A steel sphere of 5mm diameter falls in a glycerin at a terminal velocity of 0.05m/s. Assume Stoke’s law is applicable, determine (a) dynamic viscosity of glycerin, (b) drag force and (c) coefficient of drag. Take sp wt of steel and glycerin as 75 KN/m³ and 12.5 KN/m³ respectively.

Fluid Mechanics Problem Solution

Problem Statement

A steel sphere of 5mm diameter falls in a glycerin at a terminal velocity of 0.05m/s. Assume Stoke’s law is applicable, determine (a) dynamic viscosity of glycerin, (b) drag force and (c) coefficient of drag. Take sp wt of steel and glycerin as 75 KN/m³ and 12.5 KN/m³ respectively.

Given Data

Diameter of ball (D) 5 mm = 0.005 m
Specific weight of steel (γsteel) 75 KN/m³
Specific weight of glycerin (γgly) 12.5 KN/m³
Terminal velocity (V) 0.05 m/s
Density of glycerin (ρgly) 12500/9.81 = 1274.2 kg/m³
Volume of sphere (1/6)πD³ = (1/6)π(0.005)³ = 6.545×10⁻⁸ m³

Solution Approach

When an object falls through a fluid and reaches terminal velocity, the net force acting on it is zero. At this point, the weight of the object is balanced by the drag force and the buoyant force. We’ll use this principle along with Stoke’s law to determine the required parameters.

Calculations

(a) Dynamic Viscosity of Glycerin

Step 1: At terminal velocity, the forces are balanced:

Weight of sphere (W) = Drag force (FD) + Buoyant force (FB)

Therefore, Drag force:

FD = W – FB

Step 2: Calculate the weight and buoyant force:

FD = γsteel × Volume – γgly × Volume
FD = (75000 – 12500) × (1/6)π(0.005)³
FD = 62500 × 6.545×10⁻⁸
FD = 0.00409 N

Step 3: Apply Stoke’s law to find the dynamic viscosity:

FD = 3πμDV
0.00409 = 3π × μ × 0.005 × 0.05
μ = 0.00409 / (3π × 0.005 × 0.05)
μ = 0.00409 / 0.002356
μ = 1.7 N·s/m²

Dynamic Viscosity (μ) = 1.7 N·s/m² = 1.7 Pa·s

(b) Drag Force

The drag force was already calculated in part (a):

FD = 0.00409 N

Drag Force (FD) = 0.00409 N

(c) Coefficient of Drag

Step 1: First, we need to verify if Stoke’s law is applicable by calculating the Reynolds number:

Re = ρVD/μ
Re = (12500/9.81) × 0.05 × 0.005 / 1.7
Re = 1274.2 × 0.05 × 0.005 / 1.7
Re = 0.18

Step 2: Since Re < 0.2, Stoke's law is valid. For Stoke's flow, the coefficient of drag is:

CD = 24/Re
CD = 24/0.18
CD = 133

Coefficient of Drag (CD) = 133

Detailed Explanation

Terminal Velocity and Force Balance

When an object falls through a fluid, it initially accelerates due to gravity. As its velocity increases, the drag force also increases until it balances the net gravitational force (weight minus buoyancy). At this point, the object stops accelerating and falls at a constant speed called the terminal velocity.

Stoke’s Law

Stoke’s law applies to spherical particles moving through a viscous fluid at low Reynolds numbers (typically Re < 0.2). It states that the drag force is directly proportional to:

  • The fluid’s dynamic viscosity (μ)
  • The radius of the spherical object
  • The velocity of the object

The mathematical form is FD = 6πμrV, which is equivalent to FD = 3πμDV where D is the diameter.

Reynolds Number

The Reynolds number (Re) is a dimensionless quantity that predicts the flow pattern of fluids around objects. It is defined as:

Re = ρVD/μ

where ρ is the fluid density, V is the velocity, D is the characteristic length (diameter in this case), and μ is the dynamic viscosity.

For very low Reynolds numbers (Re < 0.2), the flow is dominated by viscous forces, and Stoke's law provides an accurate description of the drag.

Drag Coefficient

The drag coefficient (CD) is a dimensionless quantity used to quantify the drag or resistance of an object in a fluid environment. For a sphere in the Stokes flow regime, the drag coefficient is inversely proportional to the Reynolds number:

CD = 24/Re

This relatively high drag coefficient (133) indicates that viscous forces are dominant in this flow scenario, which is typical for small objects moving slowly through highly viscous fluids like glycerin.

Applications

Understanding the behavior of particles in viscous fluids has numerous applications:

  • Pharmaceutical industry: drug delivery systems and particle separation
  • Environmental engineering: sedimentation processes and pollutant transport
  • Chemical engineering: design of reactors and separation equipment
  • Biomedical applications: understanding cell movement in body fluids
  • Geological processes: sedimentation and erosion mechanisms

Significance of Results

The calculated dynamic viscosity of glycerin (1.7 Pa·s) is consistent with its known high viscosity at room temperature, which can range from 0.95 to 1.5 Pa·s depending on purity and temperature. This high viscosity explains why the steel sphere falls at such a slow terminal velocity despite the substantial density difference between steel and glycerin.

The verification that Re = 0.18 confirms that Stoke’s law is indeed applicable in this scenario, validating our approach and results.

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