Bernoulli Equation Paradox Imagine a funnel with an extremely wide mouth, filled with water. Attached to the stem is a water pump that pumps water upwards (into the funnel). Water is being pumped in at the rate of Q. The cross sectional area of the stem is A.
The Bernoulli Equation:
$$P_m + \frac12\rho v_m^2 + \rho gZ_m = P_s + \frac12\rho v_s^2 + \rho gZ_s$$
where the subscripts are $m$ = mouth and $s$ = stem (expulsion)
If we assume that the surface of the water is rising at an extremely slow rate due to the wide mouth, then $v_m = 0$. Setting $Z_m = 0$ and $P_m = P_{atm}$, 
$$P_{atm} = P_s + \frac12\rho v_s^2 + \rho gZ_s$$
In the second case the water pump is replaced by a water suction pump that draws water away from the funnel (downwards) at the same rate of Q.
The Bernoulli Equation still applies.
$$P_m + \frac12\rho v_m^2 + \rho gZ_m = P_t + \frac12\rho v_t^2 + \rho gZ_t $$
where the subscripts now mean $m$ = mouth and $t$ = stem(suction)
Similarly, the surface of the water at the mouth is lowering at an extremely slow rate, so the same assumptions apply ($v_m = 0, \,Z_m = 0,\, P_m = P_{atm}$),
$$P_{atm} = P_t + \frac12\rho v_t^2 + \rho gZ_t$$
Hence,
$$P_s + \frac12\rho v_s^2 + \rho gZ_s = P_t + \frac12\rho v_t^2 + \rho gZ_t$$
Since $Z_s = Z_t$ (the water level from the end of the stem to the surface of the water), then $v_s = v_t$ (Since Q is the same, so the fluid velocity is the same),
Then, 
$$P_s = P_t$$
However, this is not true because $P_t$ is definitely lower than $P_s$, as the dynamic pressure of $P_t < P_s$ due to suction as compared to expulsion.
In fact, the definition of pitot pressure is the sum of its static and dynamic pressure:
$$P_{pitot} = P_{static} + P_{dynamic}$$
$$P_s = \rho gZ + \frac12\rho v^2$$
$$P_t = \rho gZ - \frac12\rho v^2$$ 
Dynamic pressure is negative as suction occurs in the second case, reducing the pressure. From here, it is clear that,
$P_t < P_s$. However the Bernoulli equation proves that $P_t = P_s$.
What seems to be the error in the application of the Bernoulli equation? Any help is appreciated!
Update:
I'm aware that the Bernoulli equation can be derived from the conservation of energy,
$$P_1V_1 + \frac12 mv_1^2 + mgZ_1 = P_2V_2 + \frac12 mv_2^2 +  mgZ_2$$
Dividing by its mass yields the Bernoulli equation. This appears to be the solution to the paradox - because the mass is not a constant, since we are removing or adding water to the control volume.
However, in a situation where the funnel mouth is full and water simply overflows if water is pumped into the funnel, or where we constantly refill the water that has been sucked out of the funnel, then wouldn't the abovementioned paradox still hold?
My question is: in the situation where the water level in the funnel mouth is being held constant by overflow or replenishment, can we still assume that the total head at that point to be zero and unchanging?
 A: Your answer is correct on the basis of your model, which neglects the resistance to flow through the funnel (Bernoulli's equation assumes inviscid flow). If that resistance were included you'd find that $P_t < P_s$.
[Addendum in response to comments:]
As a result of the balance of forces and momentum change expressed by Bernoulli's equation, the body of water in the funnel can move up or down with equal ease, like an immersed object with neutral buoyancy. The difference between the two cases is in the work done by, or on, the pump, which for this purpose can be idealised as a piston. This work corresponds to the rate of increase of gravitational potential energy.
If the level of the upper surface of the water is allowed to rise in the first case and fall in the second (rather than being held constant by overflow or replenishment) Bernoulli's equation can be applied with the instantaneous value of the height $Z_m$.
In practice there will be a pressure loss for upward flow at the point where the funnel starts to expand, and a somewhat lower pressure loss for downward flow at the same point. This will tend to make $P_t < P_s$. But it's interesting to note that if the funnel were inverted this effect could produce a situation where $P_t > P_s$.
A: The following three statements are all roughly equivalent;


*

*Dynamic pressure can NEVER be negative.

*Bernouillis equation is valid whichever direction the flow goes in, because $v^2$ is the same.

*In a world where energy is never dissipated (which is one in which fluids are always inviscid) a movie of your experiment would make equal sense played backward or forward.


Imagine that the experiment is set up with the pump turned off and the water all at rest. Now turn the pump on. There will be a transient phase during which the flow accelerates, but once the flow has settled down (lets allow overflow) we have a situation to which Bernouilli applies. Same situation if the pump is set to suck at a similar rate. The steady conditions in either case are identical with regard to pressure and velocity, apart from the direction. 
The difference between the two experiments will be EXTERNAL. The pump works by creating a difference between the inlet and outlet. That difference is not covered by Bernouillis eqn because the flow in the pump is complicated and not reversible. If the pump inlet and outlet are the same size, then the velocities there must be equal but the pressures cannot be. Whatever the difference is, it will be reversed if the flow is reversed. The movie is reversible anywhere between the top of the pump and the and the top of the funnel, but not reversible below the top of the pump.
