General Continuity Equation in Cylinderical Coordinates

In the previous article, we have discussed the continuity equation in fluid mechanics, from the law of conservation of mass. In this article, we will discuss the continuity equation of fluid-particle in the three-dimensions and also in the Polar coordinates.

Continuity Equation

As we stated the equation based on the law of conservation of mass is called the continuity equation. For a fluid flowing through a pipe at all the cross-sections, the quantity of the fluid per second is constant.

Consider two cross-sections of a pipe as shown in the following figure

Let
V₁ = Average velocity at cross-section 1-1
ρ₁ = Density at section 1-1
A₁ = Area of pipe at section 1-1
and V₂, ρ₂ A₂, are corresponding values in section, 2-2.

The rate of flow at section 1-1 = ρ₁ A₁V₁

The rate of flow at section 2-2 = ρ₂ A₂V₂

According to the law of conservation of mass from the above

Rate of flow at section 1-1 = Rate of flow at section 2-2

The above equation is applicable to compressible as well as incompressible fluids and is called Continuity Equation.

If the fluid is incompressible, then ρ₁ = ρ₂, and the above continuity equation reduces as follows

A₁V₁ = A₂V₂

* Law of conservation of mass: Law of conservation of mass states that the mass can neither be created nor destroyed but is transformed from one form to another.

Continuity Equation in Three Dimensions

Consider a fluid element of lengths dx, dy and dz in the direction of x, y and z.

Let u, v and w are the inlet velocity components in x, y and z directions respectively.

Mass of fluid entering the face ABCD per second = ρ × Velocity in x-direction × Area Of ABCD

= ρ × u × (dy × dz)

Then the mass of fluid leaving the face EFGH per second can be written as follows

= ρu × dy dz + ∂/∂x (ρu × dy dz) dx

Gain Of mass in x-direction = Mass through ABCD — Mass through EFGH per second

= (ρu × dy dz) – [(ρu × dy dz) + ∂/∂x (ρu × dy dz) dx]
= – ∂/∂x (ρu × dy dz) dx
= – ∂/∂x (ρu) × dy dz dx

Similarly, the net gain of mass in the y-direction can be written as follows

= – ∂/∂y (ρu) × dy dz dx

Similarly, the net gain of mass in the z-direction also can be written as follows

= – ∂/∂z (ρu) × dy dz dx

The total net gin of masses will be = Sum of the net gain of mass in x-direction y-direction and z-direction

As per the Law of conservation of masses, as mentioned above, the mass is neither created nor destroyed in the fluid element, the net increase of mass per unit time in the fluid element must be to the rate of increase of mass of fluid in the element.

But the mass of fluid in the element is = ρ × dx × dy × dz

The rate of increase of mass with time is = ∂/∂t (ρ × dx × dy × dz)

As per the above statement, we can Equate the two expressions of the net increase of mass per unit of time with the rate of increase of mass of fluid in the element, we get

Equation (a)

This equation (a) is the continuity equation in cartesian coordinates in its most general form.

This equation is applicable to the following types of flows:

• Steady and unsteady flow
• Uniform and non-uniform now
• Compressible and incompressible fluids

For steady flow, we know ∂ρ/∂t = 0, hence equation (a) becomes as follows

If the fluid is incompressible, then ρ is constant and the above equation (a) becomes as follows

Equation (b)

This Equation (b) is the continuity equation in three dimensions.

For a two-dimensional flow, the component w = 0 and hence continuity equation becomes as

Equation (c)

Equation (b) and Equation (c) are the continuity equation in three-dimensions and two-dimensions respectively.

Continuity Equation in Cylindrical Polar Co-ordinates

The Continuity equation in cylindrical polar coordinates (r, θ, z co-ordinates) is derived by the procedure given below.

Let us consider a two-dimensional incompressible flow field.

The two-dimensional polar coordinates are r and θ.

Consider a fluid element ABCD between the radii r and r + dr as shown in the above figure

The angle subtended by the element at the centre is dθ.

The components of the velocity V are u_r in the radial direction and u_θ in the tangential direction.

The sides of the element are having the lengths as follows

Side AB = r dθ
Side BC = dr
Side DC = (r + dr) dθ
Side AD = dr

The thickness of the element perpendicular to the plane of the paper is assumed to be unity.

The fluid flow is in Radial Direction(r- direction)

Now Consider the flow in the radial direction

Mass of fluid entering the face AB per unit time = ρ × Velocity in r-direction × Area

= ρ × u_r × (AB × 1)
= ρ × u_r × (r dθ × 1)
= ρ u_r r dθ

Mass Of fluid leaving the face CD per unit time

The term containing (dr)² is very small and has been neglected

Gain of mass in r-direction per unit time

This is written in this form because (r × dθ × dr × 1) is equal to the volume of the element.

The fluid flow is in θ-Direction

let us now consider the fluid flow is in θ-Direction

Gain in mass in θ-direction per unit time = (Mass through BC – Mass through AD) per unit time

= [ρ × Velocity through BC × Area – ρ × Velocity through AD × Area]

The total gain in fluid mass per unit time

Equation (d)

But the mass of the fluid element = ρ × Volume of the fluid element

= ρ × [rdθ × dr × 1]
= ρ × rdθ × dr

The rate of increase of fluid mass in the element with time

equation (e)

( ∴ (rdθ × dr × 1) is the volume of element and is a constant quantity)

From the Law of Conservation of mass, the mass is neither created nor destroyed in the fluid element, hence net gain of mass per unit time in the fluid element must be equal to the rate of increase of mass Of fluid in the element.

Hence equating the two expressions given by equations (d) and (e), we get

Equation (f)

This equation (f) is the Continuity Equation in Cylindrical Polar Co-ordinates for a two-dimensional flow.

For steady-state flow, we know ∂ρ/∂t = 0, hence equation (f) becomes as follows

equation (g)

This equation (g) represents the continuity equation in polar coordinates for two-dimensional steady incompressible flow.

Example problems on Continuity Equation

Let us solve an example problem to determine whether any given velocity equation represents a possible physical fluid.

Problem Statement: Verify whether the following velocity components represent the physically possible flow.
u_r = r sin θ, u_θ = 2r cos θ

Answer:

Given data
u_r = r sin θ
u_θ = 2r cos θ

For a physically possible flow, the continuity equation,

should be satisfied

u_r = r sin θ

Multiplying the above equations by r we get

ru_r = r² sin θ

Let us differentiate the above equation with respect to r we get

∂/∂r (ru_r) = ∂/∂r (r² sin θ)
∂/∂r (ru_r) = 2r sin θ

u_θ = 2r cos θ

Let us differentiate the above equation with respect to r we get

∂/∂r (u_θ) = ∂/∂r (2r cos θ)
∂/∂r (u_θ) = – 2r sin θ

Summing the above two equations we get

∂/∂r (ru_r) + ∂/∂r (u_θ) = 2r sin θ – 2r sin θ = 0

Hence the continuity equation is satisfied.

Hence the given velocity components represent a physically possible flow.

Let us know if you have any doubts in this regard in the comment section below.

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