An idealized velocity field is given by the formula V = 4txi - 2t^2yj + 4xzkZ

ruigE

ruigE

Answered question

2021-09-21

An idealized velocity field is given by the formula V=4tξ2t2yj+4xzk Is this flow field steady or unsteady? Is it two or three dimensional? At the point (x,y,z)=(1,1,0), compute (a) the acceleration vector.

Answer & Explanation

Nathaniel Kramer

Nathaniel Kramer

Skilled2021-09-22Added 78 answers

So,
V=4tiξ2t2yj+4xzk 
(x,y,z)=(1,1,0) 
Since the three velocity components are all nonzero and the velocity field is a function of time, t, the flow field is unstable. We have this fluid movement because
u=4tx 
v=2t2y 
w=4xz 
a) Using the equation provided by, we can get the acceleration vector. 
a=dV dt =(dU dt )i+(dv dt )j+(dw dt )k 
Where: 
ax=du dt =du dt +udu dx +vdu dy +wdu dz  
ay=dv dt =dv dt +udv dx +vdv dy +wdv dz  
az=dw dt =dw dt +udw dx +vdw dy +wdw dz  
After that, by separating the velocity components according to time, we have
du dt =d dt (4tx)+(4tx)d dx (4tx)+(2ty2y)d dy (4tx)+(4xz)d dx (4tx) 
=4x+4tx4t2t2y0+4xz0 
 

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Linear algebra

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?