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The inlet angle of runner blades of a Francis turbine is 90°. The blades are so shaped that the tangential component of velocity at blade outlet is zero. The flow velocity remains constant through out the blade passage and is equal to half of the blade velocity at runner inlet. The blade efficiency of the runner is
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- 25%
- 50%
- 80%
- 89%
Correct Option: D
Given data:
Bi = 90°
Vwo = 0
Vfi = Vfo = | |
2 |
Blade efficiency of the runner,
ηH = | = | ∴ Vwi = ui | ||
gH | gH |
By energy balance equation,
ρQgH = ρQVwiUi + | |
2 |
gH = Ui² + | Ui² + | ||
2 | 8 |
gH = | Ui² | |
8 |
∴ ηH = | = | = 0.8888 | ||
(9/8)Ui² | 9 |
= 88.88% ≈ 98%