Question 2.10: Poiseuille Flow Considering steady laminar fully-developed (......
Poiseuille Flow
Considering steady laminar fully-developed (unidirectional) flow of an incompressible fluid in a pipe of length “L” and constant cross sectional area (radius r_0), i.e., Poiseuille flow. Establish reduced forms of the N–S equations and solve for the axial velocity u(r).
| Concept | Assumptions | Sketch |
| Reduced N-S eq’ns based on the postulates: \mathrm{\vec{v}=(u,0,0) }, where u = u(r) only; and \nabla\mathrm{p\Rightarrow -\frac{\partial p}{\partial x} \hat = \frac{\Delta p}{L} =\frac{p_1-p_2}{L} }=¢ | As stated, i.e., |  |
| • ∂ / ∂t = 0 (steady-state) | ||
| • \bf \vec{v} \Rightarrow \rm u (unidirectional) | ||
| • ∂ / ∂x = 0 (fullydeveloped) | ||
| • ∂ / ∂θ = 0 (axisymmetric) |
• Check continuity: \nabla\cdot\vec{\mathrm{v}}=0 can be reduced to \frac{\partial \mathrm{u}}{\partial \mathrm{x}} =0.
• The Equation of Motion (Eq. (2.21)) reduces to:
\mathsf{ρ}\left[{\frac{\partial\vec{\mathbf{v}}}{\partial t}}+(\vec{\mathbf{v}}\cdot\nabla)\vec{\mathbf{v}}\right]=-\nabla\mathrm{p}+{\mu}\nabla^{2}\vec{\mathbf{v}}+{\mathsf{ρ}}{\vec{\mathbf{g}}} (2.21)
0=-\nabla{\rm{p}}+\mu\nabla^{2}\vec{\rm{v}} (E.2.10.1)
or with the postulates and App. A
0=(\frac{\Delta{\mathrm{p}}}{{\rm L}})+\mu\frac{1}{\rm r}\frac{{\rm d}}{{\rm d r}}({\rm r}\frac{\mathrm{d}{u}}{{\rm d r}}) (E.2.10.2)
subject to the boundary conditions: \rm {u(r=r_0 )=0} (no slip) and at r = 0, du / dr = 0 (axisymmetry). Double integration and invoking the two boundary conditions yields:
\mathrm{u}(\mathrm{r})={\frac{1}{4\mu}}({\frac{\Delta\mathrm{p}}{\mathrm{L}}})(\mathrm{r}_{0}^{2}-\mathrm{r}^{2})=\mathrm{u}_{\mathrm{max}}\!\left[1-({\frac{\mathrm{r}}{\mathrm{r}_{0}}})^{2}\right] (E.2.10.3)
Comments: The solution (E.2.10.3) is the base case for steady laminar internal flows with numerous applications in terms of industrial pipe flows, such as pipe networks, heat exchangers and fluid transport, as well as idealized tubular flows in biomedical, chemical, environmental, mechanical and nuclear engineering. Equation (E.2.10.3) not only provides the parabolic velocity profile u(r) but can also be used to calculate the volumetric flow rate Q (Eq. (2.5)), wall shear stress τ_{\rm {w}} (Eq. (2.19)), pressure drop Δp (either via Eq. (2.5) or using Eq. (2.19)), head loss h_f (cf. Eq. (2.31b)), etc. Section 3.2 and Part B further illustrate the use of the Poiseuille flow solution (E.2.10.3).
{Q}=\int_{A}{\vec{\mathrm{v}}}\cdot{{\mathrm{d}}}\mathrm{\vec{A}}=\int_{A}\mathrm{v_{n}}{\mathrm{d}}\mathrm{A}={\bar{\mathrm{v}}}\mathrm{A} (2.5)
\vec{\vec{\tau }} =\mathrm{}\mu(\nabla{\rm \vec v}+\nabla{\rm\vec v^{tr}})=\mathrm{}\mu\vec{\vec{\dot{{\gamma }}}} (2.19a)
\mathrm{\tau }_{\mathrm{{ij}}}=\mu({\frac{\partial{\mathrm{v}}_{\mathrm{{i}}}}{\partial\mathrm{x}_{\mathrm{{j}}}}}+{\frac{\partial\mathrm{v}_{\mathrm{{j}}}}{\partial\mathrm{x}_{\mathrm{{i}}}}})=\mu\dot{\gamma _{\mathrm{{ij}}}} (2.19b)
\rm\vec{\vec{{\tau}}}=\lambda(\nabla\cdot\vec{\rm v})\vec{\vec{I}}+2\mu\vec{\vec{\varepsilon} } (2.19c)
\rm\tau_{ij}=2\mu\varepsilon _{ij} (2.19d)
\tau_{\mathrm{ij}}=\mu(\nabla{\vec{\mathrm{v}}}+\nabla{\vec{\mathrm{v}}^{|\rm{tr}}})=\mu{\dot{\gamma }}_{\mathrm{ij}} (2.19e,f)
{\frac{\mathrm{v}_{1}^{2}}{2\mathrm{g}}}+{\frac{\mathrm{p}_{1}}{\gamma}}+\mathrm{z}_{1}={\frac{\mathrm{v}_{2}^{2}}{2\mathrm{g}}}+{\frac{\mathrm{p}_{2}}{\gamma}}+\mathrm{z}_{2}+\mathrm{h}_{\mathrm{f}} (2.31)