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. In this article, derivation of continuity equation is given in a very simple, basic and easy to understand way.

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

**with a velocity**

*Δx*_{1}**at the lower end of the pipe.**

*v*_{1}At this time, the distance covered by the fluid will be->

*Δx _{1 }= v_{1}Δt*

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

*V = A _{1 }Δx_{1 }= A_{1} v_{1 }Δt*

It is known that** mass (m) = Density (ρ) × Volume (V). **So, the mass of the fluid in

**region will be->**

*Δx*_{1 }*Δm _{1}= Density × Volume*

*=> Δm _{1 }= ρ_{1}A_{1}v_{1}Δt ——–(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 ** A_{1},** mass flux will be->

*Δm _{1/Δt = }ρ_{1}A_{1}v_{1 }——–(Equation 2)*

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

*Δm _{2/Δt }= ρ_{2}A_{2}v_{2 }——–(Equation 3)*

Here, ** v_{2}** is the velocity of the fluid through the upper end of the pipe i.e. through

*Δx*_{2 }*,*in

*time and*

**Δt***is the cross-sectional area of the upper end.*

**A**_{2},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,

*ρ _{1}A_{1}v_{1 = }ρ_{2}A_{2}v_{2 }*

*——–(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,

*ρ*_{1 }=ρ_{2}.Thus, ** Equation 4 **can be now written as->

*A _{1 }v_{1 = }A_{2 }v_{2}*

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*

*This was the derivation of continuity equation.*

**Also Check: **Bernoulli’s Principle

*Keep visiting BYJU’S for more such derivations and other important physics articles.*

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