How fast can water travel? Another hollywood provoked science question

I was watching Die Hard III and the scene in the aquaduct got me thinking - How fast can water travel in that situation?

To set up the scenario, the aquaduct is 20 feet in diameter and 60 miles long. The gradient is basically level. The end of the aquaduct is dammed off by the ocean (to ensure no shortage of water for this scenario).

The dam is blown up and all debris is vapourized. Water rushes in to the aquaduct. At what speed? Could Bruce Willis and his dump truck out run the torrent of water?

Water pressure would be a factor but as this is using an open air source of water, I don’t think it is.
Gradient would be important but that has been eliminated here.

What is left?

I didn’t see the movie, but if I understand your description, there is the pressure exerted by the weight of the water that is being held back by the dam. Presumably there is a difference in height between the bottom of the aquaduct and the top of the water column being held back by the dam. This difference in height is the source of the pressure driving the water into the aquaduct. Ultimately it is gravity that is the driving force behind the weight of the water column.

Alright. That sounds reasonable. So is there a limit on how fast the water can rush through the aquaduct?

To add to the scenario, the tunnel is the same size so that the hole where the dam used to be is the same size as the rest of the pipe.

Is the hole at the top of the dam (very little pressure) or 300 feet from the top (much greater pressure)? That will determine how fast the water will flow into the pipe.

The dump truck isn’t on some kind of treadmill, is it?

Forget the dam. I’ll try to paint the scenario a little better.

The aquaduct pipe runs straight to the ocean, finally poking out of the coast at 50 feet below sea level. The aquaduct is uniformly 50 feet below sea level. While building the tunnel, a wall was built at the ocean end to prevent sea water getting in to the pipe. Now the wall is vapourized. How fast does the water come through the pipe?

If you assume the velocity of the water in the tunnel is the same as the velocity of the air leaving the tunnel, and that the area of air vent is the same as the water intake, then the velocity could be calculated using this equation.

 V=sqrt((25000)(D)(P)/(L))

 Where: V = Air velocity in feet per second
              D = Pipe inside diameter in inches
              P = Pressure loss due to air friction
              L = Length of pipe in feet

Ignoring friction, this gives us a velocity of about 4.4 ft/sec. A vehicle at 30 mph (44 ft/sec) could easily stay in front of this.

The difficulty would be in getting the wall of water to actually occur. The reality I think, is that a tunnel would fill from the bottom up.

I’m flattered Zoomer. You came out of lurking for me?
I, too, though that the tunnel would fill bottom up.

But the water will slow down the further it goes. The friction loss due to air will be negligible, but the higher viscosity of water will result in greater loses, lower pressure differential, and therefore lower velocity.

In idealized form, the water will always form an angled wall, and this angled wall will move through the tube. The angle of this wall is a function of the relative strengths of gravity pulling the water down and the pressure keeping it up as it is forced down the tube. My engineering intuition says that this can be found by finding the angle that produces a surface of constant pressure.

OK, rereading my post that “always” comment was wrong. I meant to say that I think we can idealize it to that form. After further consideration I think even that is wrong though as the boundary layer is inherently non linear.

In reality, the wall at one end of the aqueduct (actually Water Tunnel No. 3) is holding back the fresh water from a reservoir, not the ocean. (The whole point is to deliver fresh water to the city.) And it’s not, in reality, uniformly 50 feet below sea level. Instead, it runs downhill.

Here is a web site with a related question, but an answer that seems to fit:

http://www.newton.dep.anl.gov/askasci/eng99/eng99365.htm
“If the water is flowing out of a tank where the air above the water is at atmospheric
pressure as is the air outside the hole, and if the area of the surface of
the water is much larger than the area of the hole, then, by Bernoulli’s
Law, the water will leave through the hole at a speed v = sqrt (2gh), where
h is the height of the surface of the water above the hole. Notice that
this is the speed an object would get by falling through a height h.”

I know it’s always more complicated than this, but just plugging in the numbers, this is what we get:

v=sqrt(2gh)
g=32 feet per second^2
h=50 feet
2gh=3200
sqrt of 2gh=56 feet per second

*60 seconds in a minute * 60 minutes in an hour = 201600 feet per hour

divided by 5280 feet per mile = 38 miles per hour

So, assuming this simplistic approach is close, yes he could outrun it, but it’s certainly moving fast.

Also, in that web page they talk about the front of the column of water taking up about 2/3 of the diameter of the pipe.

Absolutely. I can’t solve the problem in any reasonable time any more, especially without specific parameters.

The general outline is this. The energy in a hydraulic system is usually given in terms of what is called “head.” For example water at an elevation of 10 ft. above some reference height has a head of 10 ft. When the water flows through a pipe there are losses in head as a result of friction, changes in direction and so forth. You start with the original head which is the height of the water behind the dam, subtract the loss in head from all the various loss factors and arrive at the velocity head. This is a measure of the kinetic energy of the flow which is related to the mass of the flow and the velocity squared. It is a balance. Losses can’t entirely stop the flow because if there is no flow there are no losses. So the velocity builds up until the input head = losses head + velocity head.

Thanks for all the info. Interesting.

So we can conclude that Bruce Willis could indeed outrun the water using a dumptruck.

the mental picture I get reading this has nothing to do with water :slight_smile:

Now you kids stop that or you’ll get us adults riled up.

The Bernoulli equation is the one to use in this case. It is:

P+1/2pV[sup]2[/sup]+pgZ=constant

Where P=pressure, p=density, V=velocity, g=gravity and Z=height. We take two points, the top of the reseverior (1) and the front of the water torrent (2), and set them equal to each other:

P[sub]1[/sub]+1/2pV[sub]1[/sub][sup]2[/sup]+pgZ[sub]1[/sub]=P[sub]2[/sub]+1/2pV[sub]2[/sub][sup]2[/sup]+pgZ[sub]2[/sub]

We assume that P[sub]1[/sub]=P[sub]2[/sub]=Patm becuase both ends are open to air. V[sub]2[/sub] for a large resevoir is negligable and taken to be 0. Z[sub]1[/sub] is taken to be the reference level, and is 0. That gets us:

1/2pV[sub]1[/sub][sup]2[/sup]=pgZ[sub]2[/sub]

Solving for V[sub]1[/sub] yields:

V[sub]1[/sub]=(2gZ[sub]2[/sub])^(1/2)

Plugging in numbers gets V[sub]1[/sub]=36mph, which mostly agrees with RaftPeople’s number, with the error due to the fact that s/he rounded during the calculation. The Bernoulli equation assumes four things:

[ol]
[li]Viscous effects are negligible[/li][li]The flow is steady[/li][li]The flow is incompressible[/li][li]The equation is applicale along a streamline[/li][/ol]

2,3 and 4 apply in this situation, but 1 does not. However, all viscous effects due are add in energy losses, and slow the flow. That means 36 mph is theoretically the fastest possible flow in this case. The actual flow will be less due to losses.

For shits and giggles, the two loss modes in this sort of flow are termed “major”, and “minor” losses. The major loss is from viscous effects and the interaction of the flow and the walls of the pipe. The minor loss is due to changes in geometry, in this case from the reseveroir to the pipe. Its important to note that “major”, and “minor” do not necessarily reflect the magnitude of the effect on the flow. In other words, in some cases the “major” loss might be much less than the “minor” loss. The values of the two losses are:

H[sub]L major[/sub]=f*(l/d)V[sup]2[/sup]/(2g)

Where f=friction factor, l=length of pipe, d=diameter, V=average velocity and g=gravity. The friction factor depends on the roughness of the pipe, the diameter, and another variable called the Reynolds number. Its values range from roughly .008 to .01.

H[sub]L minor[/sub]=K[sub]L[/sub]V[sup]2[/sup]/(2g)

Where V=velocity at the pipe entrance, g=gravity and a constant K[sub]L[/sub] that depends on how the pipe connects to the resevoir and range from .004 to .8.

You will have about a 16’ or so head from an infinite reservoir flowing into the open end of a 20 ft dia horizontal pipe.

The flow will be on the order 8 to maybe 10 MPH which you should be able to outrun. In a truck - - no sweat.

The current equations being used assume fully developed flow, which is obviously not the case here (the flow is not steady). As stated before, though, these effects would, for the most part, slow things down, viscous effects and what not. Also, pressure in the air of the pipe will be slightly higher thatn atmospheric while the air is being pushed ahead of the water.

Not all of them do this though. People have claculated the flow through the initial opening to get a velocity, but what we are really concerned with is the flow rate. The water will not fill the entirety of the pipe, the velocity of the portion that is filled will have to be higher.

The dynamic difficulties with this problem are a bitch. If the flow was fully developed it would be far easier, but it is not.

The various estimates I have seen of the Johnstown Flood indicated a water speed of from 12 to 15 miles per hour. (Obviously, a twisting valley covered in trees would be different than the scenario in the OP, but I am not sure that water in a tunnel would be orders of magnitude faster.

Different situation, but if you really want that water to move, you want what they call head (pressure).

Per McPhee’s Asembling California, hydraulic mining for gold might be done with 500 feet of pressure and nozzles called “water cannons”, and the water could come out of the cannon* at 120 miles per hour. Sure moved the alluvium.

*no moving parts in it, just a shaped brass tube.

Back to the 20’ aquaduct that is opened 50’ under the ocean, the pressure looks like about 21 lbs per square inch, and residential pressure can be from 35 to 100 psi (from here ).

Phooey to 21 psi, it’s a movie, make it 50 psi, and when the ocean rushes into the tunnel the side view would first look like water pouring out of a hose, turned all the way on, with no nozzle. But down the pipe a way I WAG it would slump out so the profile would have a shallow flat toe reaching after our heroes and a bulge of water behind that, then the bulge having less friction would catch up to the toe and make like a breaking wave, whooshing out and making another toe. Lather rinse repeat. (Hee hee! get it? Lather rinse repeat!!! Hee hee hee hee!)