(a) A nonviscous liquid of density p flows at speed v₀ through a horizontal pipe that expands smoothly from diameter d₀ to a larger diameter d₁ . The pressure in the narrower section is p₀. Find an expression for the pressure p₁ in the wider section.

Video transcript

Hey, everyone. Let's go through this problem. The water discharge from an industrial complex flows through a horizontal pipeline system. First, water circulates with a speed of three m per second at a pressure of two multiplied by 10 to the power of five pascals via a pipeline with a radius of three centimeters. Then the water enters a pipeline with a radius of five centimeters. The two pipelines are connected with a smooth adapting pipe system. Calculate the water's pressure in the five centimeter pipeline. And we're given four multiple choice options to choose from. Option A 1.96 multiplied by 10 to the power of five pascals. Option B two multiplied by 10 to the power of five pascals. Option C 2.4, multiplied by 10 to the power of five pascals. And option D 3.33 multiplied by 10 to the power of five pascals. So let's go over what's happening in this problem. We have a hose with some radius three centimeters and then the hose is connected to a section with a larger radius which were given as five centimeters. So R one is three centimeters, an R two is five centimeters where I'm using the subscript one to refer to the small section and the subscript two to refer to the large section. We're also told that the speed in the first part of the tube is three m per second. And the pressure in the first section of the tube is 2100, multiplied by 10 of the power of five pascals. We're looking for the pressure of the water in the five centimeter pipeline, which means we're looking for P sub two. That is unknown to us right now. That is what we want to find. Now, you can find a relationship between pressure and things like radius and flow speed at at different points along a streamline. Using the Bernoulli equation which states that in a flow, the pressure plus ha one half of the density multiplied by the flow speed squared plus the density multiplied by the gravitational acceleration multiplied by the vertical position is constant everywhere along the stream line. So we can use the Bernoulli equation to compare our variables in at sections one and two of the pipe flow. So let's write out the Bernoulli equation for this situation. So piece of one plus one half row V sub one squared plus row G Y sub one equals P sub two plus one half row V sub two squared plus row G Y sub two. And what's convenient about the Bernoulli equation is that it's up to us to define the vertical positions for the two points we're looking at, since the pipe is horizontal, we can imagine that we're looking at two points along a streamline that are at the exact same height as one another. And so it's convenient for us to define why sub one to be the same as Y sub two. This means that the potential energy terms cancel out. And the Bernoulli equation we're now left with says that P sub one plus one half row V sub one squared is equal to P sub two plus one half row V sub two squared, much smaller and simpler to work with. Now, since we're looking for the P sub two, let's algebraically solve this equation for P sub two by subtracting from both sides of the equation. The one half row B sub two squared term. So P sub two is equal to P sub one plus one half row V sub one squared minus one half row V sub two squared. And we can simplify this further by factoring out the one half row out of the two terms that contain the vs. So the right hand side of this equation is alternatively equal to a piece of one plus one half row multiplied by V sub one squared minus V sub two squared. Now we can't simplify this any further because the Bernoulli equation still contains two different flow speeds V sub one and V sub two. But the problem only gives us V sub one. So in order to solve for the pressure, we're going to have to find another relationship between the two flow speeds. And then connect that back to the Bernoulli equation. And another relationship we can use is the continuity equation which states that for an ideal fluid flow, the cross sectional area A through which the flow flows multiplied by the flow speed V is constant everywhere in the flow. Or alternatively, a one V one is equal to a two V two. So uh we should solve this continuity equation for V sub two so that we can make the V sub two term disappear from the Berne equation, which will mean that we'll only have V sub one in the equation. So then we'll be able to just plug in the value of V one given to us in the problem. So first to get going with solving this, we should note that we're not actually given the cross sectional area of either section of the tube, but we are given the radii. So recall that the formula for the cross sectional area of a circle is equal to pi multiplied by the radius squared. So let's plug that formula into the continuity equation to solve for the vs. So instead of writing a sub one, I'm gonna write pi R sub one squared multiplied by V sub one and that's equal to pi R sub two squared multiplied by V sub two. The pi is cancel out because pi is on both sides of the equation. So we're left with R sub one squared. V sub one equals R sub two squared V sub two. Like I said earlier, we'll want to solve this for V sub two so that we can make the V sub two in the beli equation disappear. So V sub two by dividing both sides of the equation by R sub two squared, we find that V sub two is equal to V sub one multiplied by R sub one squared divided by R sub two squared. So now we can plug V sub two into the Bernoulli equation switching back to blue. So we know that P sub two is equal to piece of one plus one half the density of the fluid multiplied by V sub one squared minus the square of V sub one multiplied by R sub one squared divided by R sub two squared. And that's what we're plugging into the Bernoulli equation. Now, now let's distribute the square. So it becomes piece of one plus one half row visa one squared minus V sub one squared multiplied by R sub one rates the power of four divided by R sub two rate, the power of four. If you want, you can simplify this further by factoring out the V sub one squared. So P sub one plus one half row V sub one squared multiplied by and, and then what we're left with, in the parentheses is one minus R sub one divided by R sub two raised to the power of four. And that's pretty much it. All that's left for us to do is plug into this equation. The variables that were given to us in the problem. So P sub one is two multiplied by 10 to the power of five pascals plus one half multiplied by the density. The problem doesn't give us the density but we're told it's water. So we can assume it's about 1000 kg per cubic meter multiplied by V sub one squared which is three m per second. And then we're squaring it, then multiplying that by one minus the ratio of the radii, raise the power of four, that's R one. So three centimeters divided by R two, which is five centimeters. And we don't need to do any unit conversions from centimeters to meters here because since they're being divided by each other, their units will just cancel out and it won't matter anyway. And then once you plug this into your calculator, then we should find a pressure of about 203, 917 pascals. Or if you round this to scientific notation, this is equal to about 2.4, multiplied by 10 to the power of five pascals. So this then is our answer to the problem. It's the pressure in the second half of the tube. And if we look at our multiple choice options. We can see that option c says exactly this 2.4 multiplied by 10 to the power of five pascals. And so that is then the answer to this problem. I hope this video helped you out. If you'd like more practice, please check out some of our other tutoring videos which will give you more experience with these types of problems. But that's all for now. I hope you all have a good day. Bye bye.