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The stream function, \(\psi\), for a particular flow field is given by $$ \psi=4 x+3 y $$ (a) Describe the velocity field. (b) If possible, find the potential function for the velocity field. (c) Determine the velocity at \((x, y)=(1,2)\).

Short Answer

Expert verified
(a) Velocity field: \( u = 3, v = -4 \). (b) Potential function: \( \phi = 3x - 4y + C \). (c) Velocity at (1,2): \( \mathbf{V} = (3, -4) \).

Step by step solution

01

Understanding the Stream Function

The stream function \( \psi = 4x + 3y \) can be used to determine the velocity field in two-dimensional incompressible flow. From the stream function, we can obtain the components of the velocity field: \( u \) and \( v \), where \( u \) is the velocity component in the \( x \)-direction, and \( v \) is the velocity component in the \( y \)-direction.
02

Derive Velocity Components

The velocity components from the stream function are given by: \( u = \frac{\partial \psi}{\partial y} \) and \( v = -\frac{\partial \psi}{\partial x} \). Calculate these derivatives: \( u = \frac{\partial}{\partial y}(4x + 3y) = 3 \) and \( v = -\frac{\partial}{\partial x}(4x + 3y) = -4 \). Thus, the velocity field is described as \( u = 3 \) and \( v = -4 \).
03

Check Conditions for Potential Function

A velocity field has a potential function if it is irrotational. This means the curl of the velocity field must be zero. The curl \( abla \times \mathbf{V} \) in two dimensions is \( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \). Since both \( u \) and \( v \) are constants, \( \frac{\partial v}{\partial x} = 0 \) and \( \frac{\partial u}{\partial y} = 0 \), making the curl zero. This flow is irrotational and has a potential function.
04

Derive the Potential Function

The potential function \( \phi \) satisfies \( \mathbf{V} = abla \phi \), so \( u = \frac{\partial \phi}{\partial x} \) and \( v = \frac{\partial \phi}{\partial y} \). Integrate \( u = 3 \) with respect to \( x \) to get \( \phi = 3x + g(y) \). Integrate \( v = -4 \) with respect to \( y \) to get \( \phi = -4y + h(x) \). Combining these, \( \phi(x, y) = 3x - 4y + C \), where \( C \) is a constant.
05

Determine Velocity at Specific Point

Use the velocity components obtained: \( u = 3 \) and \( v = -4 \). Since these are constants, the velocity does not change with position. At the point \((x, y) = (1, 2)\), the velocity is still \( u = 3 \) and \( v = -4 \). In vector form, \( \mathbf{V} = (3, -4) \).

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Key Concepts

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

Velocity Field
In fluid dynamics, the velocity field is a crucial concept as it describes how fluid velocities are distributed in space. This field helps us visualize and analyze the nature of fluid flow at every point in a fluid domain. The components of a velocity field in two-dimensional flow are typically denoted as \( u \) and \( v \).

The stream function \( \psi \) is a useful tool to obtain these components. Specifically, \( u \) is the velocity in the \( x \)-direction, defined as \( u = \frac{\partial \psi}{\partial y} \), and \( v \) is in the \( y \)-direction, defined as \( v = -\frac{\partial \psi}{\partial x} \).

In our case, for \( \psi = 4x + 3y \), the derivatives give us \( u = 3 \) and \( v = -4 \). This means the flow is uniform in terms of direction and magnitude, with the entire velocity field described by these constant components.
Potential Function
The potential function, \( \phi \), is another important concept in the context of fluid flow, especially for irrotational flows. A velocity field that possesses a potential function can be derived from the gradient of the potential function, \( \mathbf{V} = abla \phi \).

For our uniform velocity field, we found that it is irrotational, meaning it satisfies the condition that \( abla \times \mathbf{V} = 0 \). Thus, it has a potential function.

To find \( \phi \), we integrate the velocity components. For \( u = 3 \), integrating with respect to \( x \) gives \( \phi = 3x + g(y) \). Similarly, for \( v = -4 \), integrate with respect to \( y \) to obtain \( \phi = -4y + h(x) \). Combining these results gives the potential function \( \phi(x, y) = 3x - 4y + C \), where \( C \) is an arbitrary constant.
Irrotational Flow
An irrotational flow is characterized by having no rotation or swirling motion within the fluid. For a flow to be irrotational, the curl of the velocity field must be zero, \( abla \times \mathbf{V} = 0 \). This implies that the fluid parcels do not experience any angular momentum.

In our example, the velocity components derived from the stream function are constants: \( u = 3 \) and \( v = -4 \). Calculating the curl as \( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \) results in zero since both partial derivatives are zero. This confirms that the flow is irrotational, meaning it does not change its rotation over space and time. Irrotational flows often allow for a simpler analysis and solve using potential functions.
Two-Dimensional Incompressible Flow
In two-dimensional incompressible flow, the fluid density remains constant, and the flow is constrained to two dimensions. This simplifies the mathematical analysis, reducing the complexity of fluid dynamic equations.

Incompressibility implies that the divergence of the velocity field is zero, \( abla \cdot \mathbf{V} = 0 \), meaning there is no net inflow or outflow of fluid volume in a given space. For incompressible flow, we often use the stream function \( \psi \), which automatically satisfies this condition.
  • The continuity equation holds, reinforcing conservation of mass.
  • The streamlines, which are lines tangent to the velocity field, become simple to derive from \( \psi \).


Understanding these simplifications in stream function formulation is pivotal for analyzing and predicting fluid behavior effectively in two-dimensional settings.

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