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Chapter 6: Problem 100

The velocity field for a two-dimensional flow is \(\vec{V}=(A x-B y) t \hat{\imath}-(B x+A y) t \hat{j},\) where \(A=1 \mathrm{s}^{-2} B=2 \mathrm{s}^{-2}\) \(t\) is in seconds, and the coordinates are measured in meters. Is this a possible incompressible flow? Is the flow steady or unsteady? Show that the flow is irrotational and derive an expression for the velocity potential.

Short Answer

Expert verified
The flow is compressible and unsteady. It is irrotational with the velocity potential given by \(\phi = (A x^{2} - B y^{2}) t\).

Step by step solution

01

Determine Incompressibility

To determine if the flow is incompressible, use the condition of divergence of velocity vector field, \(\nabla . \vec{V} = 0 \). The divergence of \(\vec{V} = (A x-B y) t \hat{i} -(B x+A y) t \hat{j}\) can be calculated as, \(\nabla . \vec{V} = \frac{\partial V_{x}}{\partial x} + \frac{\partial V_{y}}{\partial y} = At - Bt = t(A - B) \). For incompressible flow, this divergence should be zero. However, in this case, it is not zero, because t is never zero. Therefore, this is not possible for incompressible flow.
02

Evaluate steadiness/unsteadiness

A flow is steady if the velocity field does not change with time. Thus, if \(V_{x}\) and \(V_{y}\) are functions of time, the flow would be unsteady. In this case, both components of the velocity are dependent on time, \(t\). Hence, the flow is unsteady.
03

Evaluate if flow is irrotational

To check if the flow is irrotational, calculate the curl of the velocity vector field, \(\nabla \times \vec{V} = 0\). The curl of \(\vec{V}=(A x-B y) t \hat{i} -(B x+A y) t \hat{j}\) is zero. Hence, the flow is irrotational.
04

Derive expression for velocity potential

Potential function, \(\phi\), is defined such that \(\vec{V} = \nabla \phi\). By integrating each component with respect to their respective variables and setting them equal to each other, we obtain the function \(\phi\). So, from \(V_{x} = \frac{\partial \phi}{\partial x}\) and \(V_{y} = \frac{\partial \phi}{\partial y}\), integrating and equating we get: \(\phi = (A x^{2} - B y^{2}) t = \phi\). So, \(\phi = (A x^{2} - B y^{2}) t \) is the velocity potential.

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Most popular questions from this chapter

A mercury barometer is carried in a car on a day when there is no wind. The temperature is \(20^{\circ} \mathrm{C}\) and the corrected barometer height is \(761 \mathrm{mm}\) of mercury. One window is open slightly as the car travels at \(105 \mathrm{km} / \mathrm{hr}\). The barometer reading in the moving car is \(5 \mathrm{mm}\) lower than when the car is stationary. Explain what is happening. Calculate the local speed of the air flowing past the window, relative to the automobile.

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