The pressure at B is a little tricky. Right at point B, the fluid velocity is equal to that within the exit tube running from B to C. But, within the tank, a few tube diameters upstream of exit B, the flow is converging rapidly toward the exit hole. And, along with the flow convergence, the fluid velocity is increasing from a velocity of basically zero a few diameters upstream, to the much higher velocity $v_{BC}$ at the exit. So right at point B, the pressure is equal to $$p_A=p_B+\rho g (-y_b)+\frac{1}{2}\rho v_{BC}^2\tag{1}$$The condition at point C is $$p_A=p_C=p_C+\rho g (-y_C)+\frac{1}{2}\rho v_{BC}^2\tag{2}$$
Or equivalently: $$\rho g (-y_C)+\frac{1}{2}\rho v_{BC}^2=0\tag{3}$$
Interestingly, if we combine Eqns. 1 and 3, we obtain:
$$p_B=p_A-\rho g(y_c-y_b)$$
So, as a result of the exit tube discharging at a lower depth than the tank exit, the pressure at point B is actually below atmospheric (i.e., suction). This is similar to what happens with a siphon.
Could you review my edited question?$^1$
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:
- set $v_a$ and therefore $h$ to zero.
- then $P_A$=$P_C$.
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.
Best Answer
The diagram and text does seem to be a bit confusing. Here's my best guess at interpreting the diagram and text: The $P$ is actually the ambient atmospheric pressure. In order for the first equation that you wrote down to make sense, the pressure $P_A$ then has to be the gauge pressure of the rubber bulb on the left (i.e, the pressure of the air in the bulb measured with respect to atmospheric pressure), and $P_B$ is the absolute pressure of the air at point B.
Now the pressure differential which determines the height of the liquid in the column is $P-P_B$. If this quantity is positive, the liquid in the column goes with respect to the level of the liquid in the surrounding container. If this quantity is negative, the liquid in the column goes down with respect to the surrounding liquid level.
From your first equation, we then have $P-P_B=(1/2){\sigma}V^2-P_A$, and so it follows that ${\rho}gH=(1/2){\sigma}V^2-P_A$, where H is the height of the liquid in the column. This seems to make sense. If the right end of the open tube is stopped up so that little or no air can flow out the right end of the tube, then the velocity V is negligible or zero and the height H is determined by just the gauge pressure pressure $P_A$ in rubber bulb. Note that in this case that squeezing the bulb makes the liquid in the column go down (i.e., H<0), as expected. If, on the other hand, the right end of the open tube is left completely open, then the right side of the equation is dominated by the $(1/2){\sigma}V^2$ term and the column of liquid rises (i.e., H>0).