Question

The inlet angle of runner blades of a Francis turbine is ${90}^o$. The blades are so shaped that the tangential component of velocity at blade outlet is zero. The flow velocity remains constant throughout the blade passage and is equal to half of the blade velocity at runner inlet. The blade efficiency of the runner is

• A. 25%
• B. 50%
• C. 80%
• D. 89%

Answer 1

The correct option is D 89%

${\eta }_{Blade}=\frac{RP/\dot{m}g}{\frac{RP}{\dot{m}g}+\frac{V{f}_2^2}{2g}}(\therefore V_2=V_{f2})$

$\therefore \frac{RP}{\dot{m}g}=\frac{V_{w1}{u}_1}{g}$

$=\frac{\frac{V_{w1}{u}_1}{g}}{\frac{V_{w1}{u}_1}{g}+\frac{V{f}_2^2}{2g}}\therefore V_{w1}=u_1$

$V_{f_2}=\frac{u_1}{2}$

${\eta }_{Blade}=\frac{\frac{u_1^2}{g}}{\frac{u_1^2}{g}+\frac{u_1^2}{8g}}=\frac{1}{1+\frac{1}{8}}$

$=\frac{8}{9}=89\%$