Question Video: Applying the Continuity Equation to a Branching Channel | Nagwa Question Video: Applying the Continuity Equation to a Branching Channel | Nagwa

# Question Video: Applying the Continuity Equation to a Branching Channel Physics • Second Year of Secondary School

Water flows smoothly into and through a primary pipe that divides into two secondary pipes. The secondary pipes change in thickness along their lengths, as shown in the diagram. The cross-sectional area of the pipe where it divides is identical to its cross-sectional area where water flows into it. Water flows at 0.25 m/s out of the secondary pipe that has the greater cross-sectional exit area. Water flows at 1.0 m/s out of the secondary pipe that has the smaller cross-sectional exit area. What is the difference in the cross-sectional areas of the secondary pipes where water enters them? Give your answer to two decimal places.

08:27

### Video Transcript

Water flows smoothly into and through a primary pipe that divides into two secondary pipes. The secondary pipes change in thickness along their lengths, as shown in the diagram. The cross-sectional area of the pipe where it divides is identical to its cross-sectional area where water flows into it. Water flows at 0.25 meters per second out of the secondary pipe that has the greater cross-sectional exit area. Water flows at 1.0 meters per second out of the secondary pipe that has the smaller cross-sectional exit area. What is the difference in the cross-sectional areas of the secondary pipes where water enters them? Give your answer to two decimal places.

In our diagram, we see a pipe that splits into two separate pipes. Water flows in here through this 1.00 square meter opening and then after some distance splits off into these two separate pipes. And for each one, we’re told the cross-sectional area of that pipe’s exit point. Along with this, we’re told the exit speeds of water from each pipe. Given all this information, what we want to do is solve for the difference in the cross-sectional areas of the two secondary pipes at the point where water enters them. We can see that point is along this dashed line here. If this pipe on top has a cross-sectional area at this point of 𝐴 one and the one on bottom has a cross-sectional entrance area 𝐴 two, we want to solve for their difference. And since 𝐴 two is visibly bigger than 𝐴 one, this will be 𝐴 two minus 𝐴 one.

To help us do this, we’ll use a relationship called the continuity equation for fluids. Whenever an incompressible fluid, like the water we have here, flows through a container, the continuity equation applies. It says that at any one cross section of the container, indicated by this subscript one, the volume of fluid that passes through that cross section in a unit of time is equal to the same volume of fluid that passes through another cross section of the container in that same unit of time. In other words, the volume flow rate of fluid through a container is constant.

The continuity equation for fluids is normally written this way. That said, we want to be careful how we apply it because the equation relates two cross sections in a container that have the same fluid flowing through them. For example, we could use a cross section here in a pipe and a cross section here. The continuity equation applies in this scenario because these cross sections are part of the same flow. Up in our diagram, though, we see that 𝐴 one and 𝐴 two are not part of the same flow path; the water that flows through one of these areas does not flow through the other.

That being the case, we can say that this cross section here at the entrance to the pipe and these two cross sections here at the exit points of the two pipes are indeed all part of the same flow trajectory. That’s because any water that enters here must exit either here through the first pipe or here through the second. We could say the same thing about this initial cross-sectional area and the sum 𝐴 one plus 𝐴 two here, but we won’t apply the continuity equation between these cross sections because we know so little about this split point where the pipe becomes two. We don’t yet know the speeds of water through either of these pipes, and we don’t know 𝐴 one or 𝐴 two.

So instead we’ll first apply the continuity equation between this cross section, where we know the area, and the cross section that includes the exit points of the two pipes. We know the areas of each of these pipes at their exit points, as well as the speed of water as it flows through them.

Now let’s recall that the continuity equation is all about a volume flow rate of fluid. To see this, we can recognize that if the area on each side of this equation is expressed in units of meters squared and the speeds are expressed in units of meters per second, then the units on either side of the continuity equation are cubic meters per second. Indeed then we’re considering the time rate of a volume of fluid as it flows through a container. Knowing that when we write out our equation between this point, let’s call this cross section cross section 𝑎, and at our second cross section along the exit points of the two pipes, we’ll call this cross section 𝑏, the continuity equation ends up looking like this.

Here, 𝐴 sub 𝑎 is the cross-sectional area of the pipe at its entrance. 𝑣 sub 𝑎 is the speed of water as it enters through that cross section. On the other side, 𝐴 sub one 𝑏 is the cross-sectional area of our topmost pipe at point 𝑏. And 𝑣 sub one 𝑏 is its exit speed there. That’s given as 0.25 meters per second. Similarly, 𝐴 sub two 𝑏 is the cross-sectional area of our second pipe at its exit point. And 𝑣 sub two 𝑏 is the speed of water as it exits that pipe, 1.0 meters per second. As we consider this whole expression, we know the value of 𝐴 sub 𝑎. That’s 1.00 meters squared. And we also know the values of 𝐴 sub one 𝑏, 𝑣 sub one 𝑏, 𝐴 sub two 𝑏, and 𝑣 sub two 𝑏. Those values are given in our diagram. 𝐴 sub one 𝑏 is 0.75 meters squared, and 𝐴 sub two 𝑏 is 0.50 meter squared, while 𝑣 sub one 𝑏 is 0.25 meters per second and 𝑣 sub two 𝑏 is 1.0 meters per second.

The only unknown in this equation then is 𝑣 sub 𝑎. And recall that this is the entrance speed of water in our pipe. Solving for 𝑣 sub 𝑎 will be helpful to us. To do that, we can divide both sides by 𝐴 sub 𝑎, which cancels that factor on the left and then substitute in for all the values on the right-hand side of this equation. 𝐴 sub one 𝑏 has a value of 0.75 meters squared. 𝑣 sub one 𝑏 is 0.25 meters per second. 𝐴 sub two 𝑏 is 0.50 meters squared. And 𝑣 sub two 𝑏 is 1.0 meters per second. Lastly, 𝐴 sub 𝑎 is 1.00 meters squared. And since it has a magnitude of one, we see it won’t change the magnitude of our result.

When we calculate this fraction, we get an exact result of 0.6875 meters per second. This is more precision than we’re justified in keeping for a final answer. But since this is an intermediate step leading to our final answer, we will keep all these digits for now. Writing this result off to the side, let’s now think of how we can use 𝑣 sub 𝑎 to solve for 𝐴 one and 𝐴 two. On our diagram, notice that at the point where the pipe splits into two separate pipes, right at that point, the cross-sectional area is equal to the cross-sectional area at cross section 𝐴. That means that the speed at which water enters these two pipes as they split off is equal to the speed at which it enters the first pipe at cross section 𝐴. In other words, at our two real cross-sectional areas of interest, 𝐴 one and 𝐴 two, the speed of the water equals 𝑣 sub 𝑎.

That’s useful because it means we can now apply the continuity equation over this section of the pipe in green. That equation says that 𝐴 one multiplied by 𝑣 sub 𝑎 equals 𝐴 sub one 𝑏 times 𝑣 sub one 𝑏. Recall that we know the values on the right-hand side of this expression. And now we also know 𝑣 sub 𝑎. Therefore, if we divide both sides by 𝑣 sub 𝑎, that speed cancels on the left. And we now have an expression for one of the areas we want to solve for. And note that we know all the values of this equation on the right. 𝐴 sub one 𝑏 is 0.75 square meters, and 𝑣 sub one 𝑏 is 0.25 meters per second.

Substituting in our known value for 𝑣 sub 𝑎, we get a result of 0.27 repeating meters squared. Just like we did for 𝑣 sub 𝑎, let’s store this result off to the side. And we’ll now apply the continuity equation over this part of the second pipe so that we can solve for the cross-sectional area 𝐴 two. On the left-hand side, we have 𝐴 two times 𝑣 sub 𝑎. And on the right, we have what we’ve called 𝐴 sub two 𝑏 times 𝑣 sub two 𝑏. Once again, dividing both sides by 𝑣 sub 𝑎 to isolate 𝐴 two, we can now substitute in the values in the right-hand side of this expression. 𝐴 sub two 𝑏 is 0.50 meters squared, and 𝑣 sub two 𝑏 is 1.0 meters per second.

Computing this fraction, we once again get a repeating decimal, this time, 0.72 repeating meters squared. This is 𝐴 two. So now we’re ready to calculate the difference between these areas. 𝐴 two minus 𝐴 one is 0.72 repeating meters squared minus 0.27 repeating meters squared, which is equal to 0.45 repeating meters squared. This isn’t our final answer though, because we’re to give our answer to two decimal places. We can note that the number 0.45 repeating is equal to 0.45454545 and so on. So to report our answer to two decimal places, we’ll look at the third decimal place. And noting that this number is less than five, our answer rounds to 0.45 square meters. This is the difference in the cross-sectional areas of the secondary pipes where water enters them.

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