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 ($v_m = 0, \,Z_m = 0,\, P_m = P_{atm}$),

$$P_{atm} = P_t + \frac12\rho v_t^2 + \rho gZ_t$$


$$P_s + \frac12\rho v_s^2 + \rho gZ_s = P_t + \frac12\rho v_t^2 + \rho gZ_t$$

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),


$$P_s = P_t$$

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?


2 Answers 2


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 $P_t > P_s$.

  • $\begingroup$ Thanks for the answer! However for $P_t$ the pressure would be lower than $P_s$ as there is a suction pressure resulting in the dynamic pressure term being negative, so $P_t < P_s$ even when the flow is inviscid (I've added this part into my original question) $\endgroup$
    – t.c
    Sep 27, 2014 at 16:43
  • 1
    $\begingroup$ I should have added that if in the first case the water level is rising at the top of the funnel (rather than overflowing) work will need to be done to raise the body of water, and this will require additional pressure at the base. The Bernoulli equation does not account for this because it assumes constant levels. In the second case the situation is reversed, resulting in a pressure difference in the expected direction. $\endgroup$
    – MartinG
    Sep 27, 2014 at 17:31
  • $\begingroup$ I'm a little confused. Work is supplied by the pump in the first case, and work is removed by the suction pump in the second case. However if we choose the control volume to simply enclose the fluid within the the funnel, and not the pump, could we ignore work and assume it is a Bernoulli equation? (Since there is no work transfer within the funnel itself) $\endgroup$
    – t.c
    Sep 27, 2014 at 18:02
  • $\begingroup$ I've now added what I hope is a clearer account in my extended answer above. $\endgroup$
    – MartinG
    Sep 28, 2014 at 9:49
  • $\begingroup$ In the situation where the funnel mouth's water level being held constant by overflow or replenishment, can we still assume that the total head at that point to be zero and unchanging? $\endgroup$
    – t.c
    Oct 1, 2014 at 15:39

The following three statements are all roughly equivalent;

  1. Dynamic pressure can NEVER be negative.
  2. Bernouillis equation is valid whichever direction the flow goes in, because $v^2$ is the same.
  3. 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.