Answer :
The equation in the form at point (3m, 2m) is [tex]$\ln y^2+y=2 x-2.62$[/tex], the magnitude of the velocity is 7.81 m/s, the direction of the velocity is 50.19°.
Consider the flow field:
u=2+y & v=2 y
Determine the streamline equation by using the following equation:
[tex]& \frac{d x}{u}=\frac{d y}{v} \\ \frac{d x}{2+y}=\frac{d y}{2 y} \\\\& \int d x=\int\left(\frac{2+y}{2 y}\right) d y \\\\& x=\ln y+\frac{y}{2}+C[/tex]
(A)
Since the stream line passes through (3,2).
[tex]$\begin{aligned}& x=\ln y+\frac{y}{2}+C \\& 3=\ln 2+\frac{2}{2}+C \\& C=2-\ln 2 \\& C=1.31\end{aligned}$[/tex]
Substitute the value of C in equation (1).
[tex]& x=\ln y+\frac{y}{2}+C \\\\& x=\ln y+\frac{y}{2}+1.31 \\\\& 2 x-1.31 \times 2=2 \ln y+y \\\\& \ln y^2+y=2 x-2.62[/tex]
Therefore, the equation of streamline equation at point (3m, 2m) is [tex]$\ln y^2+y=2 x-2.62$[/tex].
(B).
Calculate the magnitude of the velocity of a particle located at point (5m , 3m).
[tex]V & =\sqrt{u^2+v^2} \\\\& =\sqrt{(2+y)^2+(2 y)^2} \\\\& =\sqrt{(2+3)^2+(2 \times 3)^2} \\\\& =\sqrt{5^2+6^2} \\\\& =7.81 \mathrm{~m} / \mathrm{s}[/tex]
Therefore, the magnitude of the velocity of a particle located at point (5m , 3m) is 7.81m/s.
(C).
Calculate the direction of the velocity of a particle located at point (5m , 3m) measured counterclockwise from the positive x-axis is
[tex]\tan \theta & =\frac{v}{u} \\\\& =\frac{2 y}{2+y} \\\\& =\frac{6}{5} \\\\& =50.19^{\circ}[/tex]
Therefore, the direction of the velocity of a particle located at point (5m , 3m) is 50.19°.
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A flow field for a fluid is described by u=(2+y) m/s and v=(2y) m/s, where y is in meters.
A.) Determine the equation of the streamline that passes through point (3m, 2m). Write the equation in the form
x={x(y)}, where y is in meters.
B.) Find the magnitude of the velocity of a particle located at point (5m , 3m).
C.) Find the direction ? of the velocity of a particle located at point (5m , 3m) measured counterclockwise from the positive x axis.
