure rise in rotor}}{\text{Total pressure rise}} = \frac{W_1^2 - W_2^2}{(V_2^2 - V_1^2) + (W_1^2 - W_2^2)}\]
For an axial compressor with \(V_a\) constant:
\[R = \frac{W_{\theta 1}^2 - W_{\theta 2}^2}{2U(V_{\theta 2} - V_{\theta 1})}\]
Special Cases:
Energy Direction:
Adjust the parameters to build velocity triangles for a turbine or compressor stage. See how blade angles and speeds affect energy transfer. Watch how the blade profiles change based on the relative flow angles β!
Velocity triangle analysis sets the blade geometry in every machine that moves energy between a shaft and a fluid:
| Concept | Equation | Description |
|---|---|---|
| Velocity Triangle | \(\vec{V} = \vec{U} + \vec{W}\) | Vector relationship |
| Euler Turbine Equation | \(w = U_2 V_{\theta 2} - U_1 V_{\theta 1}\) | Specific work transfer |
| Power | \(\dot{W} = \dot{m}(U_2 V_{\theta 2} - U_1 V_{\theta 1})\) | Power transferred |
| Blade Speed | \(U = \omega r = 2\pi n r\) | Tangential velocity (\(n\) = rotational speed, rev/s; \(\omega = 2\pi n\), rad/s) |
| Absolute Flow Angle | \(\tan \alpha = \frac{V_\theta}{V_z}\) | Angle from axial direction |
| Relative Flow Angle | \(\tan \beta = \frac{W_\theta}{W_z}\) | Angle in rotating frame |
| Degree of Reaction | \(R = \frac{W_1^2 - W_2^2}{(V_2^2 - V_1^2) + (W_1^2 - W_2^2)}\) | Rotor pressure rise fraction |
An axial turbine stage has a blade speed U = 300 m/s and constant axial velocity Va = 200 m/s. The inlet absolute flow angle is α₁ = 60° and the outlet is α₂ = 20°. Calculate: (a) the specific work output, (b) the power for a mass flow rate of 50 kg/s, and (c) the inlet and outlet relative flow angles.
Show Solution
(a) Calculate tangential velocities:
\[V_{\theta 1} = V_a \tan\alpha_1 = 200 \times \tan(60°) = 200 \times 1.732 = 346.4 \text{ m/s}\]
\[V_{\theta 2} = V_a \tan\alpha_2 = 200 \times \tan(20°) = 200 \times 0.364 = 72.8 \text{ m/s}\]
Specific work (Euler equation):
\[w = U(V_{\theta 2} - V_{\theta 1}) = 300 \times (72.8 - 346.4) = 300 \times (-273.6) = -82,080 \text{ J/kg}\]
Work extracted = 82.08 kJ/kg (positive convention for turbine)
(b) Power output:
\[\dot{W} = \dot{m} \times W = 50 \times 82.08 = 4,104 \text{ kW} = 4.1 \text{ MW}\]
(c) Relative flow angles:
\[W_{\theta 1} = V_{\theta 1} - U = 346.4 - 300 = 46.4 \text{ m/s}\]
\[\beta_1 = \tan^{-1}\left(\frac{W_{\theta 1}}{V_a}\right) = \tan^{-1}\left(\frac{46.4}{200}\right) = 13.1°\]
\[W_{\theta 2} = V_{\theta 2} - U = 72.8 - 300 = -227.2 \text{ m/s}\]
\[\beta_2 = \tan^{-1}\left(\frac{-227.2}{200}\right) = -48.6°\]
Answer: (a) w = 82.08 kJ/kg extracted; (b) 4.1 MW; (c) β₁ = 13.1°, β₂ = −48.6°.
For the turbine in Problem 1, calculate the degree of reaction. Interpret what this value means for the pressure distribution in the stage.
Show Solution
Calculate relative velocity magnitudes:
\[W_1 = \sqrt{V_a^2 + W_{\theta 1}^2} = \sqrt{200^2 + 46.4^2} = \sqrt{40000 + 2153} = 205.3 \text{ m/s}\]
\[W_2 = \sqrt{V_a^2 + W_{\theta 2}^2} = \sqrt{200^2 + 227.2^2} = \sqrt{40000 + 51620} = 302.7 \text{ m/s}\]
Degree of reaction:
\[R = \frac{W_1^2 - W_2^2}{2U(V_{\theta 2} - V_{\theta 1})} = \frac{205.3^2 - 302.7^2}{2 \times 300 \times (72.8 - 346.4)}\]
\[R = \frac{42148 - 91627}{2 \times 300 \times (-273.6)} = \frac{-49479}{-164160} = 0.301\]
Interpretation: R = 0.3 means approximately 30% of the pressure drop occurs in the rotor (moving blades) and 70% in the stator (nozzle/guide vanes). This is closer to an impulse stage, which typically has lower reaction values. Lower reaction stages have higher relative velocities in the rotor, requiring stronger blade materials.
Answer: R = 0.301 (about 30% of the pressure drop in the rotor, 70% in the stator).
Design an axial compressor stage with 50% reaction (R = 0.5), blade speed U = 250 m/s, and axial velocity Va = 180 m/s. If the inlet flow is purely axial (α₁ = 0°), find the required outlet flow angle α₂ and the specific work input.
Show Solution
Given conditions:
α₁ = 0° → V_θ1 = 0 (purely axial inlet)
For 50% reaction with symmetric velocity triangles: β₁ = α₂ and β₂ = α₁ = 0°
From inlet conditions:
\[W_{\theta 1} = V_{\theta 1} - U = 0 - 250 = -250 \text{ m/s}\]
\[\beta_1 = \tan^{-1}\left(\frac{-250}{180}\right) = -54.2°\]
For 50% reaction: α₂ = β₁
\[\alpha_2 = 54.2°\]
\[V_{\theta 2} = V_a \tan\alpha_2 = 180 \times \tan(54.2°) = 180 \times 1.389 = 250 \text{ m/s}\]
Specific work input:
\[w = U(V_{\theta 2} - V_{\theta 1}) = 250 \times (250 - 0) = 62,500 \text{ J/kg} = 62.5 \text{ kJ/kg}\]
Verification: For 50% reaction, inlet and outlet velocity triangles are symmetric, which simplifies design and balances loading between rotor and stator.
Answer: α₂ = 54.2°, specific work input w = 62.5 kJ/kg.