---
title: "BL, displacement and momentum thickness"
subject: "Fluid Mechanics"
chapter: 8
key: "semester_3/fluid_mechanics:bl-thickness-displacement-momentum-applications"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/bl-thickness-displacement-momentum-applications
site: BCT Engineering
---

# BL, displacement and momentum thickness

Fluid Mechanics · Chapter 8: Boundary Layer theory

- Asked in: 2075 Chaitra, 6 marks
- Video: [Introductory Fluid Mechanics L19 p7 - Displacement and Momentum Thickness](https://www.youtube.com/watch?v=h_vk0ZYMBVc) — Ron Hugo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/bl-thickness-displacement-momentum-applications
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-8

## Question

Explain concept of Boundary layer thickness. Displacement thickness and Momentum thickness with their applications each.

## Short answer

Boundary layer thickness δ is the distance at which u = 0.99U, defining the viscous region. Displacement thickness δ* = ∫ (1 − u/U) dy corrects the external flow area in ducts and wind tunnels. Momentum thickness θ = ∫ (u/U)(1 − u/U) dy is used in the momentum integral equation to find wall shear stress and drag force.

## Answer

#### Boundary layer thickness ($\delta$)

It is the distance from the wall, measured normal to it, at which the velocity reaches 99% of the free-stream velocity $U$ ($u = 0.99U$).

**Application:** It marks out the region where viscous effects matter. It is used to fix the size of the region to be analysed, to place probes and decide the wall clearance in wind tunnels, and as the upper limit of the integrals that give $\delta^*$ and $\theta$.

#### Displacement thickness ($\delta^*$)

It is the distance by which the external flow is shifted away from the wall because of the velocity deficit in the boundary layer:

$$
\delta^* = \int_0^{\delta}\left(1 - \frac uU\right)dy
$$

**Application:** In wind tunnels, diffusers and ducts the walls are moved outward (or the area is enlarged) by $\delta^*$ to keep the core velocity constant. It also gives the pressure change in a duct entrance, and the effective contour of an aerofoil or body for the outer potential flow.

#### Momentum thickness ($\theta$)

It is the thickness of a fluid layer that, moving at velocity $U$, carries the momentum lost because of the boundary layer:

$$
\theta = \int_0^{\delta}\frac uU\left(1 - \frac uU\right)dy
$$

**Application:** It is used in the momentum integral equation to find the wall shear stress and the drag force on a surface: $\tau_0 = \rho U^2\,d\theta/dx$, and $F_D = \rho U^2\theta\,b$ for a plate of width $b$ ($\theta$ taken at the trailing edge).

| Thickness | Measures loss of | Typical use |
|---|---|---|
| $\delta$ | Extent of viscous region | Size of region |
| $\delta^*$ | Mass flow (discharge) | Area correction, outer flow |
| $\theta$ | Momentum | Drag, shear stress |

## Check yourself

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

1. How is boundary layer thickness defined?
   A. Distance where u = U/2
   B. Distance where shear is zero
   C. Normal distance from the wall where u = 0.99U
   D. Distance equal to the plate length
   Answer: C. It marks the edge of the region of significant viscous effects. (Key point: Boundary layer thickness)

2. What is a main application of displacement thickness?
   A. Calculating the drag force directly
   B. Enlarging duct or wind tunnel walls to keep core velocity constant
   C. Finding the Reynolds number
   D. Computing wall shear from d(theta)/dx
   Answer: B. delta* corrects the effective flow area for the boundary layer. (Key point: Displacement thickness use)

3. What does momentum thickness theta measure and what is it used for?
   A. Loss of momentum, used in the momentum integral equation to find shear and drag
   B. Loss of discharge, used to correct duct area
   C. Extent of the viscous region
   D. Thickness of the laminar sub-layer
   Answer: A. tau_0 = rho U^2 d(theta)/dx and F_D = rho U^2 theta b. (Key point: Momentum thickness)

4. Which integral defines theta?
   A. Integral of (1 - u/U) dy
   B. Integral of (u/U) dy
   C. Integral of (1 - u/U)^2 dy
   D. Integral of (u/U)(1 - u/U) dy
   Answer: D. The extra factor u/U accounts for momentum rather than mass flow. (Key point: Formula for theta)

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