Question 10.5: Find the velocity and acceleration in the channel flow of Exa...

From Eqs. 10.10a–10.10c, the velocity is defined by U = dX/dt, V = dY/dt, and W = dZ/dt. Taking derivatives of the particle path functions, we have

U=\frac{dX}{dt}=\frac{d}{dt}\left\{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)\right\}

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

V=\frac{dY}{dt}=\frac{d}{dt}(Y_0)=0       and  W=\frac{dZ}{dt}=\frac{d}{dt}(Z_0)=0

Thus the velocity vector is

V=\left[\frac{h²(p_1 − p_2)}{2\mu L} \right]\left[1-\left(\frac{Y_0}{h} \right)^2 \right]i+0j+0k

We note that the velocity is zero on the channel walls as expected, and has a maximum value of V = {[h²(p₁ − p₂)]/2µL}I + 0j + 0k on the channel centerline.

According to Eq. 10.13, the three Cartesian components of the acceleration vector are related to the particle path by second time derivatives:

A=\frac{d^2X}{dt^2}                              (10.13)

A_x=\frac{d^2X}{dt^2},          A_y=\frac{d^2Y}{dt^2},       and    A_z=\frac{d^2Z}{dt^2},

We could also use Eq. 10.12 and write the three components of acceleration in terms of a time derivative of velocity as

A=\frac{dV}{dt}                                     (10.12)

A_x=\frac{dU}{dt},        A_y=\frac{dV}{dt},      and      A_z=\frac{dW}{dt}

Applying the latter approach to the velocity just given, we find by inspection A = 0i + 0j + 0k. The Lagrangian acceleration is zero in this channel flow.