--- title: "Roughness height from velocity ratio" subject: "Hydraulics" chapter: 1 key: "semester_4/hydraulics:rough-pipe-velocity-ratio-roughness" source: https://www.bctengineer.com/q/semester_4/hydraulics/rough-pipe-velocity-ratio-roughness site: BCT Engineering --- # Roughness height from velocity ratio Hydraulics · Chapter 1: Pipe flow - Asked in: 2073 Bhadra, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/rough-pipe-velocity-ratio-roughness - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1 ## Question In a hydrodynamically rough pipe of 100 mm diameter, the ratio of velocities at 10 mm and 30 mm from the pipe wall is 0.838. Determine the average height of the wall roughness, shear stress at the wall and mean velocity of flow if velocity at 30 mm is 1.90 m/s. ## Short answer Using the logarithmic velocity difference in rough regime, u_* is found from the two velocities: u_*=0.1122 m/s. The roughness height from the velocity at 30 mm is k ≈ 1.024 mm, wall shear stress τ₀=12.6 Pa, and mean velocity V≈1.62 m/s. ## Answer **Data.** $D = 100$ mm so $R = 50$ mm. Distances from the wall: $y_1 = 10$ mm, $y_2 = 30$ mm. $u_2 = 1.90$ m/s at 30 mm; $u_1/u_2 = 0.838$. Water, $\rho = 1000$ kg/m³. Velocity distribution in a hydrodynamically rough pipe: $$ \frac{u}{u_*} = 5.75\log_{10}\frac{y}{k} + 8.5 $$ #### Shear velocity $u_1 = 0.838(1.90) = 1.5922$ m/s. Subtracting the equation at $y_1$ from the one at $y_2$ (the constants and $k$ cancel): $$ \frac{u_2 - u_1}{u_*} = 5.75\log_{10}\frac{y_2}{y_1} = 5.75\log_{10}3 = 2.7434 $$ $$ u_* = \frac{1.90 - 1.5922}{2.7434} = \frac{0.3078}{2.7434} = 0.1122\ \text{m/s} $$ #### (i) Average roughness height At $y_2 = 30$ mm: $\dfrac{1.90}{0.1122} = 16.935 = 5.75\log_{10}\dfrac{30}{k} + 8.5$ $$ \log_{10}\frac{30}{k} = \frac{16.935 - 8.5}{5.75} = 1.4669 \;\Rightarrow\; k = \frac{30}{10^{1.4669}} = 1.024\ \text{mm} $$ Check: $\dfrac{u_*k}{\nu} = \dfrac{0.1122(1.024\times10^{-3})}{10^{-6}} = 115 > 100$, so the pipe is hydrodynamically rough, as given. #### (ii) Wall shear stress $$ \tau_0 = \rho u_*^2 = 1000(0.1122)^2 = 12.59\ \text{Pa} $$ #### (iii) Mean velocity For rough pipes $\dfrac{V}{u_*} = 5.75\log_{10}\dfrac{R}{k} + 4.75$, with $R/k = \dfrac{50}{1.024} = 48.84$: $$ V = 0.1122\left[5.75\log_{10}(48.84) + 4.75\right] = 1.622\ \text{m/s} $$ **Answer:** $k = 1.024$ mm (about 1.0 mm); $\tau_0 = 12.59$ Pa (about 12.6 Pa); $V = 1.622$ m/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can u* be found by subtracting the rough-pipe log law at y1 from that at y2? A. The constant 8.5 and the roughness k cancel, leaving (u2-u1)/u* = 5.75 log(y2/y1) B. The pipe radius cancels, leaving u* = u2 - u1 C. The viscosity cancels, leaving u* = 5.75 log(y2/y1) D. The wall shear stress cancels, leaving u1/u2 = y1/y2 Answer: A. Both equations contain the same 8.5 and the same k, so subtraction leaves only the velocity difference over u* equal to 5.75 log10(y2/y1). (Key point: Finding shear velocity from two points) 2. Once u* = 0.1122 m/s is known, how is the roughness height k obtained? A. k = y2 x u2/u* B. From u2/u* = 5.75 log10(y2/k) + 8.5, solved for k C. From tau0 = rho u*^2 divided by the pipe radius D. From the mean-velocity formula, because k equals R/48.84 Answer: B. Substituting u2/u* = 16.935 at y2 = 30 mm into the rough log law gives log(30/k) = 1.4669 and k = 1.024 mm. (Key point: Roughness from the log law) 3. How is the wall shear stress calculated from the shear velocity? A. tau0 = u*/rho B. tau0 = rho u* R C. tau0 = rho u*^2 D. tau0 = mu u* Answer: C. Shear velocity is defined by u* = sqrt(tau0/rho), so tau0 = rho u*^2 = 12.59 Pa. (Key point: Wall shear stress) 4. Which relation gives the mean velocity V once k and u* are known? A. V/u* = 5.75 log(R/k) + 8.5 B. V/u* = 5.75 log(R/k) + 4.75 C. V = (u1 + u2)/2 D. V/u* = 5.75 log(u* R/nu) + 1.75 Answer: B. Averaging the log law over the pipe area gives the constant 4.75, so V = 0.1122[5.75 log(48.84) + 4.75] = 1.622 m/s. 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