In a hydroelectric dam, water falls 26.0 m and then spins a turbine to generate electricity. What is ∆U of 1.0 kg of water? Suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 54.0 MW of electricity? This is a typical value for a small hydroelectric dam.

potentialenergy= mgh=1kg*9.8N/kg*26m

massflow*g*h/time * .80=54*10^6
Flowrate=54^10^6/gh * 1/.80

Well, ∆U is basically the change in potential energy, which in this case is the change in gravitational potential energy. We can calculate that using the equation ∆U = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.

So, ∆U = 1.0 kg × 9.8 m/s² × 26.0 m = 254.8 J.

Now, let's move on to the second part of the question. We know that the dam is 80% efficient, so only 80% of the water's potential energy is converted to electrical energy. Therefore, we can calculate the total power output as follows:

Power output = Efficiency × ∆U/Time

54.0 MW = (0.80) × (254.8 J)/Time

To find the mass of water passing through the turbines each second, we need to rearrange the equation like this:

Time = (0.80) × (254.8 J)/(54.0 × 10⁶ W)

Now, let's plug in the numbers:

Time = (0.80) × (254.8 J)/(54.0 × 10⁶ W) = 3.76 × 10⁻⁵ s

Finally, we can calculate the mass of water passing through the turbines each second using the rearranged equation:

Mass = Power output/(Efficiency × ∆U/Time)

Mass = 54.0 × 10⁶ W/[(0.80) × (254.8 J)/(3.76 × 10⁻⁵ s)]

After some calculations, we find that the mass is approximately 8.76 × 10⁶ kg.

So, to generate 54.0 MW of electricity, approximately 8.76 × 10⁶ kg of water must pass through the turbines each second. And that's a lot of water! I hope they have good plumbing down there.

To find the change in potential energy (∆U) of 1.0 kg of water, we can use the formula:

∆U = m * g * ∆h

where m is the mass of water, g is the acceleration due to gravity (9.8 m/s^2), and ∆h is the change in height.

Given that the height is 26.0 m, we can calculate ∆U:

∆U = 1.0 kg * 9.8 m/s^2 * 26.0 m = 254.8 J

Therefore, the change in potential energy of 1.0 kg of water is 254.8 J.

Now, let's calculate the amount of water that needs to pass through the turbine per second to generate 54.0 MW of electricity.

First, we need to convert 54.0 MW to watts:

1 MW = 1,000,000 watts
54.0 MW = 54,000,000 watts

Given that the dam is 80% efficient in converting potential energy to electrical energy, only 80% of the potential energy of the water can be converted to electrical energy. Therefore, the power output of the dam would be:

Power output = Efficiency * ∆U/ time

Rearranging the formula, we get:

time = ∆U / (Efficiency * Power output)

Substituting the values:

time = 254.8 J / (0.80 * 54,000,000 W) = 4.168 x 10^-6 s

So, for each second, 1 kg of water can generate electrical energy for 4.168 x 10^-6 s.

To find the mass of water necessary to generate 54 MW of electricity per second, we divide the power output by the change in potential energy:

mass = Power output / ∆U

mass = 54,000,000 W / 254.8 J = 211,964 kg

Therefore, approximately 211,964 kg of water must pass through the turbines each second to generate 54.0 MW of electricity.

I forgot to include the 80% efficiency factor, but BobPursley did it right

If what you are calling ∆U includes the gravitational potential energy*, then it is -(density*g*(delta H)

= -(10^3 kg/m^3)*26 m*9.81 m/s^2
The product will be in Joules.

For P = 54*10^6 Watts, you need a mass flow rate such that
P = (mass flow rate)*(∆U)

*Usually, in thermodynamics, U represents internal energy only, which has nothing to do with elevation.