Question 11.7: In the Poiseuille flow of a constant density, constant viscos...
In the Poiseuille flow of a constant density, constant viscosity fluid in a round pipe, (Figure 11.6), the velocity field is given in cylindrical coordinates by u= vrer + vθeθ + vzez with components vr = 0, vθ = 0, and vz(r) = {[R^2_P (p1 − p2)]/4µL}[1−( r/R_P )²]. Find the pressure distribution in this flow if body forces are neglected.
The velocity field must satisfy the continuity equation, Eq. 11.4b, and the velocity field and pressure distribution must satisfy Eqs. 11.12a–11.12c. We begin by checking the continuity equation:
\begin{aligned}&\rho\left(\frac{\partial v_r}{\partial t}+v_r \frac{\partial v_r}{\partial r}+\frac{v_\theta}{r} \frac{\partial v_r}{\partial \theta}+v_z \frac{\partial v_r}{\partial z}-\frac{v_\theta^2}{r}\right) \\&\quad=\rho f_r-\frac{\partial p}{\partial r}+\mu\left(\frac{\partial^2 v_r}{\partial r^2}+\frac{1}{r} \frac{\partial v_r}{\partial r}+\frac{1}{r^2} \frac{\partial^2 v_r}{\partial \theta^2}+\frac{\partial^2 v_r}{\partial z^2}-\frac{v_r}{r^2}-\frac{2}{r^2} \frac{\partial v_\theta}{\partial \theta}\right)\end{aligned} (11.12a)
\begin{gathered}\rho\left(\frac{\partial v_\theta}{\partial t}+v_r \frac{\partial v_\theta}{\partial r}+\frac{v_\theta}{r} \frac{\partial v_\theta}{\partial \theta}+v_z \frac{\partial v_\theta}{\partial z}+\frac{v_r v_\theta}{r}\right) \\=\rho f_\theta-\frac{1}{r} \frac{\partial p}{\partial \theta}+\mu\left(\frac{\partial^2 v_\theta}{\partial r^2}+\frac{1}{r} \frac{\partial v_\theta}{\partial r}+\frac{1}{r^2} \frac{\partial^2 v_\theta}{\partial \theta^2}+\frac{\partial^2 v_\theta}{\partial z^2}-\frac{v_\theta}{r^2}+\frac{2}{r^2} \frac{\partial v_r}{\partial \theta}\right)(11.12b) \\ \\ \\ \rho\left(\frac{\partial v_z}{\partial t}+v_r \frac{\partial v_z}{\partial r}+\frac{v_\theta}{r} \frac{\partial v_z}{\partial \theta}+v_z \frac{\partial v_z}{\partial z}\right) \\=\rho f_z-\frac{\partial p}{\partial z}+\mu\left(\frac{\partial^2 v_z}{\partial r^2}+\frac{1}{r} \frac{\partial v_z}{\partial r}+\frac{1}{r^2} \frac{\partial^2 v_z}{\partial \theta^2}+\frac{\partial^2 v_z}{\partial z^2}\right)\end{gathered} (11.12c)
\frac{1}{r} \frac{\partial\left(r v_r\right)}{\partial r}+\frac{1}{r} \frac{\partial v_\theta}{\partial \theta}+\frac{\partial v_z}{\partial z}=0
Since we know vr = 0 and vθ = 0, this reduces to ∂vz/∂z = 0, which can be seen to be satisfied by inspection, since vz = vz(r). Writing only the nonzero terms in Eqs. 11.12a–11.12c, we have
0=-\frac{\partial p}{\partial r}, 0=-\frac{1}{r}\frac{\partial p}{\partial \theta }, and 0=-\frac{\partial p}{\partial z}+\mu\left(\frac{\partial^2 v_z}{\partial^2 r}+\frac{1}{r} \frac{\partial v_z}{\partial r}\right)
Thus the pressure is a function of z only. Inserting the given velocity component into the last equation and taking derivatives we find
\frac{\partial p}{\partial z}=\mu \left(\frac{ R^2 _P(p_1 − p_2)}{4\mu L} \right)\left[\left(-\frac{2}{R^2_P} \right)+\frac{1}{r}\left(-\frac{2r}{R^2_P} \right)\right]
=\mu \left(\frac{ R^2 _P(p_1 − p_2)}{4\mu L} \right)\left(-\frac{2}{R^2_P} -\frac{2}{R^2_P} \right)
which, after simplification, yields ∂p/∂z = (p2 − p1)/L. Integrating and evaluating the resulting constant of integration at z = z1, we find p(z) = p1 + [(p2 − p1)/L](z − z1). This is a linear drop in pressure down the pipe. Can you see by inspection that the no-slip, no-penetration conditions are satisfied on the pipe wall?