Air flows through the tube shown in FIGURE P14.63. Assume that air is an ideal fluid. What is the volume flow rate?

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Hey, everyone. In this problem, we have a device shown in the figure which measures the volume flow rate, Q of air through the pipe using the information given in the figure calculate Q and then we have four multiple choice options. Option A 1.8 times 10 to the power of negative four cubic meters per second. Option B 6.8 times 10 to the power of negative four cubic meters per second. Option C 1.8 cubic meters per second and option D 6.9 cubic meters per second. First, recall that the volume flow rate Q is related to the continuity equation which states that the flow rate is equal to the cross sectional area through which the fluid travels multiplied by the flow speed. V. Second, realize that the figure shows us a main tube with changing diameters and the two points of two different diameters are connected to each other through this U shaped tube containing the mercury at different heights. This will allow us to recognize the pressure difference between the two points of different diameters. Since this formula shows that we could find Q easily if we only had V. The first thing we want to do is find the flow speed. Recall that the Bernoulli equation states that for any fluid flow, any ideal fluid flow, that is the pressure plus one half of the density multiplied by the square of the flow speed plus the density multiplied by the gravitational acceleration. G multiplied by the horizon. The vertical position Y is constant. Since the Bernoulli equation includes the flow speed V, we can try using this equation to solve for the flow speed and then plug it into the continuity equation here. So the first thing we're going to want to do is label the two different points in the diagram where we're going to be comparing the Bernoulli equation, you can label it however you want, it doesn't really matter. But I'm going to call the point with a smaller diameter A and I'm going to call the point with a larger diameter B. So whenever I use the A subscript, I'm referring to variables that only matter necessarily at this point in the middle of the diagram with the smaller diameter. So the Bernoulli equation for this meter will look like this. It'll look like P sub A, the pressure at point A plus one half the density of the fluid, which the diagram tells us is air, we have air flowing through the main tube. So row sub air multiplied by the flow speed at point A squared plus the density of the air again, multiplied by the gravitational acceleration. G multiplied by the Y position, the vertical position of the flow at point A Y. So that, and this is going to be equal to the Bernoulli equation. But with the B subscripts, so this is equal to P sub B, the pressure at point B plus one half the density of air multiplied by the square of the speed at B plus the density of the air multiplied by G again, multiplied by the Y position at point B. Something useful to recall is that since it is up to us to define the Y positions for the point we're measuring, I'm going to say that the vertical position of A is the same as the vertical position of B. So Y sub A equals Y sub B. That means that these row G Y terms can cancel out since they're both equal and we can sub subtract them both away from the equation causing this equation to be more simplified. So I'm gonna rewrite this as P sub A plus one half row A va squared. And this is equal to P sub B plus one half row, suba V sub B squared. We've now reached a point where we can't do too much more simplifying as of yet due to the variables that we don't know because we don't know what the pressures are P sub A and P sub B. But as we mentioned earlier, and as you can tell from the diagram, we can determine the pressure difference between points A and B by looking at the levels of mercury in the youtube. The diagram tells us that the surfaces of the mercury are offset by a height difference of 4.5 centimeters. Then we can use that using the pressure equation to tell us what the pressure difference is between points A and B recall that the pressure difference due to a height is equal to the density of the fluid that has the height difference. In this case, the mercury H G multiplied by the gravitational acceleration multiplied by that height difference. And in this case, I'm going to draw on the diagram, the height difference I'm calling H which is given to us in the diagram as 4.5 centimeters. Everything we need for this part is given to us in the diagram. The density of the mercury is given as 13, kg per cubic meter. The gravitational acceleration G is just as as usual 9.81 m per second squared. And the height difference is 4.5 centimeters which in order to keep our units consistent, we're gonna convert to meters. So 4.5 times 10 to the power of negative two m. And if we put this into a calculator, then we find that there's a pressure difference here of about pascals. So now that we've found the pressure difference, let's go back to the Bernoulli equation that we've been setting up but algebraically rewrite it so that we have a term that can account for this pressure difference. I am going to subtract P sub A from both sides of the equation. So that on one side of the equation, we have a term that just says P sub B minus P sub A since that's going to be a pressure difference and the same one that we found so doing some basic algebraic moving around that's one half row, suba multiplied by V sub A squared minus one half row, sub A air squared or rows of air multiplied by V sub B squared. And the one half and the row sub air are the same. So I'm gonna factor them out. So one half row suba multiplied by and then in parentheses V sub A squared minus V sub B squared. And from here on forward instead of writing piece sub B minus P sub A instead, I'm just going to write pascals since we found that to be equal to it. So 6004 past Kells is then equal to one half row suba multiplied by V sub A squared minus V sub B squared. The density of error is given to us in the problem. So I'm going to continue to algebraically rewrite this to solve for this difference in the squares of the speed. So I'm going to get rid of this half by multiplying both sides of the equation by two. And then I'm going to divide both sides of the equation by row sub air. And what we find from that is that V sub A squared minus V sub B squared is equal to two multiplied by 6004 pascals divided by the density of the air. So row sub air and the diagram tells us that the density of air is 1.28 kg per per cubic meter. So I'm gonna write that out two times pascals divided by 1. kg per cubic meter. And if we put that into a calculator, we find a value of 9381 m squared per second squared. And so that is what the difference between VA squared and V B squared is. So we're getting closer, we've now found a relationship between the two different speeds. But in order to actually answer the problem and find the volume flow rate, we still need to actually find a numerical value for at least one of the speeds. And in order to do that, we're going to need another formula giving us another relationship between V sub A and V sub B or in order for us to find one of the two speeds. And luckily for us, we do have another relationship between fluid flow speeds using the continuity equation which as I mentioned all the way back at the beginning of the video is equal to the cross sectional area through which the fluid moves multiplied by the flow speed. And more specifically what the continuity equation states is that this value is constant throughout a flow. So in other words, a sub A multiplied by the sub A is going to be equal to a sub B multiplied by V sub B. So we, we'll solve this and find a relationship between Visa A and Visa B. Since we're looking at a tube that has circular cross sections, let's recall that the formula for the coro for the surface area of a spheres cross section or a circles cross section is equal to pi multiplied by the diameter of the circle squared divided by four. So for instead of writing a sub A and A sub B, instead, I'm going to use the formula for the surface area of a circle. So that becomes pi D sub A squared over four multiplied by V sub A. This is equal to pi multiplied by D sub B squared divided by four multiplied by V sub B. And since these PS and divided by fours are on both sides of the equation, they can cancel out and just give us the equation D sub A squared multiplied by V sub A equals D sub B squared multiplied by V sub B. Finally, I'm just going to solve for one of these V S so that we have an equation of just one V in relation to the other. I'm going to solve this for V sub A so V sub A and then divide both sides of the equation by D sub A squared. It's equal to D sub B squared divided by D sub A squared all multiplied by V sub B. In order to simplify this even further, I'm going to plug in the numerical values for D sub B and D sub A as given to us in the problem. In the, in the figure more specifically, the figure tells us that D sub B, the diameter of the, of the tube at point B is equal to three centimeters. So I'm going to write 3.0 and I'm gonna convert to meters to keep our units consistent 3.0 times 10 to the power of negative two m. It's squared all divided by D sub A squared, which if you look at the diagram at point A, the diameter of the tube is three millimeters. I'm gonna divide this by 3.0 again, converting from millimeters to meters times 10 to the power of negative three m. All that is squared and all that is multiplied by Visa B and doing some basic math of a calculator or whatever you'd like. This simplifies down quite nicely to 100 multiplied by V sub B. So now that we have this neat little relationship between V sub A and V sub B, we can plug this in for V sub A in our current form of what we found from the Bernoulli equation, the V sub A squared minus V sub B squared formula. So instead of writing V sub A, I'm going to write 100 multiplied by V sub V and that is minus V sub B squared and that is equal to m squared per second squared. Now, we can see that V sub B is the only unknown in this equation. So we can solve this equation algebraically for V sub B. And then once we have that V, we can put it into the continuity equation to find the volume flow rate of the fluid of the air. So first let's factor out the Visa Bs. So actually, I'm going to distribute the squares first. So that's 100 squared multiplied by V sub B squared minus Visa B squared. That equals the same numerical value from before 9381 m squared per second squared. Then we can factor out algebraically the V sub B is squared. So that's V sub B squared multiplied by in parentheses, 100 squared minus one. And that's equals to 9381 m square per second squared. And then we just divide both sides of the equation by 100 squared minus one. So V sub B squared is equal to 9381 m squared per second. Squared divided by 100 squared minus one. And all that's left for us to do to find V sub B is take the square root of both sides of the equation. So V sub B is equal to the square root of 9381 m squared per second squared, all divided by 100 squared minus one. And if we were to put that into a calculator, and we find a value for the speed of about 0.969 m per second. So now we have found the sub B. So now let's go back to the continuity equation that we pulled from earlier. We mentioned all the way back to the beginning of the problem. So the flow rate, the volume flow rate is equal to the cross sectional area multiplied by the flow speed. And we just found the flow speed at point A. So let's use the cross sectional area at point B as well. So A sub B equal uh Q is equal to A A B multiplied by VA B. Once again, recall that the formula for the surface area of a circle is equal to pi multiplied by the square of the diameter divided by four. And then to this, we're multiplying A V sub B. So this is pi multiplied by the square of D sub B which as we discussed earlier was three centimeters or three times 10 of the power of negative two m. Squared divided by four and then multiplied by 0.969 m per second. And if we were to put that into a calculator, then we find a value, final, final volume flow rate of about 6. multiplied by 10 to the power of negative four cubic meters per second. And since that is a volume flow rate, that is the answer to what we've been trying to find. So this should be our answer to the problem. And if we look at our multiple choice options, recall that option B was indeed 6.8 times 10 to the power of negative four cubic meters per second. So that means that option B is our answer to this problem. Thank you for watching. I hope this video helped you out if you have questions or you'd like more practice, check out some of our other videos and hopefully you'll get more practice with these sorts of problems. But I hope this helped you out and that's all for now.

Air flows through the tube shown in FIGURE P14.63. Assume that air is an ideal fluid. What is the volume flow rate?

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Key Concepts

Ideal Fluid

Bernoulli's Principle

Volume Flow Rate