Problem Statement
The pressure difference (∆P) in a pipe of diameter (D) and length (L) due to viscous flow depends on the velocity of fluid (V), viscosity (µ) and density (ρ). Using Buckingham's π theorem, show that ∆P=(µVL)/D² · f(Re) where Re=ρDV/μ is Reynold's number. (Take D, V and ρ as repeating variables)
Given Data
Solution Approach
To solve this problem, we'll apply Buckingham's π theorem, which states that a physical relationship between n variables that involve k fundamental dimensions can be rewritten as a relationship between (n-k) dimensionless π groups.
We have 6 variables and 3 fundamental dimensions, so we expect 3 dimensionless π groups. We'll form these π groups using the repeating variables D, V, and ρ, and then manipulate them to obtain the required form.
Calculations
Dimensional Analysis
Step 1: Identify the dimensional formula for each variable:
| Pressure difference (∆P) | ML-1T-2 |
| Pipe diameter (D) | L |
| Pipe length (L) | L |
| Fluid velocity (V) | LT-1 |
| Fluid viscosity (µ) | ML-1T-1 |
| Fluid density (ρ) | ML-3 |
These dimensional formulas will be used to determine the exponents for each repeating variable in our π terms.
Step 2: Form the first π term with ∆P and the repeating variables:
π1 = Da1 · Vb1 · ρc1 · ∆PFor π1 to be dimensionless, we need:
[L]a1 · [LT-1]b1 · [ML-3]c1 · [ML-1T-2] = [M0L0T0]Expanding and grouping by dimensions:
M: c1 + 1 = 0L: a1 + b1 - 3c1 - 1 = 0
T: -b1 - 2 = 0
Solving the system of equations:
From T: b1 = -2From M: c1 = -1
From L: a1 + (-2) - 3(-1) - 1 = 0
a1 - 2 + 3 - 1 = 0
a1 = 0
Therefore:
π1 = D0 · V-2 · ρ-1 · ∆P = ∆P / (ρV2)Step 3: Form the second π term with L and the repeating variables:
π2 = Da2 · Vb2 · ρc2 · LFor π2 to be dimensionless, we need:
[L]a2 · [LT-1]b2 · [ML-3]c2 · [L] = [M0L0T0]Expanding and grouping by dimensions:
M: c2 = 0L: a2 + b2 - 3c2 + 1 = 0
T: -b2 = 0
Solving the system of equations:
From T: b2 = 0From M: c2 = 0
From L: a2 + 0 - 0 + 1 = 0
a2 = -1
Therefore:
π2 = D-1 · V0 · ρ0 · L = L/DStep 4: Form the third π term with μ and the repeating variables:
π3 = Da3 · Vb3 · ρc3 · μFor π3 to be dimensionless, we need:
[L]a3 · [LT-1]b3 · [ML-3]c3 · [ML-1T-1] = [M0L0T0]Expanding and grouping by dimensions:
M: c3 + 1 = 0L: a3 + b3 - 3c3 - 1 = 0
T: -b3 - 1 = 0
Solving the system of equations:
From T: b3 = -1From M: c3 = -1
From L: a3 + (-1) - 3(-1) - 1 = 0
a3 - 1 + 3 - 1 = 0
a3 = -1
Therefore:
π3 = D-1 · V-1 · ρ-1 · μ = μ/(ρDV)We recognize that π3 = μ/(ρDV) = 1/Re, where Re is the Reynolds number.
Step 5: According to Buckingham's π theorem, we can write:
f(π1, π2, π3) = 0Or equivalently:
π1 = Φ(π2, π3)Substituting our π terms:
∆P/(ρV2) = Φ(L/D, μ/(ρDV))Step 6: To obtain the desired form ∆P=(µVL)/D² · f(Re), we manipulate the equation:
∆P/(ρV2) = Φ(L/D, μ/(ρDV))Multiply both sides by ρV²:
∆P = ρV2 · Φ(L/D, μ/(ρDV))Let's rearrange the function to express it in terms of Re:
∆P = ρV2 · Φ(L/D, 1/Re)We can rewrite this as:
∆P = ρV2 · (L/D) · Ψ(Re)Where Ψ is a new function. Now we have:
∆P = ρV2 · (L/D) · Ψ(Re)We can further manipulate this to get:
∆P = μV · (L/D²) · (ρDV/μ) · Ψ(Re) ∆P = (μVL/D²) · Re · Ψ(Re)Define f(Re) = Re · Ψ(Re):
∆P = (μVL/D²) · f(Re)∆P = (μVL/D²) · f(Re), where Re = ρDV/μ
Detailed Explanation
Significance of the Result
The result ∆P = (μVL/D²) · f(Re) is a general form of the pressure drop equation for pipe flow. It shows that:
- Pressure drop is directly proportional to fluid viscosity (μ), velocity (V), and pipe length (L)
- Pressure drop is inversely proportional to the square of pipe diameter (D²)
- The function f(Re) accounts for the flow regime (laminar or turbulent) based on the Reynolds number
Special Cases
For fully developed laminar flow in a circular pipe, f(Re) = 64/Re, and the equation reduces to the Hagen-Poiseuille equation:
∆P = (μVL/D²) · (64/Re) = 64μ²VL/(ρD³V²) = 32μVL/D²For turbulent flow, f(Re) is more complex and typically determined empirically using the Moody diagram or various friction factor correlations like Colebrook-White equation.
Applications in Engineering
This dimensional analysis result is fundamental in:
- Designing piping systems for fluid transport
- Sizing pumps based on pressure drop requirements
- Analyzing flow control devices
- Optimizing pipe networks for energy efficiency
- Scale modeling and similarity analysis in fluid dynamics research
Power of Dimensional Analysis
This problem demonstrates the power of dimensional analysis and Buckingham's π theorem in:
- Reducing the number of variables in a complex physical problem
- Identifying the key dimensionless groups that govern the phenomenon
- Establishing general relationships without solving detailed differential equations
- Facilitating experimental design and data correlation
Practical Implications
Understanding the pressure drop relationship has practical implications:
- Doubling the pipe diameter reduces pressure drop by a factor of 16 (from the D² term)
- Doubling the flow velocity increases pressure drop by at least a factor of 2 (more in turbulent flow)
- Pressure drop scales linearly with pipe length
- The transition from laminar to turbulent flow (governed by Re) significantly affects pressure drop behavior
This dimensional analysis approach provides a powerful framework for understanding and predicting fluid flow behavior in pipes without solving the complex Navier-Stokes equations directly.