Concept regarding Venturi Tube-Bernoulli application I was recently studying applications of Bernoulli Equation and came across the Venturi tube. This is diagram I have used to analyse the venturimeter. I understand how we obtain the first equation using bernoulli theorem which is
$$P_1 - P_2 =(1/2)ρ(v_2^2 - v_1^2)      \tag{1}$$
and also the continuity equation
$$A_1 v_1 = A_2 v_2. \tag{2}$$
However, I am unable to process how to obtain the following third equation
$$ P_1 - P_2 = ρgh $$
where $h$ is difference in the heights of the liquid level in the two tubes and $ρ$ is density of fluid.
Lets say the atmospheric pressure at the top of each tube is $P$. Now since the fluid in the two VERTICAL tubes are at rest and not moving, their velocities are 0. Hence, if I proceeded by applying Bernoulli equation.
$$P_1 +ρ(v_1^2)/2 = ρg(h_1) + P     \tag{3}$$  and
$$P_2 + ρ(v_2^2)/2 = ρg(h_2) + P     \tag{4}$$
Here $P_1$ and $P_2$ are the pressures at the points in the tube and constriction respectively and the points are at the SAME HORIZONTAL LEVEL.
Subtracting (3) and (4) and even using (1)  does not yield me
$$P_1 - P_2 = ρgh $$
rather gives me $0 = ρg(h)$ which makes no sense whatsoever.
What am I missing here and how do I obtain the right result ? 
 A: 
Lets say the atmospheric pressure at the top of each tube is . Now since the fluid in the two VERTICAL tubes are at rest and not moving, their velocities are 0. Hence, if I proceeded by applying Bernoulli equation.


1+ρ(21)/2=ρ(ℎ1)+     (3)

and

2+ρ(22)/2=ρ(ℎ2)+     (4)

3 and 4 are incorrect. Bernoulli theorem goes as:
1/2ρ v ^2 + ρgh + P = constant
Your equation 3 and 4 does not equate the LHS to constant but to Pressure which is incorrect.
Note that P + ρgh1 = P1 in equation 3 which gives us velocity as zero.
A: You cannot apply Bernoulli's theorem through the two lateral pressure taps (where the vertical tubes are linked to the main tube). In general, it is assumed that the pressure is continuous and the speed is clearly discontinuous at this point. (To justify the continuity of the pressure, one would have to look in detail at the nature of the flow around the hole).
For a unidirectional flow, we can show that the pressure varies as in statics in a direction perpendicular to the flow (We prove this by projecting the Euler equation perpendicular to the flow).
So you can write, as for a static fluid $P_1=P+\rho g h_1$ and $P_2=P+\rho g h_2$
A: Let the atmospheric pressure be $P_0$
Therefore
$$P_1=P_0+\rho gh_1 \quad \text{and}\quad P_2=P_0+\rho gh_2$$
Subtracting we get the required equation
$$P_1-P_2=\rho gh$$
