Newton's Law of Viscosity

Viscosity is a measurement of the resistance of a fluid to flow or deformation. It measures how much fluid resists the motion of sliding layers past one another. There is a mathematical approach under Newton's Law of Viscosity to describe this resistance as involving force and flow.

1.0Statement and Explanation of Newton’s Law of Viscosity 

Newton's Law of Viscosity states that shear stress () is directly proportional to the related shear strain rate, which is defined as the velocity gradient between two layers. Another way to say this is the force that must be used to move one layer of fluid over another is proportional to the speed with which the velocity changes in the fluid. Mathematically, it can be written as: 

$F=-\eta \frac{du}{dy}$

Here:

• F = shear stress (force per unit area (A) applied tangentially to the fluid),
• η = dynamic viscosity of the fluid,
• ​dudy = velocity gradient, which represents the rate at which the velocity of fluid changes with respect to the distance from the surface (i.e., the rate of shear strain).
• The given formula is known as Newton’s law of viscosity Formula, which shows that for the Newtonian type of fluid, the relationship between shear stress and rate is linear. 
• The negative sign indicates the direction of the shear force relative to the flow.

What is the Formula for Viscosity? 

The above-mentioned formula can also be used for deriving the formula for viscosity. It shows how the viscosity or internal resistance of fluids is related to the shear stress and the shear rate: 

$F=\frac{\tau }{\frac{du}{dy}}$

• The SI unit of dynamic viscosity - is pascal-seconds (Pa–s) while the CGS unit of viscosity is poise (P).
• 1 poise (P) = 0.1 Pascal – second 

2.0Types of Fluids: 

Newtonian Fluids:

These fluids follow Newton's Law of Viscosity. The viscosity is constant and it is independent of the shear rate. The shear stress is always directly proportional to the shear rate. Examples of Newtonian fluids comprise water, air, and any other gas.

Non-Newtonian Fluids:

These fluids do not obey Newton's Law of Viscosity. Their viscosity varies with changes in the shear rate or external factors like temperature. Such compounds include ketchup, blood, paints, and some polymer solutions.

3.0Solved Examples

Problem 1: Two fluids, A and B, are flowing between parallel plates. Fluid A has a dynamic viscosity of 0.4 Pa.s, and fluid B has a viscosity of 0.8 Pa.s. If the shear rate in fluid A is 10 s⁻¹, calculate the shear rate for fluid B when the shear stress is the same in both fluids.

Solution: Let the Dynamic viscosity of fluid A A=0.4Pa.s

Let the Dynamic viscosity of fluid B B\eta_{B}=0.8Pa.s

The shear rate of fluid $A\frac{d{u}_A}{dy}=10s^{-1}$

According to the question, the shear stress is the same for both fluids. Hence,

$FA=FB$

${\eta }_A\frac{d{u}_A}{dy}={\eta }_B\frac{d{u}_B}{dy}$

$0.4\times 10=0.8\times \frac{d{u}_B}{dy}$

$\frac{d{u}_B}{dy}=\frac{0.4\times 10}{0.8}=5s^{-1}$

Problem 2: A fluid flows through a pipe with a radius of 0.01 m. The pressure difference between the two ends of the pipe is 500 Pa. The velocity gradient in the pipe is 150 s⁻¹. Find the viscosity of the fluid. 

Solution: Let the radius of the pipe be r = 0.01m 

Pressure difference be P = 500Pa

Velocity Gradient $\frac{du}{dy}=150s^{-1}$

Here, we need shear stress for using Newton’s law of viscosity formula, which for pipe flow is given by: 

$F=\frac{\Delta P}{r}$

$F=\frac{500}{0.01}=50000Pa$

Now, using the formula of viscosity:

$\eta =\frac{F}{\frac{du}{dy}}$

$\eta =\frac{500000}{150}=333.33Pa.s$

Problem 3: A fluid flows between two parallel plates, with a distance of 0.02 m between them. The velocity at the lower plate is 0.3 m/s, and the velocity at the top plate is 0.5 m/s. If the dynamic viscosity of the fluid is 0.04 Pa.s, find the shear stress on the fluid.

Solution: Let the distance between plates by dy = 0.02m

Velocity at the lower plate u_1,= 0.3m/s

Velocity at the top plate u₂ = 0.5m/s

Dynamic viscosity of fluid = 0.04 Pa.s 

Velocity gradient can be calculated as:

$\frac{du}{dy}=\frac{u_2-u_1}{dy}$

$=\frac{0.5-0.3}{0.02}=\frac{0.2}{0.02}$

$=10m/s^2$

Now, by using Newton’s law, let’s calculate shear stress:

$F=\eta \frac{du}{dy}$

$F=0.04\times 10=0.4Pa$