Parsons turbine stage work and efficiency

Given: $D=1.0$ m, $N=3000$ rpm, $\alpha_1=\phi=20^\circ$, $V_1=250$ m/s, $m=10$ kg/s. For a Parsons turbine the moving blades are the mirror image of the fixed blades, so the velocity triangles are symmetrical: $V_{r2}=V_1$, $V_{r1}=V_2$ and the exit angle of moving blades $\phi=\alpha_1=20^\circ$.

Blade speed

$$u=\frac{\pi DN}{60}=\frac{\pi\times1.0\times3000}{60}=157.1\ \text{m/s}$$

Inlet triangle (leaving fixed blades, entering moving blades)

$$V_{w1}=V_1\cos20^\circ=234.9\ \text{m/s},\qquad V_{f1}=V_1\sin20^\circ=85.5\ \text{m/s}$$ $$V_{r1}=\sqrt{(V_{w1}-u)^2+V_{f1}^2}=\sqrt{77.8^2+85.5^2}=115.6\ \text{m/s}$$

(a) Inlet angle of moving blades

$$\tan\theta=\frac{V_{f1}}{V_{w1}-u}=\frac{85.5}{77.8}\ \Rightarrow\ \theta=47.7^\circ$$

Outlet triangle

 Inlet (moving blade)          Outlet
      V1   Vf                      V2   Vf
        \  |  Vr1                   \   |   Vr2 = V1
   20 deg\ |/                         \  |  /
   -------*----> u                ----*-----> u  (Vr2 backward)

$V_{r2}=V_1=250$ m/s at $\phi=20^\circ$. Exit whirl (backward):

$$V_{w2}=V_{r2}\cos20^\circ-u=234.9-157.1=77.8\ \text{m/s}$$

Exit velocity of steam: $V_2=\sqrt{V_{w2}^2+V_f^2}=115.6$ m/s, (equal to $V_{r1}$, as expected).

(b) Power

Work per kg of steam:

$$W=u\,(V_{w1}+V_{w2})=157.1\times312.8=49.13\ \text{kJ/kg}$$ $$P=mW=10\times49.13=491\ \text{kW}$$

(c) Blade efficiency

Energy supplied $=$ work done $+$ kinetic energy carried away by steam leaving the stage $\left(\frac12V_2^2\right)$:

$$\frac12V_2^2=\frac{115.6^2}{2}=6.69\ \text{kJ/kg}$$ $$\eta_b=\frac{W}{W+\frac12V_2^2}=\frac{49.13}{49.13+6.69}=88.0\%$$

The kinetic energy leaving this stage is not wasted when the next stage follows; it is used again by the next fixed blades.

Answer: (a) $\theta=47.7^\circ$; (b) $P=491$ kW (49.13 kJ/kg); (c) $\eta_b=88.0\%$.