---
title: "Pipe passing over barrier between reservoirs"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:flexible-pipe-over-barrier-height"
source: https://www.bctengineer.com/q/semester_4/hydraulics/flexible-pipe-over-barrier-height
site: BCT Engineering
---

# Pipe passing over barrier between reservoirs

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2073 Magh, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/flexible-pipe-over-barrier-height
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

Two reservoirs are joined by a sharp-ended flexible pipe 100 mm diameter and 36 m long. The ends of the pipe differ in level by 4 m; the surface level in the upper reservoir is 1.8 m above the pipe inlet while that in the lower reservoir is 1.2 m above the pipe outlet. At the position 7.5 m horizontally from the upper reservoir the pipe is required to pass over a barrier. Assuming that the pipe is straight between its inlet and the barrier and that f = 0.04, determine the greatest height to which the pipe may rise at the barrier if the absolute pressure in the pipe is not to be less than 40 kPa. Consider all losses. (Take atmospheric pressure = 101.3 kPa). [Figure: upper reservoir A with water 1.8 m above the pipe inlet; the pipe rises to a barrier X at horizontal distance 7.5 m, height h; lower reservoir with water 1.2 m above the pipe outlet; pipe ends differ in level by 4 m; difference between the two water surfaces 4.6 m]

## Short answer

With all losses, velocity is 2.382 m/s. Energy balance to the barrier, allowing a minimum absolute pressure of 40 kPa, gives a maximum pipe rise of 6.47 m above its inlet (4.67 m above the upper reservoir surface).

## Answer

**Data.** $D = 0.1$ m, $L = 36$ m, $f = 0.04$. Upper water surface is 1.8 m above the inlet; lower water surface is 1.2 m above the outlet; ends differ in level by 4 m, so the difference of the water surfaces is $H = 4 + 1.8 - 1.2 = 4.6$ m. Barrier at 7.5 m horizontally from the upper reservoir. Minimum absolute pressure 40 kPa; $p_a = 101.3$ kPa. All losses: entrance $0.5\dfrac{V^2}{2g}$, exit $1.0\dfrac{V^2}{2g}$, friction.

#### Step 1: velocity in the pipe
$$
H = \left(0.5 + f\frac{L}{D} + 1.0\right)\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.04\left(\frac{36}{0.1}\right) = 14.4
$$
$$
4.6 = 15.9\,\frac{V^2}{2g} \;\Rightarrow\; \frac{V^2}{2g} = 0.2893\ \text{m},\quad V = 2.382\ \text{m/s}\quad(Q = 18.71\ \text{l/s})
$$

#### Step 2: energy equation from the upper surface to the barrier X
Take the pipe inlet level as datum. Let the barrier be at height $h_X$ above the inlet. Since the pipe is straight from the inlet to the barrier, its length is $l = \sqrt{7.5^2 + h_X^2}$.

Absolute pressure heads: $\dfrac{p_a}{\gamma} = \dfrac{101.3}{9.81} = 10.326$ m; minimum $\dfrac{p_X}{\gamma} = \dfrac{40}{9.81} = 4.077$ m.
$$
\frac{p_a}{\gamma} + 1.8 = h_X + \frac{p_X}{\gamma} + \frac{V^2}{2g}\left(1 + 0.5 + f\frac{l}{D}\right)
$$
$$
h_X = 10.326 + 1.8 - 4.077 - 0.2893\left(1.5 + 0.4\,l\right)
$$
Iterating ($h_X$ then $l$): $l = 9.90$ m, so the bracket is $5.46$ and
$$
h_X = 6.47\ \text{m above the pipe inlet}
$$

**Answer:** The pipe may rise to about 6.47 m above its inlet at the barrier (about 4.67 m above the upper reservoir water level).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the head difference between the two water surfaces?
   A. 4 m
   B. 4.6 m = 4 + 1.8 - 1.2
   C. 3.4 m
   D. 1.8 m
   Answer: B. Upper surface is 1.8 m above the inlet and lower surface is 1.2 m above the outlet, with 4 m between ends. (Key point: Driving head)

2. How is V found?
   A. 4.6 = f L/D x V^2/2g
   B. 4.6 = V^2/2g
   C. 4.6 = (0.5 + 14.4 + 1.0) V^2/2g, so V^2/2g = 0.2893 m
   D. 4.6 = 14.4 V^2/2g only
   Answer: C. All losses (entrance, friction over 36 m, exit) use up the 4.6 m, giving V = 2.382 m/s. (Key point: Velocity)

3. Why is the pipe length to the barrier l = sqrt(7.5^2 + h^2)?
   A. The pipe is vertical
   B. The pipe is straight from inlet to barrier, so it is the hypotenuse
   C. The pipe length equals 7.5 m
   D. The pipe length equals h
   Answer: B. The straight pipe forms the hypotenuse, so friction depends on h and the equation is solved by iteration. (Key point: Barrier energy equation)

4. What fixes the highest barrier position?
   A. Maximum velocity
   B. Exit loss
   C. Reservoir size
   D. Absolute pressure not less than 40 kPa at the barrier
   Answer: D. The energy equation from the upper surface to the barrier with p_X/gamma = 4.077 m gives h = 6.47 m. (Key point: Pressure limit)

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