# Have I understood the Bernoulli equation wrong?

I just want to solve for the speed of flow of a column of fluid of height $h$ down a vertical smooth pipe. The pressure on the fluid is $P_{atm}$ everywhere. The pressure difference due to gravity is $pgh$, so at the bottom of the column the equation gives $C = \frac{1}{2}pv_0^2 + pgh_0 + P_{atm}$ and at the top of the column $C = \frac{1}{2}pv^2 + pg(h_0+h) + P_{atm}$ so $\frac{1}{2}v_0^2 = \frac{1}{2}v^2 + 2gh$, meaning that fluid at the bottom is traveling faster than fluid at the top. This can't be true, what have I got wrong?

• "The pressure on the fluid is $P_{atm}$ everywhere. The pressure difference due to gravity is pgh" How is it both the same pressure everywhere and a pressure difference? It's not really clear what you mean. – JMac Apr 7 '17 at 21:02
• sorry, i should say energy density due to pressure is atm. energy density due to gravity and so on – ggmate Apr 7 '17 at 22:40
• dimensionally, gravitational potential provides a pressure, but it's like a negative pressure i suppose, as is kinetic energy. (F/A = F.ds/A.ds) (you can check this using the vector definition) – ggmate Apr 7 '17 at 22:54

I think what you have in mind is an isolated column of inviscid fluid in free fall down a pipe. The friction between fluid and pipe must be zero if the fluid is to be truly in free fall; and once in free fall the enclosing pipe is really unnecessary.

Now you must be aware that gravitational effects become zero for any object in free fall. Hence $g$ should not appear anywhere in your equation (equivalently, set $g=0$ in your equation). Thus contradiction dispapears: if $P_{top}=P_{bottom}$ then we must have $v_{top}=v_{bottom}$ (for a fluid column in free fall).

• ah, the bernoulli equation only works for steady flow – ggmate Apr 8 '17 at 7:27
• There is a more general form of Bernoulli equation for unsteady flows. My point was that, freely falling body in a uniform gravitational field is equivalent to an identical body far removed from any gravitating bodies. Therefore in a freely falling fluid column, there cannot be a gravity-induced pressure gradient. – Deep Apr 9 '17 at 3:58

The pressure at the top has to be less than the pressure at the bottom. It can't be atmospheric at both locations. If it is atmospheric at both locations, the fluid will be in free fall, and will lose contact with the pipe wall as its velocity increases on the way down, while its cross section decreases. So the final equation you derived relating the velocities at both locations is actually correct.

• but this is what i wanted to do (calculate free fall of a liquid). Surely in free fall everything falls at the same rate – ggmate Apr 7 '17 at 22:42
• No. In freefall everything accelerates at the same rate. But the portion falling for longer (and therefore further down the pipe) has greater speed. If the pipe is full of fluid, then it won't be in freefall. – BowlOfRed Apr 7 '17 at 22:47
• it all falls at the same time? – ggmate Apr 7 '17 at 22:55
• What? No. It's like a stream of water falling from a spigot. The parcels just exiting the spigot are falling slowly, and they speed up as they fall, while the diameter of the stream decreases with distance from the spigot.. – Chet Miller Apr 8 '17 at 1:30

If the fluid is in a pipe of constant cross section, then if it's incompressible the entire column will have to travel at the same velocity.

This constraint (constant cross section == constant velocity) has to be taken into account. Bernoulli's equation assumes there is no such constraint - so you cannot blindly apply it in this situation.

• Bernoulli's equation assumes there is no such constraint. Meaning it can cotradicth the constraint in some cases? – ggmate Apr 7 '17 at 21:32
• i think i know what you're saying though, the velocity of the fluid is not necessarily known if we just know height and density from bernoulli's equation – ggmate Apr 8 '17 at 7:32