A hydroelectric power plant uses spinning turbines to transform the kinetic energy of moving water into electric energy with 80% efficiency. That is, 80% of the kinetic energy becomes electric energy. A small hydroelectric plant at the base of a dam generates 50 MW of electric power when the falling water has a speed of 18 m/s . What is the water flow rate—kilograms of water per second—through the turbines?

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Accepted Answer

Hey, everyone. So in this problem, we're working with power in the work energy theorem. Let's see what they're telling us. We have a water turbine at a dam using kinetic water kinetic energy to turn a shack connected to an impeller that pumps water from that same dam, We're told to assume a 90% efficiency in converting the kinetic energy of the water into the kinetic energy of the shaft. That's further explained by 90% of the kinetic energy is successfully used to pump the water. A typical water pump pushes water at a rate of 22. kW. If the velocity of the water hitting the turbine is 20 m/s, we are asked to determine the flow rate of the water hitting the turbines that facilitates pumping water at that 22.5 kilowatt value. So our multiple choice answers here are a 25 100 kg per second. B 62.5 kg per second. C 31.3 kg per second or d 125 kg per second. So the first thing we can do in this problem is recall that power is equal to work over time. And using the work energy theorem in this problem, we can recall that work is equal to delta K or the change in kinetic energy. So that looks like one cat M V F squared minus one half M B I S square. Our initial velocity of water is zero because it is starting at rest. And so our delta K simplifies 21 half M V F squared. When we plug that into our work, our power equation, we are left with M V F squared over two, multiplied by delta T. Now they're asking us to find the flow rate and that flow rate is going to be M over delta T. So if we look at everything else in this equation, we have what we need to solve. So we have power and we have our final velocity. The trick here is to realize that the power of the pump is not going to be the same as the power of the shaft because of that efficiency rating. So the power of our pump is going to be 0.9 or 90% times the power of the shaft which is Given to us as 22.5 kW. So that equals 25 watts. Now, we can rearrange this equation to solve for mass over delta T and that equals two P sub P the pump over V F squared. So M over Delta T, it's gonna be two Times 25 kW Over VF which is given to us in the problem is m/s squared. And we plug that in and we get kilograms per second. So that is our final answer for this problem. And that aligns with answer choice D that's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Kinetic Energy

Power and Efficiency

Flow Rate