Question

5. Imagine the same system described in the previous problem, but now focus on fluid particle A located on the x-axis at x =x4 at t = 0, At some later time t, that fluid particle moved with the flow to some new location xx(figure below). The flow is symmetric about the x-axis, so the fluid particle remains on the x-axis at all times. Write the analytical expression describing thex- location of the fluid particle as a function of time in terms of its initial location x and constant:s Uo and n. Fluid particle at some later time t The velocity field is given by: v-(u, v) = (u, + nx)i-nyj Also note that the x-component of the velocity is can be described by: dx particle dt Fluid particle at time , = 0

Answer 1

we start our answer with the available information with us. we have velocity field and we need to find the position of a fluid particle as a function of time t.

velocity field has two component one is in x-direction i.e. (U_o + nx), this component changes the position of a fluid particle in the x-direction. so we use this velocity component to get the position of a particle in the x-direction.

we also know that the rate of change displacement in any direction gives velocity in that direction.

so,

in x-direction, $V_{x-direction}= \frac{dx_ {particle}}{dt}$

$U_o +nx= \frac{dx}{dt}$

integrating both sides, at t=0, x=x_A

$\int_{0}^{t} dt = \int_{x_{A}}^{x}\frac {dx}{U_o +nx}$

$t = \frac{\ln (U_o+nx) - \ln (U_o+nx_{A})}{n}$

$nt =\ln\frac{(U_o+nx)}{(U_o+nx_{A})}$

$e^{nt} =\frac{(U_o+nx)}{(U_o+nx_{A})}$

$x=\frac{e^{nt} (U_o+nx_{A}) -U_o}{n}$

Hence this is our final answer which shows the position of a fluid particle on the x-axis at x distance from initial position x_A at time t.