---
title: "Parabolic velocity profile in laminar pipe flow"
subject: "Hydraulics"
chapter: 1
key: "semester_4/hydraulics:laminar-parabolic-velocity-profile"
source: https://www.bctengineer.com/q/semester_4/hydraulics/laminar-parabolic-velocity-profile
site: BCT Engineering
---

# Parabolic velocity profile in laminar pipe flow

Hydraulics · Chapter 1: Pipe flow

- Asked in: 2080 Chaitra, 4 marks
- Video: [Viscous flow in circular pipe](https://www.youtube.com/watch?v=_25UvS19ke4) — TECHNICAL CLASSES
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/laminar-parabolic-velocity-profile
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1

## Question

Prove that velocity distribution in the case of laminar flow in circular pipe is parabolic.

## Short answer

A force balance on a fluid cylinder gives τ = -(dp/dx)r/2. Using Newton’s law and integrating with no-slip at the wall yields u = -(1/(4μ)) (dp/dx) (R² - r²). This is a parabolic distribution with maximum at the axis.

## Answer

Consider steady laminar flow in a horizontal circular pipe of radius $R$. Take a coaxial cylinder of fluid of radius $r$ and length $L$.

```text
        <-- tau (on surface)
   p1 ->  ____________  <- p2
         |   cylinder |  radius r, length L
   ------|------------|-------- axis
         |____________|
```

**Force balance (no acceleration):** pressure force on the ends = shear force on the curved surface.
$$
(p_1 - p_2)\pi r^2 = \tau\,(2\pi r L) \quad\Rightarrow\quad \tau = \frac{(p_1 - p_2)}{L}\frac{r}{2} = -\frac{dp}{dx}\frac{r}{2}
$$
So the shear stress varies linearly from zero at the axis to $\tau_0$ at the wall.

**Newton's law of viscosity.** With $y = R - r$, $\tau = \mu\dfrac{du}{dy} = -\mu\dfrac{du}{dr}$:
$$
-\mu\frac{du}{dr} = -\frac{dp}{dx}\frac{r}{2} \quad\Rightarrow\quad \frac{du}{dr} = \frac{1}{2\mu}\frac{dp}{dx}\,r
$$
Integrate:
$$
u = \frac{1}{4\mu}\frac{dp}{dx}\,r^2 + C
$$
Boundary condition (no slip): $u = 0$ at $r = R$, so $C = -\dfrac{1}{4\mu}\dfrac{dp}{dx}R^2$.
$$
\boxed{u = -\frac{1}{4\mu}\frac{dp}{dx}\left(R^2 - r^2\right)}
$$

This is the equation of a parabola in $r$ ($u$ falls as $r^2$), with its maximum at the axis:
$$
u_{max} = -\frac{1}{4\mu}\frac{dp}{dx}R^2, \qquad \frac{u}{u_{max}} = 1 - \frac{r^2}{R^2}
$$
The mean velocity is $V = u_{max}/2$, so the velocity distribution is parabolic. (For an inclined pipe, replace $p$ by the piezometric pressure $p + \gamma z$.)

## Check yourself

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

1. From the force balance on a coaxial cylinder, how does shear stress vary across the pipe?
   A. Constant
   B. Linearly from zero at the axis to maximum at the wall
   C. Parabolically with maximum at the axis
   D. Zero at the wall
   Answer: B. tau = (-dp/dx) r/2 is proportional to r. (Key point: Shear distribution)

2. Which condition fixes the integration constant?
   A. u = u_max at r = R
   B. du/dr = 0 at r = R
   C. u = 0 at r = R (no slip)
   D. u = V at r = 0
   Answer: C. No slip at the wall gives C = -(1/4mu)(dp/dx)R^2. (Key point: Boundary condition)

3. What is the ratio of mean velocity to maximum velocity in laminar pipe flow?
   A. 1
   B. 2/3
   C. 0.85
   D. 1/2
   Answer: D. u/u_max = 1 - r^2/R^2 and V = u_max/2. (Key point: Profile result)

## Related in Hydraulics

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