Question 10.3: Find the inverse particle paths for the channel flow of Examp...

The process begins by inspecting the particle paths. As given earlier, these are

X=X_0+\left[\frac{h²(p_1 − p_2)}{2\mu L} \right]\left[1-\left(\frac{Y_0}{h} \right)^2 \right](t-t_0),     Y = Y₀,         and          Z = Z₀

We immediately note that two of the inverse functions are available by inspection as

Y₀ = Y                and          Z₀ = Z

Isolating the X₀ variable in the remaining function, we have

X_0=X-\left[\frac{h²(p_1 − p_2)}{2\mu L} \right]\left[1-\left(\frac{Y_0}{h} \right)^2 \right](t-t_0)

We are not finished, however, since the identity variable Y₀ occurs in the right-hand side. We must replace this identity variable, since the inverse function is only allowed to be a function of (X, Y, Z, t). In general, the final step in developing an inverse function is to get rid of any identity variables, and this can be difficult if highly complex functions are involved. We do it here by substituting for Y₀ using the previously determined inverse Y₀ = Y. The final component of the inverse is then

X_0=X-\left[\frac{h²(p_1 − p_2)}{2\mu L} \right]\left[1-\left(\frac{Y}{h} \right)^2 \right](t-t_0)

To find the initial position of the indicated particle, we substitute the current position X = (2h)i + (0)j + 0k and current time t = t₀ +4µL/[h(p₁ − p₂)] into the inverse functions to obtain

X_0=2h-\left[\frac{h²(p_1 − p_2)}{2\mu L} \right]\left[1-\left(\frac{0}{h} \right)^2 \right]\left[\frac{4 \mu L}{h(p_1-p_2)} \right]=2h-2h=0

Y₀ = Y = 0       and        Z₀ = Z = 0

The initial position of this particle is X₀ = 0i + 0j + 0k as stated.