--- title: "Inclined smooth pipe with wall shear limit" subject: "Hydraulics" chapter: 1 key: "semester_4/hydraulics:smooth-pipe-inclined-wall-shear-limit" source: https://www.bctengineer.com/q/semester_4/hydraulics/smooth-pipe-inclined-wall-shear-limit site: BCT Engineering --- # Inclined smooth pipe with wall shear limit Hydraulics · Chapter 1: Pipe flow - Asked in: 2076 Baisakh, 8 marks - Video: [Laminar and Turbulent flow - 5 Numerical on Laminar flow in circular pipe](https://www.youtube.com/watch?v=1O7YzmuQ0vQ) — Learn Engineering - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/smooth-pipe-inclined-wall-shear-limit - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1 ## Question A straight smooth pipe 100 mm diameter and 60 m long is inclined at 10° to the horizontal. A liquid of relative density 0.9 and kinematic viscosity 120 mm²/s is to be pumped through it into a reservoir at the upper end where the gauge pressure is 120 kPa. The pipe friction factor f is given by 64/Re for laminar flow and by $0.32\,Re^{-1/4}$ for turbulent flow when Re < $10^5$. Determine (a) the maximum pressure at the lower, inlet, end of the pipe if the mean shear stress at the pipe wall is not to exceed 200 Pa; (b) the corresponding rate of flow. ## Short answer The flow is turbulent. Equating wall shear stress to the turbulent friction relation yields V ≈ 6.96 m/s, giving Q = 54.7 L/s. Using the force balance along the pipe, the maximum inlet gauge pressure is about 692 kPa. ## Answer **Data.** $D = 0.1$ m, $L = 60$ m, inclination 10°, $\rho = 900$ kg/m³ (R.D. 0.9), $\nu = 120$ mm²/s $= 1.2\times10^{-4}$ m²/s, outlet gauge pressure $p_2 = 120$ kPa, maximum mean wall shear $\tau_0 = 200$ Pa. #### Which flow regime? Mean wall shear in terms of $f$: $\tau_0 = \dfrac{f\rho V^2}{8}$. - If laminar: $\tau_0 = \dfrac{8\mu V}{D}$, i.e. $V = \dfrac{\tau_0 D}{8\mu} = \dfrac{200(0.1)}{8(900)(1.2\times10^{-4})} = 23.15$ m/s, giving $Re = 19290$, which is not laminar. So the flow is turbulent. - Turbulent: $f = 0.32\,Re^{-1/4}$ (valid since $Re < 10^5$). #### (b) Flow rate $$ \tau_0 = \frac{\rho V^2}{8}(0.32)\left(\frac{\nu}{VD}\right)^{1/4} = 0.04\,\rho\left(\frac{\nu}{D}\right)^{1/4} V^{7/4} $$ $$ 200 = 0.04(900)\left(\frac{1.2\times10^{-4}}{0.1}\right)^{1/4} V^{1.75} = 6.700\,V^{1.75} $$ $$ V = \left(\frac{200}{6.700}\right)^{1/1.75} = 6.963\ \text{m/s} $$ Check: $Re = \dfrac{6.963(0.1)}{1.2\times10^{-4}} = 5803 < 10^5$, so the formula applies; $f = 0.32Re^{-1/4} = 0.03666$ and $\dfrac{f\rho V^2}{8} = 200.0$ Pa. $$ Q = \frac{\pi}{4}D^2V = \frac{\pi}{4}(0.1)^2(6.963) = 0.05469\ \text{m}^3/\text{s} = 54.69\ \text{l/s} $$ #### (a) Maximum inlet pressure Force balance on the liquid in the pipe along the incline (steady, uniform flow): $$ (p_1 - p_2)\frac{\pi D^2}{4} = \tau_0\,\pi D L + \rho g\,\frac{\pi D^2}{4}L\sin\theta $$ $$ p_1 - p_2 = \frac{4\tau_0L}{D} + \rho g L\sin\theta = \frac{4(200)(60)}{0.1} + 900(9.81)(60)\sin10^\circ $$ $$ p_1 - p_2 = 480000 + 91988 = 571988\ \text{Pa} $$ $$ p_1 = 120\,000 + 480000 + 91988 = 691988\ \text{Pa} \approx 692\ \text{kPa (gauge)} $$ **Answer:** (a) $p_{1,max} \approx 692$ kPa gauge; (b) $Q = 54.7$ l/s ($V = 6.96$ m/s). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the flow shown to be turbulent? A. Laminar formula gives V = 23.15 m/s, Re = 19290, which is not laminar B. Re is exactly 2000 C. The pipe is inclined D. Because f = 64/Re Answer: A. Assuming laminar flow leads to Re far above 2000, so the assumption fails. (Key point: Regime check) 2. Which relation links mean wall shear to velocity? A. tau_0 = f rho V^2 / 8 B. tau_0 = f rho V^2 C. tau_0 = rho g h D. tau_0 = mu V Answer: A. tau_0 = f rho V^2/8 with f = 0.32 Re^(-1/4) gives V = 6.96 m/s. (Key point: Wall shear relation) 3. What does the net pressure force on the liquid balance? A. Wall shear force only B. Weight only C. Wall shear force plus weight component along the incline D. Momentum flux difference only Answer: C. (p1 - p2) pi D^2/4 = tau_0 pi D L + rho g (pi D^2/4) L sin(theta). (Key point: Force balance) 4. What is the maximum gauge inlet pressure? A. About 120 kPa B. About 572 kPa C. About 692 kPa D. About 480 kPa Answer: C. p1 = 120000 + 480000 + 91988 = about 692 kPa. (Key point: Inlet pressure) ## Related in Hydraulics - [Laminar oil flow in inclined pipe](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Alaminar-inclined-oil-upward) — Hydraulics, chapter 1 - [Derive laminar velocity in inclined pipe](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Alaminar-inclined-pipe-velocity-derivation) — Hydraulics, chapter 1 - [Pipe diameter to halve pumping cost](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apetrol-pipe-diameter-pumping-cost) — Hydraulics, chapter 1 - [Viscosity of oil from laminar pipe flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahorizontal-oil-pipe-laminar-viscosity) — Hydraulics, chapter 1 - [Pressure to drive oil up inclined pipe](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aviscous-fluid-upslope-pressure) — Hydraulics, chapter 1 - [Pipe over hill with cavitation limit](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apipe-over-hill-cavitation-limit) — Hydraulics, chapter 2 ## Related in other subjects - [Cross wall design, 120 mm slab, 3.0 m](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Ainterior-cross-wall-120mm-2-5m-slab) — Concrete Technology and Masonry Structure, chapter 8, asked 2× - [Brick and mortar for 230 mm wall (100 kN/m)](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Awall-eccentric-230mm-100kn-5m) — Concrete Technology and Masonry Structure, chapter 8, asked 7× - [Model liquid viscosity, Froude and Reynolds](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Amodel-liquid-viscosity-froude-reynolds) — Fluid Mechanics, chapter 10 - [Cross wall design, 120 mm slab, 3 m wide](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Ainterior-cross-wall-120mm-3-0m-slab) — Concrete Technology and Masonry Structure, chapter 8, asked 4×