---
title: "Oil flow between fixed plates problem"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_4/fluid_mechanics_mechanical:oil-flow-between-plates-numerical"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/oil-flow-between-plates-numerical
site: BCT Engineering
---

# Oil flow between fixed plates problem

Fluid Mechanics · Chapter 6: Viscous Effects

- Asked in: year not given, 6 marks
- Video: [Numerical | Flow of Viscous Fluid Through pipe | Flow through Parallel Plate | FMHM | 3141906 | GTU](https://www.youtube.com/watch?v=DoTPucG-JTc) — krunal khiraiya
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/oil-flow-between-plates-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-6

## Question

Oil of viscosity 0.12 Pa s and specific gravity 0.9 flows steadily in laminar motion between two fixed horizontal parallel plates 10 mm apart. The pressure drops by 20 kPa over a length of 5 m. For a plate width of 1 m, find (a) the maximum velocity, (b) the discharge, (c) the mean velocity, (d) the shear stress at the wall, (e) the velocity at 2.5 mm from a plate, (f) the head loss over 5 m, and check that the flow is laminar.

## Short answer

Using pressure gradient 4000 Pa/m, u_max = h^2|dp/dx|/(8μ) = 0.417 m/s; Q = (h^3|dp/dx|/(12μ))×b = 2.78 L/s; ū = 0.278 m/s; τ_w = (h/2)|dp/dx| = 20 Pa; u(2.5 mm) = 0.3125 m/s; h_f = Δp/(ρg) = 2.27 m; Re = 20.8, laminar.

## Answer

#### Given data
$\mu = 0.12\ \text{Pa s}$, $\rho = 900\ \text{kg/m}^3$, $h = 0.01\ \text{m}$, $L = 5\ \text{m}$, $\Delta p = 20\,000\ \text{Pa}$, $b = 1\ \text{m}$.

Pressure gradient magnitude: $\left|\dfrac{dp}{dx}\right| = \dfrac{20\,000}{5} = 4000\ \text{Pa/m}$

#### (a) Maximum velocity
$$
u_{max} = \frac{h^2}{8\mu}\left|\frac{dp}{dx}\right| = \frac{(0.01)^2 \times 4000}{8 \times 0.12} = 0.4167\ \text{m/s}
$$

#### (b) Discharge
Per unit width:

$$
q = \frac{h^3}{12\mu}\left|\frac{dp}{dx}\right| = \frac{(0.01)^3 \times 4000}{12 \times 0.12} = 2.778\times10^{-3}\ \text{m}^2/\text{s}
$$

For $b = 1\ \text{m}$: $Q = 2.778\times10^{-3}\ \text{m}^3/\text{s} = 2.78\ \text{L/s}$.

#### (c) Mean velocity
$$
\bar u = \frac{q}{h} = \frac{2.778\times10^{-3}}{0.01} = 0.2778\ \text{m/s} \quad (= \tfrac23 u_{max})
$$

#### (d) Wall shear stress
$$
\tau_w = \frac{h}{2}\left|\frac{dp}{dx}\right| = 0.005 \times 4000 = 20\ \text{Pa}
$$

#### (e) Velocity at $y = 2.5\ \text{mm}$
$$
u = \frac{1}{2\mu}\left|\frac{dp}{dx}\right|(hy - y^2) = \frac{4000}{0.24}\left(0.01 \times 0.0025 - 0.0025^2\right) = 0.3125\ \text{m/s}
$$

#### (f) Head loss
$$
h_f = \frac{\Delta p}{\rho g} = \frac{20\,000}{900 \times 9.81} = 2.27\ \text{m of oil}
$$

#### Check for laminar flow
Taking the plate gap as the characteristic length:

$$
Re = \frac{\rho \bar u h}{\mu} = \frac{900 \times 0.2778 \times 0.01}{0.12} = 20.8 \ll 2000
$$

So the flow is laminar.

**Answer: $u_{max} = 0.417\ \text{m/s}$, $Q = 2.78\ \text{L/s}$, $\bar u = 0.278\ \text{m/s}$, $\tau_w = 20\ \text{Pa}$, $u(2.5\ \text{mm}) = 0.3125\ \text{m/s}$, $h_f = 2.27\ \text{m}$**

## Check yourself

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

1. Delta p = 20 kPa over 5 m gives dp/dx = 4000 Pa/m. Which formula gives u_max for h = 10 mm, mu = 0.12 Pa s?
   A. u_max = h^2 |dp/dx| / (12 mu) = 0.278 m/s
   B. u_max = h^2 |dp/dx| / (8 mu) = 0.417 m/s
   C. u_max = h |dp/dx| / (2 mu) = 0.167 m/s
   D. u_max = h^3 |dp/dx| / (8 mu) = 0.417 m/s
   Answer: B. u_max = (0.01)^2 x 4000 / (8 x 0.12) = 0.4167 m/s. (Key point: Pressure gradient and max velocity)

2. What is the wall shear stress and how is it found?
   A. tau_w = h |dp/dx| = 40 Pa
   B. tau_w = h/2 x |dp/dx| = 20 Pa
   C. tau_w = mu u_max / h = 5 Pa
   D. tau_w = Delta p = 20 000 Pa
   Answer: B. tau_w = (h/2)|dp/dx| = 0.005 x 4000 = 20 Pa. (Key point: Wall shear stress)

3. How is laminar flow confirmed here?
   A. Re = rho u_bar h / mu = 20.8, much below 2000
   B. Re = rho u_max h / mu = 5000 which exceeds 2000
   C. Because the plates are fixed
   D. Because head loss is 2.27 m
   Answer: A. Using the gap as length and mean velocity 0.278 m/s gives Re = 20.8, far below 2000. (Key point: Laminar check)

## Related in Fluid Mechanics

- [Laminar flow between fixed parallel plates](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Alaminar-flow-between-fixed-parallel-plates) — Fluid Mechanics, chapter 6
- [Laminar oil flow in a pipe](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Alaminar-oil-pipe-pressure-drop) — Fluid Mechanics, chapter 6
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