Derive by the method of dimensions, an expression for the maximum velocity u for fluid flow through a circular tube. Assume that u depends on (i) the radius of the tube R, (ii) vis... Derive by the method of dimensions, an expression for the maximum velocity u for fluid flow through a circular tube. Assume that u depends on (i) the radius of the tube R, (ii) viscosity of the fluid η, (iii) pressure gradient (P/ℓ), ℓ is the length of the tube.
Understand the Problem
The question is asking us to derive an expression for maximum velocity in fluid flow through a circular tube using dimensional analysis, considering factors like tube radius, fluid viscosity, and pressure gradient.
Answer
$$ u = k \cdot \frac{P}{\ell \eta} $$
Answer for screen readers
The derived expression for maximum velocity is: $$ u = k \cdot \frac{P}{\ell \eta} $$
Steps to Solve
- Identify the Variables Define the variables that affect the maximum velocity $u$:
- Tube radius $R$ (length)
- Viscosity of the fluid $\eta$ (mass/time length)
- Pressure gradient $\frac{P}{\ell}$ (force/length$^2$)
- Write the Dimensional Formulae Establish the dimensional formula for each variable:
- Viscosity $\eta$ has dimensions $[M][L]^{-1}[T]^{-1}$
- Pressure gradient $\frac{P}{\ell}$ has dimensions $[M][L]^{-1}[T]^{-2}$
- Tube radius $R$ has dimensions $[L]$
-
Establish the Relationship Using Dimensional Analysis Assume that maximum velocity $u$ can be expressed as: $$ u = k \cdot R^x \cdot \eta^y \cdot \left(\frac{P}{\ell}\right)^z $$ where $k$ is a constant.
-
Determine the Dimensions of the Left Side The left side, maximum velocity $u$, has dimensions $[L][T]^{-1}$.
-
Write the Dimensional Relationship Substituting the dimensions into the equation gives: $$ [L][T]^{-1} = k \cdot [L]^x \cdot [M][L]^{-1}[T]^{-1}^y \cdot [M][L]^{-1}[T]^{-2}^z $$
-
Simplify the Right Side This results in: $$ [M]^{y+z} [L]^{x-y-1-z} [T]^{-y-2z} $$
-
Equate Powers of Dimensions Set powers of each dimension equal to those on the left side:
- For $M$: $y + z = 0$
- For $L$: $x - y - 1 - z = 1$
- For $T$: $-y - 2z = -1$
- Solve the Equations From $y + z = 0$, we can write $z = -y$.
Substituting into the other equations:
- $x - y + 1 + y = 1 \implies x + 1 = 1 \implies x = 0$
- $-y - 2(-y) = -1 \implies -y + 2y = -1 \implies y = -1$
Thus, $z = 1$.
- Final Expression The derived expression for maximum velocity is: $$ u = k \cdot \frac{P}{\ell} \cdot R^{0} \cdot \eta^{-1} $$ Hence, we simplify to: $$ u = k \cdot \frac{P}{\ell \eta} $$
The derived expression for maximum velocity is: $$ u = k \cdot \frac{P}{\ell \eta} $$
More Information
This result indicates that the maximum velocity in fluid flow through a circular tube is proportional to the pressure gradient and inversely proportional to the viscosity of the fluid. The constant $k$ depends on factors not specified in the problem.
Tips
- Miscounting the dimensions when equating powers of $M$, $L$, and $T$.
- Ignoring the dependency of other parameters when deriving the relationships; ensure all given factors are considered.
AI-generated content may contain errors. Please verify critical information
