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For low-speed (laminar) steady flow through a circular pipe, as shown in Fig. P1.12, the velocity \(u\) varies with radius and takes the form $$u=B \frac{\Delta p}{\mu}\left(r_{0}^{2}-r^{2}\right)$$ where \(\mu\) is the fluid viscosity and \(\Delta p\) is the pressure drop from entrance to exit. What are the dimensions of the constant \(B ?\)

Short Answer

Expert verified
The dimension of \( B \) is dimensionless, \([1]\).

Step by step solution

01

Identify Given Variables and Constants

We have the velocity expression: \( u = B \frac{\Delta p}{\mu}(r_0^2 - r^2) \). Here, \( \Delta p \) is the pressure drop, \( \mu \) is the fluid viscosity, \( r_0 \) and \( r \) are radii. Our task is to determine the dimensions of the constant \( B \). We know that velocity \( u \) has dimensions \([L][T]^{-1}\).
02

Determine Dimensions of Known Quantities

The pressure \( \Delta p \) has dimensions of force per unit area, \([M][L]^{-1}[T]^{-2}\). The dynamic viscosity \( \mu \) has dimensions of \([M][L]^{-1}[T]^{-1}\). The radius \( r_0^2 - r^2 \) has dimensions of \([L]^2\).
03

Express Equation in Terms of Dimensions

Substitute the dimensional formulas into the velocity expression: \[ [L][T]^{-1} = B \frac{[M][L]^{-1}[T]^{-2}}{[M][L]^{-1}[T]^{-1}}([L]^2) \]
04

Simplify the Equation

Simplifying the right-hand side gives: \[ \frac{[M][L]^{-1}[T]^{-2}}{[M][L]^{-1}[T]^{-1}}([L]^2) = [L][T]^{-1} \] Thus, the simplified equation is: \[ [L][T]^{-1} = B [L][T]^{-1] \]
05

Solve for Dimensions of B

Since \( [L][T]^{-1} = B [L][T]^{-1] \), \( B \) must be dimensionless, meaning it has dimensions of \([1]\).

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Key Concepts

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

Fluid Viscosity
When we're diving into the world of fluid dynamics, fluid viscosity is a crucial concept. Fluid viscosity essentially describes a fluid's resistance to flow. Imagine trying to stir a pot of honey compared to a pot of water. Honey has a higher viscosity than water, which means it resists flow more strongly. In this exercise, viscosity (represented as \( \mu \)) is a central part of determining the flow characteristics in the given problem.

In mathematical terms, viscosity can be thought of as the "thickness" of a fluid. That is why higher viscosity means a thicker, more resistant fluid flow, while lower viscosity indicates a thinner flow. Viscosity is represented in dimensions as \([M][L]^{-1}[T]^{-1}\) (mass per length per time), which links how a force is needed over a certain area to move the fluid at a specific speed.

Understanding viscosity is essential for dealing with various engineering problems, as it affects how fluids are pumped, spread in channels, or interact with solid surfaces.
Pressure Drop
Pressure drop is another key factor when analyzing laminar flow and fluid mechanics in general. It denotes the difference in pressure from one point in the fluid to another. In the context of our exercise, it's the drop from the entrance to the exit of the pipe, denoted as \( \Delta p \).

A pressure drop occurs because of friction within the fluid and also between the fluid and the pipe walls. The pressure essentially 'pushes' the fluid through the pipe, so any drop indicates energy loss, mostly due to this friction. In dimensional terms, pressure is described as force per unit area, \([M][L]^{-1}[T]^{-2}\).

Calculating pressure drop is vital in systems where maintaining certain flow rates or conditions are critical, like in HVAC systems or in medical devices where specific fluid velocities are needed for function. Knowing the pressure drop allows engineers to design systems that can handle necessary flow rates without compromising efficiency or performance.
Dimensional Analysis
Dimensional analysis is a valuable tool in engineering and physics, allowing us to understand and predict relationships between different physical quantities. In this exercise, it was used to determine the dimensions of the constant \( B \) in the velocity equation.

The core idea of dimensional analysis is to represent physical quantities in terms of their basic dimensions, such as mass \([M]\), length \([L]\), and time \([T]\). This approach ensures that equations are dimensionally consistent, meaning that both sides of any equation must have the same dimensions.

In our case, starting from the equation \( u = B \frac{\Delta p}{\mu}(r_0^2 - r^2) \), we express everything in terms of dimensions and simplify step-by-step. Since the velocity \( u \) is \([L][T]^{-1}\), all parts of the equation must equate to this.

Dimensional analysis not only checks for dimensional consistency but often provides insight into the nature of physical relationships and is fundamental in scaling models, converting units, and even deriving new formulas. It's all about making sure the math aligns with the physical world.

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Most popular questions from this chapter

Algebraic equations such as Bernoulli's relation, Eq. (1) of Ex. \(1.3,\) are dimensionally consistent, but what about differential equations? Consider, for example, the boundary-layer \(x\) -momentum equation, first derived by Ludwig Prandtl in 1904 : $$\rho u \frac{\partial u}{\partial x}+\rho v \frac{\partial u}{\partial y}=-\frac{\partial p}{\partial x}+\rho g_{x}+\frac{\partial \tau}{\partial y}$$ where \(\tau\) is the boundary-layer shear stress and \(g_{x}\) is the component of gravity in the \(x\) direction. Is this equation dimensionally consistent? Can you draw a general conclusion?

Yaws et al. [34] suggest the following curve-fit formula for viscosity versus temperature of organic liquids: $$\log _{10} \mu \approx A+\frac{B}{T}+C T+D T^{2}$$ with \(T\) in absolute units. (a) Can this formula be criticized on dimensional grounds? ( \(b\) ) Disregarding \((a)\), indicate analytically how the curve-fit constants \(A, B, C, D\) could be found from \(N\) data points \(\left(\mu_{i}, T_{i}\right)\) using the method of least squares. Do not actually carry out a calculation.

A gas at \(20^{\circ} \mathrm{C}\) may be considered rarefied, deviating from the continuum concept, when it contains less than \(10^{12}\) molecules per cubic millimeter. If Avogadro's number is 6.023 E23 molecules per mole, what absolute pressure (in Pa) for air does this represent?

Do some reading and report to the class on the life and achievements, especially vis-à -vis fluid mechanics, of (a) Evangelista Torricelli \((1608-1647)\) (b) Henri de Pitot \((1695-1771)\) \((c)\) Antoine Chézy \((1718-1798)\) \((d)\) Gotthilf Heinrich Ludwig Hagen \((1797-1884)\) \((e)\) Julius Weisbach \((1806-1871)\) \((f)\) George Gabriel Stokes \((1819-1903)\) \((g)\) Moritz Weber \((1871-1951)\) (h) Theodor von Kármán ( \(1881-1963\) ) (i) Paul Richard Heinrich Blasius \((1883-1970)\) (j) Ludwig Prandtl (1875-1953) (k) Osborne Reynolds \((1842-1912)\) ( \(l\) ) John William Strutt, Lord Rayleigh \((1842-1919)\) \((m)\) Daniel Bernoulli \((1700-1782)\) \((n)\) Leonhard Euler \((1707-1783)\).

A disk of radius \(R\) rotates at an angular velocity \(\Omega\) inside a disk- shaped container filled with oil of viscosity \(\mu,\) as shown in Fig. P1.54. Assuming a linear velocity profile and neglecting shear stress on the outer disk edges, derive a formula for the viscous torque on the disk.

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