 Consider a cylindrical tank draining as a result of gravity through a five millimeter hole in the bottom. If the tank starts with a depth of water of 40 centimeters, determine the depth of the water in the tank after, I don't know, like 12 seconds. I will point out that when I made this PDF I didn't actually include enough information to actually answer the question without an assumption that you would have to make, which would simplify it too much. So I'm going to specify the top diameter. I'm going to say that this tank is actually a diameter of, I don't know, let's make it a nice even increment, let's call it 50 millimeters. So everyone in agreement, that is part of the problem statement itself. That's not an assumption that we're making. Totally written out here. I'll even switch to black here. And also the tank diameter is 50 millimeters. There, look, it's in the problem statement and everything. So this time around, we don't have the luxury of treating this as a steady state problem. But I do have the luxury of assuming incompressible flow, because for our purposes, we're going to be treating any liquid as an incompressible fluid. So I'm going to start a list of assumptions over here. And I'm going to lead off the list of assumptions with the assumption of incompressible flow. The next, I recognize that I'm going to be using my Reynolds transport theorem simplified for mass, that conservation of mass equation is dm dt of our system is equal to d dt of the integral of the control volume that's density with respect to volume plus the integral across the control surface, which is density times velocity vector with respect to area. And then I will simplify dm dt as zero for control volume. In the previous problem, we had been able to simplify this second term as being zero because it was steady state. This time I can't treat it as steady state. So I will have to include it for now. So we have zero is equal to d dt of the integral of density with respect to volume plus the integral across the control surface of density times velocity vector with respect to area. But I can simplify it a little bit by bringing out the density because I've assumed incompressible flow. Then dividing all three terms by density yields zero is equal to then I have d dt of the integral of the volume plus the integral across the control surface of the velocity vector with respect to area. And I should have indicated that was a vector. So we have a changing volume here. And the volume is changing with respect to time. So it's going to be easiest to explore that first. We have a pretty good grasp on how to quantify the velocity of the fluid across the orifice here. The only orifice that's relevant is going to be the outlet. Because my control volume is encompassing all the water in the tank. And as it goes down, my control volume is just shrinking. So there's no moving inlet or anything. So we will set this aside for now and focus on the new more difficult part, which is the volume changing. And I will describe my dy term. So as the infinitesimal step down, dy, then I can describe d volume, which is this volume as the cross sectional area times dy. So I can say d volume is equal to cross sectional area. And I will call that area of tank area top at for now area of the tank times dy. Then I'm saying zero is equal to d dt times the integral of at dy plus the integral across the control surface of velocity vector with respect to area. And then I am going to recognize that my area on the top doesn't change with respect to time. So I'm bringing that out. So I have a top times dy dt, which is something that we can handle here. It's not as new and scary as it seems. Then on the right, let's treat this as uniform flow, because I was described a velocity. And that velocity well that's the same as average should really establish some state points here. I'm calling this two. The outlet is two that top is top abbreviated with the t perfectly clear. So when I get rid of this vector, I'm running that as uniform flow that would have the velocity vector and the area of vector in the same direction, which means that I'm writing this as average velocity state two times area state two. Again, area at state two is the area of a circle with a diameter of five millimeters. Okay, that's definitely not confusing. So I have zero is equal to at times dy dt plus the velocity, which is a function of y times area. So I'll write this as dy dt by bringing this over to the left and dividing by a top at which point I would have negative a two divided by a t times v bar two. And then I will substitute in my equation that I was given for the average velocity state two, which by the way is just conservation of energy here, you can even simplify for a streamline and use Bernoulli's principle at which point you're saying your velocity at the top is going to be very small and your pressure change there is zero. But we're describing the velocity at state two as the square root of two times gravity times y and as dy dt. So I will break apart my radical right that is negative a two over a t times the square root of two times gravity times y to the one half power. Then I can group together my y terms and my t terms and get this in a little bit more of a convenient format. So I'm saying dy over y to the one half is equal to negative a two over a top times square root of two times gravity times dt. And I am going to start a new column. And actually, let's just start a whole new page here, because we're probably going to need more than that. Not anyway. The whole new page and the equation that is starting off the new page is this one. So we are going to be integrating y to the negative one half. Because as you guys know, I cannot do any sort of integration that is in the power rule. And I will write this a little bit more cleanly. So hopefully it's easier to follow. Now what are we integrating across here, we are integrating from the initial position of y, which I will call y not because I mean, why not to y at the time that we defined earlier, which was 12 seconds. So I'm going to call that y, we're integrating from y not to y. And that would allow us to describe a function for y is a function of t. So then on my integral on the right hand side, where I have negative a two over a top times the square root of two times gravity dt. Here we are integrating from a time of zero to a time of t, so that we can come up with an equation as a function of t. Okay. And then that's going to be see, we add one, that becomes y to the one half power, and then we have two, so two times y to the one half power, because that would be divided by negative one half plus one, which is divided by one half, which is multiplied by two. Does that make sense? And we are evaluating that from zero to y. And then on the right hand side, a two, a top two and gravity are all constants, so they all come out. And we have negative a two over a t times the radical two times gravity times the integral of dt from zero to time, which means that we're just left with time, because the time at zero is zero. Okay. And should I have indicated this in more steps? Yeah, maybe. I'll just pad this out with a little bit more work. So hopefully it's easier for people who are just looking at the PDF. So that's drop bracket, one over negative one half plus one times y to the negative one half plus one divided from y not to y. And then on the right, we have negative a two over a top times the square root of two times gravity times the integral of dt from zero to t, which simplifies down to just t. Okay, hopefully that's easier to follow. Then when we plug in, this is not zero to y, this is y not. When we plug in y and we subtract y not, and this is going to be the quantity two times y to the one half power minus two times y not to the one half power, I can factor out the two and which one I have two times the quantity, y to the one half power minus y not to the one half power is equal to negative a two over a top times the square root of two times gravity, which I really should have just copied and pasted by now times time. Why not is a constant. So I'm going to bring that over to the right hand side. And I'm going to say two, that's divided by two divided by two add y not to the one half power and I have y to the one half power is equal to would be y not to the one half power divided by two. No, excuse me, is equal to y not to the one half power because I already divided by two plus negative a two divided by two times a t times the square root of two times gravity times hot. Then I want to square everything. So that would be y on the left. And then squaring everything on the right, I can't just take y not plus the quantity squared, I have to square the entire quantity, that would be y not to the one half power plus negative a two over two times a top times the square root of two times gravity times time quantity squared. And then I could simplify that a little bit if I wanted to, I could, I could factor out y not and have y not out front, but it would be one minus a two over two times a two. Yeah, that wouldn't be super convenient, because I still have the y not term here. So I'm going to have to keep track of that square. Anyway, y as a function of time then is the quantity y not to the one half power plus that's okay. I'm just going to make a simplification right that has minus one half times the quantity a two over a time that's excuse me a top times square root of two times gravity times time. And by leaving it this way, we don't have to foil anything. So a two over a top, the area of a circle could be written as pi over four times diameter squared. So I can write this as a two is equal to pi over four times diameter two squared and a top as pi over four times diameter top squared. Therefore, that proportion would be pi over four times d two squared divided by pi over four times d top squared, the pi is over four, excuse me, the pi over fours would cancel. So that'd be left with d two over d top quantity squared. So I'm going to make that substitution here. And then that's pretty much as far as I can go substitution wise. So this is why not the one half power minus one half times d two over d top squared times the square root of two times gravity times time quantity squared. And then we are going to evaluate this at a. Okay, let me start over on a new page. Just so that we have some room to work. So this equation, I will bring to a new page. And I will write that as evaluated at time equals 12 seconds and gravity is 9.81 meters per second squared, we should list that as an assumption. d two was the diameter at the bottom, which was five millimeters. The top I had made up was 50 millimeters. And why not is 40 centimeters. So I'm going to have a quantity in centimeters to the one half power minus one half times the unitless proportion because the millimeters will cancel squared times the square root of two times gravity gravity is going to be in length per time squared. So the square root of that would be linked to the one half power over time times time. So the time dimension is going to cancel. So I'm going to be left with length to the one half minus length to the one half quantity squared. So the result inside of the parentheses is going to be in length to the one half power. And then when I square it, I will be left with length. It is what I want. So I will have square root of 40 centimeters to the one half power minus one half times d two over d top squared. Why am I doing that? I wanted to plug in numbers now. So dt was five over 50 quantity squared times the square root of two times 9.81 meters per second squared multiplied by seconds, that would be 12 seconds entire quantity to the second bar. And then I want centimeters to the one half power to come out of the radical. So when I take two times 9.81 meters per second squared, I have to multiply by the conversion from meters to centimeters within that radical. So I'm going to add in here. One meter is 100 centimeters, meters cancels meters I'm left with centimeters to the one half power over seconds when I'm done taking the square root. Suddenly me with centimeters to the one half power quantity squared, which will leave me with centimeters. Perhaps that would be the advantage of factoring out why not. But I wouldn't have to worry about units. People will try that. We'll see how this goes. Okay calculator, if you would come back, we have quantity 40 to the one half power minus 0.5 times the quantity five over 50 squared times the square root of two times 9.81 times 100. Not sure why I put so many parentheses there times 12. And then we are taking that entire thing to the second bar. We get a syntax error as is tradition. So I will just add parentheses until it stops breaking. Okay, I will subtract parentheses until it stops breaking. How about that? How about that? That's not actually what I want. Calculator. Why you got to be difficult? Ah, that's why that was there. Okay. So one more here for good measure. Nope. One more here for good measure and we get 13.4461 centimeters. So why evaluated at a time of 12 seconds yields an answer of 13.4461, which I will round to 45. Because that's a little bit more of a reasonable quantity. And if you heard me rambling earlier about factoring out why not. My logic here is if you really didn't want to hurt your brain about the units, it might be easier to factor out why not. So if we here I will open a new page, if we establish an alternate universe here where we hadn't yet calculated an answer, then I will take this quantity and I'll show you how that would work. This is just an algebra preference thing. It can be confusing to try to keep track of units inside of a radical. So if I were to Okay, let's let's actually back up. Why squared you remember was why not the one half power minus this entire quantity in here one half times d two over d top quantity squared times a square road of two times gravity times time. So if I were to factor out why not squared, I would have one, excuse me, factor out why not to the one half power. I would have one why not to the one half power times one minus one over two times why not to the one half power. And that would be multiplied by t two over dt square times the square road of two times gravity times time. And the advantage of that is the quantity inside of that parenthetical statement here, I'll switch that to a bar so it matches. Okay, the quantity inside of that bar is just going to be a unitless proportion. So I'm taking a unit multiplying by why not one half power, which once I square everything is just going to be why not times that entire quantity square. So that might be easier to keep track of especially if I were to bring this why not term inside of the radical. So just because we're going down this rabbit hole already, guys, why not to the one half power times one minus one half times d two over dt squared. So this is a unitless proportion, which one squared is still a unitless proportion. And then I'm multiplying by the square root of two times gravity divided by why not, because why not to the one half power is square root of why not sauce divided by the square root of why not combine the radicals. So now I would just be left with times squared square root of which is time in the denominator times time, which is unitless. And then when I square that entire quantity, excuse me, this, this wasn't right. This was one half. So when I square everything now, I'm left with why to the one half power. No, John, what are you doing? When I square everything, I'm left with y times one minus one half or that neater one minus one half times the quantity d two over d top squared times the quantity square root of two times gravity over why not times time quantity squared. So the only real advantage of that would be the fact that you wouldn't have to worry about your unit conversion inside of that parenthetical statement, it would just be whatever you plug in for the unit on why not is what you get out over here. We would just have to make sure that the units canceled over here, which is still meters per second squared divided by centimeters. So you'd have to convert centimeters to meters to get the the length dimension to cancel within the radical. So I guess it's not even that much easier, but potato potato, whatever you prefer. And you know what, while we're, you know, just going down a rabbit hole of stuff that we didn't necessarily need to calculate in order to answer this question. What if I were to pose the question, how long would it take for this to empty? How long would it take for the tank to empty? Well, I mean, we would just plug in a y value of zero and solve for t, we could do that math together. But since that's a little bit outside the scope of what I'm actually asking for right now, and I don't want to have to do all the algebra just to talk through that answer, let me jump to a computer. The easiest way for me to show that would be Wolfram Alpha. So I will pop you over to a new Chrome tab. And we will type y is equal to 40 centimeters times one minus 0.5 times five over 50 squared times square root of two times 9.81. Put that in parentheses just for fun. Divide it by 40 and then I have to make the meters cancel. So that'd be 9.81 meters per second squared divided by quantity and centimeters. So I need to cover that to meters, which would be divide by 100. The meters and meters are canceled, leaving me with a second squared in the denominator of a radical, which when I multiply by time, cancels, which is what I want. Times, let's just call it x for now, because x is going to give it to us. So we're saying y is equal to 40 times this entire quantity as a function of x. So we have a parabola shape. And of course, we're only actually considering the time period from zero to the point at which the tank is empty. So we could see that visually as being somewhere between 20 seconds and 30 seconds. But if I were to plug in zero, I think Wolfram Alpha will just interpret that as solve for x and it did. So the amount of time it would take for us to empty the tank would be 28.56 seconds. But we could have done that algebra manually. And is that neat that we can just see the plot of the height of the water with respect to time? And I will also point out that that velocity at the exit was a simplification made by the Bernoulli principle. So as a result, those six assumptions required to use the Bernoulli principle are already applied behind the scenes in the generation of the problem. So that's something else to bear in mind. Let's see if I can plot this thing from zero to 28.569. So that's the water level as a function of time. As the height of the water gets lower, the velocity at the outlet gets lower. So the slope is decreasing or perking up here. Neat. I think we can consider that problem solved.