Problem Statement
Determine the viscosity of a liquid having a kinematic viscosity of 6 stokes and a specific gravity of 2.0.
Given Data
- Kinematic Viscosity, \(\nu = 6 \, \text{stokes}\)
- Specific Gravity, \(S.G. = 2.0\)
Solution
1. Convert Units to a Consistent System (SI)
First, we convert the kinematic viscosity from stokes to m²/s and calculate the density from the specific gravity.
Kinematic Viscosity Conversion (1 stoke = 10⁻⁴ m²/s):
Density Calculation (Density of water \(\approx 1000\) kg/m³):
2. Calculate the Dynamic Viscosity (\(\mu\))
Dynamic viscosity (\(\mu\)) can be found by rearranging the definition of kinematic viscosity (\(\nu = \mu/\rho\)).
3. Convert Dynamic Viscosity to Poise
The CGS unit for dynamic viscosity is the Poise (1 N·s/m² = 10 Poise).
The viscosity (dynamic) of the liquid is \( \mu = 1.2 \, \text{N s/m}^2 \) or \( 12 \, \text{Poise} \).
Explanation of the Relationship
Dynamic Viscosity (\(\mu\)) is the fundamental property of a fluid that measures its internal resistance to shearing forces. It's often thought of as the fluid's "thickness."
Kinematic Viscosity (\(\nu\)) relates this "thickness" to the fluid's "heaviness" (density). It is defined as the ratio of the dynamic viscosity to the density: $$ \nu = \frac{\mu}{\rho} $$ This calculation simply reverses the definition. By knowing the kinematic viscosity (a measure of flowability under gravity) and the density, we can determine the fluid's intrinsic, absolute resistance to shear, which is the dynamic viscosity.
Physical Meaning
The result tells us that the liquid is quite viscous. With a dynamic viscosity of 12 Poise (1.2 N·s/m²), it is about 1200 times more viscous than water at room temperature.
This problem highlights how two fluids can have the same kinematic viscosity (flow similarly under gravity) but vastly different dynamic viscosities if their densities are different. In this case, we are given a fluid that is twice as dense as water. To have a kinematic viscosity of 6 stokes, its dynamic viscosity must be substantial to overcome its own "heaviness." This calculation essentially "removes" the effect of density to find the fluid's true internal friction.
