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An incompressible and irrotational flow field has a potential function, \(\phi\), that can be described by the relation $$ \phi=y^{6}-15 x^{2} y^{4}+15 x^{4} y^{2}-x^{6} $$ Determine an expression for the stream function that can be used in plotting the streamlines.

Short Answer

Expert verified
The stream function is \(\psi = -6xy^5 + 20x^3y^3 - 6x^5y\).

Step by step solution

01

Understanding Irrotational Flow and Potential Function

In an irrotational flow field, the velocity components can be derived from the gradient of a potential function \(\phi\). Here, \(\phi\) is given by the relation \(\phi=y^{6}-15 x^{2} y^{4}+15 x^{4} y^{2}-x^{6}\).
02

Derive Velocity Components

We obtain the velocity components from the potential function by finding the partial derivatives: \( u = \frac{\partial \phi}{\partial x} \) and \( v = \frac{\partial \phi}{\partial y} \).- Compute \( u = \frac{\partial }{\partial x} (y^{6} - 15x^{2}y^{4} + 15x^{4}y^{2} - x^{6}) = -30xy^{4} + 60x^{3}y^{2} - 6x^{5} \).- Compute \( v = \frac{\partial }{\partial y} (y^{6} - 15x^{2}y^{4} + 15x^{4}y^{2} - x^{6}) = 6y^{5} - 60x^{2}y^{3} + 30x^{4}y \).
03

Understand Stream Function Components

In an incompressible and irrotational flow, the stream function \(\psi\) is related to the velocity components as follows: \( u = \frac{\partial \psi}{\partial y} \) and \( v = -\frac{\partial \psi}{\partial x} \).
04

Integrate to Find Stream Function

To find the stream function \(\psi\), solve the following partial differential equations: - From \( u = \frac{\partial \psi}{\partial y} \): \(\psi = \int (-30xy^{4} + 60x^{3}y^{2} - 6x^{5}) \, dy \). - From \( v = -\frac{\partial \psi}{\partial x} \): \(\psi = -\int (6y^{5} - 60x^{2}y^{3} + 30x^{4}y) \, dx \).
05

Solve for \(\psi\) from Both Integrals

- Solving \(\int (-30xy^{4} + 60x^{3}y^{2} - 6x^{5}) \, dy = -6xy^{5} + 20x^{3}y^{3} - 6x^{5}y + f(x) \). - Solving \(-\int (6y^{5} - 60x^{2}y^{3} + 30x^{4}y) \, dx = -6xy^{5} + 20x^{3}y^{3} - 6x^{5}y + g(y) \). - Consistency: \( f(x) \) and \( g(y) \) are constants for specific integration choice or can be merged as zero because of the boundary conditions.
06

Final Expression for Stream Function

The consistent form of the stream function \(\psi\) based on both integrations is:\[ \psi = -6xy^{5} + 20x^{3}y^{3} - 6x^{5}y \]

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Key Concepts

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

Potential Function
In fluid mechanics, a potential function, often denoted by \( \phi \), represents a scalar field whose gradient matches the flow velocity of an irrotational flow. This concept is particularly useful when dealing with incompressible and irrotational fluids. In the given exercise, the potential function is \( \phi = y^{6} - 15x^{2}y^{4} + 15x^{4}y^{2} - x^{6} \). This function essentially outlines the energy per unit mass at each point in the field.

For an irrotational flow, the condition \( abla \times \mathbf{V} = 0 \) holds, meaning the curl of the velocity field is zero. Consequently, we can express the velocity field as the gradient of our potential function:
  • \( u = \frac{\partial \phi}{\partial x} \)
  • \( v = \frac{\partial \phi}{\partial y} \)
By differentiating the given \( \phi \), we obtain the velocity components as explained further below, showcasing how the flow behaves according to energy variations in the system.
Stream Function
For incompressible and irrotational flows, aside from potential functions, we also use stream functions, denoted as \( \psi \). The stream function is another scalar field from which the velocity components can be derived. Its lines represent streamlines that illustrate the flow pattern of the fluid.

The beauty of stream functions in two-dimensional flows is they inherently satisfy the condition for incompressibility. They are related to the velocity components as follows:
  • \( u = \frac{\partial \psi}{\partial y} \)
  • \( v = -\frac{\partial \psi}{\partial x} \)
The task is to integrate these relationships to find \( \psi \). Starting with the expressions for \( u \) and \( v \) derived from the potential function, we perform integrations:
- For \( u \), integrate \( \psi = \int (-30xy^{4} + 60x^{3}y^{2} - 6x^{5}) \, dy \).
- For \( v \), consider \( \psi = -\int (6y^{5} - 60x^{2}y^{3} + 30x^{4}y) \, dx \).
The solutions must be consistent, leading to a stream function \( \psi = -6xy^{5} + 20x^{3}y^{3} - 6x^{5}y \). This expression describes the flow's streamline structure, where every constant value of \( \psi \) represents a specific streamline.
Velocity Components
Velocity components in a flow field indicate the rate and direction of flow at any point. In a two-dimensional irrotational flow, these components can be directly derived from the potential function, \( \phi \). The components are defined as part of differentiating the potential function:

  • \( u = \frac{\partial \phi}{\partial x} = -30xy^{4} + 60x^{3}y^{2} - 6x^{5} \)
  • \( v = \frac{\partial \phi}{\partial y} = 6y^{5} - 60x^{2}y^{3} + 30x^{4}y \)
These components are essential to understand how the flow velocity behaves at any given point in the field. The component \( u \) encodes horizontal movement, while \( v \) captures vertical movement of fluid elements.

By using these expressions, we uncover the complete velocity field of the flow. Such detail aids in analyzing flow characteristics, unraveling kinetic patterns based on space variation, which is critical for plotting streamlines and understanding fluid behavior.

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Most popular questions from this chapter

Two-dimensional unsteady flow of an incompressible fluid occurs in the \(x y\) plane. When polar coordinates are used, the \(\theta\) -component of the velocity is given by $$ v_{\theta}=-\frac{C \sin \theta}{r^{2}} $$ where \(C\) is a constant. Determine the required functional form of the \(r\) -component of the velocity.

Measurements in a flow field indicate that the velocity components are \(u=6 \mathrm{~m} / \mathrm{s}\) and \(v=2 \mathrm{~m} / \mathrm{s}\) at a location where \(x=3 \mathrm{~m}\) and \(y=4 \mathrm{~m}\). Express the given location in polar coordinates \((r, \theta)\) and determine the \(r\) and \(\theta\) components of the velocity, which are commonly represented by \(v_{r}\) and \(v_{\theta}\).

The two-dimensional velocity field of an incompressible fluid is given by \(u=3 y(4 x+\) 1), \(v=-6 y^{2}+x^{2}\). Determine the analytic expression for the stream function and describe how you would use this stream function to plot the streamlines of the flow.

SAE 30 oil at \(20^{\circ} \mathrm{C}\) flows between two horizontal \(0.50-\mathrm{m}\) -wide parallel plates separated by \(30 \mathrm{~mm}\). The length of the top and bottom plates in the direction of flow is \(2 \mathrm{~m}\), the bottom plate is stationary, the top plate moves at \(3 \mathrm{~cm} / \mathrm{s}\), an adverse pressure gradient of \(700 \mathrm{~Pa} / \mathrm{m}\) is applied between the plates, and the flow is laminar between the plates. Determine the flow rate between the plates and the force that must be applied to move the top plate.

A flow field closely approximates that of a free line vortex with a vortex strength \(\Gamma\). The pressure far from the vortex center is \(p_{\infty}\), and the density of the fluid is \(\rho\). (a) Determine the pressure at any distance \(r\) from the vortex center in terms of \(r, p_{\infty}\), \(\rho\), and \(\Gamma\). (b) If the fluid is air with a density of \(1.225 \mathrm{~kg} / \mathrm{m}^{3}\), the pressure far from the influence of the vortex is \(101 \mathrm{kPa}\), and the strength of the vortex is \(3000 \mathrm{~m}^{2} / \mathrm{s}\), determine the radial distance from the vortex center to where the pressure is \(10 \%\) less than the undisturbed pressure of \(101 \mathrm{kPa}\).

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