Consider a burette (like the one in a chemistry lab) with uniform cross-section and open at both ends so that they are exposed to the atmosphere. Once we open the stopcock/plug (P in the diagram) the water filled inside it will begin to flow down. The speed of the water throughout should be the same as the cross sectional area(A in diagram)is same throughout.But if an analysis using Bernoulli's equation is done involving the top portion(T) and the bottom portion(T) of the water column ( calculation in diagram ), the result is absurd.

What is the mistake in my reasoning? Isn't Bernoulli's equation supposed to be based on conservation of energy ? I believe that Bernoulli's equation is used for a specific instance of time only where the energy remains constant but here the energy changes once the stopcock is opened.

Rough diagram and calculation

  • $\begingroup$ The form of the Bernoulli equation you employed applies to a steady state flow (or approximately steady state), and this is not a steady state flow. The velocity is changing with time. There is a form of the Bernoulli equation you can use that includes non-steady state flow. $\endgroup$ Jun 17 '19 at 13:17
  • $\begingroup$ Could you provide a link to the non steady flow Bernoulli's equation. $\endgroup$
    – UnoWindJTS
    Jun 17 '19 at 15:54
  • $\begingroup$ Google "transient Bernoulli." $\endgroup$ Jun 17 '19 at 17:00

This is the misconception on the "area".

The continuity equation reads: \begin{align} v_{i} A_{i} \; = \; v_{f} A_{f} \end{align}

  • $v_{i}$ and $v_{f}$ denotes the initial, and final velocity of the fluid patch.
  • $A_{i}$ and $A_{f}$ denotes the initial and final cross section area of the fluid patch, NOT the burette cross section area!

Without using Bernoulli's equation, we should be able to deduce that $v_{f} > v_{i}$. The reason is that the particles are accelerating due to gravity.


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