Continuity Equation: Overview, Questions, Preparation

• The tube has only one entry and one exit.
• The fluid is non-viscous.
• The flow is steady and incompressible. 

A continuity equation is useful when flux could be defined. It means any effect that travels through a surface or a substance. 

Weightage of the Topic

The topics form part of the unit of Properties of Bulk Matter that carries a weightage of about 20 marks. 

The continuity equation is the product of flow velocity and flow area. It is the relationship between the pipe's area and the speed of the fluid along with the pipe, which is considered to be constant. The equation of the continuity is as follows:

R = A v = constant

here, the volume flow rate is denoted as R,

flow area is denoted as A,

the velocity of the flow is denoted as V

Assumptions of the continuity equation

There are some of the assumptions of continuity equation which are listed as below:

1. It is assumed that the entry and exit of the tube are single. One tube can enter and exit from a single tube.
2. It is assumed that the flow of the fluid is steady
3. The flow of the fluid is not compressible
4. It is also assumed that the flow of the fluid in the tube is non-viscous.

Let us consider that the flow of the fluid is in the pipe for a short amount of time. This short amount is denoted as Δt, and the distance it covers is denoted as Δx1, and the velocity is v1.

Now, the distance covered by the fluid is

Δx1 = v1 Δt

At the end of the pipe, the fluid will flow into the pipe, and the volume will be:

V = A1 Δx1 = A1v1 Δt

We all know that the mass is equal to the product of density and volume and can be written as:

M = P x V

So, the mass of the fluid in the region Δx1 will be:

Δm1 = Density x volume

Δm1 = p1A1v1 Δt      (equation 1) 

Let us now calculate the mass flux at the end of the pipe.

Mass flux can be described as the per-unit fluid passing through the pipe. For the end of the pipe A1, mass flux is:

Δm1 / Δt = p1A1v1      (equation 2)

The mass flux at the starting of the pipe will be:

Δm2 / Δt = p2A2v2     (equation 3)

Now let us equate the equation 2 and 3, we will get:

p1A1v1 = p2A2v2   (equation 4)

In general, it can be written as 

P A v = constant  

From equation 4, we can say that,

A v = constant

R is the flow rate of volume, and by integrating it in the equation, we will get:

R = A v = constant

This process derived the continuity equation.

The chapter of 'Properties of fluids’ holds a weightage of 6 marks in total. It consists of one objective type question (1 mark), one very short question (2 marks), one short question (3 marks).

1. A piece of copper, bearing a rectangular cross-section of 15.2mm x 19.1mm, is pulled in tension with 44,500 N force, producing only elastic deformation. Ascertain the resultant strain.  

Solution: 

Given,

Length of the copper piece, 

l = 19.1mm = 19.1 x 10-3m

Breadth of the copper piece, 

b = 15.2mm = 15.2 x 10-3m

Area of the piece of copper, 

A = l x b

= 19.1 x 10-3 x 15.2 x 10-3

= 2.9 x 15.2 x 10-4m2

Force of tension applied on the copper, F = 44500N

Modulus of the copper’s elasticity, = 41*109 N/m2

Modulus of elasticity of the copper, = stress / strain

= (F/A) / Strain

Therefore, Strain = F/A

= 44500 / (2.9 x 10-4 x 42 x 109 )

= 3.65 x 10-3

2. A steel rope having a radius of 1.5cm holds a chairlift at a ski area. If the maximum stress is not supposed to be over 108 N m-2, what is the maximum weight that the steel cable can support?

Solution: 

Given, 

Radius of the steel cable = 15.cm = 0.015m

Maximum stress allowed =  108 N m-2

Maximum stress = Maximum Force/Area of cross section

Therefore, maximum force = maximum stress x area of cross section 

= 108x (0.015)2

= 7.065 x 104 N

Therefore, the maximum load the cable can bear is 7.065 x 104 N

3. The tube of a spray pump has a cross-section of 8.0 sq. cm. one end of which contains 40 fine holes of diameter 1.0mm. If the flow of liquid inside the tube is 1.5m min-1, what is the ejection speed of the liquid through the holes?

Solution: 

Cross section area, A = 8 sq. cm. = 8 x 10-4m2

Number of holes, n = 40  

Diameter of each hole = 1mm = 1 x 10-3m

Radius of each hole = d / 2 = 0.5 x 10-3m

Area of cross section of each hole = r2 = (0.5 x 10-3)2 m2

Total area of 40 holes A2 = n * a =40 * (0.5 * 10-3)2m2

= 31.41 x 10-6m2

Speed of the liquid flow inside the tube = 1.5m/min = 0.025m/s

Speed of liquid ejection through holes = V2

According to the Law of Continuity 

A1V1 =A2V2

V2 =A1V1 / A2 

= ( 8 x 10-4 x 0.025) 31.61 x 10-6

= 0.633 m /s

Therefore, the speed of ejection is 0.633 m /s

Q: What do you mean by the continuity equation?

Q: Write one assumption of the continuity equation.

Q: What is A in the continuity equation?

Q: What is the primary application of the continuity equation?

Q: Who discovered the continuity equation?