A velocity field is given by \(\mathbf{V}=x \hat{\mathbf{i}}+x(x-1)(y+1) \hat{\mathbf{j}}\) where \(u\) and \(v\) are in \(f t / s\) and \(x\) and \(y\) are in feet. Plot the streamline that passes through \(x=0\) and \(y=0 .\) Compare this streamline with the streakline through the origin.

Short Answer

Expert verified
The streamline and streakline that pass through the origin are given by \( y = e^{x^2/2 - x/2} - 1 \)

Step by step solution

01

Identify the velocity components

In this exercise, the components of the velocity vector are specified as \( u = x \) and \( v = x(x-1)(y+1) \). This implies that the velocity of the particle in the x-direction is proportional to the position \(x\), and the velocity in the y-direction is both x and y dependent.
02

Streamline Calculation

The equation of a streamline is given by \( dy/dx = v/u \). Substituting \( u \) and \( v \) into this equation gives us \( dy/dx = x(x-1)(y+1) / x \). Simplifying this equation gives \( dy/dx = (x-1)(y+1) \). This differential equation can now be solved to find the streamline pattern.
03

Solve the Differential Equation

Separate the variables to solve the differential equation obtained in Step 2. We obtain \( dy / (y + 1) = (x - 1) dx \), by integrating both sides we get \( \ln(y + 1) = x^2/2 - x /2 + C \). Taking the exponential of both sides yields the streamline equation \( y + 1 = e^{x^2/2 - x/2 + C} \), where the constant \( C \) can be determined by the condition that the streamline passes through the origin, i.e., when \( x = 0 \), \( y = 0 \). This gives \( e^{C} = 1 \), hence \( C = 0 \). With this, the streamline equation finally becomes \( y = e^{x^2/2 - x/2} - 1 \).
04

Streakline Calculation

For steady flows, the streaklines coincide with the streamlines. Therefore, the streakline will also be given by \( y = e^{x^2/2 - x/2} - 1 \).

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