If the resistance to motion of a sphere through a fluid (R) is a function of the density (ρ), viscosity (µ) of the fluid, and the radius (r) and velocity (u) of the sphere, develop a relationship of R using Buckingham's π theorem.

Fluid Mechanics Problem Solution

Problem Statement

If the resistance to motion of a sphere through a fluid (R) is a function of the density (ρ), viscosity (µ) of the fluid, and the radius (r) and velocity (u) of the sphere, develop a relationship of R using Buckingham’s π theorem. (Take u, r and ρ as repeating variables and take the dimension of shear stress for R)

Given Data

Resistance to motion R [ML^-1T^-2] (Dimension of shear stress)

Fluid density ρ [ML^-3]

Fluid viscosity μ [ML^-1T^-1]

Sphere radius r [L]

Sphere velocity u [LT^-1]

Repeating variables u, r, and ρ

Solution Approach

To solve this problem, we’ll apply Buckingham’s π theorem, which helps derive dimensionless groups from a set of physical variables. The steps are:

Identify the number of independent variables and fundamental dimensions
Determine the number of π terms using n – k, where n is the number of variables and k is the number of fundamental dimensions
Select repeating variables that contain all the fundamental dimensions
Form π terms and solve for the exponents using dimensional homogeneity
Express the relationship between the dimensionless groups

Calculations

Applying Buckingham’s π Theorem

Step 1: Express the functional relationship and analyze dimensions:

f₁(R, ρ, μ, r, u) = 0

Total number of variables: n = 5
Number of fundamental dimensions: k = 3 (M, L, T)
Number of π terms: n – k = 5 – 3 = 2

We can express this as:

f(π₁, π₂) = 0

Using the repeating variables u, r, and ρ as specified.

Step 2: Determining the first π term with R:

π₁ = u^a1 r^b1 ρ^c1 R

For π₁ to be dimensionless, we need:

[M⁰L⁰T⁰] = [LT^-1]^a1 [L]^b1 [ML^-3]^c1 [ML^-1T^-2]

This gives us the following system of equations by comparing exponents:

For M: c1 + 1 = 0

For L: a1 + b1 – 3c1 – 1 = 0

For T: -a1 – 2 = 0

Solving these equations:

From T equation: a1 = -2

From M equation: c1 = -1

Substituting into L equation: -2 + b1 + 3 – 1 = 0

Solving for b1: b1 = 0

Therefore, the first π term is:

π₁ = u^-2 r⁰ ρ^-1 R = R/(ρu²)

Step 3: Determining the second π term with μ:

π₂ = u^a2 r^b2 ρ^c2 μ

For π₂ to be dimensionless, we need:

[M⁰L⁰T⁰] = [LT^-1]^a2 [L]^b2 [ML^-3]^c2 [ML^-1T^-1]

This gives us the following system of equations:

For M: c2 + 1 = 0

For L: a2 + b2 – 3c2 – 1 = 0

For T: -a2 – 1 = 0

Solving these equations:

From T equation: a2 = -1

From M equation: c2 = -1

Substituting into L equation: -1 + b2 + 3 – 1 = 0

Solving for b2: b2 = -1

Therefore, the second π term is:

π₂ = u^-1 r^-1 ρ^-1 μ = μ/(urρ)

Step 4: Establishing the functional relationship between the π terms:

f(π₁, π₂) = 0

f(R/(ρu²), μ/(urρ)) = 0

We can rewrite this as:

R/(ρu²) = f(μ/(urρ))

R = ρu² f(μ/(urρ))

Detailed Explanation

Physical Interpretation

The final relationship R = ρu² f(μ/(urρ)) provides valuable insight into the resistance of a sphere moving through a fluid:

The term ρu² represents the dynamic pressure or inertial force per unit area.
The dimensionless group μ/(urρ) is the reciprocal of the Reynolds number (Re = ρur/μ), which characterizes the ratio of inertial forces to viscous forces.

Therefore, we can rewrite the relationship as:

R = ρu² f(1/Re)

This means that the resistance coefficient (R/(ρu²)) is a function of the Reynolds number, a fundamental result in fluid mechanics.

Practical Significance

This relationship has several important implications:

Scaling: It allows engineers to predict the resistance of geometrically similar bodies of different sizes moving at different speeds in various fluids.
Flow regimes: At low Reynolds numbers (viscous flow), resistance is dominated by viscous effects. At high Reynolds numbers (turbulent flow), resistance is primarily due to inertial effects.
Drag coefficient: The function f(1/Re) represents the drag coefficient, which varies with the Reynolds number and has been extensively studied experimentally.

Stokes’ Law

For very low Reynolds numbers (Re ≪ 1), the resistance follows Stokes’ law:

R = 6πμru

This can be written in our dimensionless form as:

R/(ρu²) = 6π/(ρur/μ) = 6π/Re

Which confirms that f(1/Re) = 6π×(1/Re) for very small Reynolds numbers.

High Reynolds Number Behavior

For high Reynolds numbers (Re > 1000), the resistance becomes proportional to ρu²r², and the function f(1/Re) approaches a constant value of approximately 0.2-0.5, depending on surface roughness and other factors.

Buckingham’s π Theorem in Engineering

This problem demonstrates the power of Buckingham’s π theorem in engineering analysis:

It reduces the number of variables that need to be studied experimentally
It provides a framework for dimensional analysis and scaling
It helps identify the key dimensionless parameters that govern physical phenomena
It enables engineers to design experiments more efficiently and interpret results more effectively

Applications

The relationship derived in this problem is fundamental to many practical applications:

Design of particles and spherical objects in fluid systems
Settling velocity calculations in sedimentation processes
Drug delivery systems using microparticles in the bloodstream
Aerodynamic design of sports balls (golf, tennis, etc.)
Meteorology for predicting the fall of raindrops and hailstones
Bubble formation and rise in chemical reactors and natural systems

If the resistance to motion of a sphere through a fluid (R) is a function of the density (ρ), viscosity (µ) of the fluid, and the radius (r) and velocity (u) of the sphere, develop a relationship of R using Buckingham’s π theorem.