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Chapter 4: Problem 10

(a) A fluid is said to be incompressible if its density \(\rho\) is a constant (that is, is independent of \(x, y, z\), and \(t)\). Use the continuity equation to show that the velocity \(\mathbf{v}\) of an incompressible fluid satisfies the equation \(\nabla \cdot \mathbf{v}=0\). (b) If \(\nabla \times \mathbf{v}=0\), the fluid flow is said to be irrotational. Show that for an incompressible fluid undergoing irrotational flow, $$ \nabla^{2} \phi=0 $$ where \(\phi\), a scalar function called the velocity potential, is so defined that \(\mathbf{v}=\nabla \phi\).

Short Answer

Expert verified
For an incompressible fluid, \( \nabla \cdot \mathbf{v} = 0 \). If it's irrotational, \( \nabla^2 \phi = 0 \).

Step by step solution

01

Understand the Continuity Equation

The continuity equation for fluid dynamics states that the rate of change of density \( \rho \) in a fluid plus the divergence of the product of velocity \( \mathbf{v} \) and density is zero: \[ \frac{\partial \rho}{\partial t} + abla \cdot (\rho \mathbf{v}) = 0. \] For an incompressible fluid, \( \rho \) is constant, therefore \( \frac{\partial \rho}{\partial t} = 0 \). This simplifies the equation to \( abla \cdot \mathbf{v} = 0 \).
02

Definition of Incompressible Fluid Condition

For a fluid to be incompressible, its density \( \rho \) must be constant, indicating that any variation or change in volume will not alter the density. By substituting \( \frac{\partial \rho}{\partial t} = 0 \) into the continuity equation, since \( \rho \) is not dependent on time or space, this directly leads to the condition \( abla \cdot \mathbf{v} = 0 \) as the velocity field must adapt to maintain constant density.
03

Understanding Irrotational Flow

Irrotational flow implies that the curl of the velocity field \( \mathbf{v} \) is zero, denoted as \( abla \times \mathbf{v} = 0 \). This suggests that the velocity field can be expressed as the gradient of a scalar function \( \phi \), or \( \mathbf{v} = abla \phi \).
04

Derive the Laplace's Equation for Velocity Potential

Given \( \mathbf{v} = abla \phi \) and knowing that the fluid flow is incompressible, \( abla \cdot \mathbf{v} = 0 \), we substitute to get \( abla \cdot abla \phi = 0 \). This simplifies to \( abla^2 \phi = 0 \), which is known as Laplace's equation, indicating that the velocity potential \( \phi \) is a harmonic function.

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Key Concepts

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

Incompressible Fluid
In fluid dynamics, the term **incompressible fluid** refers to a fluid whose density \( \rho \) is constant under the assumption that it doesn鈥檛 change over time or across different points in space. This condition simplifies many equations in fluid dynamics, as it implies that the fluid's volume does not vary whether it's moving or at rest.
Understanding incompressible fluids is crucial for analyzing many real-world scenarios, such as water in pipes or air under typical atmospheric conditions.
  • Key Point: When a fluid is incompressible, its density \( \rho \) remains constant.
  • No matter how the fluid moves, the density or compactness of particles within it does not change.
This characteristic leads to the concept of the **continuity equation**, which simplifies greatly under incompressibility, helping us comprehend how velocity fields act simply because the equation \( abla \cdot \mathbf{v} = 0 \) must hold true.
Continuity Equation
The **continuity equation** in fluid mechanics ensures the conservation of mass within a fluid flow. It accounts for how fluid flows in and out of a given section, maintaining balance.
This balance is like tracking money in a bank account鈥攚here the inflow equals the outflow to keep a stable balance. For fluid flows, this translates to:\[\frac{\partial \rho}{\partial t} + abla \cdot (\rho \mathbf{v}) = 0.\]
  • The first term represents changes in fluid density \( \rho \) over time.
  • The second term, \( abla \cdot (\rho \mathbf{v}) \), captures how density changes with the flow's divergence.
For an **incompressible fluid**, where the density \( \rho \) is constant, the time derivative is zero (\( \frac{\partial \rho}{\partial t} = 0 \)). This simplifies the equation to:\[abla \cdot \mathbf{v} = 0.\]This result tells us that the fluid, despite moving, maintains its mass consistency. Fluid flows adapt such that there is no net change in fluid volume within a flow region.
Irrotational Flow
In fluid dynamics, **irrotational flow** describes a scenario where the fluid has no rotation at any point. This means the fluid elements do not spin around their own axes. Mathematically, this is expressed by having the curl of the velocity \( \mathbf{v} \) be zero:\[abla \times \mathbf{v} = 0.\]
  • Implication: There is no spinning motion or vorticity within the fluid flow.
  • Mathematically, such a flow can be represented by a gradient of a potential function: \( \mathbf{v} = abla \phi \).
The velocity potential \( \phi \) is a scalar field that uniquely defines the irrotational flow. For **incompressible and irrotational flows**, the function \( \phi \) satisfies the **Laplace鈥檚 equation**:\[abla^2 \phi = 0.\]This equation is central to understanding such flows, as it defines a situation where the potential is "harmonic," keeping the flow efficient and continuous without internal whirls. This understanding aids in solving many practical fluid problems by simplifying the equations we have to deal with.

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Most popular questions from this chapter

(a) Starting with the divergence theorem, derive the equation $$ \iint_{s} \hat{\mathbf{n}} \cdot(u \nabla v) d S=\iiint_{V}\left[u \nabla^{2} v+(\nabla u) \cdot(\nabla v)\right] d V $$ where \(u\) and \(v\) are scalar functions of position and \(S\) is a closed surface enclosing the volume \(V\). This is sometimes called the first form of Green's theorem. (b) If \(\nabla^{2} u=0\) use the first form of Green's theorem to show that $$ \iint_{S} \hat{\mathbf{n}} \cdot(u \nabla u) d S=\iiint_{V}|\nabla u|^{2} d V $$ where \(|\nabla u|^{2}=(\nabla u) \cdot(\nabla u)\). (c) Use the first form of Green's theorem to show that $$ \iint_{s} \hat{\mathbf{n}} \cdot(u \nabla v-v \nabla u) d S=\iiint_{V}\left(u \nabla^{2} v-v \nabla^{2} u\right) d V $$ This is the second form of Green's theorem.

Here is a "proof" that there is no such thing as magnetism. One of Maxwell's equations tells us that $$ \nabla \cdot \mathbf{B}=0 $$ where \(\mathbf{B}\) is any magnetic field. Then using the divergence theorem, we find $$ \iint_{S} \mathbf{B} \cdot \hat{\mathbf{n}} d S=\iiint_{V} \nabla \cdot \mathbf{B} d V=0 $$ Because B has zero divergence, we know (see Problem III-24) there exists a vector function, call it \(\mathbf{A}\), such that $$ \mathbf{B}=\nabla \times \mathbf{A} $$ Combining these last two equations, we get $$ \iint_{s} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Next we apply Stokes' theorem and the preceding result to find $$ \oint_{C} \mathbf{A} \cdot \hat{\mathbf{t}} d s=\iint_{S} \hat{\mathbf{n}} \cdot \nabla \times \mathbf{A} d S=0 $$ Thus we have shown that the circulation of \(\mathbf{A}\) is path independent. It follows that we can write \(\mathbf{A}=\nabla \psi\), where \(\psi\) is some scalar function. Since the curl of the gradient of a function is zero, we arrive at the remarkable fact that $$ \mathbf{B}=\nabla \times \nabla \psi=0 $$ that is, all magnetic fields are zero! Where did we go wrong? [Taken from G. Arfken, Amer. J. Phys., 27, 526 (1959).]

(a) Find the charge density \(\rho(x, y, z)\) that produces the electric field $$ \mathbf{E}=g(\mathbf{i} x+\mathbf{j} \mathbf{y}+\mathbf{k} z) $$ where \(g\) is a constant. (b) Find an electrostatic potential \(\Phi\) such that \(-\nabla \Phi\) is the field \(\mathbf{E}\) given in (a). (c) Verify that \(\nabla^{2} \Phi=-\rho / \epsilon_{0}\)

Suppose you find a solution of Laplace's equation that satisfies certain boundary conditions. Is this solution unique or are there others? This problem will answer that question in certain simple cases. Consider the region of space completely enclosed by a surface \(S_{0}\) and containing in its interior objects \(1,2,3, \ldots\) (two of which are pictured in the diagram). Suppose that \(S_{0}\) is maintained at a constant potential \(\Phi_{0}\), object no. 1 at \(\Phi_{1}\), object no. 2 at \(\Phi_{2}\), and so on. Then in the charge-free region \(R\) enclosed by \(S_{0}\) and between the objects, the potential must satisfy Laplace's equation $$ \nabla^{2} \Phi=0 $$ and the boundary conditions $$ \Phi=\left\\{\begin{array}{c} \Phi_{0} \text { on } S_{0} \\ \Phi_{1} \text { on } S_{1} \\ \Phi_{2} \text { on } S_{2} \\ \vdots \end{array}\right. $$ The following steps will guide you through a proof that \(\Phi\) is unique. (a) Assume that there are two potentials \(u\) and \(v\), both of which satisfy Laplace's equation and the boundary conditions listed earlier. Form their difference \(w=u-v\). Show that \(\nabla^{2} w=0\) in \(R\). (b) What are the boundary conditions satisfied by \(w\) ? (c) Apply the divergence theorem to $$ \iint_{S} \hat{\mathbf{n}} \cdot(w \nabla w) d S $$ where the integration is carried out over the surface \(S_{0}+S_{1}+S_{2}\) \(+\cdots\), and show thereby that $$ \iiint_{V}|\nabla w|^{2} d V=0 $$ where \(V\) is the volume of the region \(R\). (d) From the result of (c) argue that \(\nabla w=0\) and that this, in turn, means \(w\) is a constant. (e) If \(w\) is a constant, what is its value? (Use the boundary conditions on \(w\) to answer this.) What does this say about \(u\) and \(v ?\) (f) The uniqueness proof outlined in (a) to (e) involves specifying the value of the potential on various surfaces. Might we have specified a different kind of boundary condition and still proved uniqueness? If so, in what way or ways would the proof and the result differ from those given above?

(a) Calculate \(\mathbf{F}=\nabla f\) for each of the following scalar functions: (i) \(f=x y z\). (ii) \(f=x^{2}+y^{2}+z^{2}\). (iii) \(f=x y+y z+x z\). (iv) \(f=3 x^{2}-4 z^{2}\). (v) \(f=e^{-x} \sin y\). (b) Verify that $$ \oint_{C} \mathbf{F} \cdot \hat{\mathbf{t}} d s=0 $$ for one or more of the functions \(\mathbf{F}\) determined in part (a) choosing for the curve \(C\) : (i) the square in the \(x y\)-plane with vertices at \((0,0),(1,0)\), \((1,1)\), and \((0,1)\). (ii) the triangle in the \(y z\)-plane with vertices at \((0,0),(1,0)\), and \((0,1)\). (iii) the circle of unit radius centered at the origin and lying in the \(x z\)-plane. (c) Verify by direct calculation that \(\nabla \times \mathbf{F}=0\) for one or more of the functions \(\mathbf{F}\) determined in part (a).
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