---
title: "Pipe over hill with cavitation limit"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:pipe-over-hill-cavitation-limit"
source: https://www.bctengineer.com/q/semester_4/hydraulics/pipe-over-hill-cavitation-limit
site: BCT Engineering
---

# Pipe over hill with cavitation limit

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2072 Asoj, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/pipe-over-hill-cavitation-limit
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

Two reservoirs are connected by a pipe 1000 m long of diameter 300 mm. The pipe passes over a hill whose height is 5 m above the level of water in the upper reservoir. The difference in water levels in the two reservoirs is 13 m. If the absolute pressure of water anywhere in the pipe is not allowed to fall below 1.2 m of water to prevent cavitations, calculate the length of pipe in the portion between the upper reservoir and the hill summit; and also the discharge through the pipe. Assume the reservoirs are open to the atmosphere having atmospheric pressure of 760 mm of mercury. Take friction factor, f = 0.032 and neglect bend losses.

## Short answer

Discharge is 0.1093 m³/s (109.3 L/s). The maximum pipe length from the upper reservoir to the summit is about 309 m, determined by the minimum absolute pressure of 1.2 m water at the hilltop.

## Answer

**Data.** $L = 1000$ m, $D = 0.3$ m, $f = 0.032$, level difference $H = 13$ m, summit 5 m above the upper reservoir level, minimum absolute pressure 1.2 m of water. Atmospheric pressure $= 760$ mm Hg $= 0.76(13.6) = 10.336$ m of water. Head losses are taken as pipe friction only (bend losses neglected; entry and exit losses would change the result by less than 1%).

#### Discharge
$$
H = f\frac{L}{D}\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.032\left(\frac{1000}{0.3}\right) = 106.67
$$
$$
\frac{V^2}{2g} = \frac{13}{106.67} = 0.12187\ \text{m} \;\Rightarrow\; V = 1.546\ \text{m/s}
$$
$$
Q = \frac{\pi}{4}(0.3)^2(1.546) = 0.1093\ \text{m}^3/\text{s} = 109.3\ \text{l/s}
$$

#### Length of pipe from the upper reservoir to the summit
The critical point is the summit S, where absolute pressure head is 1.2 m. Energy equation from the upper reservoir surface (datum, atmospheric) to the summit:
$$
\frac{p_a}{\gamma} = 5 + 1.2 + \frac{V^2}{2g} + f\frac{l_1}{D}\frac{V^2}{2g}
$$
$$
10.336 = 6.2 + 0.12187 + f\frac{l_1}{D}(0.12187) \;\Rightarrow\; f\frac{l_1}{D}\,(0.12187) = 4.0141
$$
$$
l_1 = \frac{4.0141\times 0.3}{0.032\times 0.12187} = 308.8\ \text{m}
$$

**Answer:** The summit may be at most about 309 m of pipe from the upper reservoir; the discharge is $0.1093$ m³/s ($109.3$ l/s).

## Check yourself

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

1. How is the discharge found?
   A. From 13 = f L/D x V^2/2g for the whole pipe
   B. From the hill height
   C. From the pressure at the summit
   D. From 5 m head
   Answer: A. The level difference equals friction loss over the full 1000 m, giving V = 1.546 m/s and Q = 0.1093 m3/s. (Key point: Discharge)

2. How is atmospheric pressure converted to metres of water?
   A. 760 mm Hg = 0.76 m of water
   B. 760 mm Hg = 13.6 m
   C. 760 mm Hg = 0.76 x 13.6 = 10.336 m of water
   D. 760 mm Hg = 7.6 m
   Answer: C. Mercury is 13.6 times denser than water. (Key point: Atmospheric head)

3. Where does pressure become critical, and what equation gives l1?
   A. At the lower reservoir; energy from lower tank
   B. At the pipe outlet; momentum equation
   C. At the inlet; Bernoulli with no losses
   D. At the summit; p_a/gamma = 5 + 1.2 + V^2/2g + f l1/D x V^2/2g
   Answer: D. The summit has the highest elevation and lowest pressure; solving gives l1 = 308.8 m. (Key point: Critical point)

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