Calculating the flow rate in a non-ideal siphon? When calculating flow rate in a siphon, the simplifying assumption is usually made that the velocity head at the input is effectively zero. This requires that the cross-sectional area at the input be much larger than at the output. But what if I don't want to make that assumption? What if the input has a relatively small area?
Using Bernoulli's principal, I started by stating that $$\frac{v_{in}^2} {2g}+h=\frac{v_{out}^2}{2g}$$
After substituting velocity in terms of flow rate and area, and rearranging to solve for flow rate I got the equation $$Q=A_{in}A_{out}\sqrt{\frac{2gh}{A_{in}^2-A_{out}^2}}$$
Either I've made a mistake, or I'm not understanding what this equation is telling me. It appears that the flow rate would approach infinity as $A_{in}$ gets smaller and approaches $A_{out}$. Did I make a mistake somewhere, or am I still making a simplifying assumption that would affect the flow rate when $A_{in}$ and $A_{out}$ are close to each other?
Edit: I'm going to add my step-by-step calculations to make it easier to spot a mistake.
$$\frac{v_{in}^2}{2g}+h=\frac{v_{in}^2}{2g}$$
$$v_{in}^2+2gh=v_{out}^2$$
$$v_{in}=\frac{Q}{A_{in}},v_{out}=\frac{Q}{A_{out}}$$
$$\frac{Q^2}{A_{in}^2}+2gh=\frac{Q^2}{A_{out}^2}$$
$$Q^2(\frac{1}{A_{out}^2}-\frac{1}{A_{in}^2})=2gh$$
$$Q^2(\frac{A_{in}^2-A_{out}^2}{A_{in}^2A_{out}^2})=2gh$$
$$Q^2=\frac{2ghA_{in}^2A_{out}^2}{A_{in}^2-A_{out}^2}$$
$$Q=A_{in}A_{out}\sqrt{\frac{2gh}{A_{in}^2-A_{out}^2}}$$
 A: Your calculation is correct. So is your conclusion.
What is happening here is that Bernoulli's equation conserves energy. The work done on the fluid by gravity as it falls through height $h$ equals its increase in kinetic energy, which is proportional to $v^2$ :
$$mgh=\frac12 m(v_{out}^2-v_{in}^2)=m\frac{v_{out}+v_{in}}{2}(v_{out}-v_{in})=m \bar{v} \Delta v$$ 
For a fixed value of $h$, if the increase in velocity of the fluid $\Delta v$ is small then the average speed between input and output $\bar{v}$ must be correspondingly high. In the limit that $\Delta v \to 0$ the average speed $\bar{v} \to \infty$ and so does the volume flow rate $Q=A\bar{v}$.
Bernoulli's equation does not take into account frictional mechanisms which transform the streamline kinetic energy of the fluid into thermal energy - such as viscosity and turbulence. These losses usually increase as $\bar{v}$ increases : a greater fraction of the loss in gravitational potential energy is dissipated as thermal energy, with very little increase in kinetic energy, until an equilibrium is reached. It is then possible to have $\Delta v \approx 0$ while $\bar{v} \ll \infty$.
Another point to note is that if $A_{in} \approx A_{out}$ then the hydrostatic head $h$ will not be constant in the above equation (nor in your equation for $Q$). It will be falling close to the speed $\bar{v}$. So $\Delta v$ can become small without $\bar{v}$ becoming large, because $h$ is falling at the same time as $\Delta v$.
