Problem Statement
A metallic sphere of sp.gr. 8.0 falls in an oil of density 800 kg/m³. The diameter of the sphere is 10mm and it attains a terminal velocity of 50mm/s. Find the viscosity of the oil in Poise.
Given Data
Solution Approach
When a sphere falls through a viscous fluid and reaches terminal velocity, three forces are in equilibrium:
- Weight of the sphere (downward)
- Buoyant force (upward)
- Drag force due to viscosity (upward)
For a sphere moving slowly through a viscous fluid, we can apply Stokes’ law to find the viscosity of the oil.
Calculations
Force Balance Analysis
Step 1: At terminal velocity, the weight of the sphere equals the sum of buoyant force and drag force:
Where μ is the dynamic viscosity of the oil, and the drag force is given by Stokes’ law: FD = 3πμDV.
Step 2: Substituting the given values:
Step 3: Simplifying:
Step 4: Solving for μ:
Step 5: Converting to Poise (1 Pa·s = 10 Poise):
Step 6: Verifying that Reynolds number is less than 0.2 for Stokes’ law validity:
Since Re = 0.05 < 0.2, Stokes' law is valid for this problem.
Viscosity of the oil = 78.48 Poise
Detailed Explanation
Terminal Velocity Principle
When an object falls through a fluid, it eventually reaches a constant velocity called terminal velocity. At this point, the downward gravitational force is exactly balanced by the upward forces (buoyancy and drag).
Stokes’ Law
Stokes’ law describes the drag force exerted on spherical objects moving through a viscous fluid:
Where:
- μ is the dynamic viscosity of the fluid
- D is the diameter of the sphere
- V is the velocity of the sphere relative to the fluid
This law applies when the Reynolds number is less than 0.2, which indicates laminar flow conditions.
Force Analysis
The three forces acting on the falling sphere are:
- Weight force: FW = ρs × g × Volume = ρs × g × (1/6)πD³
- Buoyant force: FB = ρo × g × Volume = ρo × g × (1/6)πD³
- Drag force: FD = 3πμDV
Units Conversion
The viscosity calculated in SI units is 7.848 Pa·s (or N·s/m²). The unit Poise is a CGS unit of dynamic viscosity:
- 1 Poise = 0.1 Pa·s
- Therefore, 7.848 Pa·s = 78.48 Poise
Applications
This method of determining fluid viscosity is the principle behind the falling ball viscometer, which is used in various industries:
- Oil and petroleum industry for quality control
- Food processing for consistency measurement
- Pharmaceutical industry for medication viscosity testing
- Chemical industry for product development and quality assurance
Importance of Reynolds Number
The Reynolds number (Re = 0.05) being well below 0.2 confirms that we are in the creeping flow regime where Stokes’ law applies. This is characterized by:
- Dominant viscous forces over inertial forces
- Streamline, laminar flow around the sphere
- Negligible wake formation behind the sphere
Understanding fluid viscosity is essential in many engineering applications, including lubrication systems, hydraulic machinery, and fluid transport systems.
