# Delay to transmit pressure in water

1. Sep 12, 2010

### lba

Hi ;)

I'm new on the forum

I have a question in fluid mechanics:

A tube of lenght 1500 meters has water inside at 1 bar of pressure.
A time t=0s we put 100 bars of pressure in a side close to the left end of the tube. The pressure will be at the other side at time t=1s because it's the speed of sound in water. So, an end of the tube has 100 bars and another end has 1 bar. The tube move to the left during 1 second. If the tube move it can give energy. Where come from this energy ? Put pressure in the tube can be recover because pressure it's only potential energy, so after 1s we can depress at right the tube and recover energy (in this case the tube move to left again...). If someone can explain to me ?

Thanks

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2. Sep 12, 2010

### Staff: Mentor

Welcome to PF.
Agreed.
Well I don't know - it's your thought experiment. What are you doing to the tube to make it move? What are you doing to it to suddenly pressurize one end?

You're inputting energy but not telling us how, then asking us how it is being input!

3. Sep 13, 2010

### lba

Hi russ_watters,

The tube has 100 bars at end and 1 bar at another end. The difference of pressure move the tube during one second only, no ? If I put on a side of a surface 100 bars and 1 bar to another side, the surface move, no ?

I open suddently a door and the pressure which is in a big container pass to the tube. Maybe we can consider the time for open the door extremely fast ?

It's a pump that create pressure in a big container with water inside, consider the container 1000 m^3 of water. The pump press water.

If the tube don't move it's not a problem with energy, but if the tube move we can press a spring for example so this give energy to the spring. Where the energy come from ?

I consider no loss in the system, it's a theorical study, for find where come from the energy that move the tube, if it move and can give energy...Consider the tube in steel and consider the tube indeformable. Imagine the system which bring the pressure move like the tube if it's a problem.

We can consider:

A/ Figures 2 and 2-1

At t=0s, we open the door, at left side we have 100 bars, at right side we have 1 bar. A force exist which give energy to a system (spring, hydraulic cylinder, etc.).
At t=1s the pressure it's the same at 2 ends of the tube, we close the door.
The container has lost pressure, but the tube has win pressure, it's the same energy ?

B/ figure 2-2

At t=1s we close the door and we can open a right door for depress the tube at right, we can depress the water so we can recover this energy. If a system can depress very fast and recover all energy (in theory), the tube will have less than 100 bars at right and 100 bars at left during 1s. The tube move to the left too.

I'm french, sorry if my english is not perfect. On figures: POMPE=PUMP.

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4. Sep 13, 2010

### Staff: Mentor

You said the tube moves to the left - did you mean it moves to the right?

Yes, if you create a pressure wave inside the tube moving from left to right, the tube will move from left to right: but not much. The pressure is applied to the water, not the tube so unless the water is confined to the tube by a valve or a sharp elbow at the end, the tube will move very little.
A valve opening can create the effect you are describing, yes (though it is more typical with a closing valve than an opening one). This effect is called the water hammer effect: http://en.wikipedia.org/wiki/Water_hammer
the energy to move the water and the tube comes from the pump.
You can't consider a no loss situation, as it is the loss that creates the motion of the tube (unless the tube is closed at the other end).

5. Sep 13, 2010

### lba

For me the tube moves from the right to the left ! Like the figure show. The two ends of the tube are CLOSED. It's very important. I don't show it on figure and I thought it was logical (for me ... lol). If water was uncompressible no water move in the tube and the 100 bars pressure will be at the same time at 2 ends of the tube. The tube has 0.01*0.01*1500=150 l of water inside. Like water is compressible and like the difference of pression is 99 bars, the water put inside the tube when the door is open: 99*100000/(2.2*1000000000)*150=0.675 l . It's for that I question me about the source of the energy, because the container lost energy of compression but the tube win energy of compression, like this law is linear, in theory I win that I lost and if the tube move (in this case the difference of pressure is 99 bars, the force = 990 N). The volume of container V=1000m^3 = constant, the volume of the tube Vt=150l = constant. We transfer pressure from container to tube, the total volume is V+Vt = constant. We have only transmit energy from one volume to another. The right side has 1 bar of pressure at t=0s, when the tube move it increase the pressure at right but not like the force of 990 N, I will be surprising if this force is the same than at left. This pressure at right is created by acceleration, so the water in the tube will have a speed, this speed is kinetics energy and can be recover. If the tube move and can press a spring for example, I don't know where come from this energy ?

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6. Sep 15, 2010

### lba

When the water come from the container to move to the tube, the energy give by water is 0.01*0.01*99*100000*6.75=6682 J. I think this energy can be recover by a system fix on the ground. We can insert for example a piston or a turbine in the tube near the left side and recover the energy. In my example with a tube of lenght of 1500 meters, the acceleration must be very high for recover energy from the tube. But we can take a tube of 15000 meters with a tube finer.

Maybe my system is a little too complicated for calculations. Don't use container if it's a problem. We can use a tube with a piston on the side near the left. Like this, we pressure water with the piston, this energy is only a potential energy and can be recover at least 50% easily. The pressure at left side is more than right side. In this case where come from the energy ? If the lenght of the tube is 15000 meters (with 150l of water inside), we have 10 seconds for equilibrate pressure. In 10s we can move the tube to d=100*a/2. The energy is give by E=F*ds=F*s, if the tube have a speed before pressure the piston, the energy increase...
Cycle:
1/ Move at speed 40m/s the tube with water inside, cost energy 1/2*150*40^2=120000 J
2/ Press piston and recover energy from the tube (external system), the force is 99N (the tube is 150l => surface = 0.00001 m^2) and the distance is in 10 s of 400 meters. We recover 99*400=39600 J
(The speed of the tube is fix during recover energy). We need to press the piston need: 6682 J
3/ Decelerate the tube to 0m/s. This give 120000 J. Depressure the water in the tube: recover 3341 J

Where come from the 39600 J - 3341 J ?

Maybe you see where the energy come from ? I don't reach to have efficiency=1.

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7. Sep 16, 2010

### lba

Just an up of the message. I think the second system easy to understand, tell me ? Where is my error ?

Someone can help me ? ;)

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8. Sep 16, 2010

### Bob S

If you close a valve at the end of a pipe suddenly, you get a hydraulic pressure surge because of the moving water. This sometimes leads to a hydraulic pressure oscillation called "water hammer". See equations in

http://en.wikipedia.org/wiki/Water_hammer

Bob S

9. Sep 16, 2010

### lba

OK, thanks, Russ tell me that too for the first system. With the second system a tube with a piston in a side, we have the same oscillation ? If yes how is the oscillation at left and at right ? is there a delay ? I don't understand. Don't forget the tube is close at all sides. No water move in the tube. It's only a tube with water inside at 1 bar at start. The water don't move in the tube, the tube and the water have the same speed in my last example. There is only 0.675 l which move with the difference of pressure. The "water hammer" is not the momentum of water due to the speed ? Here it's not that, I only pressure the water at left, maybe is there an oscillation, but if the piston move for have 100 bars constant at left, how the oscillation move ? For show what I think about this system, see the second figure, the piston press only local water for have 100 bars always at left, like a wave move to the right at speed of 1500 m/s we move pistons too in the same time, like that the pressure at left is always 100 bars and I don't see how a wave move to the right ? Maybe you can imagine a tube of lenght of 1500m or 15000m with a diameter of Xmm with water inside, we change the diameter from X to Y with Y<X but with the start of change at left and the change of diameter move at speed of 1500m/s (like a wave). I know if there isn't a wave or something else there is a big problem with energy and CG but I don't find what ?

My problem is: there is a difference of pressure during 1 or 10 s (1500 or 15000m). Maybe an oscillation exist, if yes, the oscillation move at right after left, after right after left, etc. Without loss the oscillation don't move nothing but the delay of transmit pressure do it. Sure there are losses in water but are you sure the oscillation compensate all the delay of transmit pressure ?

Another thing, the CG never move without loss weight. With the difference of pressure, the delay move the CG ? I thought it come back with the oscillation but with all systems can't move the CG, never...So what's wrong in my calculations ? We can put pressure at left during x seconds move to the left and before the wave come to the right put pressure at right, the waves in water cancel themselves but the CG has moved ? Maybe you think water move to the right but what's happen if the change of diameter of the tube is faster than the speed of water (1500m/s) ?

With the 3° figure. If we take a tube of 1500 m with a square section of 0.01 cm * 0.01 cm we have 150l of water inside and difference of volume of 0.675 l with 100b of difference. We need to move one side of the tube of 4.54 e-5 m = 0.045 mm at a speed > 1500m/s. Move one side of 4.54e-5m will reduce the volume of 0.675l {1500*0.01*(0.01-4.54e-5)=0.150-0.000675}. The figure show a big difference of section is only for show it, in reality the modification of section is very small. For it's works we need to recover energy from the system in other case the acceleration caused problem at right side. In last example, acceleration = 6.67 m/s^2, if we press a spring during movement and change acceleration for 1 m/s^2 in this case we recover energy and move the CG without lost weight because pressure at right end is P=F/S=m*a/S. Or put the tube on a weight plate for add the total weight.

I know it's stupid, it's not possible to move CG and take more energy than we give, but I can't find my error. If someone can explain to me ? thanks

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10. Sep 19, 2010

### lba

I have study "water hammer" and I don't find 0 for the momentum of the tube. See the figure for whatch the tube, please. I concern only about momentum for this message. This is only a piston that move in a close volume with water inside.

Datas:

The tube is close at all side, 37.5m^3 of water are in the tube
Length of tube = 15000m
Surface tube (square section of 0.05m side) = 0.0025 m²
Difference of pressure reach at final = 100 bars
Volume of water press = 170 l
Speed in water = 1500m/s
Length of the piston = 300m
Side Surface of piston = 0.05 * 0.011 (depth) m²
The piston move in water of 0.011m
Delay for reach the other side = 9.7 s (the piston is 300m: 10-0.3)
System that support the tube = 10000 tons (like this I can omit the pressure at right because I don't know calculate it ...maybe you can help me ?)
The piston move in water with an acceleration of 2.2 m/s²
The piston move during 0.1s
The speed of piston at 0.1s is 0.22m/s
The pressure at start is 1 bar all in tube (t=0s)
The pressure at left when the piston has moves (t=0.1s) is around 4840 bars (I do approximations because I don't know how calculate this) For this I consider the volume of water press by the piston like this: the volume at start is (300+75)*0.05*0.05 and it is at end : (300+75)*0.05*(0.05-0.011), I add 75m because the piston move during 0.1s.
The pressure at t=10s is 100 bars

So for the momentum:

To the right :
Momentum of right pressure due to the global acceleration = very few because the system which support the tube is 10000 tons (0.37 bar, this force reduce the start force, but it's only 100N), I omit it, can I ?
Momentum Wave speed = very few (170*0.22 kg*m/s), I omit it, can I ?
The momentum of water is 170 kg * 1500 m/s = 255000 kg*m/s, is that ? Is that the "water hammer" ?

To the left:
The momentum of force with the delay is (4850+100)/2*100000*0.0025 *9.7 = 6e6 kg*m/s

Like there is a big difference I have a problem with raisoning... can you help me ?

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