Streakline of 2D Flow from (1,1) at t=0

[2clients]

Homework Statement

A 2D flow is given by u = v/(1+t) , v = 1

Find the equation of the streakline at t=0 formed by releasing dye from the point (1,1) at t= 0

Homework Equations

u = dx/dt = v/(1+t) = 1/(1+t) -> (eq. 1)

v = dy/dt = 1 -> (eq. 2)

The Attempt at a Solution

Integrating and solving for x, I get x = ln(1+t)+c ->(eq 1a)

Integrating and solving for y, I get y = t + c -> (eq 2a)

If x = 1 and t = Tao , (1a) becomes 1 = ln(1+tao) + c

If y = 1 and t = Tao, (2a) becomes 1 = tao + c

I'm unsure if this is correct and what the next steps are. I do understand the concept of the streakline- that it's the time history of a fluid particle, like the movement of an entire trace of smoke from a chimney. I realize it is like the pathline, but we set t=0 to be now and tao to be some time in the past for the streakline.

Thank you very much for any help.

 

[mark726]

Hi there,

Your approach looks correct so far. To find the equation of the streakline, we need to solve for the constant 'c' in equations (1a) and (2a) using the given initial conditions.

From equation (1a), we can rearrange to get c = 1 - ln(1+tao). Similarly, from equation (2a), we can rearrange to get c = 1 - tao. Since both equations must equal the same constant 'c', we can set them equal to each other and solve for tao.

1 - ln(1+tao) = 1 - tao
ln(1+tao) = tao
e^(ln(1+tao)) = e^tao
1+tao = e^tao
tao = W(1) where W is the Lambert W function

Therefore, the equation of the streakline is x = ln(1+t) + 1 - W(1), y = t + 1 - W(1).

Hope this helps! Let me know if you have any other questions.