 Incompressible water flows at a rate of 15,000 gallons per minute through a hydroelectric dam, dropping 133 feet and losing 17 feet of head due to friction along the way. Determine the theoretical maximum power output of the turbine. I'm going to recognize that this is a conservation of energy problem. We know that because we were told to find power. The presence of power implies to us that we're going to be using the conservation of energy in the same way that the presence of forces implied to us that we were going to be using the conservation of momentum. That doesn't necessarily mean that the conservation of energy is the only equation we're applying, but it's definitely going to be involved because it's going to be what gets us to our answer. Furthermore, I recognize that I can't simplify this all the way down to a Bernoulli principle problem because applying the Bernoulli principle requires that we make six big assumptions, including but not limited to assuming no friction, which we can't do, and assuming no shaft work, which we can't do. Therefore, I can't use Bernoulli's principle. I have to use the full conservation of energy simplification that we made from the Reynolds transport theorem. Let's start by running that out. To be able to apply this equation, I must be describing a control volume, which means that we have to establish a control volume. I have a couple of options here. I could assume that my control volume is something like this, at which point I would have an inlet and an outlet that didn't have a substantial change in potential energy, but do have a change in either enthalpy or kinetic energy. And that increase in enthalpy and kinetic energy on the inlet side comes from the conversion of potential energy up here into either enthalpy or kinetic energy depending on the conditions before it enters the turbine. So we would have to analyze that conversion before we even started our control volume. Instead, it would be smarter for us to define our control volume like this. This way we are accounting for the potential energy inside of our control volume. We aren't performing a conversion just to perform an analysis on the result of the conversion. We are handling everything all at the same time. This is better too because the difference in potential energy is what we actually have. It's what's driving the conversion of energy in the first place. So it's better to just start with the information we have as much as possible. At this point, I have an inlet, which I will call state 1, and an outlet, which I will call state 2. I can make a couple of simplifications and assumptions for this. I will start by assuming incompressible flow so the density of our water isn't assumed to change anywhere in the problem. Then I will assume steady state because it doesn't appear that time matters in this problem. Regardless of when you look at the turbine, there is 15,000 gallons per minute flowing through it, and a potential energy change, some friction, head loss, and a power output, none of which are affected by time. So steady state is going to be how we are going to analyze this problem. Then I'm going to make a couple of simplifications about the parameters of the problem itself. First of all, we recognize that if we wanted to represent P1, we would have to really work our way from atmospheric pressure down through this height of a fluid and from atmospheric pressure down across this height of a column of fluid. But because we don't know how high the river is at state 1 and state 2, we can't actually figure out how P1 and P2 could vary relative to one another. But what we can say with a reasonable amount of confidence here is that that change in potential energy that is driving the pressure increase from here and here and here and here is already accounted for because presumably the 133 feet is actually from this to this. So I've already accounted for the difference in height of the water itself. I don't actually need to account for any minor changes in elevation relative to the inlet and outlet of the water. I mean, this would be the same problem, for example, if we considered our control volume as going down out here and we had like a big old reservoir of water that filled this entire space and there's a pipe here, like it's still a pressure difference between one and two. It's just that we've defined our inlet up here as opposed to right here. Does that logic make sense? So the elevation difference is what's driving the pressure difference. Therefore, the pressure difference doesn't really matter. I'm going to get rid of all these scribbles. We're saying P1 is relatively close to P2. Similarly, I'm going to assume that the change in kinetic energy is negligibly small for exactly the same reasons. A, we don't have enough information to be able to describe how the velocity is affected by conditions at the inlet or at the outlet. But if we had any change in kinetic energy, it would presumably be caused by our change in elevation. The potential energy is probably converted to enthalpy and kinetic energy before it's converted into shaft work. So by accounting for our change in potential energy, we've already accounted for any changes that it makes along the way. Does that distinction make sense? You could also come at this assumption by thinking about the inlet and outlet pipe. If we had the same mass flow rate at state 1 and state 2, and we've assumed incompressible flow, therefore the density at 1 and 2 is the same, that means we must have the same volumetric flow rate at state 1 and state 2. That same volumetric flow rate is 15,000 gallons per minute. Then I could describe the volumetric flow rate as the average velocity times cross-sectional area. And if the inlet and outlet areas are relatively close to one another, that means the average velocity must be relatively close to one another. Therefore, the change in velocity isn't particularly relevant to this problem. We're saying v1 and v2 are pretty much the same, close enough for our purposes, so as to be able to account for that energy somewhere else. So so far, because p1 and p2 are the same, we can get rid of this term, and because v1 and v2 are the same, we can get rid of this term. That leaves me with a change in elevation, a friction head, a pump head, and a turbine head. We need the turbine head and the friction head, but the pump head doesn't make sense for this problem. We have no opportunities for a pump to be operating, therefore we can get rid of our pump head. Now we just have z1 minus z2 is equal to friction head plus turbine head. Turbine work and power is what I'm actually looking for, and the turbine head is going to be what gets me there. So I will rewrite this as the turbine head is equal to z1 minus z2 minus the friction head, which would be 133 feet minus 17 feet, which would be 117. So that 117 feet represents the head that's left over for our turbine to actually produce power from. And I could calculate that number, I mean I could compute it, we did it in our heads, at least I hope we did successfully, but the turbine head doesn't really matter in and of itself, it's just a stepping stone to get to the actual answer. And instead of writing out the number, let's consider how we get from the turbine head to the turbine power. We recognize that our head term is defined as a work specific work divided by gravity. So I could say the turbine head is equal to the specific work of the turbine divided by gravity, and I know the specific work is related to power by mass flow rate. Therefore, if I rearrange this equation to write it in terms of specific work of the turbine, I could say the specific work of the turbine is equal to the turbine head times gravity, which means the power of the turbine is equal to the mass flow rate of water times the turbine head times gravity. And we don't have mass flow rate, we have volumetric flow rate, so I will recognize that I can write mass flow rate in terms of volumetric flow rate by involving density. Density times volumetric flow rate is our mass flow rate. So density of water times the volumetric flow rate of water, which because we're in fluids, I will write as a Q instead of the volumetric flow rate symbol I use for thermal, and then turbine head and gravity. We know this, we were given it. We know the turbine head because we know all three of these variables up here. We can assume gravitational acceleration, and we can assume a temperature of water and then look up that density of water. So I will add to our assumptions, gravity is about 9.81 meters per second squared, which because we have imperial units is probably more convenient to write in terms of feet per second squared. So I will write 32.2 feet per second squared. And we will assume the water is at standard temperature and pressure, at which point we are using 20 degrees Celsius or 68 degrees Fahrenheit as our temperature and one atmosphere for our pressure when we're looking at the conditions of water. So if we jump into the appendix of our textbook, we can get the density of water from table a one in imperial units, 20 degrees Celsius or 68 degrees Fahrenheit is going to give us a density for water. This is table a one by the way of 1.937 slugs per cubic foot. So I will jump back into the iPad. And we will say from table a one at standard temperature and pressure, density of water is 1.937 slugs per cubic foot. And now we have everything we need. It's just a matter of plugging everything in all together. And in an effort to try to make this easier to follow, I will write out one more symbolic step. Now we know everything. Density of water, volumetric flow rate, the elevation difference, the friction head and gravity. It's time to plug in some numbers for that. I will need a big old horizontal line. And then I plug in about 1.937 slugs per cubic foot. And then I'm going to multiply by 15,000 and good gallons per minute. He said hoping it was indeed 15,000 and it is. And then I multiply by the difference in elevation between Z1 and Z2. That was 133 feet. This is my Imperial unit boys by the way. 133 minus our friction head, which was 17 length unit 17 feet. Yeah. And that is feet. And then I multiply by 32.2 feet per second squared. And then it's time for some unit conversions for that. I will start with gallons to cubic feet and minutes and seconds. Actually, let me rephrase. We want a power in horsepower. So let's start at our destination and work backwards. That's what I always recommend as opposed to trying to figure out what route gets you to your destination. If you start there and break it apart into its components and get the components to cancel, you're going to have an easier time. So I have my conversion factor sheet, a couple of ways to write horsepower, including feet times pounds of force per second. That's probably going to be the most convenient. So I will say one horsepower is equal to 550 feet times pound force per second. One horsepower is 550 feet times pound force per second. And I'm out of horizontal line. Of course I am. So feet, feet, cancel so far. Then I'm going to write pounds of force in terms of slugs. Remember our definition of the slug means one pound of force is one slug times one feet per second squared. And then slugs cancel slugs. And second squared cancel second squared and feet and feet cancel. And pounds of force and pounds of force cancel. Then if I convert minutes to seconds, I will get rid of the time units and gallons to cubic feet. I will get rid of the volume units. One minute is 60 seconds. And one gallon is a number of cubic feet that is conveniently available on this sheet somewhere. One gallon is one, excuse me, 0.13368, 0.13368, 0.13368, 0.13368 cubic feet. And then gallons cancel gallons and cubic feet cancel cubic feet. Leaving me with horsepower. Cool beans. So if we pop up the calculator and wake it up, we can start some computation. I'm going to take one, I'm going to wrap it in parentheses just for good measure. 1.937 times 15,000 times 133 minus 17 times 32.2 times 0.13368 divided by 550 times 60. That was all the denominator. Then I get 439.63, 439.63 horsepower. And that represents the maximum power that could be generated by the turbine because we are converting energy perfectly. We're saying all of the change in elevation except for that that's lost for friction is going directly into turbine power output. In reality, there's going to be some other conversion losses along the way, not the least of which is going to be the energy lost due to the conversion process from rotational energy to electrical energy. And that electrical energy is going to be lost in its transmission because not all of the energy produced is actually making it to wherever it's going. I mean, if we talked about this turbine powering, say my house, I live in South Dakota, and a lot of our power over here on the eastern side of South Dakota comes from the big hydroelectric dam in Peer, that energy that's produced at the turbines in Peer is being transmitted to me and a lot of it is being lost in the transmission process. So I mean, this is how much power we're getting out of it, but we can't actually make use of all of it, nor is all of it actually going to make it there in the first place. Depending on where you're measuring the power, you're going to have a lower number than that. And also, you're probably not going to be able to take advantage of all of the elevation change being converted into power in the first place.