--- title: "Pressure to drive oil up inclined pipe" subject: "Hydraulics" chapter: 1 key: "semester_4/hydraulics:viscous-fluid-upslope-pressure" source: https://www.bctengineer.com/q/semester_4/hydraulics/viscous-fluid-upslope-pressure site: BCT Engineering --- # Pressure to drive oil up inclined pipe Hydraulics · Chapter 1: Pipe flow - Asked in: 2077 Chaitra, 2+3+3 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/viscous-fluid-upslope-pressure - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1 ## Question What is the pressure needed to drive a viscous fluid upslope through a 12 cm diameter pipe? The length of the pipe is 50 m, slope is 20 deg. At the end of the pipe, the pressure is 1 bar. Oil of specific gravity 0.85 and viscosity 0.01 kg/m-s, and the target flow rate is 0.25 cubic meter per minute. (i) What is the pressure needed to drive the flow if there were no slope? (ii) The supplier is out of 12 cm pipes and your manager wonders if multiple 8 cm diameter pipes can be used side by side to achieve the same total flow rate at the same driving pressure, determine how many 8 cm pipes are needed. Round up to nearest integer. [Figure: pipe of length 50 m inclined at 20° to the horizontal, driving pressure P at the lower end, 1 bar at the upper end] ## Short answer (i) Without slope, pressure needed ≈ 100.41 kPa (friction drop ~409 Pa). (ii) On the 20° slope, required pressure ≈ 243.0 kPa. For multiple 8 cm pipes, flow scales as D⁴, so n = (12/8)⁴ = 5.06, thus 6 pipes are required. ## Answer **Data.** $\rho = 850$ kg/m³ (s.g. 0.85), $\mu = 0.01$ Pa s, $D = 0.12$ m, $L = 50$ m, slope 20°, outlet pressure 1 bar $= 10^5$ Pa, $Q = 0.25$ m³/min $= 0.004167$ m³/s. The oil is viscous, so Hagen-Poiseuille (laminar) flow is assumed. #### (i) Pressure needed if there were no slope Mean velocity and Reynolds number: $$ V = \frac{Q}{\pi D^2/4} = 0.3684\ \text{m/s}, \qquad Re = \frac{\rho V D}{\mu} = 3758 $$ (This is above 2000, so the flow is really in the critical zone. The laminar formula is used because the oil is described as viscous and the flow is steady. A turbulent estimate would give a friction drop of only about 1 kPa, so the friction loss is small in either case.) Friction pressure drop (Hagen-Poiseuille): $$ \Delta p_f = \frac{128\mu L Q}{\pi D^4} = \frac{128(0.01)(50)(0.004167)}{\pi(0.12)^4} = 409.3\ \text{Pa} $$ Pressure at the inlet without slope: $P = 10^5 + 409.3 = 100409.3$ Pa $\approx 100.41$ kPa. #### Pressure needed on the actual slope Extra pressure to lift the oil through the height $L\sin 20^\circ = 17.10$ m: $$ \Delta p_z = \rho g L \sin 20^\circ = 850(9.81)(50)(0.3420) = 142597\ \text{Pa} $$ $$ P = 10^5 + 409.3 + 142597 = 243006\ \text{Pa} \approx 243.0\ \text{kPa} $$ #### (ii) Number of 8 cm pipes in parallel At the same driving pressure the friction drop per pipe is the same, and in laminar flow $Q \propto D^4$ (the slope term is the same for any pipe size). Flow per 8 cm pipe compared with the 12 cm pipe: $$ \frac{Q_8}{Q_{12}} = \left(\frac{8}{12}\right)^4 = 0.1975 $$ $$ n = \frac{Q_{12}}{Q_8} = \left(\frac{12}{8}\right)^4 = 5.0625 \;\Rightarrow\; n = 6 \text{ pipes (rounded up)} $$ Check with 6 pipes: each carries $Q/6 = 0.000694$ m³/s, $V = 0.138$ m/s, $Re = 939$ (laminar), and the friction drop is $345.4$ Pa, which is less than 409.3 Pa, so the driving pressure is not exceeded. **Answer:** (i) $P \approx 100.41$ kPa without slope ($\Delta p_f = 409$ Pa); on the 20° slope $P \approx 243.0$ kPa; (ii) 6 pipes of 8 cm diameter. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the inlet pressure on the slope consist of? A. Outlet pressure + friction drop + rho g L sin(theta) B. Outlet pressure only C. Friction drop only D. Outlet pressure - friction drop Answer: A. P = 1 bar + 409 Pa friction drop + 142.6 kPa to lift oil through 17.1 m. (Key point: Pressure components) 2. Which formula gives the laminar friction pressure drop? A. 128 mu L Q / (pi D^4) B. 64 mu L Q / D^2 C. f rho V^2 / 2 D. rho g Q L Answer: A. Hagen-Poiseuille: delta p = 128 mu L Q/(pi D^4). (Key point: Friction drop) 3. Which term vanishes when there is no slope? A. The friction drop B. The outlet pressure C. The elevation term rho g L sin(theta) D. The flow rate Answer: C. Without slope the pressure is 1 bar plus friction only, about 100.41 kPa. (Key point: No slope case) 4. Why does the 8 cm pipe carry (8/12)^4 of the flow of the 12 cm pipe at the same pressure? A. Q is proportional to D^2 in laminar flow B. Q is proportional to D^4 in laminar flow at the same pressure drop C. Q is independent of D D. Q is proportional to D Answer: B. From Hagen-Poiseuille Q is proportional to D^4, so n = (12/8)^4 = 5.06, rounded up to 6. 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