---
title: "Plane Poiseuille flow from Navier-Stokes"
subject: "Applied Computational Fluid Dynamics"
chapter: 2
key: "semester_7/applied_computational_fluid_dynamics:plane-poiseuille-flow-numerical"
source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/plane-poiseuille-flow-numerical
site: BCT Engineering
---

# Plane Poiseuille flow from Navier-Stokes

Applied Computational Fluid Dynamics · Chapter 2: Governing Equations and Numerical Foundations

- Asked in: year not given, 8 marks
- Video: [Fluid Mechanics 2 | C2-L9 | fully developed plane Couette-Poiseuille flow](https://www.youtube.com/watch?v=RKuF08x4Mo8) — المقررات المفتوحة - Open Courses
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/plane-poiseuille-flow-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-2

## Question

Starting from the Navier-Stokes equation, derive the velocity profile for steady, fully developed laminar flow of an incompressible fluid between two stationary parallel plates separated by a distance $h$. Oil of viscosity 0.08 Pa·s and density 870 kg/m³ flows between two plates 6 mm apart under a pressure gradient $dp/dx = -1500$ Pa/m. Find the maximum velocity, the flow rate per metre width, the mean velocity, the wall shear stress and the Reynolds number based on the gap.

## Short answer

Derivation yields the parabolic profile u = −(1/(2μ))(dp/dx) y (h−y). For oil data: maximum velocity 0.0844 m/s, flow rate per metre width 3.375×10⁻⁴ m²/s, mean velocity 0.0563 m/s, wall shear stress 4.5 Pa, and Reynolds number 3.67 (laminar).

## Answer

#### Derivation
Assumptions: steady, incompressible, fully developed ($\partial u/\partial x=0$), $v=w=0$, no body force effect along x, infinitely wide plates ($\partial/\partial z=0$).

```text
   y=h  ======================  fixed plate
         ->    ->     ->
         ->   -->    --->   u(y)
         ->    ->     ->
   y=0  ======================  fixed plate
```

The x-momentum equation reduces to
$$
0=-\frac{dp}{dx}+\mu\frac{d^2u}{dy^2}\ \Rightarrow\ \frac{d^2u}{dy^2}=\frac1\mu\frac{dp}{dx}
$$
(the y-momentum equation shows that $p$ depends only on $x$). Integrate twice:
$$
u=\frac{1}{2\mu}\frac{dp}{dx}y^2+C_1y+C_2
$$
No-slip: $u=0$ at $y=0$ gives $C_2=0$; $u=0$ at $y=h$ gives $C_1=-\dfrac{1}{2\mu}\dfrac{dp}{dx}h$. So
$$
u=-\frac{1}{2\mu}\frac{dp}{dx}\,y\,(h-y)
$$
The profile is parabolic. Maximum velocity at $y=h/2$:
$$
u_{max}=-\frac{h^2}{8\mu}\frac{dp}{dx}
$$
Flow rate per unit width $q=\int_0^hu\,dy=-\dfrac{h^3}{12\mu}\dfrac{dp}{dx}$, so mean velocity $\bar u=q/h=\tfrac23u_{max}$. Wall shear stress $\tau_w=\mu\,\dfrac{du}{dy}\Big|_{0}=-\dfrac h2\dfrac{dp}{dx}$ in magnitude.

#### Numerical
Data: $\mu=0.08$ Pa·s, $\rho=870$ kg/m³, $h=0.006$ m, $G=-dp/dx=1500$ Pa/m.

$$
u_{max}=\frac{G h^2}{8\mu}=\frac{1500\times(0.006)^2}{8\times0.08}=0.0844\ \text{m/s}
$$
$$
q=\frac{Gh^3}{12\mu}=\frac{1500\times(0.006)^3}{12\times0.08}=3.375\times10^{-4}\ \text{m}^2/\text{s per m width}
$$
$$
\bar u=\frac qh=\frac{3.375\times10^{-4}}{0.006}=0.05625\ \text{m/s}\quad(=\tfrac23\times0.0844)
$$
$$
\tau_w=\frac{Gh}{2}=\frac{1500\times0.006}{2}=4.5\ \text{Pa}
$$
$$
Re=\frac{\rho\bar uh}{\mu}=\frac{870\times0.05625\times0.006}{0.08}=3.67
$$
$Re=3.67$ is far below the transition value for channel flow (of the order of 1000), so the flow is laminar and the assumption is valid.

**Answer:** $u_{max}=0.0844$ m/s; $q=3.375\times10^{-4}$ m²/s per m; $\bar u=0.0563$ m/s; $\tau_w=4.5$ Pa; $Re=3.67$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For fully developed flow between stationary plates, the x-momentum equation reduces to:
   A. d^2u/dy^2 = (1/mu) dp/dx
   B. du/dy = 0
   C. dp/dx = 0
   D. u du/dx = -dp/dx / rho
   Answer: A. Convective terms vanish, so pressure gradient balances viscous shear: 0 = -dp/dx + mu d^2u/dy^2. (Key point: Reduced equation)

2. The velocity profile is:
   A. Linear
   B. Uniform
   C. Parabolic, u = -(1/2mu)(dp/dx) y (h - y)
   D. Logarithmic
   Answer: C. Integrating twice with no-slip at y = 0 and y = h gives a parabola with maximum at y = h/2. (Key point: Velocity profile)

3. The mean velocity is related to the maximum velocity by:
   A. u_mean = 0.5 u_max
   B. u_mean = u_max
   C. u_mean = 1.5 u_max
   D. u_mean = (2/3) u_max
   Answer: D. q = G h^3/(12 mu) gives u_mean = G h^2/(12 mu) = (2/3) u_max. (Key point: Mean velocity relation)

4. With G = 1500 Pa/m, h = 6 mm, mu = 0.08 Pa.s, u_max is:
   A. 0.0563 m/s
   B. 0.0844 m/s
   C. 0.844 m/s
   D. 0.169 m/s
   Answer: B. u_max = G h^2/(8 mu) = 1500 x 3.6e-5/0.64 = 0.0844 m/s. (Key point: Maximum velocity)

5. The wall shear stress is:
   A. 4.5 Pa from G h/2
   B. 9 Pa from G h
   C. 2.25 Pa from G h/4
   D. 0.0844 Pa
   Answer: A. tau_w = G h /2 = 1500 x 0.006 / 2 = 4.5 Pa. (Key point: Wall shear stress)

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