Engineering, 27.02.2020 17:59 angienunez2983

Consider the following steady, incompressible, two- dimensional velocity field: ecV=(u, v)=(4.35+0.656x)eci+(−1.22−0.656y)e cj. Generate an analytical expression for the flow streamlines and draw several streamlines in the upper-right quadrant from x = 0 to 5 and y= 0 to 6.

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Answer from: oliviaberta91

An analytical expression for the flow streamlines is;

y = ( $\frac{0.656x + 4.35}{0.656}$ ) $e^{0.656C}$

We are given the velocity field as;

v^ = (u, v) = (4.35 + 0.65x)i^ + (-1.22 - 0.65y)j^

Therefore, the x and y components respectively are;

u = (4.35 + 0.656x)

v = (-1.22 - 0.656y)

The streamline function is written as;

dy/dx = v/u = (-1.22 - 0.656y)/(4.35 + 0.656x)

Rearranging, we have;

dy/(-1.22 - 0.656y) = dx/(4.35 + 0.65x)

To get the streamline equation, we need to integrate both sides to get;

∫dy/(-1.22 - 0.656y) = ∫dx/(4.35 + 0.656x)

⇒ [(1/0.656)(In (0.656y + 1.22)] = [(1/0.656)(In 0.656x + 4.35)] + C

C is a constant

Multiply through by 0.656 to get;

In(0.656y + 1.22) = In(0.656x + 4.35) + 0.656C

Rearranging, we have;

In(0.656y + 1.22) - In(0.656x + 4.35) = 0.656C

using property of logarithms, we have;

In [(0.656y + 1.22)/(0.656x + 4.35)] = 0.656C

⇒ (0.656y + 1.22)/(0.656x + 4.35) = $e^{0.656C}$

⇒ (0.656y + 1.22) = (0.656x + 4.35) × $e^{0.656C}$

⇒ 0.656y = (0.656x + 4.35) $e^{0.656C}$

y = ( $\frac{0.656x + 4.35}{0.656}$ ) $e^{0.656C}$

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Consider the following steady, incompressible, two- dimensional velocity field: ecV=(u, v)=(4.35+0.6...

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