---
title: "Velocity profile near smooth boundary"
subject: "Hydraulics"
chapter: 1
key: "semester_4/hydraulics:prandtl-smooth-boundary-profile"
source: https://www.bctengineer.com/q/semester_4/hydraulics/prandtl-smooth-boundary-profile
site: BCT Engineering
---

# Velocity profile near smooth boundary

Hydraulics · Chapter 1: Pipe flow

- Asked in: 2068 Bhadra, 2+6 marks
- Video: [Hydrodynamically Smooth & Rough Boundaries | Turbulent Flow | Smooth & Rough Boundaries in Turbul...](https://www.youtube.com/watch?v=MR0FSvWMHBM) — Professor Nidhi Trivedi
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/prandtl-smooth-boundary-profile
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1

## Question

Explain Prandtl mixing length theory. Starting from the expression for turbulent shear stress derive the velocity distribution for turbulent flow near hydrodynamically smooth boundaries in the form $\frac{u}{u_*} = 5.75\log_{10}\left(\frac{u_* y}{\nu}\right) + 5.5$.

## Short answer

Prandtl's mixing length theory gives τ_t = ρ(κy)²(du/dy)². With constant τ₀ near the wall, du/dy = u*/(κy). Integrating and matching to the laminar sublayer at y = 11.6 ν/u* (where u = 11.6 u*) yields u/u* = (1/κ) ln(u* y/ν) + 5.5. With κ=0.4, 1/κ=2.5, converting ln to log₁₀ gives 5.75 log₁₀(u* y/ν) + 5.5.

## Answer

#### Prandtl's mixing length theory
In turbulent flow, lumps of fluid are moved across layers by eddies. Prandtl assumed a lump keeps its momentum over a distance $l$ (mixing length) before mixing. The velocity fluctuations are then $u' \approx v' \approx l\,\dfrac{du}{dy}$, and the turbulent shear stress is
$$
\tau_t = -\rho\overline{u'v'} = \rho\,l^2\left(\frac{du}{dy}\right)^2
$$
Near a wall the mixing length increases with distance from it: $l = \kappa y$ with $\kappa \approx 0.4$ (von Karman's constant).

#### Velocity distribution near a smooth boundary
Let $u_* = \sqrt{\tau_0/\rho}$. Close to the wall the shear stress is practically constant, $\tau = \tau_0$, and the viscous stress is negligible outside the laminar sub-layer:
$$
\tau_0 = \rho\kappa^2y^2\left(\frac{du}{dy}\right)^2 \;\Rightarrow\; \frac{du}{dy} = \frac{u_*}{\kappa y}
$$
Integrating:
$$
u = \frac{u_*}{\kappa}\ln y + C \quad\cdots(1)
$$

**Constant from the laminar sub-layer.** In the laminar sub-layer of thickness $\delta'$, $\tau_0 = \mu u/y$, so $u = \dfrac{\tau_0y}{\mu} = \dfrac{u_*^2y}{\nu}$. At its edge $y = \delta'$, experiments give $\delta' = \dfrac{11.6\nu}{u_*}$, so
$$
u_{\delta'} = \frac{u_*^2}{\nu}\cdot\frac{11.6\nu}{u_*} = 11.6\,u_*
$$
Applying (1) from $y = \delta'$ to $y$:
$$
u - 11.6u_* = \frac{u_*}{\kappa}\ln\frac{y}{\delta'} = \frac{u_*}{\kappa}\ln\frac{u_*y}{11.6\nu}
$$
With $\kappa = 0.4$:
$$
\frac{u}{u_*} = 11.6 + 2.5\ln\frac{u_*y}{\nu} - 2.5\ln 11.6 = 2.5\ln\frac{u_*y}{\nu} + 11.6 - 6.13
$$
Convert $\ln$ to $\log_{10}$: $2.5\ln x = 2.5(2.303)\log_{10}x = 5.75\log_{10}x$:
$$
\boxed{\frac{u}{u_*} = 5.75\log_{10}\left(\frac{u_*y}{\nu}\right) + 5.5}
$$
(the exact constant 5.47 is rounded to 5.5, a value that agrees with Nikuradse's measurements).

## Check yourself

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

1. Near a wall, how does Prandtl take the mixing length?
   A. Constant across the section
   B. l = kappa y with kappa about 0.4
   C. l = kappa/y
   D. Equal to the pipe diameter
   Answer: B. The mixing length grows with distance from the wall, l = 0.4 y, where 0.4 is von Karman's constant. (Key point: Mixing length assumption)

2. Setting tau = tau0 and l = kappa y, what does integration lead to?
   A. u proportional to y^2
   B. u = tau0 y / mu
   C. u = (u*/kappa) ln y + C
   D. u = constant
   Answer: C. tau0 = rho kappa^2 y^2 (du/dy)^2 gives du/dy = u*/(kappa y), which integrates to a logarithm. (Key point: Logarithmic law)

3. How is the constant C fixed?
   A. By setting u = 0 at y = R
   B. By setting u = u_max at the axis
   C. By setting u = 0 at y = 0 in the log law
   D. By matching to the laminar sub-layer edge y = 11.6 nu/u*, where u = 11.6 u*
   Answer: D. In the sub-layer u = u*^2 y/nu, so at delta' = 11.6 nu/u* the velocity is 11.6 u*, which fixes the constant. (Key point: Laminar sub-layer matching)

4. How does 5.5 arise from 11.6 - 2.5 ln 11.6?
   A. 11.6 - 6.13 = 5.47, rounded to 5.5
   B. 11.6 + 6.13
   C. 11.6 x 0.4
   D. 11.6/2.5
   Answer: A. 2.5 ln 11.6 = 6.13, so the constant is 5.47, rounded to 5.5 in agreement with Nikuradse. (Key point: Constant 5.5)

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