Problem Statement
The space between two square flat parallel plates is filled with oil. Each side of the plate is 60 cm. The thickness of the oil film is 12.5 mm. The upper plate, which moves at 2.5 metre per sec requires a force of 98.1 N to maintain the speed. Determine:
- The dynamic viscosity of the oil in poise
- The kinematic viscosity of the oil in stokes if the specific gravity of the oil is 0.95
Given Data
- Each side of square plate = 60 cm = 0.60 m
- Area, A = 0.6 × 0.6 = 0.36 m²
- Thickness of oil film, dy = 12.5 mm = 12.5 × 10⁻³ m
- Velocity of upper plate, u = 2.5 m/sec
- Change of velocity between plates, du = 2.5 m/sec
- Force required on upper plate, F = 98.1 N
- Specific gravity of oil, S = 0.95
Solution
1. Calculate Shear Stress (τ)
2. Calculate Dynamic Viscosity (μ)
\( \tau = \mu \frac{du}{dy} \)
\( 272.5 = \mu \times \frac{2.5}{12.5 \times 10^{-3}} \)
\( \mu = \frac{272.5 \times 12.5 \times 10^{-3}}{2.5} = 1.3635\, \text{Ns/m}^2 \)
Convert to poise (1 Ns/m² = 10 poise):
\( \mu = 1.3635 \times 10 = 13.635\, \text{poise} \)
3. Calculate Kinematic Viscosity (ν)
\( \rho = S \times \rho_{water} = 0.95 \times 1000 = 950\, \text{kg/m}^3 \)
Kinematic viscosity:
\( \nu = \frac{\mu}{\rho} = \frac{1.3635}{950} = 0.001435\, \text{m}^2/\text{sec} \)
Convert to stokes (1 m²/sec = 10⁴ stokes):
\( \nu = 0.001435 \times 10^4 = 14.35\, \text{stokes} \)
- Dynamic viscosity (μ) = 13.635 poise
- Kinematic viscosity (ν) = 14.35 stokes
Explanation
1. Shear Stress Calculation:
The shear stress of 272.5 N/m² represents the force per unit area required to maintain the motion of the upper plate. This value is derived from the applied force (98.1 N) distributed over the plate area (0.36 m²).
2. Dynamic Viscosity:
The dynamic viscosity of 13.635 poise indicates the oil's resistance to flow under shear stress. This value is calculated using Newton's law of viscosity, relating shear stress to the velocity gradient (du/dy).
3. Kinematic Viscosity:
The kinematic viscosity of 14.35 stokes represents the oil's viscous forces relative to its inertial forces. This value is obtained by dividing the dynamic viscosity by the oil's density (950 kg/m³).
Physical Meaning
1. Dynamic Viscosity Significance:
A dynamic viscosity of 13.635 poise indicates a moderately viscous oil. This property determines how easily the oil flows under shear stress and is crucial in lubrication applications.
2. Kinematic Viscosity Importance:
The kinematic viscosity of 14.35 stokes is important for understanding how the oil's viscosity changes with temperature and pressure. It's particularly relevant in applications involving fluid flow and heat transfer.
3. Practical Implications:
These viscosity values suggest this oil would be suitable for applications requiring moderate lubrication and thermal stability. The relatively high viscosity indicates good load-bearing capacity but may require more energy to maintain flow.
4. Measurement Context:
The parallel plate configuration used in this problem is a classic method for viscosity measurement, simulating real-world conditions where fluids are subjected to shear between moving surfaces.
