Question Video: Calculating Cross-Sectional Area Change Using the Continuity Equation for Fluids | Nagwa Question Video: Calculating Cross-Sectional Area Change Using the Continuity Equation for Fluids | Nagwa

Question Video: Calculating Cross-Sectional Area Change Using the Continuity Equation for Fluids Physics • Second Year of Secondary School

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Water flows smoothly through a pipe at a speed of 1.66 m/s. The water then flows smoothly from the first pipe into and through a second pipe, at a speed of 1.22 m/s. What is the ratio of the cross-sectional area of the first pipe to that of the second pipe?

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Video Transcript

Water flows smoothly through a pipe at a speed of 1.66 meters per second. The water then flows smoothly from the first pipe into and through a second pipe at a speed of 1.22 meters per second. What is the ratio of the cross-sectional area of the first pipe to that of the second pipe?

Weโll start by drawing a diagram. We have two pipes. The first pipe has a cross-sectional area that we will call ๐ด one, and the second pipe has a cross-sectional area that we will call ๐ด two. These pipes are connected so that any fluid that passes through the first pipe also passes through the second pipe. In this case, we are told that water flows through the pipes, flowing through the first pipe with a speed that we will call ๐ฃ one and through the second pipe with a speed that we will call ๐ฃ two.

The question asked us to calculate the ratio of the cross-sectional area of the first pipe to that of the second pipe. So we must calculate ๐ด one divided by ๐ด two. The question tells us that water flows through the first pipe with a speed of 1.66 meters per second. So ๐ฃ one is equal to 1.66 meters per second. The question goes on to tell us that the water flows through the second pipe with a speed of 1.22 meters per second. So ๐ฃ two is equal to 1.22 meters per second. And these are both in SI units, so we donโt need to worry about converting any of them before substituting them into any equations.

We will answer this question using the continuity equation for fluids, which states that a fluidโs density multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the fluidโs speed is constant. This means that we can write the waterโs density multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the waterโs speed in the first pipe is equal to the waterโs density multiplied by the cross-sectional area of the pipe it is flowing in multiplied by the speed of the water in the second pipe.

Because the fluid is water, we can say that it is incompressible, meaning that its density is constant. So the density of the water in the first pipe is equal to the density of the water in the second pipe. And we will call this overall density ๐. And we can simplify our equation by dividing both sides by ๐, where we see that the ๐โs on the left and on the right cancel, leaving us with ๐ด one multiplied by ๐ฃ one is equal to ๐ด two multiplied by ๐ฃ two.

And we would like to rearrange this to get an expression for ๐ด one divided by ๐ด two. We will start by dividing both sides of the equation by ๐ฃ one, where we see that the ๐ฃ ones on the left cancel. Next, we will divide both sides of the equation by ๐ด two, where we see that the ๐ด twos on the right cancel. And this gives us our expression of the ratio of the cross-sectional area of the first pipe to that of the second pipe. ๐ด one divided by ๐ด two is equal to ๐ฃ two divided by ๐ฃ one.

We can now substitute unknown values of ๐ฃ one and ๐ฃ two into this equation. ๐ด one divided by ๐ด two is equal to 1.22 meters per second divided by 1.66 meters per second. Evaluating this expression gives ๐ด one divided by ๐ด two is equal to 0.735 to three decimal places. The ratio of the cross-sectional area of the first pipe to that of the second pipe is equal to 0.735 to three decimal places.

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