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. $$\frac{v_{in}^2}{2g}+h=\frac{v_{in}^2}{2g}$$ $$v_{in}^2+2gh=v_{out}^2$$ $$v_{in}=\frac{Q}{A_{in}},v_{out}=\frac{Q}{A_{out}}$$ $$\frac{Q^2}{A_{in}^2}+2gh=\frac{Q^2}{A_{out}^2}$$ $$Q^2(\frac{1}{A_{out}^2}-\frac{1}{A_{in}^2})=2gh$$ $$Q^2(\frac{A_{in}^2-A_{out}^2}{A_{in}^2A_{out}^2})=2gh$$ $$Q^2=\frac{2ghA_{in}^2A_{out}^2}{A_{in}^2-A_{out}^2}$$ $$Q=A_{in}A_{out}\sqrt{\frac{2gh}{A_{in}^2-A_{out}^2}}$$
