Parsons reaction turbine blade efficiency

Degree of reaction

The degree of reaction $R$ is the ratio of the enthalpy drop in the moving blades to the enthalpy drop in the whole stage:

$$R = \frac{h_2 - h_3}{h_1 - h_3} = \frac{\Delta h_{rotor}}{\Delta h_{stage}}$$

$R = 0$ for impulse; $R = 0.5$ for a Parsons turbine, where fixed and moving blades are identical and mirror images of each other.

Velocity diagrams (50% reaction)

Because fixed and moving blades are identical: $\alpha_1 = \beta_2$ and $\beta_1 = \alpha_2$, so $V_{r2} = C_1$ and $C_2 = V_{r1}$.

 Inlet                   Outlet
     C1                     Vr2=C1   C2=Vr1
    /|\                       \      /|
   / | \ Vr1                   \    / | Ca
  /a1|b1\                       \b2/  |
 +---U---+                      +--U--+

Work done per kg

$$C_{w1} = C_1\cos\alpha_1, \quad V_{w2} = V_{r2}\cos\beta_2 = C_1\cos\alpha_1$$ $$w = U(C_{w1} + V_{w2} - U) = U(2C_1\cos\alpha_1 - U)$$

Blade efficiency

Energy supplied = enthalpy drop in the fixed blades ($C_1^2/2$, entry velocity neglected) + enthalpy drop in the moving blades $(V_{r2}^2 - V_{r1}^2)/2$. This equals the work done plus the exit kinetic energy, $w + C_2^2/2$. With $C_2^2 = V_{r1}^2 = C_1^2 - 2UC_1\cos\alpha_1 + U^2$:

$$\text{Supply} = w + \frac{C_2^2}{2} = \frac{C_1^2}{2} + UC_1\cos\alpha_1 - \frac{U^2}{2}$$ $$\eta_b = \frac{2U(2C_1\cos\alpha_1 - U)}{C_1^2 + 2UC_1\cos\alpha_1 - U^2} = \frac{2\rho(2\cos\alpha_1 - \rho)}{1 + 2\rho\cos\alpha_1 - \rho^2}, \quad \rho = \frac{U}{C_1}$$

Condition for maximum

Putting $d\eta_b/d\rho = 0$ gives $\rho = \cos\alpha_1$, i.e. $U = C_1\cos\alpha_1$.

$$\boxed{\eta_{b,max} = \frac{2\cos^2\alpha_1}{1 + \cos^2\alpha_1}}$$

For $\alpha_1 = 20^\circ$, $\eta_{b,max} = 2(0.883)/(1.883) = 0.938$. This is higher than the single-stage impulse maximum $\cos^2\alpha_1 = 0.883$, and the optimum blade speed is about twice as high, so reaction stages are used in multistage turbines at moderate speed.