Air flows through the tube shown in FIGURE P14.62 at a rate of...

Question

Air flows through the tube shown in FIGURE P14.62 at a rate of 1200 cm³/s. Assume that air is an ideal fluid. What is the height h of mercury in the right side of the U-tube?

Accepted Answer

Hello everyone. Let's go through this problem. Consider a mono meter formed with a U shaped tube containing liquid of density 13.6 g per cubic centimeter and connected to a frustra by the upper end of the tube. As shown in the figure, air flows inside the frustum at a rate of 900 cubic centimeters per second. Air is assumed to be an ideal fluid. Determine the difference in height between the two vertical columns of the manometer. We're given a diagram showing the U tube and the thrust and the diagram gives us the diameter of the thrust at the two points connected to each of the arms of the U tube. The diagram also gives us the volume flow rate of the fluid. And lastly, we have four multiple choice options to choose from. Option A 2.1 centimeters, option B 5.4 centimeters, option C 7.7 centimeters and option D centimeters. Now because we're looking for the height difference of the liquid in the youtube recall that the pressure difference that's related to a height difference is equal to the density of the liquid multiplied by the gravitational acceleration. G multiplied by the height difference delta H we're trying to solve for delta H. So solving this algebraically for delta H by dividing both sides of the equation by row G, we can see that the height difference is going to be equal to the pressure difference of the liquid divided by the density of that liquid multiplied by the gravitational acceleration. G. The problem does give us the density of the liquid but we don't have the pressure difference. So we're gonna wanna find that first. Now the pressure difference of the liquid is going to be the same as the pressure difference at the two arms of the denominator of the air. Because that pressure difference of the air flow is going to be what causes the height difference of the liquid. So the first thing we should do is compare the pressures at the two different diameters described to us. And we can do this using the Bernoulli equation which states that the pressure at some point in the streamline of fluid plus the density one half of the density multiplied by the square of the speed plus the density multiplied by the gravitational acceleration multiplied by the y position at a point on the streamline is constant. So we can use the Bernoulli equation to compare the variables we have at two different points along a, a streamline of a flow. So for the purpose of this problem, I'm going to declare that the subscript one will refer to variables affecting the smaller diameter, that diameter of 1.5 millimeters. And I'm going to use the subscript two to refer to variables affecting the larger diameter of 3.5 centimeters. So let's use the Bernoulli equation to compare the variables at those two different points. So P sub one, the pressure diameter one plus one half the density multiplied by V sub one squared plus the density multiplied by G multiplied by the vertical location of the first point we're analyzing Y sub one is equal to P sub two plus one half row V sub two squared plus row G and then Y sub two. Now it's useful to know is that it's up to us to define the vertical positions of the points we're looking at. So the way this problem is set up, it is totally possible for us to assume that Y sub one and why sub two are the same? Because that'll simplify our math. Because if Y sub two and Y sub one are equal, then that means that we can just cancel out the row G Y terms since they'll both be equal to each other. And we can simplify the equation by subtracting them out. So the new equation is P sub one plus one half row V one squared equals P sub two plus one half Rome Visa two squared. And we're going to solve this for the pressure difference so that we can substitute it in to the delta H equation that we discovered earlier. So I'm gonna subtract from both sides of the equation P sub one and then get that on its own. So P sub two minus P sub one is then equal to and then subtracting from both sides of the equation. One half row of the sub two squared. This becomes one half row V sub one squared minus one half row V sub two squared. And then we can factor out to the one half row simplify it. So the right hand side of the equation then becomes one half row in parentheses, V sub one squared minus V sub two squared. Now we can use this equation to solve for the pressure difference. But in order to do that, we also, we first need to know what the actual speeds of the flow at the two different diameters. And we can figure that out using the continuity equation. So recall that the volume flow rate of a fluid flow is equal to the cross sectional area that the fluids flowing through multiplied by the flow speed. We can use this equation to solve for the flow speed at the two different diameters. Keeping in mind the fact that the continuity equation tells us that the volume flow rate is constant throughout the Rustom. So in other words, Q sub one is going to be equal to a sub one vis sub one and Q sub two is equal to a sub two V sub two. But since Q is always constant Q sub one and Q sub two are just going to be equal to one another. So if we want to solve for V sub one, we can divide both sides of the equation by a sub one. So V sub one is equal to the volume flow rate of the air divided by the cross sectional area. The volume flow rate is given to us in the diagram as 900 cubic centimeters per second. So we're going to have to convert this into cubic meters per second. So converting it into cubic meters per second, that becomes 900 multiplied by 10 to the power of negative six cubic meters per second divided by the cross sectional area that the fluids flowing through. And that's a circular path. So recall that the area of a circle is equal to pi multiplied by the square of the radius. So in the context of for us here, that's going to be pi multiplied by the radius of the circle at one. So the radius of circle one is 1.5 millimeters. So 1.5 multiplied by and then we got to convert it to meters. So multiplied by 10 of the power of negative three m squared. And then if we were to put this into our calculator, then we find a speed of about 127. m per second. Now let's do the exact same thing to find V sub two. So V sub two is going to be equal to the volume flow rate divided by the cross sectional area two. And it's gonna be equal to the exact same flow rate. 900 multiplied by 10 to the power of negative six cubic meters per second, all divided by the area of the diameter at that point. So pi multiplied by and then the radius at circle two is 3.5 centimeters. So that's 3.5, multiplied by 10 to the power of negative two m squared. This time we're converting from centimeters to meters. And then if we put that into a calculator, then we find a flow rate of about 0.23, 386 m per second. Now let's take these values for V sub one and V sub two and put them into the delta P formula. We found earlier define the change in pressure between the two different points. So going back to the earlier equation or delta P delta P is equal to one half multiplied by the density of the air multiplied by V sub one squared. So that is 127.324 m per second squared minus the two squared. So that's 0.23386 m per second squared. And then for, for row sub air for the density of the air recall that the density of air has a value of about one 0.28 kilograms per cubic meter. If we put that value in for ros of air and put all of this into a calculator, then we find a value for the change in pressure of about 10,000 0.3 pascals. So now that we have the pressure difference between the two different points in the air stream line, That's the same as the pressure difference between the two different arms in the lower liquid stream. So now we can apply this to the delta H formula that we found all the way back in the beginning of the problem where delta H is equal to the pressure difference, which we just found delta P all divided by the density of the lower liquid multiplied by the gravitational acceleration G. Now let's put our values in. So delta P is 10, 0.3 pascals. And this is being divided by the density of the liquid, which is given to us in the problem as 13.6 g per cubic centimeter. As usual, we need to keep our units consistent. So I'm gonna write this in kilograms per cubic meter as 13,600 kg per cubic meter multiplied by the dent by the gravitational acceleration 9.81 m per second squared. And if we put this into a calculator, then we find a height difference of approximately 0.77 m, which can also be written in centimeters as being about 7. centimeters. This then is our answer to the problem. And if we look at the options we were given above, we can see that option C does indeed say 7.7 centimeters. So option C is the answer to this problem. And that is it for this video. I hope this video helped you out. If you feel like you need more practice, please check out some of our other videos which will give you more experience with these types of problems, but that's all for now. Hope you all have a lovely day. Bye bye.

Air flows through the tube shown in FIGURE P14.62 at a rate of 1200 cm³/s. Assume that air is an ideal fluid. What is the height h of mercury in the right side of the U-tube?

Key Concepts

Bernoulli's Principle

Continuity Equation

Hydrostatic Pressure

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