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9. (1Opts) gas station being build in Orangevale along Greenback Lane: As part of the construction; large spherical tank will be buried to hold gasoline: If the ta...

Question

9. (1Opts) gas station being build in Orangevale along Greenback Lane: As part of the construction; large spherical tank will be buried to hold gasoline: If the tank has radius of SM and Is buried meter below the surface; how much work will be required pump all the gasoline out of the tank when it is full? You may use 750 as the density of gasoline:

9. (1Opts) gas station being build in Orangevale along Greenback Lane: As part of the construction; large spherical tank will be buried to hold gasoline: If the tank has radius of SM and Is buried meter below the surface; how much work will be required pump all the gasoline out of the tank when it is full? You may use 750 as the density of gasoline:



Answers

A gasoline storage tank in the shape of a right circular cylinder, with height $10 \mathrm{~m}$ and radius $8 \mathrm{~m}$, is full of gasoline. How much work is required to pump all the gasoline over the top of the tank? (The density of gasoline is $\left.\rho=720 \mathrm{~kg} / \mathrm{m}^{3} .\right)$

Okay. What we want to do is we want to determine how much work oh is done? Yeah, by filling a pipe and a water tank with water. And we're going to be focused on water. And so one we know the density of water is equal to 62 0.4 pounds per cubic feet. Okay. Um and so the pipe and tank are represented by this figure right here. Um We have a pipe that is 30 this is ground level, so the pipe is 30 ft um tall. Um and then goes into a spherical tank. The pipe has a radius of half a foot. Um And of course the spherical tank has a radius of 10 ft. Okay, So what we're gonna do is we're gonna kind of do these pieces. So we know that the total work is done by um filling one. The combined work of first filling the pipe and then feeling the tank. So in order to fill the pipe and we're going to do that. That pipe that were done by filling the pipe. Remember for fluids work is equal to well for everything, work is equal to force times distance and of course at force um for fluids is the volume times that um density. So what we have here is first of all, it's going to be the density and then the volume is going to be that pie times radius squared times why? Or delta Y. Because we take off a little slab and then the distance is why, because we're going this entire wide length, right? So each time we take off kind of a slab here and that would be delta Y. So that would be pi r squared delta Y. For each of those little slabs on this pipe. So work is equal to your density, times pi times the integral. And I guess we could put for four out here um times that. Why. Dy and we're going from zero 2 30 feet. So that's how we determine this. So they're still going to be um your density over eight times y squared. And we're going to evaluate it at zero and at 30 or 30 and at zero I guess. So the work done by the to fill the pipe is then equal to 60 two point 4/8 times that pie. Um And then we're going to do times 30 squared. So this is going to be equal to about, let's see 22,000 0 45 foot pounds of work. Okay, so that was the easy part. Now we have to think about the um sphere and of course the sphere, the y value the y value is going to go from 30 because it starts here at 30 and then it goes 2 50 right? And so if we look at a cross sectional area of that, so we've just kind of looked out at a cross sectional area of that slab than what we have is once again that cross sectional area is going to be a circle to that cross sectional area is going to be equal to pi r squared. Now the problem is getting the radius. So are any time we look at that cross sexual area that are is going to be equal to the square root because it's going to be um kind of like the distance um is going to be 10 squared because that would be your X value minus. And now you're why value is actually going to be you're 40 minus your y squared. Yeah, so basically it's going to be a circle where it's gonna be x squared plus Y squared equals r squared. And now we're gonna find that are right. Um So this would be um then equal to the square root and let's go ahead and multiply all this out. We have negative Y squared plus 80. Why minus 1500? Okay so now the work done by it, so that change of work done by to fill the sphere now is equal to your density times, pi r squared times delta. Why? Because we're taking those little slabs out of there. Um And then it's going to be times why as well. Okay because we're raising it Y unit. So my work and my by my sphere is actually going to be equal to my density of the water times pi and now we're only going from 30 2 50. And so this will be actually equal to negative Y que plus and that should be an 80 in there, 80 y squared minus 1500. Why do I? Okay so when we integrate this we should have 62.4 because that's the density of water pie and then this would be negative 1/4. Why do the fourth plus 80/3 Y cubed minus 1500 over two y squared. And we're going to evaluate at 50 and then at 30. And so we should get should get about 10 million 455,000 220 foot pounds of work. And then to find the total amount of work we just add those two together. And so I believe that's going to be 10,000 400 and 77,000 274 foot pounds of work. Let's make sure I did that, yep. And there

Okay, we have a cylindrical tank that is four m in diameter. And just to note that four m is going to be consistent all the time and then it is six m high. Now if you're four m in diameter that does mean that you have a radius of two m. So we're asked for how much work is required To pump half the water over the top. So think of that top half. So really if we just would have had a tank that was 3m high we would be pretty much doing the same work. So I'm gonna start with that idea here and then we're gonna talk about some options on different ways that you can integrate this. So if we consider that we have are integral we're gonna do our density of water of 1000. We have to multiply that by 9.8 and then we need our cross sectional area. Remember you multiply that um 1000 kg per meter cubed. You multiply it by a volume but that's going to be a cross sectional area with a thickness of dX. So we can say pi times two squared and it's consistent. We don't need to use a variable and then we have our D. X. Piece. So now the last piece can remember it's M. G. H. And so now we have a height. So this is where the 0 to 3 comes in, it really hasn't coming up to this point, it's the 0 to 3 and this high component that really kind of needs to go hand in hand. So if you consider for a second, if I put a three minus X here, if I put a zero in, I'd be at this point here on either drawing and everything would have to get pumped up three m. By the time I'm putting a three in I just be pumping zero m. So there's definitely other arrangements that these guys could have think about if you wanted to create the top zero and then you could go from negative 3 to 0. But will you be putting in is a negative X. That still makes it. So that when you put in the negative three it goes up three and when you put up the zero it goes and zero, another way to do, it would be 3 to 6 and then you would do a six minus X. So again, if you put a three in it would be going up three but if you put a six in it would be going up zero. So all of these arrangements work exactly the same. So if you want to try a different one than I do, you can see that they're all equivalent. Okay, so let's go ahead and bring all of those constants out front and there is a bunch of them and then we can also go ahead and integrate. So the three becomes a three x. And then our X goes up a power. So we have a negative one half X squared and our brackets will say put in a three and then minus putting in a zero. Of course once we put in the zero we see that piece zeros out. So let's put in the three. So we're gonna have our constant out front and then we'll have nine minus a half of nine or nine over to. And we end up with um the amount of work Being the 176,400 jewels.

Okay. What we have here is we have a cylindrical tank um with a radius of four ft in height. The tin loops 10 ft, not 10 10 ft but 10 ft. Um And we want to it is filled um from top to bottom with oil that has a density of £50 per cubic feet. And we want to determine the work done on pumping over all of the um all of the oil over the edge of the tank. Um And so what we need to do is um I kind of think of a slab, cut these into slabs with a thickness of delta Y. And of course, um the volume of each slab is pi r squared and then delta Y. So this is going to be 16 pie delta Y. And so we know that the work is equal to four some distance and the force is actually going to be the weight of the um weight of each of the slabs of oil. And so that is going to be the density times that volume, which is 16 pie, delta Y. And then work is force times distance. And we're moving these from a distance of 10 minus why? So if y is down here, the first lab is down here, we have to move it all of 10 ft. If the slab is this last one appeared then we're just moving it zero. So um so we know that work is equal to the integral and now we're moving that fluid 10 0 to 10, 0 to 10 of that density of this guy right here. So we have the density, A man might pull out. Well let's just put this in here time. 16 pie times 10 minus. Why do I here we go. And so work is equal to um and then let's do our density is 15 so 50. So we have 50 times 16 pie integral from 0 to 10 of 10 minus Y. Dy. So this is going to be equal to um 50 time 16 pie times 10 y minus one half. Why squared. And we're gonna evaluate at 10 and at zero. And so when we do that, um we should get 100 and 25,000 664 ft pounds of work.

All right, we've got a question. Here we have a fuel tank that is an upright cylinder burried so that it's circular top of 10 ft beneath. The ground level tank has released 5 ft and it's 15 ft high. And although the current oil level is on Lee 6 ft deep, we want to calculate the work required to pump all the oil to the surface. The oil weighs £50. Perfect. Cute. Alright, so we'll call the liquid, which is being pumped out out of height. D of h you know the area and be calculated as pi over r squared. Hey, that circular area. And then our volume would be you would just be changed. An area multiplied by the change in height. Excuse me. Area multiplied by the change in heights you have T V is equal to the area multiplied the change. All right, now we can calculate the weight off that layer. Um, excuse me, The liquid is what I'm referring to when I say the layer. So, um, we know that the vault Excuse me? The weight of that would just be the density multiplied by the volume. Alright, that is your weight And so then you could say, Ah, that your like to expand it out its density Excuse me, Rho pi r squared d of h all right. And we can then calculate our, um, work, respect the height, you know, put a dash here, distinguish between weight and work. We have a church one h two and then we plug in our row pi r squared d of H And so our height to we know would be 25 ft because it's already 10 ft underground. And then there's another 15 ft. How high it ISS. And then if we were going to calculate our, um, lower limit, we would take a these 15 as the height and we'll subtracted by, um who told the current oil level is on Lee 6 ft deep. So we'll take the 15 and we will subtracted by our six. And so we'll get nine and then we have nine plus 10. So we'll get 19 as our lower limit the H 1 19. It's too close to 25. And then when you go ahead and salt with this integral, you can pull out the row pie and the r squared. You just have the integral from 19 25 or D of h. So that equals two Rho pi r squared, inch squared. Excuse me. Eight square or two, and that will go from 19 to 25. So you plug in your row density, which we know 50. Hi. And then your radius here would be five were told. Then you have 25 squared over two minus 19. According to you, plug that into your handy dandy calculator, Uh, like plugging into my calculator so you will get 518,000 362.79 pounds. Perfect. Right. And that will be your final answer there. All right, well, I hope that clarifies the question there. Thank you so much for watching.


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