Consider the following experiment: Rising a water in a straw.
Legends:
A: a point on the top end of the straw.B: a point at the boundary between air and water.C: a point on the water surface
Applying Bernoulli's equation for the air in column AB:
\begin{align}
P_A +\tfrac{1}{2}\rho_a v_a^2 &= P_B \tag{1}\\
\end{align}
I am assuming that the hydrostatic pressure due to the air column and the air speed at B are too small and neglected.
Applying Bernoulli's equation for the air in column BC:
\begin{align}
P_B + \rho_w g h &= P_C \tag{2}\\
\end{align}
I am assuming that water speed at both B and C are too small and neglected.
Eliminating $P_B$ we have
\begin{align}
P_A +\tfrac{1}{2}\rho_a v_a^2 + \rho_w g h &= P_C \tag{3}\\
\end{align}
and because $P_A=P_C$ (atmospheric pressure), the final equation becomes
\begin{align}
\tfrac{1}{2}\rho_a v_a^2 + \rho_w g h &= 0 \tag{4}\\
\end{align}
Contradictory
The left hand side cannot be zero. What does it mean? Which part of my calculation is wrong?
Best Answer
In your eqn. $2$, $P_C$ is known to be $P_a$ by Pascal's law. $P_C$ is therefore always constant i.e. $\forall$ values of $h$ or $v_a$.
Also, in your eqn. $3$, when you try to determine what $P_C$ should be, you did the following:
So when $v_a$ and $h$ are zero, the equality holds, but not otherwise. So when you ultimately solve $h$ for a non-zero $v_a$, you can't use the equality you just derived - $P_A$ must be treated as an unknown. Eqn. $4$ therefore never follows.
So let's say you set $P_C=P_a$. How do you solve your eqn. $3$? What is $P_A$? One must look to your eqn. $1$.Due to initial conditions, the RHS of your eqn. $1$, can be determined to be $P_a$.
But this is always $P_B$ right? So is $P_B$ always at atmospheric pressure? This doesn't make any sense.
It took me a while to understand, but this a deeper mistake. A conceptual one. The Bernoulli eqn., in my opinion can not be applied to points $A$ and $B$ i.e. $$P_A+\rho_a v_{a,A}^2+\rho_a g h_A \ne P_B+\rho_a v_{a,B}^2+\rho_a g h_B \tag {0}$$
such that $v_{a,A}$ is the air velocity at $A$ with which one blows.
Mathematically, this is reflected in the fact that while the LHS of eqn. $0$ is always $P_a$ because of the initial condition, the RHS approximated to just $P_B$ obviously isn't - there is definitely a pressure drop.
Physically its because the LHS and RHS of eqn. $0$ represent different flows - the horizontal cloud of blown air that moves over the straw and the air within the straw, resp. One is clearly blowing, while the other is stagnant (at equilibrium). These flows have $A$ as the common point. The Bernoulli equation is derived by assuming a flow element conserving its energy as it moves within the flow stream. This assumption doesn't hold when two different flows merge.
So here's a correct derivation
using your notation.(all heights measured wrt. $C$)
At $A$,
$$P_A+\rho_a g h_{A}+\rho_a v_{a,A}^2/2=const.=c \tag{1}$$
This equation is taking $A$ as part of the 'blow flow'. As the blown air moves over the straw's cross-section, there is hardly any change in its height. Hence eqn. $1$ reduces to
$$P_A+\rho_a v_{a,A}^2/2=const.=c' \tag{2}$$
and, as stated earlier, with the initial condition $v_{a,A}= 0,P_A=P_a$, becomes
$$P_A+\rho_a v_{a,A}^2/2=P_a \tag{3}$$
At $A$, the following is also satisfied
$$P_A+\rho_a g h_A+\rho_a v_{a,A}^2/2 =P_B+\rho_a g h_{B}+\rho_a v_{a,B}^2/2 \tag {4}$$
This eqn. treats $A$ and $B$ as part of the 'stagnant flow', the air column inside the straw. In this way, its consistent with the refutal in eqn. ($-1$).
Since, the air column is stagnant $v_{a,A},v_{a,B}=0$. Also since $\rho_a$ is pretty small to cause gauge pressure at $B$ we can ignore it, but lets keep it anyways. Eqn. $4$ then reduces to
$$P_A =P_B+\rho_a g (h-H) \tag {5}$$
where I have taken $h_A=H, h_B=h$.
From Pascal's law, $P_C=P_a$ therefore
$$P_C=P_B+\rho_w g h=P_a\tag{6}$$
Substituting $P_A$ from eqn. $3$ and $P_B$ from eqn. $6$ into eqn. $5$ and solving for $h$ gives $$ \begin{align} h&=\frac{\rho_a}{\rho_w-\rho_a}\left(\frac{v_{a,A}^2}{2g} -H\right)\tag{7}\\ &\approx\frac{\rho_a}{\rho_w}\frac{ v_{a,A}^2}{2g}\tag{8}\\ &\approx \frac{v_a^2}{20000} m \tag{in SI units} \end{align} $$
This expression seems too small to enable suction by blowing over a straw. I am not sure what the problem is. e.g. blowing at say $10\,ms^{-1}$ only generates $5\,mm$ of lift. Maybe using another straw to blow instead of blowing directly would increase the speed.
$^1$ I re-wrote the answer to stress more on your query - the contradiction - and also to avoid notational confusion, and some corrections.