Question 4.SP.3: A two-dimensional flow field is given by u=2 y, v=x (a) Sk......
A two-dimensional flow field is given by u=2 y, v=x (a) Sketch the flow field. (b) Derive a general expression for the velocity and acceleration (x and y are in units of length L ; u and v are in units of L / T) (c) Find the acceleration in the flow field at point A(x=3.5, y=1.2).
(a) Velocity components u and v are plotted to scale, and streamlines are sketched tangentially to the resultant velocity vectors. This yields the following general picture of the flow field:
(b)
V=\left(u^{2}+v^{2}\right)^{1 / 2}=\left(4 y^{2}+x^{2}\right)^{1 / 2} \quad \text { ANS }Eq. (4.23 a) :
a_{x}=u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}=2 y(0)+x(2)=2 xEq. (4.23b)
a_{y}=u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=2 y(1)-x(0)=2 y \begin{aligned}& a=\left(a_{x}^{2}+a_{y}^{2}\right)^{1 / 2}=\left(4 x^{2}+4 y^{2}\right)^{1 / 2}=2\left(x^{2}+y^{2}\right)^{1 / 2} \text { ANS }\end{aligned}(c) \operatorname{At} A(3.5,1.2),
\left(a_{A}\right)_{x}=2 x=2(3.5)=7.00 \mathrm{~L} / T^{2} ; \quad\left(a_{A}\right)_{y}=2 y=2(1.2)=2.40 \mathrm{~L} / \mathrm{T}^{2}and \quad a_{A}=\left[\left(a_{A}\right)_{x}^{2}+\left(a_{A}\right)_{y}^{2}\right]^{1 / 2}=\left[(7.00)^{2}+(2.40)^{2}\right]^{1 / 2}=7.40 \mathrm{~L} / T^{2} \quad ANS
Rough check. Imagine a velocity vector at point A. This vector would have a magnitude approximately midway between that of the adjoining vectors, or V_{A} \approx 4 L / T. The radius of curvature of the sketched streamline at A is roughly 3 L. Thus \left(a_{A}\right)_{n} \approx 4^{2} / 3 \approx 5.3 L / T^{2}. The tangential acceleration of the particle at A may be approximated by noting that the velocity along the streamline increases from about 3.2 L / T, where it crosses the x axis, to about 7.8 L / T at B. The distance along the streamline between these two points is roughly 4 L. Hence a very approximate value of the tangential acceleration at A is
\left(a_{A}\right)_{t}=V \frac{\partial V}{\partial s} \approx 4\left(\frac{7.8-3.2}{4}\right) \approx 4.6 L / T^{2}Vector diagrams of these roughly-computed normal and tangential acceleration components plotted below allow us to compare them with the true acceleration as given by the analytic expressions. In Chap. 14 we shall prove that the flow in this example is that of an incompressible fluid.
