Single-stage impulse turbine blade work and efficiency

Given: $V_1=900$ m/s, $\alpha=20^\circ$, $u=350$ m/s, blades symmetrical ($\theta=\phi$), $k=V_{r2}/V_{r1}=0.9$, $m=2$ kg/s.

Inlet triangle

$$V_{w1}=V_1\cos20^\circ=845.7\ \text{m/s},\qquad V_{f1}=V_1\sin20^\circ=307.8\ \text{m/s}$$ $$V_{r1}=\sqrt{(V_{w1}-u)^2+V_{f1}^2}=\sqrt{495.7^2+307.8^2}=583.5\ \text{m/s}$$

(a) Blade angles

$$\tan\theta=\frac{V_{f1}}{V_{w1}-u}=\frac{307.8}{495.7}\ \Rightarrow\ \theta=31.8^\circ,\qquad \phi=\theta=31.8^\circ$$

Outlet triangle

 Inlet                    Outlet
      V1  Vf1                  V2    Vf2
        \ |                      \   |
   20deg \|  Vr1            Vr2   \  |
 --------*---------> u    <-------*------- u
  |<-Vw1->|                |Vw2|

$$V_{r2}=0.9\times583.5=525.2\ \text{m/s},\quad V_{f2}=V_{r2}\sin\phi=277.0\ \text{m/s}$$ $$V_{w2}=V_{r2}\cos\phi-u=446.2-350=96.2\ \text{m/s (opposite to blade motion)}$$

The exit whirl is opposite to the blade motion, so it adds to the inlet whirl:

$$V_{w1}+V_{w2}=845.7+96.2=941.9\ \text{m/s}$$

(b) Force and power

$$F_t=m(V_{w1}+V_{w2})=2\times941.9=1884\ \text{N}$$ $$P=F_tu=1884\times350=659\ \text{kW}$$

(c) Axial thrust

$$F_a=m(V_{f1}-V_{f2})=2\times(307.8-277.0)=61.6\ \text{N}$$

(d) Blade efficiency

$$\eta_b=\frac{\text{work done}}{\text{kinetic energy supplied}}=\frac{2u(V_{w1}+V_{w2})}{V_1^2}=\frac{2\times350\times941.9}{900^2}=81.4\%$$

(Check: $\frac12mV_1^2=810$ kW and $P/\text{KE}=81.4\%$.)

Blade speed for maximum efficiency

For a single-stage impulse turbine with equiangular blades, $\eta_b$ is a maximum at

$$u_{opt}=\frac{V_1\cos\alpha}{2}=\frac{900\times\cos20^\circ}{2}=423\ \text{m/s}$$

and the maximum efficiency (with no friction) is $\cos^2\alpha=88.3\%$. The given 350 m/s is below this value, so the efficiency is lower than the maximum.

Answer: (a) $\theta=\phi=31.8^\circ$; (b) $F_t=1884$ N, $P=659$ kW; (c) $F_a=61.6$ N; (d) $\eta_b=81.4\%$; $u_{opt}=423$ m/s.