Respuesta :
Answer:
a) 0.76
b) 0.80
c) 1964 kW
Explanation:
GIVEN DATA:
[tex]\dot m = 5000 kg/s[/tex]
Assume Mechanical energy at exist is negligible
A) Take lake bottom as reference, and then kinetic and potential energy are taken as zero.
change in mechanical energy is givrn as
[tex]e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}[/tex]
= 0.491 kJ/kg
[tex]\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW[/tex]
[tex]\eta_{OVERALL} = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76[/tex]
B) [tex]\eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80[/tex]
c) [tex]\dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)[/tex]
[tex]\dot W_{shaft} = 1964 kW[/tex]