Work
Pumping a conical tank A right-circular conical tan...

Question

Work

Pumping a conical tank A right-circular conical tank, point down, with top radius 5 ft and height 10 ft, is filled with a liquid whose weight-density is 60lb/ft³. How much work does it take to pump the liquid to a point 2 ft above the tank? If the pump is driven by a motor rated at 275ft-lb/sec (1/2 hp), how long will it take to empty the tank?

Accepted Answer

Hello there, today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. An inverted right circular conical reservoir vertex down has a height of 12 ft and a top radius of 4 ft. It is filled with oil whose weight density is 58 pounds per cubic feet. How much work is required to pump all the oil up to a discharge point located 3 ft above the rim of the reservoir? OK. So it appears for this particular prompt we're asked to take all the information that is provided to us and we're asked to determine how much work will be required in order to pump all the oil up to a discharge point that is located 3 ft above the rim of the reservoir. So now that we know that we're ultimately trying to determine what the value for the work is for this specific problem. Our first step that we need to take in order to solve this problem is we need to take a moment to pause and make sure we understand the problem. So, what is happening for this particular prong? Well, we're told that the reservoir shape is an inverted right circular cone, because that's what conical means. It means we're dealing with a cone. So let's make a note that we're dealing with a cone, and the vertex of this cone is down, so I have an arrow pointing down. We have the height of our cone, so the height H of our cone, we're told is 12 ft. We're told that the top radius, which will denote as capital R, is equal to 4 ft. We're also told the weight density of the oil, which is, we're gonna denote this as gamma, and gamma is equal to 58. Pounds per cubic feet. And the discharge point is 3 ft above the rim. So essentially, once again, we are asked to find the work that is required to pump all the oil to a point that is 3 ft above the top of this cone. So with that in mind. Our 2nd step now is we need to recall and use the work formula. So, focusing on the problem that we're given, the work that is required to move a small volume DV from a height Y to a discharge point is written as, so the formula is written as DW. is equal to the weight of the oil at the height Y. Multiplied by the distance to lift, which I'll call DL. So, once again, the work that is required to move a small volume DV from a height Y to a discharge point is DW is equal to the weight of the oil at height Y multiplied by the distance to lift. So with that in mind, we now need to set up the coordinate system, so we need to let the origin be Y is equal to 0, and this will be at the vertex or the bottom of our cone, so this will be the bottom of our cone. And we need to note that the rim is going to be at Y is equal to 12. So this is our rim will be at Y is equal to 12. So that means the oil at the height Y must be lifted 12 + 3 minus Y, which will be equal to parentheses of 15 minus Y in units of feet to the discharge point. So Our third step now is we need to express the volume of a thin slice. So at the heighty, the radius lowercase r of the cone is proportional toy. So we can go ahead and write that lowercase r divided by y is equal to R divided by H, which is equal to 4 divided by 12, which is equal to 1/3. Fantastic, which will mean that lowercase r is equal toy divided by 3. So, a thin slice at a heighty of a thickness DY has a volume DV of which DV will be equal to pi multiplied by r to the 2, specifically lowercase r, multiplied by dyy which will be equal to pi multiplied by parentheses of y divided by 3 to the power of 2. Dy, which will be simply equal to pi divided by 9 multiplied by y to the power of 2 Dy fantastic. Our 4th step now is we need to set up the integral for work. So we need to solve for the work, that's our final answer that we're ultimately trying to solve for. So, the work that is needed to move this slice to the discharge point will be DW is equal to the weight multiplied by the distance, which will be equal to gamma multiplied by DV multiplied by parentheses 15 minus Y, which will be equal to, so this will be. All equal to. 58 multiplied by parentheses of pi divided by 9 multiplied by y to the power of 2. Dy multiplied by parentheses of 15 minus y. Fantastic. So, that means we could write that the total work W will be equal to, so W denotes the total work, so W is equal to the integral evaluated from Y is equal to 0 to Y is equal to 12 of. 58 multiplied by pi divided by 9 multiplied by y to the 2 multiplied by parentheses of 15 minus y. Dy will be equal to 58 pi divided by 9 multiplied by the integral evaluated from 0 to 12 of y to the power of 2 multiplied by parentheses of 15 minus ydy which will be Equal to drum roll. So it's gonna be equal to 58 pi divided by 9 multiplied by the integral evaluated from 0 to 12 of parentheses of. 15 multiplied by y to the power of 2 minus y to the power of 3 d y. Which will be equal to 58 pi divided by 9 multiplied by parentheses of 15 multiplied by the integral evaluated from 0 to 12 of y to the power of 2 DY minus the integral evaluated from 0 to 12 of y to the power of 3DY. OK. So now, our fifth step is we need to compute the definite integral. So we need to take the integral, evaluate from 0 to 12 of y to the power of 2, Dy, which will be equal to y 3 divided by 3, evaluate from 0 to 12. Which will be equal to 12 to 3 divided by 3, which will be simply equal to 576. And the integral evaluated from 0 to 12 of y to the power of 3, Dy will be equal to y to the power of 4 divided by 4, evaluated from 0 to 12, which will be equal to 12 4 divided by 4, which will be simply equal to 5,184. And our sixth and final step is we need to plug in our values and solve for our final answer the total work. So the total work W is equal to 58 multiplied by pi divided by 9 multiplied by parentheses of 15 multiplied by 576 minus 5,184. And when we plug this expression into our calculator, we will find that it's simply equal to. 22,272 high and that's it. That is our final answer, but we need a pause. We can't forget our units and the units for this particular problem will be feet pounds, so FT dash LB feet pound, or you might hear it called feet times pound. And that's it. That is our final answer. Hooray, we did it. So going to look at the problem itself and to recap what we've learned, we could safely say that our final answer for the total work is once again without a doubt. 2, no it's gonna be 22,272 ft pounds, and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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