# Derivation of Continuity Equation

Derivation of continuity equation is one of the most important derivations in fluid dynamics. The continuity equation derivation is very simple and can be understood easily if some basic concepts are known. Before checking the continuity equation derivation, it is important to know what continuity equation is and some of its related terms. In simple words, the continuity equation explains how a fluid conserves mass in its motion (conservation of mass) when it flows along a tube having a single entry and a single exit. Visit Continuity Equation to learn more about this topic and to have a better understanding of the related terms.

### Derivation of Continuity Equation:

Continuity equation represents that the product of cross-sectional area of the pipe and the fluid speed at any point along the pipe is always constant. This product is equal to the volume flow per second or simply the flow rate.

To Derive:

R = A v = constant

Where,

R= volume flow rate,

A= flow area, and

v= flow velocity

Assumption:

To derive the continuity equation, it is important to assume a tube having a single entry and a single exit.

Derivation:

Consider the following diagram: Now, consider the fluid flows for a short interval of time in the tube. So, assume that short interval of time as Δt. In this time, the fluid will cover a distance of Δx1 with a velocity v1 at the lower end of the pipe.

At this time, the distance covered by the fluid will be->

Δx1 = v1Δt

Now, at the lower end of the pipe, the volume of the fluid that will flow into the pipe will be->

V = A1 Δx1 = A1 v1 Δt

It is known that mass (m) = Density (ρ) × Volume (V). So, the mass of the fluid in Δx1 region will be->

Δm1= Density × Volume

=> Δm1 = ρ1A1v1Δt ——–(Equation 1)

Now, the mass flux has to be calculated at the lower end. Mass flux is simply defined as the mass of the fluid per unit time passing through any cross-sectional area. For the lower end with cross-sectional area A1, mass flux will be->

Δm1/Δt = ρ1A1v1 ——–(Equation 2)

Similarly, the mass flux at the upper end will be->

Δm2/Δt = ρ2A2v2 ——–(Equation 3)

Here, v2 is the velocity of the fluid through the upper end of the pipe i.e. through Δx2 , inΔttime andA2, is the cross-sectional area of the upper end.

In this, the density of the fluid between the lower end of the pipe and the upper end of the pipe remains same with time as the flow is steady. So, the mass flux at the lower end of the pipe is equal to the mass flux at the upper end of the pipe i.e. Equation 2 = Equation 3.

Thus,

ρ1A1v1 = ρ2A2v2 ——–(Equation 4)

This can be written in a more general form as->

ρ A v = constant

With equation proves the law of conservation of mass in fluid dynamics. Also, if the fluid is incompressible, the density will remain constant for steady flow. So, ρ1 2.

Thus, Equation 4 can be now written as->

A1 v1 = A2 v2

This equation can be written in general form as->

A v = constant

Now, if R is the volume flow rate, the above equation can be expressed as->

R = A v = constant