Problem Statement
Determine the specific gravity of a fluid having viscosity 0.05 poise and kinematic viscosity 0.035 stokes.
Given Data
- Viscosity, μ = 0.05 poise = 0.05/10 = 0.005 Ns/m²
- Kinematic viscosity, ν = 0.035 stokes = 0.035 × 10⁻⁴ m²/s
- Density of water = 1000 kg/m³
Solution
1. Calculate Density (ρ)
0.035 × 10⁻⁴ = 0.005/ρ
ρ = 0.005/(0.035 × 10⁻⁴)
ρ = 1428.57 kg/m³
2. Calculate Specific Gravity
S = 1428.57/1000
S = 1.42857 ≈ 1.43
- Specific gravity = 1.43
Explanation
1. Unit Conversion:
Converted viscosity from poise to Ns/m² (1 poise = 0.1 Ns/m²) and kinematic viscosity from stokes to m²/s (1 stoke = 10⁻⁴ m²/s) for consistent SI unit calculations.
2. Density Calculation:
Used the fundamental relationship between kinematic viscosity (ν), dynamic viscosity (μ), and density (ρ): ν = μ/ρ to determine the fluid density.
3. Specific Gravity:
Calculated as the ratio of fluid density to water density, showing the fluid is 43% denser than water.
Physical Meaning
1. Specific Gravity Value:
A specific gravity of 1.43 indicates this is likely a heavy oil or similar viscous fluid, denser than water.
2. Viscosity Values:
The given viscosity values (0.05 poise dynamic, 0.035 stokes kinematic) are typical for lubricating oils or hydraulic fluids.
3. Practical Implications:
This fluid would sink in water and would require more energy to pump than less dense fluids. The relatively high viscosity suggests good lubricating properties but higher flow resistance.