We explain what the continuity equation is, its formula, applications, examples and we propose exercises to solve

**What is the continuity equation?**

The **continuity equation**, for an incompressible fluid, establishes that the total mass of a fluid flowing through a tube, without losses or gains, remains constant. In other words, the mass remains unchanged as the fluid moves.

An incompressible fluid is one whose density remains approximately constant while it flows. For example, water is a liquid considered incompressible under standard conditions of pressure and temperature.

There is a mathematical way of expressing the conservation of mass, in the continuity equation, given by:

A1∙ v1 = A2∙ v2

Where v1 and v2 represent the velocity of the fluid in two sections of a pipe, while A1 and A2 are the respective cross-sectional areas.

The product of the cross-sectional area and the velocity is called the *flow *and the continuity equation implies that, throughout the entire length of the pipe, the flow rate is constant. Flow is also known as *volume flow rate*is understood by carefully observing the previous expression, whose dimensions are volume per unit of time.

**Formula**

In the image above there is a pipe with two sections of different diameters and at the same height, although they could be at different heights without representing a problem.

In the wider section 1, the cross-sectional area is A1 and the fluid moves with velocity v1, while in the narrower section 2, the cross-sectional area is A2 and the velocity of the fluid is v2.

A piece of mass Δm1 (green) moves through section 1 in a time Δt. During this period, the portion Δm2 (red) travels through section 2. Since the fluid is incompressible, its density is the same at all its points, so starting from the definition of density:

Here ρ is the density, m is the mass, and V is the volume. According to this, the mass Δm1 is equal to:

Δm1 = ρ∙V1

Where the volume V1 is the product between the cross-sectional area and the distance Δx1:

Δm1 = ρ∙ (A1 ∙ Δx1)

But since:

Then the portion of mass Δm1 can be written, in terms of velocity and time Δt as:

Δm1 = ρ∙ A1 ∙ Δx1 = ρ∙A1 ∙ (v1 ∙Δt)

Similarly, the portion Δm2 that flows at the same time through section 2 is written:

Δm2 = ρ∙A2 ∙ Δx2 = ρ∙A2 ∙(v2 ∙Δt)

By conservation of mass:

Δm1 = Δm2

AND:

ρ∙A1 ∙ v1 ∙Δt = ρ∙A2 ∙ v2 ∙Δt

Since Δt and ρ cancel, it results:

A1 ∙ v1 = A2 ∙ v2

**flow Q**

The product of the cross-sectional area A and the velocity of the fluid v is called the flow rate and is denoted Q. It is equal to the volume of fluid per unit time through the pipe, or volume flow rate:

Being V the volume and Δt the time interval. The unit of flow in the International System of Units is m3/s, although cubic feet /s, gallons/min, gallons/s to liters/s and gallons/s are common. Some of the most used conversion factors are the following:

1 m3/s = 264.172 gal/s 1 L/s = 0.001 m3/s 1 ft3/s = 0.0283168 m3/s 1 L/s = 0.264172 gal/s 1 m3/s = 15850.3 gal/min

Note that, as the cross section of the tube decreases, the speed of the fluid increases, and vice versa, if the cross section increases, then the speed decreases so that the flow rate remains constant.

**Applications and examples**

The continuity equation is used in the analysis of fluid flow, in combination with Bernoulli’s equation, which takes into account the variations of the velocity of the fluid in the different sections, as well as the pressure changes and the effect of the height.

**Example 1**

In the familiar garden hose, when the water comes out normally, the jet has a certain range, but if you put your finger on the hose outlet, decreasing the outlet orifice, the jet’s range is greater.

Here the continuity equation is fulfilled, since, as the area of the outlet nozzle decreases, the velocity of the jet increases so that the product area by velocity remains constant.

**Example 2**

Another example where the continuity equation is evident is the stream of water that narrows as it falls, due to the increase in the speed of the water during the fall.

In this way, the flow is kept constant, as long as the jet continues to flow in a laminar regime, that is, the water falls gently without turbulence or eddies.

**solved exercises**

**Exercise 1**

Water circulates through a pipe 20 cm in diameter. Knowing that the flow is 2000 L/s, find the velocity of the water in the pipe.

It is convenient to express everything in units of the International System. First of all, the cross-sectional area of the pipe is calculated, remembering that the radius is half the diameter:

A = π∙ (D/2)2

D=20cm=0.2m

Therefore, the area is:

A = π∙ (D/2)2 = A = π∙ (0.2 m /2)2 = 0.0314 m2.

The flow is expressed in m3/s with the help of the appropriate conversion factor:

Q = 2000 L/s = 2 m3/s

From the formula Q = A ∙ v, we find the velocity with which the fluid circulates through the pipe:

**Exercise 2**

There is a pipe of variable cross section through which water flows. At a certain point, the cross-sectional area is 0.070 m2 and the speed of the water is 3.50 m/s. Calculate:

a) The speed of the water at another point in the pipe whose cross-sectional area is 0.105 m2.

b) The volume of water discharged from an open end in 1 hour.

The continuity equation is used, equating the flow of the first point with the flow of the second. The flow is:

Q = A ∙ v

By continuity:

Q1=Q2

A1 ∙ v1 = A2 ∙ v2

Now substitute the data supplied by the statement:

A1 = 0.070 m2 v1 = 3.50 m/s A2 = 0.105 m2 v2 =?

And v2 is cleared:

**solution b**

Since the flow rate is also the volume per unit time, we have that:

Therefore, the volume V is:

V = Q∙Δt = (A∙v) Δt

The flow Q can be calculated with the data from point 1 or point 2, since it is the same at both points:

Q = A1 ∙ v1 = 0.070 m2 ∙ 3.50 m/s = 0.245 m3 / s

Knowing that 1 hour = 3600 s, the volume of water discharged is:

V = Q∙Δt = (0.245 m3 / s) × (3600 s) = 882 m3

In 1 hour, 882 m3 of water are discharged through the pipeline.