12.8 Steam & Gas Turbines: Impulse, Reaction & Velocity Diagrams

Impulse Versus Reaction

All turbines extract work by turning a fluid stream, but they differ in where the pressure drop occurs.

The reason a reaction turbine requires full admission is that a pressure difference exists across the moving blades. If only part of the annulus were fed, gas would leak circumferentially through the unfed passages. An impulse turbine has no such pressure difference, so partial admission is permissible — which is why the first, high-pressure stage of a large steam turbine is usually an impulse stage.

Degree of Reaction

$R = \frac{\text{Enthalpy drop in moving blades}}{\text{Total enthalpy drop in the stage}}$

The 50% reaction case is important: the fixed and moving blade rows have the same profile, so the velocity triangles at inlet and outlet are mirror images. This symmetry simplifies both manufacture and analysis.

Velocity Triangles

The velocity diagram is the central analytical tool. Three velocities appear at each of inlet and outlet:

These are related by the vector triangle $\vec{V} = \vec{u} + \vec{V_r}$. Two component directions matter:

• $V_w$, the whirl (tangential) component, which produces torque;
• $V_f$, the flow (axial) component, which carries the mass through.

Angles: $\alpha$ is the nozzle angle at inlet, $\beta$ the blade inlet angle, and the corresponding outlet angles are usually written $\phi$ and $\theta$.

Work and Efficiency for an Impulse Stage

From the momentum equation applied tangentially, the work per unit mass is

$W = u\left(V_{w1} + V_{w2}\right)$

where the whirl components are added when the outlet whirl is in the opposite direction to the inlet whirl, which is the usual case. In terms of relative velocities and blade angles:

$W = u\,V_{r1}\left(\cos\beta + k\frac{V_{r2}}{V_{r1}}\cos\theta\right)$

where $k = V_{r2}/V_{r1}$ is the blade velocity coefficient, less than one because of friction. For smooth, symmetrical blading, $k \approx 0.9$ to $0.95$ and $\theta = \beta$.

Blade or diagram efficiency

$\eta_b = \frac{\text{Work done on blades}}{\text{Kinetic energy supplied}} = \frac{2u(V_{w1}+V_{w2})}{V_1^2}$

Defining the blade speed ratio $\rho = u/V_1$ and taking symmetric, frictionless blades:

$\eta_b = 4\rho\left(\cos\alpha - \rho\right)$

Differentiating with respect to $\rho$ and setting to zero gives the optimum:

$\boxed{\rho_{\text{opt}} = \frac{\cos\alpha}{2}}, \qquad \eta_{b,\max} = \cos^2\alpha$

So a nozzle angle of 20 degrees gives $\rho_{\text{opt}} = 0.47$ and a maximum blade efficiency of $\cos^2 20^\circ = 0.883$. Reducing the nozzle angle improves efficiency but reduces the flow area, so the practical range is 16 to 22 degrees.

At the optimum, the absolute exit velocity is axial, meaning the leaving whirl is zero and no kinetic energy is wasted in swirl.

Reaction Stage: Parsons Turbine

For 50% reaction with symmetric blading, the corresponding results are

$W = u\left(2V_1\cos\alpha - u\right)$

$\eta_b = \frac{2\rho\left(2\cos\alpha - \rho\right)}{1 + 2\rho\cos\alpha - \rho^2}$

with optimum at

$\rho_{\text{opt}} = \cos\alpha, \qquad \eta_{b,\max} = \frac{2\cos^2\alpha}{1+\cos^2\alpha}$

Note the contrast with the impulse case: the optimum blade speed ratio is $\cos\alpha$ rather than $\cos\alpha/2$, so a reaction stage runs at roughly twice the blade speed of an impulse stage for the same steam velocity. For $\alpha = 20$ degrees the maximum blade efficiency is $2(0.883)/(1.883) = 0.938$, higher than the impulse value — reaction staging is intrinsically more efficient, which is why the low-pressure end of a large turbine is reaction blading.

Compounding

A single impulse stage expanding steam from boiler to condenser pressure would produce an impractically high steam velocity and hence a dangerously high blade speed. Compounding distributes the drop across several stages.

For a Curtis stage with $n$ moving rows, the optimum blade speed ratio is

$\rho_{\text{opt}} = \frac{\cos\alpha}{2n}$

and the work is distributed across the rows in the ratio $(2n-1) : (2n-3) : \ldots : 1$. For a two-row Curtis wheel, the first row does three times the work of the second — a favourite examination result.

Losses and Efficiencies

The reheat factor in a multi-stage turbine is the ratio of the sum of the individual stage isentropic drops to the overall isentropic drop. Because the constant-pressure lines on an enthalpy-entropy chart diverge, this ratio is always greater than one, typically 1.03 to 1.06. It means a multi-stage turbine recovers part of each stage's friction loss as useful enthalpy in the next stage.

Gas Turbines

A gas turbine drives its own compressor, and the compressor absorbs a large fraction of the turbine output. The work ratio is

$\text{Work ratio} = \frac{W_T - W_C}{W_T}$

and it is typically only 0.3 to 0.5. That low value is the defining characteristic of the machine: a modest fall in turbine or compressor isentropic efficiency causes a disproportionately large fall in net output, which is why gas turbines were impractical until component efficiencies reached about 85%.

For the Brayton cycle, the pressure ratio for maximum net specific work is

$r_{p,\text{opt}} = \left(\frac{T_{\max}}{T_{\min}}\right)^{\gamma/2(\gamma-1)}$

which differs from the pressure ratio for maximum thermal efficiency — the latter rises indefinitely with pressure ratio. Real designs compromise between the two.

Blade cooling matters because turbine inlet temperature directly sets efficiency and output, but exceeds the melting point of the blade alloys. Modern practice combines directionally solidified or single-crystal nickel superalloys, internal convective cooling passages, film cooling through surface holes, and thermal barrier coatings.