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

4.326 MW

Step-by-step 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
`$K_S = 17.5 \ mm = 12 * 10^(-2)$` m

`$K_S = (QV^2)/(V^2)=(QN^2)/((√(2gH))^2) $`

`$K_S= (30 * N^2)/((√(2 * 9.81 * 200))^2)$`

`$N^2=(12 * √(2* 9.81 * 200))/(30)$`

`$N=160$`

`$u_2=u_1=(2 \pi N)/(60) =(2 \pi * 160)/(60)$`

= 16.75 m/s

Head loss (H) =
`$(v_2^2)/(2g)+(V_(\omega_1)v_1)/(g)$`

`$7=((0.75)^2)/(2 * 9.81)+(V_(\omega_1) * 16.75)/(9.81)$`

`$V_(\omega_1) = 10.13 \ m/s$`

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

`$=1000 * 30 * (10.13 * 16.75)$`

= 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