Bernoulli's equation not constant?

[n0083]

Suppose there are two tanks, tank A and tank B, of equal size and both are very large.
Suppose the bottom of tank A is at an elevation that is higher than the top of tank B.
Suppose there is a very small tube relative to the size of the tanks that connects the bottom of tank A to the bottom of tank B.
Suppose tank A and tank B are both half full of water.
Suppose the tube is completely full of water.

Take point 'a' as the top of the water surface in tank A and take point 'b' to be the top of the water surface in tank B.
Suppose point 'a' has an elevation that is a height, h, meters greater than point 'b'.

Then, points 'a' and 'b' are at a pressure of 1 atm and the velocity, v, of the decrease in the water level in tank A equals the velocity of the increase in the water level in tank B.

Bernoulli's Equation would seem to give the inconsistent equation.

$P_{atm} + \rho g h + \frac{\rho v^2}{2} = P_{atm} + \rho g 0 + \frac{\rho v^2}{2} \implies$
$\rho g h = 0 \implies$
$h = 0$ contradiction.

where ρ is equal to the density of water.

Where is my mistake?

Thanks to the community in helping me understand the limitations and consequences of bernoulli's equation.

 

[mfb]

The pressure difference between the bottom of tank A and the bottom of tank B will lead to water flow, which leads to friction. The water flow will reach an equilibrium where this friction exactly cancels the power gained from water flowing down a non-zero h.

 

[256bits]

$P_{atm} + \rho g h + \frac{\rho v^2}{2} = P_{atm} + \rho g 0 + \frac{\rho v^2}{2} \implies$
$\rho g h = 0 \implies$
$h = 0$ contradiction.

where ρ is equal to the density of water.

Why should ρ not be equal instead to the density if AIR.
Afterall, you seem to be calculating the difference in air pressure between surface a and b.
P_atm(a) ≠ P_Patm(b)

 

[n0083]

Thanks for your assistance.

I do not think the pressure at the bottom of tank A and tank B are the same.

I do recognize that water will flow from tank A to tank B. I am troubled with why the equation (as I've tried to apply it) provides a contradiction in an otherwise

seemingly simple example. It would seem that the pressure at the bottom of tank B would be greater than the pressure at the bottom of tank A. This would be due to

the weight of the water column in tank A plus the weight of the water column in the tube (i.e. the total water column elevation from the water line in tank A to the

bottom of tank B). However, I thought the pressure at the water line in tank A would be equal to the pressure at the water line in tank B (assuming that the extra column of air pressure

of length h, would be negligible). It is this aspect which must not be correct. That is, the pressure at water line in tank B, must not be equal to the pressure at

the water line in tank A. I assumed because both water lines were exposed to air that their pressures had to both be equal to P_atm. I realize this is incorrect.

The pressure at the water line in tank A is P_atm. The pressure at the water line in tank B, P_b, is P_atm + pgh. This could be determined from the Bernoulli Equation.

P_atm + pgh + pv^2/2 = P_b + pg0 + pv^2/2 ==>
P_atm + pgh = P_b

Because the pressure at the water line in tank B is P_atm + pgh > P_atm, then the water line in tank B will increase in elevation.

Thanks for assisting me in thinking this through.

 

[mfb]

I do recognize that water will flow from tank A to tank B. I am troubled with why the equation (as I've tried to apply it) provides a contradiction in an otherwise

Again: You forgot to include friction.

 

[rcgldr]

Let hA = height of upper surface of water above ground in tank A and hB = height of upper surface of water above ground in tank B, and let hW = height of upper surface of water above the bottom its tank for either tank.

Assume that the tube connection to tank B is sealed off. Rearranging the terms in Bernoulli's equstion and eliminating the velocity term, the pressure in the tube at height of the bottom of tank A

$P_a = P_{atm} + \rho \ g \ {hW}$

and the pressure in the tube at the height of the bottom of tank B (remember the tube is sealed off from the bottom of tank B)

$P_b = P_{atm} + \rho \ g \ ( {hW} + {hA} - {hB})$

So Pb is greater than Pa. Once the seal between the tube and the bottom of tank B is removed, the higher pressure at hB results in water flowing into to the bottom of tank B. Once a flow exists, then friction in the tube results in pressure decreasing as it flows through the tube, but the pressure at the outlet at the bottom of tank B will continue to be greater than the pressure at the bottom of tank B, and the flow continues until the height of the upper surface of the water in tank A or the tube (if tank A is emptied) equals the height of the upper surface of the water in tank B.