The velocity field of a flow is given by \(\mathbf{V}=2 x^{2} t \hat{\mathbf{i}}+[4 y(t-1)\) \(+2 x^{2}+1 j\) is where \(x\) and \(y\) are in meters and \(t\) is in seconds. For fluid particles on the \(x\) axis, determine the speed and direction of flow.

Short Answer

Expert verified
The speed of the fluid particles on the x-axis is given by \(\sqrt{(2x^{2}t)^{2}+(2x^{2}+1)^{2}\). The angle θ, indicating the direction of fluid flow, is given by \(tan^{-1}[(2x^{2}+1)/(2x^{2}t)]\).

Step by step solution

01

Understand the given information

The problem presents a velocity vector field \(\mathbf{V}=2 x^{2} t \hat{\mathbf{i}}+(4 y(t-1)+2x^{2}+1)\hat{\mathbf{j}}\). We are asked to find the speed and direction of fluid flow for fluid particles existing on the x-axis.
02

Simplify the velocity vector for particles on x-axis

For the particles on the x-axis, \(y = 0\). Substituting \(y = 0\) into the equation of \(\mathbf{V}\), we get \(\mathbf{V}=2x^{2} t \hat{\mathbf{i}}+(2x^{2}+1)\hat{\mathbf{j}}\).
03

Calculate the speed

The speed of the fluid particles is the magnitude of the velocity vector. This can be calculated using the formula \( Speed = \sqrt{V_{x}^{2} + V_{y}^{2}}= \sqrt{(2x^{2}t)^{2} + (2x^{2}+1)^{2}}\).
04

Determine the direction

The direction of the fluid particles can be indicated by finding the angle θ made with respect to the x-axis. Tan θ can be obtained by the ratio of the y-component to the x-component of the velocity vector. \( θ = tan^{-1}(V_{y}/V_{x}) = tan^{-1}[(2x^{2}+1)/(2x^{2}t)]\).

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Most popular questions from this chapter

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