(a) A cylindrical tank of radius 𝑅, filled to the top with a liq...

Question

(a) A cylindrical tank of radius 𝑅, filled to the top with a liquid, has a small hole in the side, of radius 𝓇, at distance d below the surface. Find an expression for the volume flow rate through the hole.

Accepted Answer

Hey, everyone. Let's go through this problem. A chemical manufacturing plant uses a large vertical cylindrical reservoir to store oil. The diameter of the reservoir is capital D. The reservoir is a small opening with a diameter of lower case D situated at a distance of L under the surface which is used to extract the oil, write the expression of the oil volume flow rate. Through the opening, we have four multiple choice options to choose from A through D and all four options have the same sort of general thing going on. They all start with pi multiplied by small D divided by two squared, multiplied by the square root of two G L divided by something though in all four options, the thing that's in the denominator is different. In option A, we have the big D divided by small D squared. In option B, we have one minus the big D divided by small D to the power of four. In option C that denominator is one minus small D divided by big D squared. And in option D, it's one minus small D divided by big D to the power of four. So the way the problem is described to us there is a large cylindrical reservoir and on the side of it, there is a hole with a diameter of small D and the diameter of the reservoir itself is capital D. We're told that the small D hole is located at a distance of L below the surface of the water or of the surface of the plant. And I'm going to define a new variable h to refer to the total height of the cylinder. Now, the problem asks us for the volume flow rate of the oil through the little hole. In order to do that, we're first going to need to find the standard flow speed of the regular meters per second speed of the flow. And to find that we're gonna want to find a relationship between the flow speed and the other variables we have because of the way the problem is given from the pressure and the atmosphere or the, the or just the heights of the fluid in the tube. And one way we can represent all those variables together into one equation is to use the Bernoulli equation which states that the pressure in a fluid along a streamline plus one half multiplied by the density of the fluid multiplied by the square of its speed plus the density of the fluid multiplied by the gravitational acceleration. G multiplied by the height of the flow, the height of the fluid is constant. So if we were to pick two different points in the fluid to compare to one another, we could find a relationship between all the different variables we have in the problem which we could then use to find the flow speed of the oil out the little hole. So in order for that to work, let's first label the two different points in the problem that we're going to analyze. I'm going to use the subscript one to refer to the variables related to the surface of the oil. And I'm going to use the subscript two to refer to the variables affecting the fluid just as it's about to exit the hole at the bottom. So let's now write out the Bernoulli equation. So P sub one plus one half row V, sub one squared plus row G and then the height, the, the vertical position of the point in the case of the top of the cylinder, that's just going to be H the variable I defines to be the total height of the cylinder. And this is equal to P sub two plus one half row, the sub two squared plus row G and then the height of the fluid at the hole, which is going to be the total height of the cylinder minus the L that's given to us in the problem. So in the parentheses here, I'm gonna write H minus L. Now there are a few things we can do here to simplify this. First note that at both points we're looking at there is exposure to the atmospheric pressure, which is going to be the same value in both cases. So P sub one is approximately equal to P sub two, which means those terms can cancel out. So a new, more simplified version of the equation is one half row V, one squared plus row G H equals one half row V two squared plus. And then for the final term on the right hand side of the equation, I'm going to do a distribution. So row G H minus row G L and now notice that on both sides of the equation, we have a row G H those can cancel out, which is good anyways because the problem doesn't give us the variable capital H. So that variable is not something we'd want to have in our final answer. So now our equation becomes one half row V, one squared equals one half row V two squared minus row G L. And also notice that every term of this equation includes the density row. So they can all cancel out as well. So we're left with one half V, one squared equals one half V two squared minus G L. The way things stand now, we can't simplify this any further because we don't have a relationship between V one and V two. Remember the problem wants us to find the volume flow rate through the opening. So we're gonna want to have some way to solve for V sub two. So we can find another relationship between our two speeds using the continuity equation, which states if you'll recall at the cross sectional area of travel multiplied by the flow speed is going to be constant. Or in the case of our problem, A one V one is going to be equal to A two V two. So I'm going to rewrite this equation again, except instead of writing a, I'm going to use the facts. If you'll recall at the cross sectional area of a circle is equal to pi multiplied by the diameter of the circle squared divided by four. So let's substitute that into our continuity equation. So instead I'm going to write out I D squared and I'm specifically making it the capital D because the one subscripts refer to the large diameter of the entire cylinder. So big D squared divided by four multiplied by V one and equal to the cross sectional area of the little hole. So that's pi small D squared divided by four multiplied by V two. Then we can simplify this. Now by looking at the fact the pi is cancel out. And so do the fours in the denominator. So we're left with V one big D squared equals V two small D squared. And remember we're solving for V sub two because that's the substitution we need to make. So I'm going to solve the continuity equation for V sub one, that way we'll have something to replace the V sub one here with in the Bernoulli equation that we simplified. So that the final equation we have will only be in terms of the sub two. So V sub one solving for this algebraically is going to be equal to V sub two multiplied by small D squared divided by big D squared. So now let's plug this, let's plug this continuity equation into V sub one that we had in the Bernoulli equation. So that's one half multiplied by the square of the sub one. So that's one half V sub two little D squared over big D squared. And all of this is inside another big square and that's equal to one half V sub two squared minus G L. So now let's distribute the square out. This becomes one half multiplied by V sub two squared multiplied by small D to the power of four, divided by big D to the power of four. And that's equal to one F V sub two squared minus G L. You want to solve this equation for V sub two. So I'm going to do a few things to simplify it. First, I'm gonna get these halves out of the way by multiplying both sides of the equation by two. That'll make this at least look a little easier to deal with. So it becomes V sub two squared multiplied by small deed of power of four over big D to the power of four equals the sub two squared minus two G L. Next, I'm going to put the V sub twos into one term by putting them both on the same side of the equation and then factoring them out. So let's first get the V sub twos onto one side of the equation algebraically. So I'm gonna add the, both sides of the equation to G L and then subtract the V sub two term with all the DS in it. So that means that two G L is equal to V sub two squared minus V sub two squared multiplied by small D to the power of four divided by big D to the power of four. And then we can factor out the V sub two squared. So that becomes V sub two squared multiplied by one minus deep to the power of four. Then we can actually share the parentheses here. So instead, I'm just gonna write this as small D divided by big D and put the entire D fractional in a power of four. So next, I'm going to divide both sides of the equation by the term containing the DS. So that becomes V sub two squared is equal to two G L all divided by one minus small D divided by big D raised to the power of four. And then lastly, we can solve for V sub two by taking the square root of both sides. Of the equation. So V sub two is equal to the square root of two G L divided by one minus small D divided by big D raised to the power of four. So now we have found an equation for V sub two, the speed of the water of the oil coming out of the hole. But the problem asks for the volume flow rate which according to the continuity equation tells us that we also have to multiply this by the cross sectional area through that hole. As we established earlier, the cross sectional area of that hole is equal to pi multiplied by small D, the diameter of that small hole divided by four. So all we gotta do to get the flow rate is multiply this area pi small D squared divided by four multiply that against what we found for the flow speed out the hole. So is this multiplied by the square root of two G L all divided by one minus small D divided by big D raise the power of four. And this then is the final answer to our problem. And if we look at our multiple choice options, we can see that this choice exists as option. D option D says pi multiplied by little D over two squared multiplied by the square root of two G L all divided by one minus small D divided by big D raised to the power of four. So that is our answer to this problem and that is all. I hope this video helped you out if it did and if you'd like more experience, please check out some of our other videos which will hopefully give you more experience and practice with these types of problems, but that's all for now. I hope you all have a lovely day. Bye bye.

(a) A cylindrical tank of radius 𝑅, filled to the top with a liquid, has a small hole in the side, of radius 𝓇, at distance d below the surface. Find an expression for the volume flow rate through the hole.

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