Respuesta :
Answer:
4.326 MW
Explanation:
Given :
Discharge through system = [tex]$30 \ m^3/s$[/tex]
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 [tex]$K_S = 17.5 \ mm = 12 \times 10^{-2}$[/tex] m
[tex]$K_S = \frac{QV^2}{V^2}=\frac{QN^2}{(\sqrt{2gH})^2} $[/tex]
[tex]$K_S= \frac{30 \times N^2}{(\sqrt{2 \times 9.81 \times 200})^2}$[/tex]
[tex]$N^2=\frac{12 \times \sqrt{2\times 9.81 \times 200}}{30}$[/tex]
[tex]$N=160$[/tex]
[tex]$u_2=u_1=\frac{2 \pi N}{60} =\frac{2 \pi \times 160}{60}$[/tex]
= 16.75 m/s
Head loss (H) = [tex]$\frac{v_2^2}{2g}+\frac{V_{\omega_1}v_1}{g}$[/tex]
[tex]$7=\frac{(0.75)^2}{2 \times 9.81}+\frac{V_{\omega_1} \times 16.75}{9.81}$[/tex]
[tex]$V_{\omega_1} = 10.13 \ m/s$[/tex]
The runner power obtained, [tex]$P_R = \rho Q(V_{\omega_1}u_1)$[/tex]
[tex]$=1000 \times 30 \times (10.13 \times 16.75)$[/tex]
= 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
