Question 10.2: Consider the laminar flow through a channel driven by a press...
Consider the laminar flow through a channel driven by a pressure gradient as illustrated in Figure 10.5A. The particle paths for this flow are given by
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
where 2h is the channel height, and the pressures are measured a distance L apart as shown. Consider the fluid particle initially located at X0 = x1, Y0 = 0, Z0 = 0 at time t0 as shown in Figure 10.5B. Where will this particle be at time t1 = t0 + 4µL/ [h(p1 − p2)]? Where will the particle initially located at X0 = x1, Y0 = h/2, Z0 = 0 be located at time t1?
Examining the particle paths, we see that the Y and Z coordinates of a fluid particle cannot change, but the X coordinate changes in time and depends on the initial coordinate Y0. We are asked first for the future position of a fluid particle that is initially located on the channel centerline at X0 = x1, Y0 = 0, Z0 = 0. To determine the future position, we enter the specified initial values X0 = x1, Y0 = 0, Z0 = 0 and desired time t1 = t0 + 4µL/[h(p1 − p2)] into the three components of the particle path to obtain
X=(x_1)+\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]
=(x_1)+\left[\frac{h²(p_1 − p_2)}{2\mu L} \right](1)\left[\frac{4 \mu L}{h(p_1-p_2)} \right]=x_1+2h
Y = 0 and Z = 0
Thus, the location of this fluid particle at time t1 is given by the vector
X = (x1 + 2h)i + 0j + 0k
We see that the particle has moved the equivalent of one channel height to the right in this amount of time as shown in Figure 10.5B. Note that since this flow is described by Y = Y0 and Z = Z0, the values of these two coordinates never change for any fluid particle.
The location of the second particle at time t1 is found using the same procedure to
be
X = \left(x_1+\frac{3}{2}h\right)i + \frac{h}{2}j+ok
We see that as expected, this particle remains at its initial elevations Y0 = h/2, Z0 = 0. It has moved in the X direction, but not as far as the first particle. Why is that? From the velocity profile in Figure 10.5B we see that the velocity at the initial location of the second particle is less than that at the initial location of the first. If we check for the future location of fluid particles located initially on either wall (Y0 = +h or −h), we find that these particles do not move at all. This is consistent with the no-slip, no-penetration boundary conditions on the channel walls.