Problem Statement
Find the kinematic viscosity of an oil having density 981 kg/m³. The shear stress at a point in oil is 0.2452 N/m² and velocity gradient at that point is 0.2 per second.
Given Data
- Mass density, ρ = 981 kg/m³
- Shear stress, τ = 0.2452 N/m²
- Velocity gradient, du/dy = 0.2 s⁻¹
Solution
1. Calculate Dynamic Viscosity (μ)
\( \tau = \mu \frac{du}{dy} \)
\( 0.2452 = \mu \times 0.2 \)
\( \mu = \frac{0.2452}{0.2} = 1.226\, \text{Ns/m}^2 \)
2. Calculate Kinematic Viscosity (ν)
Convert to stokes (1 m²/sec = 10⁴ cm²/s):
\( \nu = 0.00125 \times 10^4 = 12.5\, \text{cm}^2/\text{s} \)
\( 1\, \text{cm}^2/\text{s} = 1\, \text{stoke} \)
- Kinematic viscosity (ν) = 12.5 stokes
Explanation
1. Dynamic Viscosity Calculation:
The dynamic viscosity of 1.226 Ns/m² represents the oil's internal resistance to flow when subjected to shear stress. This value is obtained by dividing the shear stress (0.2452 N/m²) by the velocity gradient (0.2 s⁻¹).
2. Kinematic Viscosity Conversion:
The kinematic viscosity of 12.5 stokes indicates how easily the oil flows under gravity, considering both its viscosity and density. This is calculated by dividing the dynamic viscosity by the oil's density (981 kg/m³) and converting to stokes.
Physical Meaning
1. Practical Significance of Values:
A kinematic viscosity of 12.5 stokes suggests this oil would be suitable for applications requiring moderate flow characteristics, such as certain types of lubrication or hydraulic systems.
2. Measurement Context:
The calculation method used here is fundamental in rheology, demonstrating how basic measurements of shear stress and velocity gradient can determine a fluid's viscous properties.
3. Industrial Applications:
Knowing both dynamic and kinematic viscosity is crucial for engineers designing systems involving fluid flow, heat transfer, or lubrication, as these values affect pump selection, pipe sizing, and thermal management.
