Parsons stage: work, power and efficiency

Data: $\alpha_1 = \beta_2 = 20^\circ$, $U = 200$ m/s, $\rho = U/C_1 = 0.7$, $\dot m = 12$ kg/s. For a Parsons stage: $V_{r2} = C_1$, $C_2 = V_{r1}$, $\beta_1 = \alpha_2$.

(a) Inlet velocity triangle

$$\begin{aligned} C_1 &= U/0.7 = 285.7\ \text{m/s} \\ C_{w1} &= C_1\cos 20^\circ = 268.5\ \text{m/s}, \quad C_{a} = C_1\sin 20^\circ = 97.7\ \text{m/s} \\ V_{w1} &= C_{w1} - U = 68.5\ \text{m/s} \\ \tan\beta_1 &= C_a/V_{w1} \Rightarrow \beta_1 = 55.0^\circ \end{aligned}$$ $$V_{r1} = \sqrt{68.5^2 + 97.7^2} = 119.3\ \text{m/s} = C_2$$

(b) Work and power

$$w = U(2C_1\cos\alpha_1 - U) = 200 \times (2 \times 268.5 - 200) = 67.4\ \text{kJ/kg}$$ $$P = 12 \times 67.4 = 809\ \text{kW}$$

(c) Energy supplied (entry velocity neglected)

$$\frac{C_2^2}{2} = \frac{119.3^2}{2} = 7.12\ \text{kJ/kg}$$ $$\Delta h = w + \frac{C_2^2}{2} = 67.4 + 7.12 = 74.5\ \text{kJ/kg}$$

(d) Blade efficiency

$$\eta_b = \frac{w}{\Delta h} = \frac{67.4}{74.5} = 0.904$$

Check with the formula $\eta_b = \dfrac{2\rho(2\cos\alpha_1 - \rho)}{1 + 2\rho\cos\alpha_1 - \rho^2} = \dfrac{1.4 \times 1.1794}{1.8256} = 0.904$.

Answer: $\beta_1 = 55.0^\circ$; $w = 67.4$ kJ/kg; $P = 809$ kW; $\Delta h = 74.5$ kJ/kg; $\eta_b = 90.4\%$.