Derive siphon inlet to outlet leg ratio

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).

        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.