---
title: "Derive siphon inlet to outlet leg ratio"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:siphon-inlet-outlet-leg-ratio"
source: https://www.bctengineer.com/q/semester_4/hydraulics/siphon-inlet-outlet-leg-ratio
site: BCT Engineering
---

# Derive siphon inlet to outlet leg ratio

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2075 Bhadra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/siphon-inlet-outlet-leg-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

Derive an expression for ratio of length of inlet to outlet leg for typical siphon as follows: $\frac{l_1}{l_2} \le \frac{P_a - (Z_B - Z_A)}{(Z_B - Z_C) - P_a}$, where $P_a$ is atmospheric pressure, $Z_A$, $Z_B$, $Z_C$ are elevation of inlet, summit and outlet of syphon.

## Short answer

Applying Bernoulli and constant friction slope, the siphon’s inlet-to-outlet leg ratio must satisfy l1/l2 ≤ [Pa – (ZB – ZA)] / [(ZB – ZC) – Pa] to prevent absolute pressure at the summit from dropping below zero.

## Answer

**Setting up.** A siphon has an inlet leg $AB$ (length $l_1$) from the intake A in the upper reservoir up to the summit B, and an outlet leg $BC$ (length $l_2$) down to the outlet C. The pipe has uniform diameter, so the friction loss per unit length is the same in both legs. Neglect minor losses and the velocity head. Let $Z_A$, $Z_B$, $Z_C$ be the elevations of intake water level, summit and outlet, and $P_a$ the atmospheric pressure head (m of water).

```text
        B (summit)
       /\
      /  \
 A __/    \
 ~~~ l1    \ l2
(upper      \___ C (outlet)
 reservoir)
```

**Step 1: energy gradient.** The whole fall is used to overcome friction over the total length:
$$
H = Z_A - Z_C = h_f, \qquad S = \frac{h_f}{l_1 + l_2} = \frac{Z_A - Z_C}{l_1 + l_2}
$$

**Step 2: pressure at the summit.** Applying Bernoulli between A and B (velocity head is the same and ignored):
$$
\frac{p_A}{\gamma} + Z_A = \frac{p_B}{\gamma} + Z_B + S\,l_1 \;\Rightarrow\; \frac{p_B}{\gamma} = -\left[(Z_B - Z_A) + S\,l_1\right] \quad\text{(gauge)}
$$
The pressure at B is below atmospheric. The water column will break (cavitation, air release) if the absolute pressure falls below zero, i.e. $\dfrac{p_B}{\gamma} < -P_a$. For the siphon to work:
$$
(Z_B - Z_A) + S\,l_1 \le P_a \;\Rightarrow\; S\,l_1 \le P_a - (Z_B - Z_A) \quad\cdots(1)
$$

**Step 3: substitute the gradient.**
$$
\frac{(Z_A - Z_C)\,l_1}{l_1 + l_2} \le P_a - (Z_B - Z_A)
$$
Let $X = P_a - (Z_B - Z_A)$ and $H = Z_A - Z_C$. Then $H\,l_1 \le X\,(l_1 + l_2)$, so
$$
l_1\,(H - X) \le X\,l_2 \;\Rightarrow\; \frac{l_1}{l_2} \le \frac{X}{H - X}
$$
Now
$$
H - X = (Z_A - Z_C) - P_a + (Z_B - Z_A) = (Z_B - Z_C) - P_a
$$
$$
\boxed{\frac{l_1}{l_2} \le \frac{P_a - (Z_B - Z_A)}{(Z_B - Z_C) - P_a}}
$$

**Meaning.** For a siphon of given height the inlet leg must be shorter than the limit given by this ratio. For the denominator to be positive the outlet must be lower than $P_a$ below the summit; otherwise the siphon works for any ratio. In practice, a smaller limiting pressure (vapour pressure, about 2 to 3 m of water absolute) is used in place of zero.

## Check yourself

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

1. Why is the friction gradient the same in both legs?
   A. The pipe has uniform diameter and flow, so S = (Z_A - Z_C)/(l1 + l2)
   B. The legs have equal lengths
   C. Minor losses dominate
   D. The summit is at atmospheric pressure
   Answer: A. The whole fall Z_A - Z_C is spent on friction over the total length l1 + l2. (Key point: Energy gradient)

2. What does Bernoulli from A to B give for the summit pressure head (gauge)?
   A. -(Z_B - Z_A)
   B. -[(Z_B - Z_A) + S l1]
   C. +S l1
   D. -(Z_B - Z_C)
   Answer: B. The summit is above A by (Z_B - Z_A) and friction S l1 is lost, so the pressure is below atmospheric by that sum. (Key point: Summit pressure)

3. What is the limiting condition for the siphon to work?
   A. Gauge pressure at B equals atmospheric
   B. Summit pressure at least 2 P_a
   C. (Z_B - Z_A) + S l1 is at most P_a, so absolute pressure at B is not below zero
   D. S l1 is at least P_a
   Answer: C. The absolute pressure must not fall below zero (in practice vapour pressure), giving S l1 <= P_a - (Z_B - Z_A). (Key point: Limiting condition)

4. With X = P_a - (Z_B - Z_A) and H = Z_A - Z_C, what gives l1/l2 <= X/(H - X)?
   A. H l1 >= X l2
   B. H l2 <= X l1
   C. l1 + l2 <= X
   D. H l1 <= X (l1 + l2)
   Answer: D. Rearranging H l1 <= X(l1 + l2) gives l1(H - X) <= X l2, and H - X = (Z_B - Z_C) - P_a. (Key point: Algebra to the ratio)

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