Question 10.1: Find the equation of, and sketch, the streamline that passes...
Find the equation of, and sketch, the streamline that passes through (1, –2) for the velocity field given by
\underline{V}=xy\hat{i}-2y^2\hat{j} m/s.
Given: Velocity vector \underline{V}.
Find: Streamline through (x = 1, y = –2).
Assume: No assumptions are necessary.
By definition, a streamline is everywhere tangent to the velocity field. So, the streamline through (1, –2) is tangent to the velocity V at this point. We state this relationship mathematically as
slope \left.\frac{dy}{dx} \right|_{streamline} =\frac{v}{u}
\frac{dy}{dx}=\frac{-2y^2}{xy}=\frac{-2y}{x} .
Separating variables,
\frac{dy}{-2y}=\frac{dx}{x} .
Integrating both sides,
\int{\frac{dy}{-2y}} = \int{ \frac{dx}{x}}
\rightarrow -\frac{1}{2}\ln y=\ln x +C .
We have absorbed the constants of integration from both sides into this new constant C:
\rightarrow \ln x + \frac{1}{2}\ln y = C .
This new C is simply –C from the previous expression. To get rid of the natural logs and find a graphable function y(x), we take the exponent of the entire expression:
x\sqrt{y} =C , or
x²y = C .
This C may no longer bear much resemblance to our initial constant C, but since the product x²y must equal a constant, we might as well use C to represent that constant.
At point (1, –2), x²y = (1)² (–2) = –2, so the equation of the streamline through (1, –2) is x²y = –2.
We plot this streamline in Figure 10.9.
