Impulse stage: blade angles, work and thrust

Data: $\alpha_1 = 20^\circ$, $C_1 = 600$ m/s, $U = 250$ m/s, $k = 0.9$, $\dot m = 10$ kg/s. Symmetrical blades: $\beta_1 = \beta_2$.

(a) Inlet velocity triangle and blade angles

$$\begin{aligned} C_{w1} &= C_1\cos 20^\circ = 563.8\ \text{m/s}, \quad C_{a1} = C_1\sin 20^\circ = 205.2\ \text{m/s} \\ V_{w1} &= C_{w1} - U = 313.8\ \text{m/s} \\ V_{r1} &= \sqrt{V_{w1}^2 + C_{a1}^2} = 375.0\ \text{m/s}, \quad \tan\beta_1 = \frac{C_{a1}}{V_{w1}} \Rightarrow \beta_1 = 33.2^\circ \end{aligned}$$

Blade inlet angle = blade outlet angle = 33.2 degrees.

(b) Outlet velocity triangle

$$\begin{aligned} V_{r2} &= k V_{r1} = 0.9 \times 375.0 = 337.5\ \text{m/s} \\ V_{w2} &= V_{r2}\cos\beta_2 = 282.4\ \text{m/s (backward)}, \quad C_{a2} = V_{r2}\sin\beta_2 = 184.7\ \text{m/s} \\ C_{w2} &= V_{w2} - U = 32.4\ \text{m/s (backward, opposite to } U\text{)} \end{aligned}$$ $$C_2 = \sqrt{C_{w2}^2 + C_{a2}^2} = 187.5\ \text{m/s}$$

The absolute exit velocity is 187.5 m/s at an angle $\alpha_2 = \tan^{-1}(C_{a2}/|C_{w2}|) = 80.0^\circ$ to the blade direction.

(c) Power

$$w = U(V_{w1} + V_{w2}) = 250 \times (313.8 + 282.4) = 149.1\ \text{kJ/kg}$$ $$P = \dot m w = 10 \times 149.1 = 1491\ \text{kW}$$

(d) Diagram efficiency

$$\eta_d = \frac{w}{C_1^2/2} = \frac{149.1}{180.0} = 0.828$$

(e) Axial thrust

$$F_a = \dot m (C_{a1} - C_{a2}) = 10 \times (205.2 - 184.7) = 205\ \text{N}$$

Answer: blade angles $33.2^\circ$; exit velocity 187.5 m/s at $80.0^\circ$; power 1491 kW; $\eta_d$ = 82.8%; axial thrust 205 N.