A Kaplan turbine has an outside diameter of runner and hub diameter as 4 m and 2 m, respectively. If the velocity of flow at inlet is 8 m/s, then what will be the discharge passing through the turbine? 

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  1. 68 m3/s
  2. 6.8 m3/s
  3. 7.536 m3/s
  4. 75.36 m3/s

Answer (Detailed Solution Below)

Option 4 : 75.36 m3/s
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Detailed Solution

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Concept:

Kaplan Turbine:

  • Kaplan turbine is an Axial Flow Reaction Turbine.
  • It offers High Discharge and Low Head.
  • In this turbine water flow parallel to the axis of rotation of the shaft.

 

Discharge (Q) for a Kaplan Turbine is given by

\(Q = \frac{\pi}{4}(D_0^2 - D_h^2)V_{f1} = \frac{\pi}{4}D_0^2(1 - \frac{D_h^2}{D_o^2})V_{f1} \)

where Do = Tip Diameter, Dh = Hub Diameter,Vf1 = Flow velocity

Flow Ratio: The ratio of Velocity of Flow at the inlet (Vf1) to the velocity given \(\sqrt{2gh}\) is known as flow ratio.

Flow Ratio, \(\psi = \frac{V_{f1}}{\sqrt{2gh}}\)

where h = Head on Turbine

Calculation:

Given:

Tip Diameter, Do = 4 m, Dh = Hub Diameter, =  2 m, Vf1 = Flow velocity = 8 m/s

Discharge, \(Q = \frac{\pi}{4}(D_0^2 - D_h^2)V_{f1}\)

\(Q = \frac{\pi}{4}(4^2 - 2^2)\times 8 =75.36 ~\rm{m^3/s}\)

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