Question

The velocity components u and v in a two-dimensional flow field are given by: u = 4yt ft./s, v = 4xt ft./s, where t is time. What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?

Answer 1

Answer:

vec(a) = 16 i + 16 j

mag(a) = 22.63 ft/s^2

Explanation:

Given,

- The two components of velocity are given for fluid flow:

                                       u = 4*y ft/s

                                       v = 4*x ft/s

Find:

What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?

Solution:

- The rate of change of velocity is given to be acceleration. We will take derivative of each components of velocity with respect to time t:

                                      a_x = du / dt

                                      a_x = 4*dy/dt

                                      a_y = dv/dt

                                      a_y = 4*dx/dt

- The expressions dx/dt is the velocity component u and dy/dt is the velocity component v:

                                     a_x = 4*(4*y) = 16y

                                     a_y = 4*(4*x) = 16x

- The acceleration vector can be expressed by:

                                     vec(a) = 16y i + 16x j

- Evaluate vector (a) at x = 1 and y = 1:

                                     vec(a) = 16*1 i + 16*1 j = 16 i + 16 j

- The magnitude of acceleration is given by:

                                     mag(a) = sqrt ( a^2_x + a^2_y )

                                     mag(a) = sqrt ( 16^2 + 16^2 )

                                     mag(a) = 22.63 ft/s^2