Electrical power from turbine?

[evenstar]

Homework Statement

A small dam produces electrical power. The water falls a distance of 27m to turn a turbine. If the efficiency to produce electrical energy is 65%, at what rate must water flow over the dam to produce 780 kW of electrical energy?

Homework Equations

v² = u² + 2as

P = $$\tau$$ x $$\omega$$

The Attempt at a Solution

I've basically been at this one on and off for a few hours, and have tied myself so completely in knots that I have no idea what I'm doing anymore. Please help someone!

 

[Carid]

Consider a litre of water going over that dam in one second.
What is its mass?
How much kinetic energy will it pick up by falling 27 metres?
The energy conversion is not totally efficent so reduce that by 35%.
Now how many litres per second do I need?

 

[thomate1]

I would approach this problem by first understanding the basic principles of electrical power and how it is generated. Electrical power is the rate at which energy is transferred or converted into electricity, and it is typically measured in watts (W) or kilowatts (kW). In this problem, we are given a specific target of 780 kW of electrical energy that needs to be produced, and we are also given information about the height of the water and the efficiency of the system.

To solve this problem, we can use the equation P = \tau x \omega, where P is the power, \tau is the torque, and \omega is the angular velocity. In this case, the torque is generated by the falling water, and the angular velocity is determined by the speed at which the turbine is rotating. We can also use the equation v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

First, we need to find the torque generated by the falling water. This can be calculated by using the equation \tau = mg\Delta h, where m is the mass of the water, g is the acceleration due to gravity, and \Delta h is the height of the water. In this case, we are given the height of the water (27m) and we can assume a mass of 1kg for simplicity. Therefore, the torque is 27 x 1 x 9.8 = 264.6 Nm.

Next, we need to find the angular velocity of the turbine. This can be calculated by dividing the power (780 kW) by the torque (264.6 Nm). This gives us an angular velocity of 2961.2 rad/s.

Now, we can use the equation v2 = u2 + 2as to find the initial velocity of the water. We know that the final velocity (v) is zero, since the water stops moving once it reaches the bottom of the dam. We also know the acceleration (a) is due to gravity (9.8 m/s2) and the distance (s) is 27m. Therefore, we can rearrange the equation to solve for u, the initial velocity. This gives us u = \sqrt{2as} = \sqrt{2 x 9.8 x 27} = 23.9 m/s.

Finally