---
title: "Wall shear stress as dimensionless function"
subject: "Fluid Mechanics"
chapter: 10
key: "semester_3/fluid_mechanics:wall-shear-stress-dimensionless-pi"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/wall-shear-stress-dimensionless-pi
site: BCT Engineering
---

# Wall shear stress as dimensionless function

Fluid Mechanics · Chapter 10: Similitude and physical modeling

- Asked in: 2075 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/wall-shear-stress-dimensionless-pi
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-10

## Question

The wall shear stress $\tau_w$ in a boundary layer is assumed to be a function of stream velocity U, boundary layer thickness $\delta$, local turbulence velocity u', density $\rho$, and local pressure gradient dp/dx. Using ($\rho$, U, $\delta$) as repeating variables, rewrite this relationship as a dimensionless function.

## Short answer

τ_w/(ρU²) = φ(u′/U, (δ/(ρU²)) dp/dx). The left side is half the skin friction coefficient, right side includes turbulence intensity and dimensionless pressure gradient.

## Answer

Variables: $\tau_w, U, \delta, u', \rho, dp/dx$. So $n = 6$, $m = 3$ (M, L, T), giving $n - m = 3$ $\pi$ terms. Repeating variables: $\rho, U, \delta$ (as given).

#### Dimensions

$\tau_w = ML^{-1}T^{-2}$, $U = u' = LT^{-1}$, $\delta = L$, $\rho = ML^{-3}$, $dp/dx = ML^{-2}T^{-2}$.

#### $\pi_1$ (with $\tau_w$)

$$
\pi_1 = \tau_w\,\rho^{a}U^{b}\delta^{c}
$$

M: $1 + a = 0 \Rightarrow a = -1$. T: $-2 - b = 0 \Rightarrow b = -2$. L: $-1 - 3a + b + c = 0 \Rightarrow c = 0$.

$$
\pi_1 = \frac{\tau_w}{\rho U^2}
$$

#### $\pi_2$ (with $u'$)

It has the same dimensions as $U$, so:

$$
\pi_2 = \frac{u'}{U}
$$

#### $\pi_3$ (with $dp/dx$)

$$
\pi_3 = \frac{dp}{dx}\rho^{a}U^{b}\delta^{c}
$$

M: $1 + a = 0 \Rightarrow a = -1$. T: $-2 - b = 0 \Rightarrow b = -2$. L: $-2 - 3a + b + c = 0 \Rightarrow -2 + 3 - 2 + c = 0 \Rightarrow c = 1$.

$$
\pi_3 = \frac{\delta}{\rho U^2}\,\frac{dp}{dx}
$$

#### Result

$$
\frac{\tau_w}{\rho U^2} = \phi\left(\frac{u'}{U},\ \frac{\delta}{\rho U^2}\frac{dp}{dx}\right)
$$

The left side is half the skin friction coefficient ($C_f/2$), and the second argument is a dimensionless pressure gradient.

## Check yourself

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

1. How many pi terms?
   A. 6
   B. 2
   C. 3, since n = 6 and m = 3
   D. 9
   Answer: C. n - m = 3. (Key point: Number of pi terms)

2. What is pi1?
   A. tau_w/(rho U delta)
   B. tau_w U/rho
   C. tau_w delta/U
   D. tau_w/(rho U^2)
   Answer: D. Equating exponents gives a = -1, b = -2, c = 0. (Key point: pi1)

3. What is pi2 for u'?
   A. u'/U, since both have velocity dimension
   B. u' U
   C. u'/delta
   D. u' rho
   Answer: A. u' has the dimension of U. (Key point: pi2)

4. What is pi3 for dp/dx?
   A. (dp/dx)/(rho U^2)
   B. (delta/(rho U^2)) dp/dx
   C. rho U^2 dp/dx
   D. (dp/dx) delta U
   Answer: B. dp/dx has ML^-2T^-2, and the exponents give c = 1. (Key point: pi3)

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