Table of contents  
Derivation of Stoke’s Law  
Derivation Of Reynolds Number  
Terminal Velocity Derivation  
Derivation of Continuity Equation 
Stoke’s Law is a mathematical equation that expresses the settling velocities of the small spherical particles in a fluid medium. The law is derived considering the forces acting on a particular particle as it sinks through the liquid column under the influence of gravity. The force that retards a sphere moving through a viscous fluid is directly proportional to the velocity and the radius of the sphere, and the fluid’s viscosity.
Sir George G. Stokes, an English scientist, clearly expressed the viscous drag force F as:
Where r is the sphere radius, η is the fluid viscosity, and v is the sphere’s velocity.
From Stoke’s Law viscosity equation, we know that viscous force acting on a sphere is directly proportional to the following parameters:
Mathematically, this is represented as
Now let us evaluate the values of a, b and c.
Substituting the proportionality sign with an equality sign, we get
Here, k is the constant of proportionality, a numerical value with no dimensions.
Writing the dimensions of parameters on either side of equation (1), we get
[MLT^{–2}] = [ML^{–1}T^{–1}]^{a} [L]^{b} [LT^{1}]^{c}
Simplifying the above equation, we get
[MLT^{–2}] = M^{a }⋅ L^{–a+b+c} ⋅ T^{–a–c}… (2)
According to classical mechanics, mass, length and time are independent entities.
Equating the superscripts of mass, length and time, respectively from equation (2), we get
a = 1… (3)
–a + b + c = 1… (4)
–a –c = 2 or a + c = 2… (5)
Substituting (3) in (5), we get
1 + c = 2
c = 1 (6)
Substituting the value of (3) & (6) in (4), we get
–1 + b + 1 = 1
b = 1 (7)
Substituting the value of (3), (6) and (7) in (1), we get
The value of k for a spherical body was experimentally obtained as
6π
Therefore, the equation gives the viscous force on a spherical body falling through a liquid.
In the case of raindrops, initially, it is due to gravity that it accelerates. As the velocity increases, the retarding force also increases. Finally, when viscous force and the buoyant force is equal to the force due to gravity, the net force becomes zero, and so it does the acceleration. The raindrop then falls with a constant velocity, known as terminal velocity. Thus, in equilibrium, the terminal velocity v_{t} is given by the equation
ρ and σ are sphere and fluid mass densities, respectively.
From the equation above, we can infer that the terminal velocity depends on the square of the radius of the sphere and is inversely proportional to the viscosity of the medium.
Q: A solid metal ball is falling in a long liquid column and has attained a terminal velocity of 4 m/s. What is the viscosity of the liquid if the radius of the metal ball is r = 5 cm and its density is
(The density of liquid is 1000 𝑘𝑔/𝑚^{3} and g is 10 m/s^{2}.)
Ans:
The radius of the sphere is r = 0.05 m.
The density of the sphere is 𝜌_{s} = 8050 kg/m^{3}
The density of the liquid is 𝜌_{s}= 1000 kg/m^{3}
The terminal velocity is 4 m/s
Let the viscosity of the liquid be 𝜂.
Substituting the values in the terminal velocity equation, we get
Stokes’s law finds application in several areas such as:
Reynolds number referred to as Re is a dimensionless quantity which is used in fluid mechanics to predict the flow patterns in different fluid flow conditions. It is defined as the ratio of inertial force to viscous force within the fluid. The mathematical representation is given as:
Where,
Using Reynolds number, it becomes easy to determine whether the fluid flow is laminar or turbulent. Following are the boundary values of a circular pipe that can be used to determine the flow pattern:
The mathematical form of the Reynolds number can be derived as follows:
R_{e} = inertial force/viscous force
Where,
Following is the Reynolds number representation for the flow in a pipe:
Where,
Critical Reynold number of any fluid flow is defined when the flow pattern changes from laminar to turbulent. It is referred to as Re_{cr}. The critical Reynold number is not fixed as it varies from geometry to geometry and also is dependent on flow conditions.
For a circular pipe, the critical Reynolds number is given as Re_{cr}=2300 and for noncircular pipes it depends on the hydraulic diameter D_{h} which is defined as
Where,
A_{c} is the crosssectional area of the pipe
P is the wetted perimeter of the pipe
Hydraulic radius is defined as the crosssectional area of the channel divided by the wetted perimeter and referred as R_{H}.
Following is the table of hydraulic diameter of noncircular tubes and channels:
Example 1: A Newtonian fluid with a dynamic viscosity of 0.40 n.s.m^{2} and a specific gravity of 0.93 flows through a 28 mm diameter pipe with a velocity of 2.9 m.s^{1}.
Ans: The density can be calculated using the specific gravity like
⍴=0.93(1000kg.m^{3})
=930kg.m^{3}
The Reynolds number can be calculated using the formula as
R_{e}=(930kg.m^{3}) (2.9m.s^{1}) (28mm) (10^{3}m.mm^{1})/(0.38Ns.m^{2})
= 198 (kg.m.s^{2})/N
Since R_{e}=198 the flow is laminar.
Example 2: A person turns on the water tap in his kitchen. Water flows out from a copper pipe with 7.00 mm diameter at a velocity of 1.00 ms^{1}. The density of water in the pipe is 1000kg.m^{3} and viscosity 0.00135 Pa.s^{1}. What is the Reynolds number, and what is the type of flow in the pipe?
Ans: The diameter of the pipe=7.00 mm = 0.007 m
Using Reynolds number formula:
R_{e}=(1000kg.m^{3}) (1.00m.s^{1}) (0.007m)/0.00135 Pa.s^{1}
=(1000kg.m^{3}) (1.00m.s^{1}) (0.007m)/0.00135 kg.m^{1}.s^{1}
=5185
Since R_{e}=5185 the flow is turbulent.
Terminal velocity is defined as the highest velocity attained by an object falling through a fluid. It is observed when the sum of drag force and buoyancy is equal to the downward gravity force acting on the object. The acceleration of the object is zero as the net force acting on the object is zero.
In fluid mechanics, for an object to attain its terminal velocity, it should have a constant speed against the force exerted by the fluid through which it is moving.
The mathematical representation of terminal velocity is:
Where,
vt is the terminal velocity, m is the mass of the falling object, g is the acceleration due to gravity, Cd is the drag coefficient, 𝜌 density of the fluid through which the object is falling, and A is the area projected by the object.
Deriving terminal velocity using mathematical terms according to the drag equation as follows:
F = bv2
Where b is the constant depending on the type of drag
(integrating the equations)
Where,
After integration,
After substituting for v_{t}
Therefore, above is the derivation of terminal velocity.
According to the continuity equation, the product of the crosssectional area of the pipe and the velocity of the fluid at any given point along the pipe is constant.
Continuity equation represents that the product of crosssectional 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,
Following are the assumptions of continuity equation:
Consider the following diagram:
Equation of Continuity
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 Δx_{1} with a velocity v_{1} at the lower end of the pipe.
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 Δx_{1 }region will be:
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 crosssectional area. For the lower end with crosssectional 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 Δt time and A_{2}, is the crosssectional 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,
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
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, 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.
Following is the continuity equation in cylindrical coordinates:
Following is the continuity equation for incompressible flow as the density, ρ = constant and is independent of space and time, we get:
∇.v = 0
Following is the continuity equation in cylindrical coordinates:
102 videos411 docs121 tests


Explore Courses for NEET exam
