 Water flows at a rate of 0.6 cubic meters per second through a turbine as shown in the figure below. If the intended power output of the turbine is 60 kilowatts, determine the pressure drop across the turbine. I recognize that this is a conservation of energy problem because we were told about a power, and power implies conservation of energy in the same way that forces implied conservation of momentum. Then I recognize that I can't simplify all the way down to Bernoulli's principle because I have a shaft work. Therefore, I'm going to be applying our conservation of energy equation at some point. It might not be the only thing that I use, but it's definitely involved because it's how we can relate 60 kilowatts to something more meaningful for our analysis. Next, I will establish my control volume, and I'm going to define my control volume like this. And I will establish some assumptions, let's say over here, beginning as per usual with incompressible flow and steady state, and then incorporating, let's say, elevation change. I'm going to say state one and state two are pretty close to the same elevation. So change in z is pretty close to zero. I have no friction losses to speak of because I wasn't told about any. And no opportunity for a pump to do anything of consequence. And I also recognize that I forgot an alpha. And let's assume that we have turbulent flow. For turbulent flow, which remember for our purposes, if you aren't told enough information to deduce otherwise, it's turbulent for now. Once we start getting the tool set to be able to calculate whether the flow is going to be laminar or turbulent, we can actually involve that. But for now, if you don't know otherwise, it's turbulent. So if we're treating this as turbulent flow, turbulent, then alpha is about one. Okay. So so far, we've been able to eliminate this term and this term and this term. Can we neglect the changes in kinetic energy like we did last time? No, because we have a change in cross-sectional area. We were told a diameter at the inlet and the outlet. And that change in diameter is going to mean, because we have the same mass flow rate and incompressible flow, that we have a different velocity at states one and two. It may not be significant in the calculation of the problem, but we should include it because we can include it. So the way that we will include that information is in the form of our conservation of mass, which if we skip a couple of the simplification steps, we can write as density times A1 times V1 on average is equal to density times A2 times average velocity 2. Densities are the same because of incompressible flow. The cross-sectional area is pi over 4 times that diameter squared. Pi over 4 cancels at which point average velocity 2 is going to equal average velocity 1 times D1 over D2. Quantity squared. Now I got a little bit ahead of myself because I assumed while I was simplifying down here that I had V1, but I don't have the volumetric flow rate. So let's get rid of this. And let's leave the pi over 4 for now because the volumetric flow rate is the cross-sectional area, meaning both of these terms are the volumetric flow rate. Therefore I can write the volumetric flow rate at state 1 as 4 times Q divided by pi times diameter squared at state 1 and average velocity at state 2 as 4 times the volumetric flow rate divided by pi times diameter 2 squared. So if I wanted to, I could compute a number, plug those in numerically, but you guys know that I like to work as far as I can symbolically before plugging in any numbers. So I'm going to leave them symbolically. Then our turbine head here, remember, is the specific work over gravity. Therefore work could be written as head times gravity and power can be related to specific work by taking mass flow rate times specific work, which means power can be written as density times volumetric flow rate times specific work, which could be written as density times volumetric flow rate times Hg. So we are taking our mass flow rate, we are multiplying by our specific work, which is the head multiplied by gravity, and that power term is actually what I have, not what I'm looking for. What I need to be able to plug into the conservation of energy before this relationship to be written out, solved for the turbine head. So turbine head is going to be the power of the turbine divided by the density times volumetric flow rate times gravity. So in our conservation of energy equation, we have enough information to be able to plug in for our turbine head and our velocities, that is plug in quantities in terms of things that I know. And I can look up a density and assume a gravity, so I should be able to write P1 minus P2, which would be the pressure drop, in terms of quantities that I know. So let's do some algebra. Let's hopefully do it correctly. I will break it out into multiple steps, because we have infinite space. And why not? P1 minus P2 divided by density times gravity is equal to alpha 2 times V2 squared minus alpha 1 times V1 squared divided by density times gravity, no excuse me, 2 times gravity plus power of the turbine divided by density times volumetric flow rate times gravity. And then I will multiply both sides by gravity, get in order of the gravity term, and I will multiply both sides by density, at which point I have P1 minus P2, which is what I want. The alpha terms are 1 because of turbulent flow is equal to density over 2 times the quantity V2 squared minus V1 squared plus the power of the turbine divided by u, because again density cancels. Then I can make the substitution for our velocity, so that's going to be density over 2 times 4 squared times q squared divided by pi squared times d1 to the fourth power because it'd be d1 squared squared, so in our velocity terms we're plugging in 4 squared, q squared, pi squared, and then diameter to the fourth, and I probably could just factor that out all in one fell swoop, but I won't for now to hopefully make this easier to follow. This is 4 times q over pi times diameter 2 squared squared minus 4 times q divided by pi times diameter 1 squared squared plus the power of the turbine divided by q and then I'm going to be writing 4 times q over pi times d2 squared quantity squared is 4 squared times q squared divided by pi squared times d2 to the fourth and then I'm going to be factoring out the 4 squared, the q squared, and the pi squared, so this is going to become 4 squared times rho times q squared divided by 2 times pi squared yeah that all makes sense, times 1 over d2 to the fourth power minus 1 over d1 to the fourth power plus the power of the turbine divided by q and then q diameters, power, and volumetric flow rate are all known quantities the density is something that we can look up, so we're going to be assuming that our flow is at standard temperature and pressure here so we are going to be using from table A1 at density of water at standard temperature and pressure is 998 kilograms per cubic meter and again that's because we're assuming the density doesn't change so we're looking at the density at one condition and we're calling that about one atmosphere and 20 degrees Celsius for our purposes and it's just important that we have that listed as an assumption so at standard temperature and pressure then everything is a known quantity so P1 minus P2, which is what we're looking for, the pressure drop is a function of known quantities, so just time to do some math for that I will jump to a new page and start plugging in some numbers I'll draw a horizontal line to begin with and we have 4 squared over 2 times pi and that was pi squared then we're multiplying by 998 kilograms per cubic meter times 0.6 squared cubic meter squared which would be meters to the sixth per second squared and let me just double check that that's 0.6 cubic meters per second and it is and then we are dividing that by nothing in particular because we've already accounted for the 2 and the pi so this quantity, let me back up a second we're looking for a pressure and presumably that pressure is going to be measured in either pascals, kilopascals or megapascals I'm going to start with the assumption that it's in kilopascals calculate an answer in kilopascals and then if I need to convert it into megapascals or pascals to make it a reasonable number to express as an answer I will do that but since the problem doesn't specify a pressure unit I'm just going to start with kilopascals because I have to start somewhere so I want my first quantity including the diameters to be in kilopascals and when I convert to kilopascals which I will do by saying a kilopascal is a thousand pascals and a pascal is a Newton per square meter and a Newton is a kilogram meter per second squared Newton cancels Newton's pascals, cancels pascals, kilograms, cancels kilograms second squared cancels second squared and I now have eight meters in the numerator and four in the denominator which is good because when I take this quantity multiplied by a difference between numbers that are one over meters to the fourth power I will be left with just kilopascals so I'm multiplying this by one over I believe this was 0.4 so 0.4 to the fourth power minus one over 0.3 to the fourth power and these are both in one over meters to the fourth power now meters to the fourth and cubic meters, cancel meters to the sixth and meters squared leaving me with kilopascals for our first term and then to that quantity we are going to add power divided by volumetric flow rate that power was 60 kilowatts so 60 kilowatts divided by 0.6 cubic meters per second and we want this in kilopascals presumably so I will so I will start at kilopascals and work backwards and then break everything into their component parts kilopascal is a thousand pascals and a pascal is a Newton per square meter and a kilowatt is a thousand watts that's not what I wanted my pet is a thousand watts and a watt is a joule per second and a nope why are you doing that look it really wants me to look up good thing I'm not in a horror movie joules cancel joules watts cancel watts newtons cancel newtons pascals cancel pascals kilowatts cancel kilowatts seconds cancel seconds meters and square meters cancel cubic meters leaving me with kilopascals and I can get rid of these thousands here because they cancel so at this point everything is done except for computing a quantity so calculator you are needed once again I'm going to just arbitrarily pick some parentheses out front here and then I'm going to take four squared times 998 times 0.6 squared take that entire quantity divided by 2 times i squared which is a lot of clicks in my calculator emulator times 1000 and then I'm multiplying that quantity by 1 over 0.4 to the crap I hit the clear button instead of the care button let's start over 4 squared no parentheses John 4 squared times 998 times 0.6 squared divided by 2 times i squared times 1000 close parentheses close parentheses times 1 over 0.4 to the fourth power let's add some more parentheses just for good measure minus 1 over 0.3 to the fourth power and close parentheses is that the correct number of parentheses? no how about now? cool and then we add to that 60 divided by 0.6 and we get 75.423 let's just double check that all of that looks correct 4 squared 998 0.6 squared divided by 2 times pi squared times 1000 probably could have just written 2 in the numerator instead of 4 squared divided by 2 but what's done is done and everyone knows you can't change anything in your calculations once it's written so we get 75.423 that's a quantity in kilopascals that represents the difference in pressure between states 1 and state 2 which is the pressure drop and with that we are done with this problem