---
title: "Pressure change from displacement thickness in duct"
subject: "Fluid Mechanics"
chapter: 8
key: "semester_3/fluid_mechanics:duct-displacement-thickness-pressure"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/duct-displacement-thickness-pressure
site: BCT Engineering
---

# Pressure change from displacement thickness in duct

Fluid Mechanics · Chapter 8: Boundary Layer theory

- Asked in: 2078 Bhadra, 6 marks
- Video: [Example 7.1 - Displacement thickness in a square duct](https://www.youtube.com/watch?v=i3ECUdu3YEs) — Prof. Brendan MacDonald
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/duct-displacement-thickness-pressure
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-8

## Question

Air flows in the entrance region of a square duct, as shown. The velocity is uniform, $U_0$ = 30 m/s, and the duct is 76 mm square. At a section 0.3 m downstream from the entrance, the displacement thickness on each wall measures 0.9 mm. Determine pressure change between section 1 and 2. [Figure: duct entrance, uniform velocity $U_0$ at section 1, section 2 with $\delta_2$ = 0.9 mm; square cross-section 76 mm by 76 mm]

## Short answer

Displacement thickness reduces effective area, accelerating core flow. By continuity, U₂ = 30(76/74.2)² = 31.47 m/s. Bernoulli gives pressure drop p₁ − p₂ = ½×1.23×(31.47² − 30²) = 55.7 Pa.

## Answer

Outside the boundary layers the core flow is accelerated, because the displacement thickness reduces the effective flow area. Apply continuity to the core, then Bernoulli. The pressure falls by about **55.7 Pa**.

#### Assumptions

Air density $\rho = 1.23\ \text{kg/m}^3$. The flow in the core is frictionless and incompressible, so Bernoulli applies between sections 1 and 2.

```text
  Section 1                  Section 2
  ----------------------     ----------------------
   U0 = 30 m/s uniform  -->    core U2 (faster)
  ----------------------     = boundary layer (d* = 0.9 mm)
```

#### Continuity (effective area)

At section 2 the effective flow area is reduced by $\delta^*$ on each wall:

$$
A_{eff} = (76 - 2 \times 0.9)^2 = 74.2^2\ \text{mm}^2,\qquad A_1 = 76^2\ \text{mm}^2
$$

$$
U_0A_1 = U_2A_{eff} \;\Rightarrow\; U_2 = 30\left(\frac{76}{74.2}\right)^2 = 31.47\ \text{m/s}
$$

#### Bernoulli (horizontal duct)

$$
\begin{aligned}
p_1 - p_2 &= \tfrac12\rho\,(U_2^2 - U_0^2) \\
&= \tfrac12(1.23)(31.47^2 - 30^2) = 55.7\ \text{Pa}
\end{aligned}
$$

**Answer: $p_2 - p_1 = -55.7$ Pa, i.e. the pressure drops by about 55.7 Pa (about 5.7 mm of water) between sections 1 and 2.**

## Check yourself

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

1. Why does the core flow accelerate in the duct?
   A. The duct narrows physically
   B. Boundary layers on the walls reduce the effective flow area
   C. Air density increases
   D. Wall friction pushes the core forward
   Answer: B. Displacement thickness on each wall reduces the area available to the core, so by continuity the speed rises. (Key point: Effect of displacement thickness)

2. What is the effective flow area at section 2?
   A. (76 - 2 x 0.9)^2 mm^2
   B. (76 - 0.9)^2 mm^2
   C. (76 + 2 x 0.9)^2 mm^2
   D. 76^2 - 0.9^2 mm^2
   Answer: A. Each of the two opposite walls removes delta* from the side, giving 74.2 mm side. (Key point: Effective area)

3. Which equation then gives the pressure change between 1 and 2?
   A. Hagen-Poiseuille equation
   B. Momentum equation with wall friction only
   C. p1 - p2 = rho g (U2 - U0)
   D. Bernoulli for the core: p1 - p2 = 1/2 rho (U2^2 - U0^2)
   Answer: D. Core flow is frictionless and horizontal; result is a drop of about 55.7 Pa. (Key point: Pressure change method)

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