# Continuity Equation (Fluids): Definition, Forms & Examples

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Consider a stream of cars driving down a segment of road with no onramps or offramps. In addition, suppose the cars can’t change their spacing at all – that they are somehow kept a fixed distance apart from each other. Then, if one car in the long line changes its speed, all of the cars would be automatically forced to change to the same speed. No car could ever be going faster or slower than the car in front of it, and the number of cars passing a point on the road per unit time would be the same along all points on the road.

But what if the spacing isn’t fixed and the driver of one car steps on their brakes? This causes other cars to slow down as well and can create a region of slower moving, closely spaced cars.

Now imagine you have observers at different points along the road whose job is to count the number of cars going by per unit time. An observer at a location where the cars are moving faster counts the cars as they go by, and because of the larger spacing between the cars, still ends up coming up with the same number of cars per unit time as an observer near the traffic jam location because even though the cars move more slowly through the jam, they are more closely spaced.

The reason the number of cars per unit time passing each point along the road remains roughly constant boils down to a conservation of car number. If a certain number of cars pass a given point per unit time, then those cars are necessarily moving on to pass the next point in approximately the same amount of time.

This analogy gets at the heart of the continuity equation in fluid dynamics. The continuity equation describes how fluid flows through pipes. Just as with the cars, a conservation principle applies. In the case of a fluid, it is conservation of mass that forces the amount of fluid passing any point along the pipe per unit time to be constant so long as the flow is steady.

## What Is Fluid Dynamics?

Fluid dynamics studies fluid motion or moving fluids, as opposed to fluid statics, which is the study of fluids that are not moving. It is closely related to the fields of fluid mechanics and aerodynamics but is narrower in focus.

The word ​fluid​ often refers to a liquid or an incompressible fluid, but it can also refer to a gas. In general a fluid is any substance that can flow.

Fluid dynamics studies patterns in fluid flows. There are two main ways in which fluids are compelled to flow. Gravity can cause fluids to flow downhill, or fluid can flow due to pressure differences.

## Equation of Continuity

The continuity equation states that in the case of steady flow, the amount of fluid flowing past one point must be the same as the amount of fluid flowing past another point, or the mass flow rate is constant. It is essentially a statement of the law of conservation of mass.

The explicit formula of continuity is the following:

\rho_1A_1v_1 = \rho_2A_2v_2

Where ​ρ​ is density, ​A​ is cross-sectional area and ​v​ is the flow velocity of the fluid. The subscripts 1 and 2 indicate two different regions in the same pipe.

## Examples of the Continuity Equation

Example 1:​ Suppose water is flowing through a pipe of diameter 1 cm with a flow velocity of 2 m/s. If the pipe widens to a diameter of 3 cm, what is the new flow rate?

Solution:​ This is one of the most basic examples because it occurs in an incompressible fluid. In this case, density is constant and can be cancelled from both sides of the continuity equation. You then only need to plug in the formula for area and solve for the second velocity:

A_1v_1 = A_2v_2 \implies \pi(d_1/2)^2v_1 =\pi(d_2/2)^2v_2

Which simplifies to:

d_1^2v_1 =d_2^2v_2 \implies v_2 = d_1^2v_1/d_2^2 = 0.22 \text{ m/s}

Example 2:​ Suppose a compressible gas is flowing through a pipe. In a region of the pipe with a cross-sectional area of 0.02 m2, it has a flow rate of 4 m/s and a density of 2 kg/m3. What is its density as it flows through another region of the same pipe with a cross-sectional area of 0.03 m2 at velocity 1 m/s?

Solution:​ Applying the continuity equation, we can solve for the second density and plug in values:

\rho_2 = \rho_1 \frac{A_1v_1}{A_2v_2}=5.33 \text{ kg/m}^3

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