Question

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Derive equation of continuity for steady and irrotational flow of a perfectly mobile and incompressible fluid. What conclusion is drawn from it?​

Answer 1

Answer:

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. The continuity equation is given as:

R = A v = constant

Where,

R is the volume flow rate

A is the flow area

v is the flow velocity

Assumption of Continuity Equation

Following are the assumptions of continuity equation:

The tube is having a single entry and single exit

The fluid flowing in the tube is non-viscous

The flow is incompressible

The fluid flow is steady

Derivation

Consider the following diagram

Explanation:

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 , in Δt time and A2, 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 the 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

The 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 =ρ2.

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

This was the derivation of continuity equation.

Answer 2

Explanation:

Consider a non-viscous liquid in streamlined flow through a tube AB at the varying cross-section.

Let a1,a2 be the area of cross-section of the tube at A and B.

v1,v2 = velocity of flow of liquid at A and B.

Volume of the liquid entering per second at A=a1v1

Mass of liquid entering per second at A=a1v1ρ1

Similarly mass of liquid leaving per second at B=a2v2ρ2

Assuming, no loss of liquid in the tube and

steady flow, then the mass of liquid entering at A/sec = mass of liquid leaving at B/sec.

⇒a1v1ρ1=a2v2ρ2

Assuming liquid is incompressible,

ρ1=ρ2

⇒a1v1=a−2v2

av=constant

This is the equation of continuity.