---
title: "Laminar flow between fixed parallel plates"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_4/fluid_mechanics_mechanical:laminar-flow-between-fixed-parallel-plates"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/laminar-flow-between-fixed-parallel-plates
site: BCT Engineering
---

# Laminar flow between fixed parallel plates

Fluid Mechanics · Chapter 6: Viscous Effects

- Asked in: year not given, 8 marks
- Video: [Laminar Flow between Parallel Plates (Navier-Stokes)](https://www.youtube.com/watch?v=Cj8-vu2zsQI) — LearnChemE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/laminar-flow-between-fixed-parallel-plates
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-6

## Question

Derive the expressions for velocity distribution, shear stress distribution, maximum velocity, average velocity and discharge per unit width for steady laminar flow of a viscous incompressible fluid between two fixed parallel plates a distance $h$ apart. Also obtain the pressure drop in a length $L$.

## Short answer

Velocity distribution u = -(1/(2μ))(dp/dx)(hy - y^2). Shear stress τ = -(dp/dx)(h/2 - y), linear. Maximum velocity u_max = -(h^2/(8μ))(dp/dx). Average velocity ū = (2/3)u_max = -(h^2/(12μ))(dp/dx). Discharge per unit width q = -(h^3/(12μ))(dp/dx). Pressure drop Δp = (12μūL)/h^2.

## Answer

#### Assumptions
Steady, laminar, fully developed flow of a Newtonian incompressible fluid between two fixed wide plates, gap $h$. Flow is in the $x$-direction; $y$ is measured from the lower plate. There is no body force effect other than hydrostatic, and $dp/dx$ is constant.

```text
   y ^ ---------------------------  upper plate (u = 0)
     |        ______
     |  h   /        \   u(y)
     |      \________/
     +---------------------------> x   lower plate (u = 0)
```

#### Force balance on an element
Consider a fluid element of length $dx$, thickness $dy$ and unit width. Net force in $x$ is zero:

$$
p\,dy - \left(p + \frac{dp}{dx}dx\right)dy - \tau\,dx + \left(\tau + \frac{d\tau}{dy}dy\right)dx = 0
$$

$$
\frac{d\tau}{dy} = \frac{dp}{dx}
$$

With $\tau = \mu\,du/dy$:

$$
\mu\frac{d^2u}{dy^2} = \frac{dp}{dx}
$$

#### Velocity distribution
Integrating twice, $u = \dfrac{1}{2\mu}\dfrac{dp}{dx}y^2 + C_1y + C_2$. Boundary conditions $u = 0$ at $y = 0$ and at $y = h$ give $C_2 = 0$ and $C_1 = -\dfrac{h}{2\mu}\dfrac{dp}{dx}$.

$$
u = -\frac{1}{2\mu}\frac{dp}{dx}\left(hy - y^2\right)
$$

The profile is parabolic. The pressure falls in the flow direction, so $dp/dx < 0$ and $u > 0$.

#### Shear stress distribution
$$
\tau = \mu\frac{du}{dy} = -\frac{dp}{dx}\left(\frac{h}{2} - y\right)
$$

It is zero at the centre line and maximum at the walls: $\tau_w = \dfrac{h}{2}\left|\dfrac{dp}{dx}\right|$ (linear variation).

#### Maximum velocity
At $y = h/2$:

$$
u_{max} = -\frac{h^2}{8\mu}\frac{dp}{dx}
$$

#### Discharge and average velocity
$$
q = \int_0^h u\,dy = -\frac{1}{2\mu}\frac{dp}{dx}\left[\frac{hy^2}{2} - \frac{y^3}{3}\right]_0^h = -\frac{h^3}{12\mu}\frac{dp}{dx}
$$

$$
\bar u = \frac{q}{h} = -\frac{h^2}{12\mu}\frac{dp}{dx} = \frac23 u_{max}
$$

#### Pressure drop
For a length $L$, $dp/dx = -\Delta p/L$:

$$
\Delta p = \frac{12\mu \bar u L}{h^2} = \frac{12\mu q L}{h^3}
$$

The head loss is $h_f = \Delta p/\rho g = \dfrac{12\mu\bar u L}{\rho g h^2}$.

## Check yourself

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

1. What equation results from the force balance on the fluid element?
   A. mu d^2u/dy^2 = dp/dx
   B. mu du/dy = dp/dx
   C. d^2u/dy^2 = 0
   D. tau = dp/dx
   Answer: A. d(tau)/dy = dp/dx with tau = mu du/dy gives mu d^2u/dy^2 = dp/dx. (Key point: Governing equation)

2. For fixed plates, how do velocity and shear stress vary across the gap?
   A. Velocity linear, shear parabolic
   B. Both parabolic
   C. Velocity parabolic, shear zero at walls
   D. Velocity parabolic, shear linear, zero at centre and maximum at walls
   Answer: D. u = -(1/2mu)(dp/dx)(hy - y^2) is parabolic and tau = -(dp/dx)(h/2 - y) is linear with maximum at the walls. (Key point: Velocity and shear distribution)

3. What is the relation between average and maximum velocity?
   A. u_avg = 0.5 u_max
   B. u_avg = u_max
   C. u_avg = (2/3) u_max
   D. u_avg = 0.8 u_max
   Answer: C. u_max = h^2/(8mu)|dp/dx| and u_avg = h^2/(12mu)|dp/dx|, which is 2/3 of u_max. (Key point: Maximum and average velocity)

4. How does the pressure drop in length L depend on the gap h at fixed mean velocity?
   A. Proportional to h^2
   B. Inversely proportional to h^2 (Delta p = 12 mu u L / h^2)
   C. Inversely proportional to h^3
   D. Independent of h
   Answer: B. Delta p = 12 mu u_bar L / h^2, so a smaller gap gives a larger pressure drop. (Key point: Pressure drop)

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