Answer:
[tex]P_{out} = 508.071 kW[/tex]
Given:
efficiency of the turbine, [tex]\eta [/tex] = 65% = 0.65
available gross head, [tex]H_{G}[/tex] = 45 m
head loss, [tex]H_{loss}[/tex] = 5 m
Discharge, Q = [tex]2 m^{3}[/tex]
Solution:
The nozzle is 100% (say)
Available power at the inlet of the turbine, [tex]P_{inlet}[/tex] is given by:
[tex]P_{inlet} = \rho Qg(H_{G} - H_{loss})[/tex] (1)
where
[tex]\rho [/tex] = density of water = 997 [tex]kg/m^{3}[/tex]
acceleration due to gravity, g = [tex]9.8 m^{2}[/tex]
Using eqn (1):
[tex]P_{inlet} = 997\times 2\times 9.8(45 - 5) = 781.65 kW[/tex]
Also, efficency, [tex]\eta [/tex] is given by:
[tex]\eta = \frac{P_{out}}{P_{inlet}}[/tex]
[tex]0.65 = \frac{P_{out}}{781.648\times 1000}[/tex]
[tex]P_{out} = 0.65\times 781.648\times 1000 = 508071 W = 508.071 kW[/tex]
[tex]P_{out} = 508.071 kW[/tex]
