Lesson Explainer: The Continuity Equation for Fluids Physics • 9th Grade

In this explainer, we will learn how to calculate the rate of transfer of smoothly flowing fluids through channels with varying cross sections.

Let us start with an incompressible fluid. This means that changes in pressure have no effect on the density of the fluid. Recall that a fluid can be a liquid or a gas; in many cases, we can assume that a liquid, such as water or oil, is incompressible.

To understand what is meant by this, imagine a fluid that fills a container that has one freely moving wall. The freely moving wall is pushed down by some mass. If the fluid is compressible, more mass pushing down on the freely moving wall will cause the particles of the fluid to move closer together, increasing its density. The following diagram shows this.

If the fluid is incompressible, no matter how much mass is placed on the freely moving wall, the particles of fluid will not move closer together; the density of the fluid will not change. It is common for liquids such as water and oil to be considered incompressible fluids. This is illustrated in the following diagram.

Definition: Incompressible Fluid

An incompressible fluid is a fluid whose density is not affected by changes in pressure.

Let us consider a scenario where a fluid is flowing through a cylindrical pipe of cross-sectional area ๐ด. Although all of the particles have random velocities, we would see that, on average, they travel in one direction with velocity ๐‘ฃ. The following diagram shows a section of this pipe.

If we look at a cross section of the pipe, over ๐‘‡ seconds, some amount of fluid will pass through this cross section. We can visualize this in the following diagram, where the fluid that passes through the cross section is highlighted.

This amount of fluid takes up a section of the pipe with cross-sectional area ๐ด and length ๐‘‘. This is a volume, ๐‘‰, of ๐‘‰=๐ด๐‘‘.

Knowing that the fluid is traveling at velocity ๐‘ฃ, the length of this section is equal to ๐‘‘=๐‘ฃ๐‘‡.

Substituting this into the equation for volume, we get ๐‘‰=๐ด๐‘ฃ๐‘‡.

If we divide both sides by ๐‘‡, we get an expression for the amount of volume flowing through the cross section every ๐‘‡ seconds: ๐‘‰๐‘‡=๐ด๐‘ฃ.

This is known as the volumetric flow rate of a fluid and has units of cubic metres per second (m3/s).

We can also consider the mass of fluid that passes through the cross section, ๐‘š.

If the fluid has density ๐œŒ, the mass of the fluid that passes through the cross section is ๐‘š=๐œŒ๐‘‰.

Substituting the equation for volume, ๐‘‰=๐ด๐‘ฃ๐‘‡, into this, we get ๐‘š=๐œŒ๐ด๐‘ฃ๐‘‡.

If we divide both sides by ๐‘‡, we get an expression for the amount of mass flowing through the cross section per ๐‘‡ seconds: ๐‘š๐‘‡=๐œŒ๐ด๐‘ฃ.

This is known as the mass flow rate of a fluid and has units of kilograms per second (kg/s).

Let us work through a question that looks at the mass flow rate of a liquid.

Example 1: Calculating the Cross-Sectional Area of a Pipe from Mass Flow Rate

45 kg of a liquid with a constant density of 1โ€Žโ€‰โ€Ž055 kg/m3 flows smoothly through a 2.5 m long pipe each second. What is the cross-sectional area of the pipe? Give your answer to three decimal places.

Answer

In this question we are asked to calculate the cross-sectional area of a pipe given the amount of liquid that has traveled a certain distance in a set amount of time.

Given the mass of liquid that travels through the pipe each second, 45 kg, and the density of the liquid, ๐œŒ=1055/kgm๏Šฉ, we can calculate the volume of liquid that travels through the pipe each second: ๐‘‰=๐‘š๐œŒ๐‘‰=451055/๐‘‰=0.0427.kgkgmm๏Šฉ๏Šฉ

The cross-sectional area of the pipe, ๐ด, can then be calculated by dividing the volume of the liquid that passes through it each second by the distance the liquid travels, ๐‘‘=2.5m: ๐ด=๐‘‰๐‘‘=0.04272.5.mm๏Šฉ

The cross-sectional area of the pipe is ๐ด=0.017.m๏Šจ

Let us consider an incompressible fluid that is flowing through a pipe that changes in cross-sectional area from ๐ด๏Šง to ๐ด๏Šจ. We will see that the velocity of the fluid also changes from ๐‘ฃ๏Šง to ๐‘ฃ๏Šจ. We can look at two cross sections, one before the cross-sectional area change, at point 1, and one after, at point 2. The following diagram shows this pipe, with cross sections at point 1 and 2 highlighted.

As before, we can measure the volume of fluid that passes through each cross section, ๐‘‰๏Šง and ๐‘‰๏Šจ, over ๐‘‡ seconds. Let the length of pipe that each volume takes up be ๐‘‘๏Šง and ๐‘‘๏Šจ. We can visualize these volumes on the following diagram.

These volumes are equal to ๐‘‰=๐ด๐‘‘,๐‘‰=๐ด๐‘‘.๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ

The length of each volume can be calculated by multiplying the velocity of the fluid at that point by the time the fluid takes to cross that volume: ๐‘‘=๐‘ฃ๐‘‡,๐‘‘=๐‘ฃ๐‘‡.๏Šง๏Šง๏Šจ๏Šจ

Substituting these into our expressions for volume, we get ๐‘‰=๐ด๐‘ฃ๐‘‡,๐‘‰=๐ด๐‘ฃ๐‘‡.๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ

Now, because we are considering an incompressible fluid, we can see that the volume that passes through each cross section must be equal. Therefore, we can write ๐‘‰=๐‘‰,๐ด๐‘ฃ๐‘‡=๐ด๐‘ฃ๐‘‡.๏Šง๏Šจ๏Šง๏Šง๏Šจ๏Šจ

Finally, we can divide both sides by ๐‘‡ to give an expression relating the fluid velocity before and after a cross-sectional area change as follows: ๐ด๐‘ฃ=๐ด๐‘ฃ.๏Šง๏Šง๏Šจ๏Šจ

Importantly, the cross-sectional area multiplied by the velocity of a fluid has units of cubic metres per second (m3/s). This is the volumetric flow of a fluid, and it is constant when the fluid is incompressible.

Example 2: Using the Continuity Equation to Model The Flow of an Incompressible Fluid

Water with a speed ๐‘ฃ=1.25/๏Šงms flows smoothly through a cylindrical pipe of radius ๐‘Ÿ=0.325๏Šงm, as shown in the diagram. The water passes through the pipe in 0.955 s before smoothly flowing through a second cylindrical pipe of radius ๐‘Ÿ=0.118๏Šจm and length ๐ฟ=0.975๏Šจm.

  1. What is the length of the first pipe? Give your answer to two decimal places.
  2. What is the time interval between the water entering and leaving the second pipe? Give your answer to three decimal places.

Answer

Part 1

The first part of this question asks us to calculate the length of the first pipe given the time it takes for water to pass through the first pipe and the velocity of the water.

To calculate the length of the first pipe, ๐ฟ๏Šง, we can simply multiply the velocity of the water in the first pipe, ๐‘ฃ=1.25/๏Šงms, by the time taken to pass through the first pipe, ๐‘‡=0.955๏Šงs: ๐ฟ=๐‘ฃ๐‘‡๐ฟ=1.25/ร—0.955๐ฟ=1.19.๏Šง๏Šง๏Šง๏Šง๏Šงmssm

The length of the first pipe is 1.19 m.

Part 2

Next, the question asks us to calculate the time it takes for water to pass through the second pipe.

First, we must calculate the velocity of the water flowing through the second pipe. Recall that the continuity equation for incrompressible fluids is ๐ด๐‘ฃ=๐ด๐‘ฃ,๏Šง๏Šง๏Šจ๏Šจ where ๐ด๏Šง is the cross-sectional area of the first pipe, ๐‘ฃ๏Šง is the velocity of fluid in the first pipe, ๐ด๏Šจ is the cross-sectional area of the second pipe, and ๐‘ฃ๏Šจ is the velocity of the fluid in the second pipe.

We can calculate the cross-sectional area of the first pipe by using the equation for the area of a circle: ๐ด=๐œ‹๐‘Ÿ๐ด=๐œ‹ร—(0.325)๐ด=0.332.๏Šง๏Šจ๏Šง๏Šง๏Šจ๏Šง๏Šจmm

Similarly, we use the same equation for the second pipe: ๐ด=๐œ‹๐‘Ÿ๐ด=๐œ‹ร—(0.118)๐ด=0.0437.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจmm

The continuity equation can be rearranged by dividing both sides by the cross-sectional area of the second pipe, ๐ด๏Šจ: ๐‘ฃ=๐‘ฃ๐ด๐ด.๏Šจ๏Šง๏Šง๏Šจ

Then, the values for ๐‘ฃ๏Šง, ๐ด๏Šง, and ๐ด๏Šจ can be substituted into this equation: ๐‘ฃ=1.25/ร—0.3320.0437๐‘ฃ=9.48/.๏Šจ๏Šจ๏Šจ๏Šจmsmmms

This velocity can then be used to calculate the length of time it takes for water to pass through the second pipe, ๐‘‡๏Šจ.

To calculate ๐‘‡๏Šจ, the length of the second pipe, ๐ฟ=0.975๏Šจm, is divided by the velocity of water in the second pipe, ๐‘ฃ=9.48/๏Šจms: ๐‘‡=๐ฟ๐‘ฃ๐‘‡=0.9759.48/๐‘‡=0.103.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจmmss

The time interval between the water entering and leaving the second pipe is 0.103 s.

Now, we must consider the case when the fluid is compressible. Let us start again with the same scenario; a fluid is flowing through a pipe that changes in cross-sectional area from ๐ด๏Šง to ๐ด๏Šจ. This time, the fluid is compressible, so its density can change. We will see that the velocity of the fluid can change from ๐‘ฃ๏Šง to ๐‘ฃ๏Šจ and also that the density of the fluid can change from ๐œŒ๏Šง to ๐œŒ๏Šจ. We can visualize this scenario with a similar diagram to previous, but we notice how the density of the fluid is different at points 1 and 2.

Again, we can measure the volume of fluid that passes through each cross section, ๐‘‰๏Šง and ๐‘‰๏Šจ, over ๐‘‡ seconds. Let the length of pipe that each volume takes up be ๐‘‘๏Šง and ๐‘‘๏Šจ. We can visualize these volumes on the following diagram.

These volumes are equal to ๐‘‰=๐ด๐‘‘,๐‘‰=๐ด๐‘‘.๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ

The length of each volume can be calculated by multiplying the velocity of the fluid at that point by the time the fluid takes to cross that volume: ๐‘‘=๐‘ฃ๐‘‡,๐‘‘=๐‘ฃ๐‘‡.๏Šง๏Šง๏Šจ๏Šจ

Substituting these into our expressions for volume, we get ๐‘‰=๐ด๐‘ฃ๐‘‡,๐‘‰=๐ด๐‘ฃ๐‘‡.๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ

Previously, we equated the two volumes. In the case of a compressible fluid this is not necessarily possible because the density of the fluid can change. Instead, we must consider the mass of fluid that passes through each cross section.

The mass of fluid that passes through cross section 1 must equal the mass of fluid that passes through cross section 2. If the masses are not equal, then the particles of the fluid must be being created or destroyed somewhere in the pipe; this is not possible.

Let the mass that passes through cross section 1 be ๐‘š๏Šง and the mass that has passed through cross section 2 be ๐‘š๏Šจ. The mass that passes through each cross section is equal to the volume that passes through each of them multiplied by the density of the fluid at that point: ๐‘š=๐œŒ๐ด๐‘ฃ๐‘‡,๐‘š=๐œŒ๐ด๐‘ฃ๐‘‡.๏Šง๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ๏Šจ

We can equate these as follows: ๐œŒ๐ด๐‘ฃ๐‘‡=๐œŒ๐ด๐‘ฃ๐‘‡.๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ

We can then divide both sides by ๐‘‡ to give us an expression that relates the cross-sectional area, density, and velocity of a compressible fluid at any two points in a pipe as follows: ๐œŒ๐ด๐‘ฃ=๐œŒ๐ด๐‘ฃ.๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ

Example 3: Using the Continuity Equation to Model The Flow of a Compressible Fluid

A gas flows smoothly through a pipe. The pipeโ€™s cross-sectional area contracts from 0.075 m2 to 0.025 m2. The gas enters the pipe moving at 1.8 m/s and leaves the pipe moving at 2.0 m/s. The density of the gas as it enters the pipe is 1.4 kg/m3. What is the ratio of the density of the gas where it enters the pipe to its density where it exits the pipe? Give your answer to one decimal place.

Answer

This question asks us to calculate the ratio in density of a gas flowing through a pipe after the pipe changes cross-sectional area.

We are told that, initially, the pipe has a cross-sectional area of ๐ด=0.075๏Šง๏Šจm, and the gas flowing through it has a velocity of ๐‘ฃ=1.8/๏Šงms and a density of ๐œŒ=1.4/๏Šง๏Šฉkgm. At the point the gas exits the pipe, the pipe has a cross-sectional area of ๐ด=0.025๏Šจ๏Šจm, and the gas has a velocity of ๐‘ฃ=2.0/๏Šจms.

Recall that the continuity equation for compressible fluids is ๐œŒ๐ด๐‘ฃ=๐œŒ๐ด๐‘ฃ.๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ

We are asked to work out the ratio of the density of the gas where it enters the pipe to its density where it exits the pipe: ๐œŒ๐œŒ.๏Šง๏Šจ

We can rearrange the continuity equation to give this ratio. First, we divide both sides by ๐ด๐‘ฃ๏Šง๏Šง: ๐œŒ=๐œŒ๐ด๐‘ฃ๐ด๐‘ฃ.๏Šง๏Šจ๏Šจ๏Šจ๏Šง๏Šง

Then, we divide both sides by ๐œŒ๏Šจ: ๐œŒ๐œŒ=๐ด๐‘ฃ๐ด๐‘ฃ.๏Šง๏Šจ๏Šจ๏Šจ๏Šง๏Šง

Note that in order to calculate this ratio, we do not actually need to know ๐œŒ๏Šง or ๐œŒ๏Šจ.

We can now substitute the values given to us into the equation: ๐œŒ๐œŒ=0.025ร—2.0/0.075ร—1.8/,๏Šง๏Šจ๏Šจ๏Šจmmsmms which, to one decimal place, gives a ratio of ๐œŒ๐œŒ=0.4.๏Šง๏Šจ

To further understand why we can assume that certain fluids, such as water, are incompressible, we can calculate how much effect the tiny density changes that these fluids undergo would have on the continuity equation.

Letโ€™s imagine a fluid flowing through a pipe and being compressed by 0.001% from pressure; the pipe it is flowing through does not change in cross-sectional area. The density of this fluid before and after compression is equal to ๐œŒ=0.99999ร—๐œŒ.๏Šจ๏Šง

Using these values in the continuity equation, we get ๐œŒ๐ด๐‘ฃ=๐œŒ๐ด๐‘ฃ๐œŒ๐ด๐‘ฃ=0.99999ร—๐œŒ๐ด๐‘ฃ.๏Šง๏Šง๏Šจ๏Šจ๏Šง๏Šง๏Šง๏Šจ

Dividing both sides by ๐œŒ๐ด๏Šง, we can get an expression relating the velocity of the fluid before and after compression as follows: ๐‘ฃ=0.99999ร—๐‘ฃ.๏Šง๏Šจ

This difference is below the uncertainty of flow measurements.

This incompressibility assumption is generally applied to liquids. For example, the volume of water will only change by around 5ร—10%๏Šฑ๏Šฎ per pascal.

We can also use the continuity equation to understand what happens when pipes split into two.

Let us start by considering an incompressible fluid flowing with velocity ๐‘ฃ๏Šง into a pipe of cross-sectional area ๐ด๏Šง. The pipe splits into two smaller pipes, with cross-sectional areas ๐ด๏Šจ and ๐ด๏Šฉ respectively. The velocity of the fluid in each of these pipes is ๐‘ฃ๏Šจ and ๐‘ฃ๏Šฉ. This pipe branching into two smaller pipes can be seen in the following diagram.

We know that, over a fixed time, ๐‘‡, the mass of fluid that enters the first pipe, ๐‘š๏Šง, must be the same as the mass of fluid that leaves the two smaller pipes, ๐‘š+๐‘š๏Šจ๏Šฉ: ๐‘š๐‘‡=๐‘š๐‘‡+๐‘š๐‘‡.๏Šง๏Šจ๏Šฉ

Recall that the mass, ๐‘š, of fluid that passes through a pipe in a set time, ๐‘‡, is equal to the density of the fluid, ๐œŒ, multiplied by its velocity, ๐‘ฃ, multiplied by the cross-sectional area of the pipe, ๐ด, as follows: ๐‘š๐‘‡=๐œŒ๐ด๐‘ฃ.

Applying this to the three pipes, we get ๐‘š๐‘‡=๐œŒ๐ด๐‘ฃ,๐‘š๐‘‡=๐œŒ๐ด๐‘ฃ,๐‘š๐‘‡=๐œŒ๐ด๐‘ฃ.๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ๏Šฉ๏Šฉ๏Šฉ

Then, we can substitute these into the equation balancing the mass of fluid flowing through the pipes as follows: ๐œŒ๐ด๐‘ฃ=๐œŒ๐ด๐‘ฃ+๐œŒ๐ด๐‘ฃ.๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ

Finally, because the fluid is incompressible, we can divide both sides by ๐œŒ to give us an equation that relates the velocity of the fluid in each pipe and the cross-sectional area of each pipe as follows: ๐ด๐‘ฃ=๐ด๐‘ฃ+๐ด๐‘ฃ.๏Šง๏Šง๏Šจ๏Šจ๏Šฉ๏Šฉ

We will now work through an example question where a pipe splits into two secondary pipes.

Example 4: Applying the Continuity Equation to a Branching Channel

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.

  1. What is the difference in the cross-sectional areas of the secondary pipes where water enters them? Give your answer to two decimal places.
  2. What is the difference in the speed of water flow where water enters the two secondary pipes?
  3. What is the speed of water flow where water enters the primary pipe? Give your answer to two decimal places.

Answer

Part 1

The first part of the question asks us to calculate the difference in the cross-sectional areas of the secondary pipes where the water enters them.

Let us call the cross-sectional area of the upper secondary pipe where the water enters it ๐ด๏Šจ and the lower secondary pipe ๐ด๏Šฉ. We are aiming to calculate ๐ดโˆ’๐ด๏Šฉ๏Šจ.

We know that the total cross-sectional area of the secondary pipes where the water enters them is equal to the cross-sectional area of the primary pipe, ๐ด=1.00๏Šง๏Šจm. We can write this as ๐ด+๐ด=๐ด๐ด+๐ด=1.00.๏Šจ๏Šฉ๏Šง๏Šจ๏Šฉ๏Šจm

We also know that the velocity of the water at the very beginning of the secondary pipes in each pipe, which is equal to the velocity of the water flowing in the primary pipe, ๐‘ฃ๏Šง. We can write this as ๐‘ฃ=๐‘ฃ=๐‘ฃ.๏Šง๏Šจ๏Šฉ

At the end of the secondary pipes, the upper pipeโ€™s cross-sectional area is ๐ด=0.75๏Šช๏Šจm and the velocity is ๐‘ฃ=0.25/๏Šชms, and the lower pipeโ€™s cross-sectional area is ๐ด=0.50๏Šซ๏Šจm and the velocity is ๐‘ฃ=1.0/๏Šซms.

We can apply the continuity equation for incompressible fluids to each secondary pipe to relate the fluid velocity at the start and the end of each pipe.

Applying this to the upper pipe, we get ๐ด๐‘ฃ=๐ด๐‘ฃ.๏Šจ๏Šจ๏Šช๏Šช

Dividing both sides by ๐ด๐‘ฃ๏Šช๏Šช, we get ๐ด๐‘ฃ๐ด๐‘ฃ=1.๏Šจ๏Šจ๏Šช๏Šช

Applying this to the lower pipe, we get ๐ด๐‘ฃ=๐ด๐‘ฃ.๏Šฉ๏Šฉ๏Šซ๏Šซ

Dividing both sides by ๐ด๐‘ฃ๏Šซ๏Šซ, we get ๐ด๐‘ฃ๐ด๐‘ฃ=1.๏Šฉ๏Šฉ๏Šซ๏Šซ

Equating the expression for the upper pipe and the expression for the lower pipe, we get ๐ด๐‘ฃ๐ด๐‘ฃ=๐ด๐‘ฃ๐ด๐‘ฃ.๏Šจ๏Šจ๏Šช๏Šช๏Šฉ๏Šฉ๏Šซ๏Šซ

Dividing both sides by ๐ด๐‘ฃ๏Šฉ๏Šฉ, we get ๐ด๐‘ฃ๐ด๐‘ฃ๐ด๐‘ฃ=1๐ด๐‘ฃ.๏Šจ๏Šจ๏Šฉ๏Šฉ๏Šช๏Šช๏Šซ๏Šซ

Then, multiplying both sides by ๐ด๐‘ฃ๏Šช๏Šช gives us ๐ด๐‘ฃ๐ด๐‘ฃ=๐ด๐‘ฃ๐ด๐‘ฃ.๏Šจ๏Šจ๏Šฉ๏Šฉ๏Šช๏Šช๏Šซ๏Šซ

Knowing that ๐‘ฃ=๐‘ฃ๏Šจ๏Šฉ, these cancel each other out: ๐ด๐ด=๐ด๐‘ฃ๐ด๐‘ฃ.๏Šจ๏Šฉ๏Šช๏Šช๏Šซ๏Šซ

Substituting the known values for ๐ด๏Šช, ๐‘ฃ๏Šช, ๐ด๏Šซ, and ๐‘ฃ๏Šซ gives us ๐ด๐ด=0.25ร—0.75/1.0ร—0.5/๐ด๐ด=0.375.๏Šจ๏Šฉ๏Šจ๏Šจ๏Šจ๏Šฉmmsmms

Multiplying both sides by ๐ด๏Šฉ, we can obtain an expression relating ๐ด๏Šจ and ๐ด๏Šฉ: ๐ด=0.375๐ด.๏Šจ๏Šฉ

Substituting this value into ๐ด+๐ด=1.00๏Šจ๏Šฉ gives us 1.375๐ด=1.00,๐ด=0.727,๐ด=0.273.๏Šฉ๏Šฉ๏Šจ๏Šจ๏Šจmm

The difference in cross-sectional area between the two pipes is therefore ๐ดโˆ’๐ด=0.45,๏Šฉ๏Šจ๏Šจm so the difference in cross-sectional areas of the secondary pipes where water enters them is 0.45 m2.

Part 2

The second part of the question requires us to calculate the difference in velocity of the fluid as it enters the upper and lower secondary pipes.

As mentioned before, at the very instant the fluid enters the secondary pipes, it must have the same velocity as the fluid in the primary pipe.

This can be written as ๐‘ฃ=๐‘ฃ=๐‘ฃ.๏Šง๏Šจ๏Šฉ

Therefore, the difference between ๐‘ฃ๏Šฉ and ๐‘ฃ๏Šจ is ๐‘ฃโˆ’๐‘ฃ=0/.๏Šฉ๏Šจms

Therefore, the difference in the speed of water flow where the water enters the two secondary pipes is 0 m/s.

Part 3

The third part of the question asks us to calculate the speed of water flow where the water enters the primary pipe.

We can apply the continuity equation for incompressible fluids to the entire pipe, relating the fluid flowing out of the two secondary pipes to the fluid flowing into the primary pipe. Recall our continuity equation for a pipe that splits in two: ๐ด๐‘ฃ=๐ด๐‘ฃ+๐ด๐‘ฃ.๏Šง๏Šง๏Šช๏Šช๏Šซ๏Šซ

We can divide both sides by ๐ด๏Šง to give an expression for the velocity of the water in the first pipe: ๐‘ฃ=1๐ด(๐ด๐‘ฃ+๐ด๐‘ฃ).๏Šง๏Šง๏Šช๏Šช๏Šซ๏Šซ

Then, we can substitute our values in for ๐ด๏Šง, ๐ด๏Šช, ๐ด๏Šซ, ๐‘ฃ๏Šช, and ๐‘ฃ๏Šซ: ๐‘ฃ=11.00๏€น0.25ร—0.75/+1.0ร—0.5/๏…๐‘ฃ=0.69/๏Šง๏Šจ๏Šจ๏Šจ๏Šงmmmsmmsms

So, the velocity of fluid flowing in the primary pipe is 0.69 m/s.

Note that this answer could have been reached by applying the continuity equation for incompressible fluids to either of the secondary pipes individually, since we have already calculated the cross-sectional area of each secondary pipe at the point where liquid enters them.

Let us summarize what we have learned in this explainer in the following key points.

Key Points

  • The volumetric flow rate of a fluid flowing through a pipe is equal to the volume of fluid, ๐‘‰, that passes through a cross section of the pipe divided by the time, ๐‘‡, the cross section is measured for and is equal to the cross-sectional area of the pipe, ๐ด, multiplied by the velocity of the fluid, ๐‘ฃ: ๐‘‰๐‘‡=๐ด๐‘ฃ.
  • The mass flow rate of a fluid flowing through a pipe is equal to the mass of fluid, ๐‘š, that passes through a cross section of the pipe divided by the time, ๐‘‡, the cross section is measured for and is equal to the density of the fluid, ๐œŒ, multiplied by the cross-sectional area of the pipe, ๐ด, multiplied by the velocity of the fluid, ๐‘ฃ: ๐‘š๐‘‡=๐œŒ๐ด๐‘ฃ.
  • When an incompressible fluid is flowing through a pipe, its volumetric flow rate is constant: ๐ด๐‘ฃ=.constant
  • For any fluid flowing through a pipe, its mass flow rate must be constant. This is known as the continuity equation for fluids: ๐œŒ๐ด๐‘ฃ=.constant
  • When a primary pipe splits into several smaller pipes, the total mass flow rate of the fluid through all of the smaller pipes is equal to the mass flow rate of the fluid through the primary pipe: ๐œŒ๐ด๐‘ฃ=๐œŒ๐ด๐‘ฃ+๐œŒ๐ด๐‘ฃ+โ‹ฏ+๐œŒ๐ด๐‘ฃ.๏Šง๏Šง๏Šง๏Šจ๏Šจ๏Šจ๏Šฉ๏Šฉ๏Šฉ๏Œญ๏Œญ๏Œญ

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