---
title: "Is flow between parallel plates rotational?"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_3/fluid_mechanics:parallel-plates-rotational-check"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/parallel-plates-rotational-check
site: BCT Engineering
---

# Is flow between parallel plates rotational?

Fluid Mechanics · Chapter 4: Hydrokinematics

- Asked in: 2072 Chaitra, 3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/parallel-plates-rotational-check
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-4

## Question

Consider fully developed two-dimensional flow between two infinite parallel plates separated by distance h, with the both top and bottom plate stationary and forced pressure gradient $\frac{dP}{dx}$ driving the flow ($\frac{dP}{dx}$ is constant and negative). The flow is steady, incompressible and two-dimensional in x-y plane. The velocity components are given by $u = \frac{1}{2\mu}\frac{dP}{dx}(y^2 - hy)$; v = 0, where $\mu$ is fluid's viscosity. Is this flow rotational or irrotational?

## Short answer

The flow is rotational. Vorticity ω_z = -(1/(4μ))(dP/dx)(2y-h) is non-zero except at y=h/2; viscous shear causes rotation.

## Answer

A flow is **irrotational** if the rotation (half the vorticity) is zero everywhere: $\omega_z = \dfrac{1}{2}\left(\dfrac{\partial v}{\partial x} - \dfrac{\partial u}{\partial y}\right) = 0$.

**Given:** $u = \dfrac{1}{2\mu}\dfrac{dP}{dx}(y^2 - hy)$, $v = 0$.

$$
\frac{\partial v}{\partial x} = 0,\qquad \frac{\partial u}{\partial y} = \frac{1}{2\mu}\frac{dP}{dx}(2y - h)
$$

$$
\omega_z = \frac{1}{2}\left(0 - \frac{1}{2\mu}\frac{dP}{dx}(2y - h)\right) = -\frac{1}{4\mu}\frac{dP}{dx}(2y - h)
$$

The vorticity is $\zeta = 2\omega_z = -\dfrac{1}{2\mu}\dfrac{dP}{dx}(2y - h)$.

Since $dP/dx$ is a non-zero constant, $\omega_z$ is zero only at the mid-plane $y = h/2$ and non-zero elsewhere. Because $dP/dx < 0$, $\omega_z$ has the sign of $(2y - h)$: it is negative near the bottom plate ($y < h/2$) and positive near the top plate ($y > h/2$).

**Answer: the flow is rotational**, because $\omega_z \neq 0$ everywhere except along the centreline $y = h/2$. The rotation arises from viscous shear (the velocity gradient $\partial u/\partial y$), which is largest at the walls.

## Check yourself

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

1. Which condition makes a 2-D flow irrotational?
   A. dv/dx - du/dy = 0
   B. du/dx + dv/dy = 0
   C. du/dx = dv/dy
   D. u = v
   Answer: A. Irrotational means vorticity is zero; continuity is a separate condition. (Key point: Rotation condition)

2. For this flow what is du/dy?
   A. (1/(2 mu))(dP/dx)(y^2 - hy)
   B. (1/(2 mu))(dP/dx)(2y - h)
   C. 0
   D. dP/dx
   Answer: B. Differentiating y^2 - hy gives 2y - h; dv/dx = 0. (Key point: Evaluating)

3. What is the conclusion?
   A. Irrotational everywhere
   B. Irrotational at the walls
   C. Rotational, vorticity is zero only at y = h/2
   D. Rotational only at the mid-plane
   Answer: C. 2y - h is zero only at the centreline; viscous shear causes rotation. (Key point: Conclusion)

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