---
title: "Hagen-Poiseuille flow in a circular tube"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_4/fluid_mechanics_mechanical:hagen-poiseuille-derivation-circular-tube"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/hagen-poiseuille-derivation-circular-tube
site: BCT Engineering
---

# Hagen-Poiseuille flow in a circular tube

Fluid Mechanics · Chapter 6: Viscous Effects

- Asked in: year not given, 8 marks
- Video: [Fluid Mechanics 12.5 - Hagen - Poiseuille Flow, Viscous flow in Circular Pipes](https://www.youtube.com/watch?v=LfBDyeaaAoA) — Rasim Guldiken
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/hagen-poiseuille-derivation-circular-tube
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-6

## Question

Derive the velocity distribution for steady laminar flow of a Newtonian fluid in a circular pipe of radius $R$. Hence obtain the expressions for maximum velocity, average velocity, discharge (Hagen-Poiseuille equation), wall shear stress, head loss and the friction factor in terms of Reynolds number.

## Short answer

For laminar pipe flow, velocity u = (Δp/(4μL))(R^2 - r^2). Maximum u_max = Δp R^2/(4μL). Average V = u_max/2 = Δp R^2/(8μL). Discharge Q = πΔp R^4/(8μL) (Hagen-Poiseuille). Wall shear stress τ_w = Δp R/(2L). Head loss h_f = 32μVL/(ρgD^2). Friction factor f = 64/Re.

## Answer

#### Assumptions
Steady, fully developed, laminar, incompressible flow of a Newtonian fluid in a horizontal circular pipe of radius $R$ and length $L$. Velocity is axial only and depends on $r$ only.

```text
        r
        ^   _____________________________
   R    |  /      ->  ->   ->
        | /   ->  ->  ->   ->   -> u(r)
   0 ---+--------------------------------->  x
        | \   ->  ->  ->   ->
        |  \_______________________________
           p1                       p2
```

#### Force balance on a cylinder of radius r
Pressure force $= (p_1 - p_2)\pi r^2 = \Delta p\,\pi r^2$. Shear force on the cylindrical surface $= \tau\,2\pi r L$. For equilibrium:

$$
\tau = \frac{\Delta p\, r}{2L}
$$

This is linear in $r$: zero at the axis, maximum at the wall.

#### Velocity distribution
With $\tau = -\mu\,du/dr$ (velocity decreases as $r$ increases):

$$
\frac{du}{dr} = -\frac{\Delta p}{2\mu L}r
$$

Integrating, with $u = 0$ at $r = R$:

$$
u = \frac{\Delta p}{4\mu L}\left(R^2 - r^2\right)
$$

The profile is a paraboloid.

#### Maximum velocity (at $r = 0$)
$$
u_{max} = \frac{\Delta p R^2}{4\mu L}
$$

#### Discharge
$$
Q = \int_0^R u\,2\pi r\,dr = \frac{\pi \Delta p}{2\mu L}\left[\frac{R^4}{2} - \frac{R^4}{4}\right] = \frac{\pi \Delta p R^4}{8\mu L} = \frac{\pi \Delta p D^4}{128\mu L}
$$

This is the **Hagen-Poiseuille equation**.

#### Average velocity
$$
\bar V = \frac{Q}{\pi R^2} = \frac{\Delta p R^2}{8\mu L} = \frac{u_{max}}{2}
$$

#### Wall shear stress
$$
\tau_w = \frac{\Delta p R}{2L} = \frac{\Delta p D}{4L} = \frac{8\mu\bar V}{D}
$$

#### Head loss
$$
h_f = \frac{\Delta p}{\rho g} = \frac{32\mu \bar V L}{\rho g D^2}
$$

#### Friction factor
Equating with the Darcy-Weisbach equation $h_f = \dfrac{f L \bar V^2}{2 g D}$:

$$
f = \frac{64\mu}{\rho \bar V D} = \frac{64}{Re}
$$

The result is valid only for laminar flow, $Re < 2000$. The head loss is directly proportional to the mean velocity and independent of the pipe roughness.

## Check yourself

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

1. From the force balance on a cylinder of radius r, how does shear stress vary?
   A. Parabolic, maximum at axis
   B. Constant across the section
   C. Linear in r: zero at axis, maximum at wall (tau = Delta p r / 2L)
   D. Maximum at axis, zero at wall
   Answer: C. Pressure force Delta p pi r^2 balances tau 2 pi r L, giving tau = Delta p r/(2L). (Key point: Shear stress distribution)

2. What is the velocity distribution and the ratio of mean to maximum velocity?
   A. Parabolic u = (Delta p/4 mu L)(R^2 - r^2); mean = u_max/2
   B. Linear; mean = u_max/2
   C. Parabolic; mean = 2u_max/3
   D. Logarithmic; mean = 0.8 u_max
   Answer: A. Integrating gives a paraboloid with u_max = Delta p R^2/(4 mu L) and V_bar = Delta p R^2/(8 mu L) = u_max/2. (Key point: Velocity profile and ratio)

3. Which gives the discharge in a laminar circular pipe?
   A. Q = pi Delta p D^2/(32 mu L)
   B. Q = pi Delta p D^4/(128 mu L)
   C. Q = pi Delta p D^3/(64 mu L)
   D. Q = Delta p D^4/(8 mu L)
   Answer: B. Q = pi Delta p R^4/(8 mu L) = pi Delta p D^4/(128 mu L), proportional to D^4. (Key point: Hagen-Poiseuille equation)

4. How is f = 64/Re obtained and what is its range?
   A. From the Moody chart for any Re
   B. Equating h_f = 32 mu V L/(rho g D^2) with Darcy-Weisbach; valid for laminar flow, Re < 2000
   C. From roughness; valid for Re > 4000
   D. By assuming turbulent shear stress
   Answer: B. Equating the Hagen-Poiseuille head loss with f L V^2/(2gD) gives f = 64/Re, independent of roughness, for laminar flow only. (Key point: Friction factor)

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