Answer:
a) 75%
b) 82%
Explanation:
Assumptions:
[tex]\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}[/tex]
Properties: The density of water [tex]\delta = 1000 kg/m^3[/tex]
Conversions:
[tex]165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\[/tex]
Analysis:
Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:
[tex]e_{mech_{in}} - e_{mech_{out}} = gh - 0[/tex]
Then;
[tex]gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}[/tex]
gh = 0.491 kJ/kg
[tex]\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg[/tex]
= 1559 kW
Therefore; the overall efficiency is:
[tex]\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}[/tex]
[tex]= \dfrac{1166 \ kW}{1559 \ kW}[/tex]
= 0.75
= 75%
b) mechanical efficiency of the turbine:
[tex]\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}[/tex]
thus;
[tex]\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%[/tex]
