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The volume rate of flow, \(Q,\) through a pipe containing a slowly moving liquid is given by the equation \\[ Q=\frac{\pi R^{4} \Delta p}{8 \mu \ell} \\] where \(R\) is the pipe radius, \(\Delta p\) the pressure drop along the pipe, \(\mu\) a fluid property called viscosity \(\left(F L^{-2} T\right),\) and \(\ell\) the length of pipe. What are the dimensions of the constant \(\pi / 8 ?\) Would you classify this equation as a general homogeneous equation? Explain.

Short Answer

Expert verified
The constant \( \frac{\pi}{8} \) is dimensionless. Yes, the equation is homogeneous.

Step by step solution

01

Analyze the Dimensions of Each Variable

First, let's determine the dimensions for each of the variables. Using the given information:- The quantity \( R \) is a radius, so its dimension is \( [L] \) (length).- The pressure drop \( \Delta p \) has dimensions of force per unit area, thus \( [F][L]^{-2} \).- Viscosity \( \mu \) has dimensions \( [F][L]^{-2}[T] \).- The length of the pipe \( \ell \) has dimensions \( [L] \).- The volume flow rate \( Q \) is volume per unit time, so it has dimensions \( [L]^3[T]^{-1} \).
02

Express Dimensions of the Given Equation

The equation for volume flow rate is given by:\[Q = \frac{\pi R^4 \Delta p}{8 \mu \ell}.\]Substitute the dimensions of the variables:- \( R^4 \) yields \( [L]^4 \).- \( \Delta p \) gives \( [F][L]^{-2} \).- \( \mu \) yields \( [F][L]^{-2}[T] \).- \( \ell \) is \( [L] \).Thus, the equation becomes:\[[L]^3[T]^{-1} = \frac{\pi [L]^4 [F][L]^{-2]}{8 [F][L]^{-2}[T][L]}.\]
03

Simplify and Determine Dimensions of \( \frac{\pi}{8} \)

Simplify the dimensional equation:\[[L]^3[T]^{-1} = \frac{\pi [L]^4[F][L]^{-2]}{8[F][L]^{-1}[T]}\]\[= \frac{\pi [L]^2}{8[T]}\]To maintain dimensional consistency, the dimensions of \( \frac{\pi}{8} \) must be dimensionless, as all other dimensions on both sides of the equation are equivalent. Therefore, \( \frac{\pi}{8} \) is dimensionless.
04

Classify the Equation as Homogeneous or Not

A homogeneous equation is one where all terms have the same dimensions. In our case, the equation simplifies dimensionally to \[[L]^3[T]^{-1} = [L]^3[T]^{-1},\]which means the dimensional expression is consistent and all terms have the same dimensions. Therefore, this equation is homogeneous.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dimensional Analysis
Dimensional analysis is a critical technique used to understand the dependencies and relationships between various physical quantities by examining their dimensions. Dimensions are represented using basic entities such as length \([L]\), mass \([M]\), time \([T]\), etc. This helps in checking the consistency of physical equations and deducing relationships between different variables.
In the context of the volume flow rate equation, we analyze each component of the equation by its dimension to ensure they lead to a coherent and consistent formula. For example, with the given flow rate equation, we break down each variable based on its physical representation:
  • The pipe radius \(R\) has a dimension of \([L]\).
  • The pressure drop \(\Delta p\) is force per unit area, expressed as \([F][L]^{-2}\).
  • Viscosity \(\mu\) captures fluid's resistance to flow and is described dimensionally as \([F][L]^{-2}[T]\).
  • The pipe length \(\ell\) is naturally a length, so \([L]\).
By breaking down the equation into dimensions and ensuring that both sides match, we confirm its validity and dimensionless nature of the constant \(\frac{\pi}{8}\). This ensures the entire equation is dimensionally consistent, a hallmark of dimensional analysis.
Viscosity
Viscosity is a fundamental property of fluids that characterizes their internal friction when in motion. It's what makes honey flow more slowly than water. It inhibits fluid motion, causing energy loss within the fluid due to frictional forces acting between fluid layers.
When we talk about viscosity in the context of the volume flow rate equation, we denote it by \(\mu\), which is critical in determining how easily a fluid can be moved through a pipe. In dimensional terms, viscosity has the dimensions of \([F][L]^{-2}[T]\), aligning with the idea that it's a force-related property affecting fluid flow.
Its role in the volume flow rate equation reflects how forcefully the fluid can be moved through a pipe of a certain dimension, bridging the relationship between the applied pressure difference and resulting flow rate. Understanding this viscosity relationship is important for engineering applications, such as designing piping systems to maximize efficiency and minimize losses due to friction.
Homogeneous Equations
A homogeneous equation in physics and engineering is one where each term has the same dimensional composition. Such equations reflect a consistent relationship across all variables and their dimensions, ensuring the entire equation is dimensionally balanced and correctly formulated.
In checking whether the flow rate equation is homogeneous, we need to ensure that every term on both sides of the equation embodies the same dimensional makeup. Upon simplifying the dimensional expression of the flow rate equation, we derived \[ [L]^3[T]^{-1} = [L]^3[T]^{-1} \] from both sides, confirming that they indeed match. This dimensional consistency signifies the equation's homogeneity.
Ensuring homogeneity is crucial for verifying that physical equations accurately reflect the systems they represent, thereby allowing for correct application and predictions in real-world scenarios.

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