--- title: "Hose siphon discharge and minimum pressure" subject: "Hydraulics" chapter: 2 key: "semester_4/hydraulics:hose-siphon-flow-and-minimum-pressure" source: https://www.bctengineer.com/q/semester_4/hydraulics/hose-siphon-flow-and-minimum-pressure site: BCT Engineering --- # Hose siphon discharge and minimum pressure Hydraulics · Chapter 2: Simple pipe flow problems and solution - Asked in: 2070 Bhadra, 8 marks - Video: [How a siphon (syphon) works](https://www.youtube.com/watch?v=xmZNUqcSI94) — Fluids Explained - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/hose-siphon-flow-and-minimum-pressure - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2 ## Question Liquid (s.g. = 0.6, $\nu = 5.0 \times 10^{-7}$ m²/s) is drawn from a tank through a hose of inside diameter 25 mm (see figure). The relative roughness for the hose is 0.0004. Calculate the volumetric flow and the minimum pressure in the hose. The total length of hose is 9 m and the length of hose to point A is 3.25 m. Neglect minor losses at head entrance. [Figure: hose inlet 1 m below the liquid surface in the tank; top point A of the hose 1.5 m above the liquid surface; hose outlet 5 m below point A] ## Short answer Volumetric flow is 1.46 L/s (0.00146 m³/s). The minimum gauge pressure at the highest point A is -17.95 kPa. ## Answer **Data.** s.g. 0.6 so $\rho = 600$ kg/m³, $\nu = 5.0\times10^{-7}$ m²/s, $D = 0.025$ m, $k/D = 0.0004$, total hose length $L = 9$ m, length to the highest point A $= 3.25$ m. From the figure: A is 1.5 m above the liquid surface; the outlet is 5 m below A, so it is $5 - 1.5 = 3.5$ m below the liquid surface. The hose discharges freely to the atmosphere. #### Volumetric flow Energy equation from the tank surface to the outlet (entrance loss neglected, velocity head at the outlet kept): $$ 3.5 = \frac{V^2}{2g}\left(1 + f\frac{L}{D}\right) = \frac{V^2}{2g}\left(1 + f\frac{9}{0.025}\right) = \frac{V^2}{2g}\left(1 + 360f\right) $$ $f$ depends on $Re = VD/\nu$ and $k/D = 0.0004$ (Colebrook-White), so iterate: 1. Assume $f = 0.02$: $\dfrac{V^2}{2g} = \dfrac{3.5}{8.2} = 0.427$ m, $V = 2.89$ m/s, $Re = 1.4\times10^5$. 2. Colebrook: $f \approx 0.0189$ and the result converges (3 iterations) to $$ f = 0.01891,\qquad \frac{V^2}{2g} = \frac{3.5}{1 + 360(0.01891)} = 0.4483\ \text{m},\quad V = 2.966\ \text{m/s},\quad Re = 148284 $$ $$ Q = \frac{\pi}{4}(0.025)^2(2.966) = 0.001456\ \text{m}^3/\text{s} = 1.456\ \text{l/s} \;(87.3\ \text{l/min}) $$ #### Minimum pressure (at the top point A) Energy equation from the tank surface (gauge pressure 0) to A, 1.5 m above the surface, with friction over 3.25 m: $$ 0 = 1.5 + \frac{p_A}{\gamma} + \frac{V^2}{2g}\left(1 + f\frac{3.25}{0.025}\right), \qquad f\frac{3.25}{0.025} = 2.458 $$ $$ \frac{p_A}{\gamma} = -1.5 - 0.4483(3.458) = -3.050\ \text{m of liquid} $$ $$ p_A = \rho g\left(\frac{p_A}{\gamma}\right) = 600(9.81)(-3.050) = -17954\ \text{Pa} \approx -17.95\ \text{kPa (gauge)} $$ This is about 83.3 kPa absolute, above the vapour pressure of the liquid unless it is very volatile. **Answer:** $Q = 1.46$ l/s ($0.001456$ m³/s, $V = 2.97$ m/s); minimum pressure at A $= -17.95$ kPa gauge. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What head drives the flow through the hose? A. 5 m, the drop from A B. 3.5 m, the outlet being 5 - 1.5 below the tank surface C. 1.5 m D. 1 m Answer: B. The outlet is 5 m below A, and A is 1.5 m above the surface, so the net fall is 3.5 m. (Key point: Driving head) 2. Why is the flow found by iteration? A. The density is unknown B. The head is unknown C. f depends on Re (hence on V) via Colebrook-White, while V depends on f D. Because the hose length is unknown Answer: C. 3.5 = V^2/2g (1 + 360 f) with f a function of Re and k/D, so V and f are iterated: V = 2.966 m/s. (Key point: Why trial is needed) 3. Where is the minimum pressure and why? A. At the outlet, since it is lowest B. At the tank surface C. At the inlet D. At the top point A, as it is highest and has accumulated friction Answer: D. A is 1.5 m above the surface, so pressure falls there to -3.05 m of liquid, about -17.95 kPa gauge. (Key point: Minimum pressure location) 4. How is -3.050 m of liquid converted to pressure? A. Multiply by rho g with rho = 600, giving -17.95 kPa B. Multiply by 1000 x 9.81 C. Divide by rho D. Multiply by 9.81 only Answer: A. The liquid has s.g. 0.6, so p = 600 x 9.81 x (-3.050) = -17954 Pa. 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