Video Transcript
Water with a speed 𝑣 one equals 1.25 meters per second flows smoothly through a cylindrical pipe of radius 𝑟 one equals 0.325 meters as shown in the diagram. The water passes through the pipe in 0.955 seconds before smoothly flowing through a second cylindrical pipe of radius 𝑟 two equals 0.118 meters and length 𝐿 two equals 0.975 meters. What is the length of the first pipe? What is the time interval between the water entering and leaving the second pipe?
Let’s start with the first part of this question, which asks “What is the length of the first pipe?” We are told that water flows through the first pipe with a speed of 𝑣 one. If we were to observe the water from the moment it entered the first pipe to the moment it leaves the first pipe, a certain time interval would have passed. And we will call this time interval Δ𝑇 one. And this is the time taken for the water to pass through the first pipe. We know that the distance that something travels is equal to its speed multiplied by the time it travels at that speed for. So the distance that the water travels, which is 𝐿 one, is equal to its speed, which is 𝑣 one, multiplied by the time interval Δ𝑇 one.
The question tells us that 𝑣 one is equal to 1.25 meters per second, and the water passes through the pipe in 0.955 seconds. So Δ𝑇 one is equal to 0.955 seconds. And these are both in SI units, so we don’t need to convert them before we substitute them into our equation for 𝐿 one. Substituting them in gives us 𝐿 one is equal to 1.25 meters per second multiplied by 0.955 seconds. Evaluating this gives 𝐿 one is equal to 1.19 meters to two decimal places. The length of the first pipe is equal to 1.19 meters to two decimal places.
What is the time interval between the water entering and leaving the second pipe?
Just as we did for the first pipe, we can observe the second pipe for an amount of time, from the moment that water enters the second pipe to the moment that water leaves the second pipe. We will call this time interval Δ𝑇 two. And this can be calculated from the length of the second pipe, 𝐿 two, and the speed of the water flowing through it, 𝑣 two. We know that the time taken for something to travel a certain distance is equal to that distance divided by the speed that something is traveling at. Or, in this case, the time taken for the water to pass through the second pipe is equal to the length of the second pipe, 𝐿 two, divided by the speed of the water through the second pipe, 𝑣 two. However, 𝑣 two is unknown, so you must first calculate 𝑣 two. We will keep a note of our equation for Δ𝑇 two over here on the right.
We can calculate the speed of the water in the second pipe using the continuity equation for fluids, which states that a fluid’s density multiplied by the cross- sectional area of the pipe it is flowing in multiplied by the fluid’s speed is constant. This means that the water’s density multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the water’s speed in the first pipe is equal to the water’s density multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the water’s speed in the second pipe. Because the fluid in the pipe is water, we say that its density is constant. This is because water can be assumed to be incompressible. This means that the density of the water in the first pipe is equal to the density of the water in the second pipe, which we will say is equal to an overall density that we will call 𝜌.
And we can use this to simplify our equation. We will divide both sides by 𝜌, where we see that the 𝜌’s on the left and on the right cancel, leaving us with the cross- sectional area of the first pipe multiplied by the speed of the water in the first pipe is equal to the cross-sectional area of the second pipe multiplied by the speed of the water in the second pipe. And we can rearrange this for the speed of the water in the second pipe, 𝑣 two, by dividing both sides of the equation by the cross- sectional area of the second pipe, 𝐴 two, where we see that the 𝐴 twos on the right cancel. And this gives us our expression for the speed of the water in the second pipe. We can write this a bit more neatly. 𝑣 two is equal to 𝐴 one divided by 𝐴 two multiplied by 𝑣 one.
Because the pipes are cylindrical, we can write the cross-sectional area of the first pipe, 𝐴 one, is equal to 𝜋 multiplied by the radius of the first pipe squared, which is 𝑟 one squared. Similarly, the cross-sectional area of the second pipe, 𝐴 two, is equal to 𝜋 multiplied by the radius of the second pipe squared, which is 𝑟 two squared. And 𝐴 one divided by 𝐴 two is equal to 𝜋𝑟 one squared divided by 𝜋𝑟 two squared. And we see that these 𝜋’s cancel, meaning 𝐴 one divided by 𝐴 two is equal to 𝑟 one squared divided by 𝑟 two squared. And we can substitute this back into our equation for 𝑣 two. So 𝑣 two is equal to 𝑟 one squared divided by 𝑟 two squared multiplied by 𝑣 one.
The question tells us that 𝑟 one is equal to 0.325 meters and 𝑟 two is equal to 0.118 meters. And we already know that 𝑣 one is equal to 1.25 meters per second. These are all in SI units. So we can carry on and substitute them into our equation for 𝑣 two, giving us 𝑣 two is equal to 0.325 meters squared divided by 0.118 meters squared multiplied by 1.25 meters per second. Evaluating this gives us 𝑣 two is equal to 9.48 meters per second to two decimal places.
We can now use 𝐿 two is equal to 0.975 meters given to us by the question and 𝑣 two is equal to 9.48 meters per second that we calculated to calculate Δ𝑇 two. Again, these are all in SI units, so we don’t need to worry about converting any of them before substituting them into our equation for Δ𝑇 two. Δ𝑇 two is equal to 0.975 meters divided by 9.48 meters per second. Evaluating this gives us Δ𝑇 two is equal to 0.103 seconds to three decimal places. The time interval between the water entering and leaving the second pipe is equal to 0.103 seconds to three decimal places.
