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Question 5.10: The turbine in Fig. 5–24 is used in a small hydroelectric pl......

The turbine in Fig. 5–24 is used in a small hydroelectric plant. If the discharge through the 0.3-m-diameter draft tube at B is 1.7 m³/s, determine the amount of power that is transferred from the water to the turbine blades. The frictional head loss through the penstock, turbine, and draft tube is 4 m.

fig 5-24
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Fluid Description.   This is a case of steady flow. Here viscous friction losses occur within the fluid. We consider the water to be incompressible, where γ = 9810 N/m³.
Control Volume.   A portion of the reservoir, along with water within the penstock, turbine, and draft tube, is selected to be the fixed control volume. The average velocity at B can be determined from the discharge.
Q = V_BA_B;                          1.7  m^3/s = V_B[\pi(0.15  m^2)]

                                                      V_B = 24.05  m/s
Energy Equation.   Applying the energy equation between A (in) and B (out), with the gravitational datum set at B, we have

\frac{p_A}{\gamma}  +  \frac{V_A^2}{2g}  +  z_A  +  h_{pump}  =  \frac{p_B}{\gamma}  +  \frac{V_B^2}{2g}  +  z_B  +  h_{turbine}  +  h_{L}

 

0 + 0 + 60  m + 0 = 0 +\frac{(24.05  m/s)^2}{2(9.81  m/s^2)} + 0 + h_{turbine} + 4  m

h_{turbine} = 26.52 m
As expected, the result is positive, indicating that energy is supplied by the water (system) to the turbine.
Power.   Using Eq. 5–17, the power transferred to the turbine is therefore
\dot{W_s} = Q\gamma h_s = (1.7  m^3/s)(9810  N/m^3)(26.52  m)

= 442 kW
By comparison, the power lost due to the effects of friction is
\dot{W_L} = Q\gamma h_L  = (1.7  m^3/s)(9810  N/m^3)(4  m) = 66.7  kW

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