--- title: "Mean velocity law for rough pipes" subject: "Hydraulics" chapter: 1 key: "semester_4/hydraulics:rough-pipe-log-law-mean-velocity" source: https://www.bctengineer.com/q/semester_4/hydraulics/rough-pipe-log-law-mean-velocity site: BCT Engineering --- # Mean velocity law for rough pipes Hydraulics · Chapter 1: Pipe flow - Asked in: 2070 Bhadra, 8 marks - Video: [Velocity Distribution for Turbulent Flow in Pipe | Turbulent Flow in Pipe | Turbulent Flow](https://www.youtube.com/watch?v=fcal3FN6gHk) — Professor Nidhi Trivedi - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/rough-pipe-log-law-mean-velocity - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1 ## Question Show that for turbulent flow in rough pipes $\frac{V}{V_*} = 5.75\log\left(\frac{R}{K}\right) + 4.75$, where V = mean velocity, $V_*$ = shear velocity, R = radius of pipe, K = average height of surface protrusions. ## Short answer Start from the rough-pipe log law u/u* = 5.75 log(y/K) + 8.5. Integrate over the cross-section: V = (2/R²)∫₀ᴿ u (R-y) dy. The integration adds a constant -3.75, giving V/V* = 5.75 log(R/K) + 8.5 - 3.75 = 5.75 log(R/K) + 4.75. ## Answer **Start.** For a hydrodynamically rough pipe the velocity at distance $y$ from the wall, from Prandtl's mixing length theory and Nikuradse's experiments, is $$ \frac{u}{u_*} = 5.75\log_{10}\frac{y}{K} + 8.5 \quad\cdots(1) $$ where $u_* = V_* = \sqrt{\tau_0/\rho}$ is the shear velocity and $K$ is the roughness height. It is assumed to hold over the whole section. Mean velocity is $$ V = \frac{1}{\pi R^2}\int_0^R u\,2\pi r\,dr $$ Put $y = R - r$, so $r = R - y$ and $dr = -dy$: $$ V = \frac{2}{R^2}\int_0^R u\,(R - y)\,dy $$ **Split (1)** as $\dfrac{u}{u_*} = 5.75\log_{10}\dfrac{R}{K} + 8.5 + 5.75\log_{10}\dfrac{y}{R}$. The first two terms are constant, and for them $\dfrac{2}{R^2}\displaystyle\int_0^R (R-y)\,dy = 1$. For the last term use $\log_{10}x = 0.4343\ln x$ and let $\eta = y/R$: $$ \frac{2}{R^2}\int_0^R (R - y)\ln\frac{y}{R}\,dy = 2\int_0^1 (1 - \eta)\ln\eta\,d\eta $$ Using $\int_0^1 \ln\eta\,d\eta = -1$ and $\int_0^1 \eta\ln\eta\,d\eta = -\dfrac14$: $$ 2\left[-1 - \left(-\frac14\right)\right] = 2\left(-\frac34\right) = -\frac32 $$ Hence the area-weighted average of $5.75\log_{10}(y/R)$ is $$ 5.75(0.4343)\left(-\frac32\right) = 2.5\left(-\frac32\right) = -3.75 $$ **Result.** $$ \frac{V}{V_*} = 5.75\log_{10}\frac{R}{K} + 8.5 - 3.75 $$ $$ \boxed{\frac{V}{V_*} = 5.75\log_{10}\left(\frac{R}{K}\right) + 4.75} $$ (The same integral applied to the smooth-pipe law gives $\dfrac{V}{V_*} = 5.75\log_{10}\dfrac{V_*R}{\nu} + 1.75$.) The equation also gives $f$ for rough pipes: since $V_* = V\sqrt{f/8}$, $\dfrac{1}{\sqrt f} = 2.03\log_{10}\dfrac{R}{K} + 1.68$, close to the von Karman-Prandtl form. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which point-velocity law is integrated to get the mean velocity in a rough pipe? A. u/u* = 5.75 log(u* y/nu) + 5.5 B. u/u* = 5.75 log(y/K) + 8.5 C. u = (dp/4 mu L)(R^2 - r^2) D. u/u* = 2.5 log(R/K) + 4.75 Answer: B. For a hydrodynamically rough boundary the velocity follows u/u* = 5.75 log10(y/K) + 8.5, assumed valid over the whole section. (Key point: Starting log law) 2. Why is V = (2/R^2) integral of u (R - y) dy? A. Because V is the average of u over y only B. Because u is constant over the section C. Because the area element 2 pi r dr is rewritten with r = R - y, dividing by pi R^2 D. Because shear stress varies linearly with y Answer: C. V = (1/pi R^2) integral u 2 pi r dr; putting r = R - y gives the weighting (R - y) which reflects the larger area near the wall. (Key point: Area-weighted mean) 3. The area-weighted average of ln(y/R) is 2 integral (1 - eta) ln eta d eta. What is it? A. -1/2 B. -3/4 C. +3/2 D. -3/2 Answer: D. Using the integral of ln eta = -1 and of eta ln eta = -1/4, 2[-1 + 1/4] = -3/2. (Key point: Value of the integral) 4. How does the constant 4.75 arise? A. 8.5 - 3.75, since 5.75 x 0.4343 x (-3/2) = -3.75 B. 8.5 + 3.75 C. 5.5 - 1.75 D. 8.5 - 5.75 Answer: A. The log(y/R) term averages to -3.75, which reduces the constant 8.5 to 4.75. 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