Question 10.6: Laminar steady fluid flow through a round pipe, known as Poise...

First note that the particle paths indicate that all fluid particles move straight down the pipe without changing their radial or angular positions. Thus, the particle paths coincide with the streamlines shown in Figure 10.10. The velocity is calculated by using Eqs. 10.17, with the result

V_r=\frac{dR}{dt}=\frac{d}{dt}(R_0)=0,       V_\theta =R\frac{d \theta}{dt}=R\frac{d}{dt}(\theta_0)=0

and

V_z=\frac{dZ}{dt}=\frac{d}{dt}\left\{Z_0+\left[\left(\frac{R^2_p(p_1 − p_2)}{4\mu L}\right)\left(1-\left(\frac{R_0}{R_p} \right)^2\right)\right](t-t_0)\right\}

=\left(\frac{R^2_p(p_1 − p_2)}{4\mu L}\right)\left(1-\left(\frac{R_0}{R_p} \right)^2\right)

Thus,

V = 0e_r + 0e_θ + \left(\frac{R^2_p(p_1 − p_2)}{4\mu L}\right)\left(1-\left(\frac{R_0}{R_p} \right)^2\right)e_z

Note that the no-slip, no-penetration conditions are satisfied for a particle at R₀ = R_P, i.e., on the pipe wall, and that the single nonzero velocity component points in the flow direction as expected. The maximum velocity, which occurs for a particle on the centerline, is V_max = 0e_r + 0e_θ + {[R^2_P (p₁ − p₂)]/4µL}e_z.

The acceleration may be calculated using Eq. 10.18 or 10.19. Applying Eq. 10.18 to the preceding velocity vector, we obtain

A_r=\frac{dV_r}{dt}-\frac{V^2_{\theta}}{R}=\frac{d}{dt}(0)-\frac{(0)^2}{R}=0,         A_\theta =\frac{d V_{\theta}}{dt}+\frac{V_rV_{\theta}}{R}=\frac{d}{dt}(0)+\frac{(0)(0)}{R}=0

and

A_z=\frac{d}{dt} \left[\left(\frac{R^2_p(p_1 − p_2)}{4\mu L}\right)\left(1-\left(\frac{R_0}{R_p} \right)^2\right)\right]=0

Thus we have A = 0e_r + 0e_θ + 0e_z. There is no Lagrangian acceleration in Poiseuille flow.