Question

4. A hydropower installation is to be located where the downstream water-surface elevation is 150 m below the water-surface elevation in the reservoir. The 1.5-m-diameter concrete-lined penstock is 300 m long and has an estimated roughness height of 17.5 mm (ks). When the flow rate through the system is 30 m3 /s, the combined head loss in the turbine and draft tube is 7.5 m, and the average velocity in the tailrace is 0.60 m/s. Estimate the power that can be extracted from the system.

Answer 1

Answer:

4.326 MW

Explanation:

Given :

Discharge through system = $30 \ m^3/s$

Gross head, H = 150 m

Diameter of the pipe = 1.5 m

Length of penstock = 300 m

Head loss due to turbine and draft tube = 7.5 m

Average velocity installed tail race = 0.60 m/s

The power exerted on the system is $K_S = 17.5 \ mm = 12 \times 10^{-2}$ m

$K_S = \frac{QV^2}{V^2}=\frac{QN^2}{(\sqrt{2gH})^2}$

$K_S= \frac{30 \times N^2}{(\sqrt{2 \times 9.81 \times 200})^2}$

$N^2=\frac{12 \times \sqrt{2\times 9.81 \times 200}}{30}$

$N=160$

$u_2=u_1=\frac{2 \pi N}{60} =\frac{2 \pi \times 160}{60}$

                         = 16.75 m/s

Head loss (H) = $\frac{v_2^2}{2g}+\frac{V_{\omega_1}v_1}{g}$

       $7=\frac{(0.75)^2}{2 \times 9.81}+\frac{V_{\omega_1} \times 16.75}{9.81}$

       $V_{\omega_1} = 10.13 \ m/s$

The runner power obtained, $P_R = \rho Q(V_{\omega_1}u_1)$

                                                     $=1000 \times 30 \times (10.13 \times 16.75)$

                                                     = 5.090 MW

So the power exerted by the shaft is up to 85% of the runner power due to mechanical losses = 0.85 x 5.090

                               = 4.326 MW