In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g
1 answer:
The electric output of the plant is 48.19 MW
First we need to calculate the water power, it is given by the formula
WP=ρQgh
Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head
Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW
Now the overall efficiency of the hydroelectric power plant is given as
η=
Plugging the values in the above equation
0.84=EP/57.38
EP=48.19 MW
Therefore, the electric output of the plant is 48.19 MW.
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Answer
given,
constant speed of cart on right side = 2 m/s
diameter of nozzle = 50 mm = 0.05 m
discharge flow through nozzle = 0.04 m³
One-fourth of the discharge flows down the incline
three-fourths flows up the incline
Power = ?
Normal force i.e. Fn acting on the cart
v is the velocity of jet
Q = A V
v = 20.37 m/s
u be the speed of cart assuming it to be u = 2 m/s
angle angle of inclination be 60°
now,
F n = 2295 N
now force along x direction
Power of the cart
P = F x v
P = 1987.52 x 20.37
P = 40485 watt
P = 40.5 kW
