36 replies to this topic

[samayaraj]

In your first case, the pipe is horizontal. The head above mid point is zero. Hence the mid point pressure will same as pipe pressure. After rotating the pipe to 90 deg from its mid point, the head acting in the mid point is half of total length of pipe. So the mid point pressure is p2 + rho x g x h/2 where as h is the total length of pipe. So how the mid point pressure will be same in both the cases?

[jamese]

The sketch is correct, apart from in the example I gave the length of the pipe was 100m

 

The average pressure in your sketched example, and my example (when the pipe is 100m long) increases as the pipe is moved from horizontal to vertical

 

How does the average pressure (and hence the total energy) increase as the pipe is moved from horizontal to vertical, if no energy is added to the system?

Or is it just that we are getting this extra pressure for free, where did this energy come from?

 

If pressure at the top of the pipe drops to 95 bar and the base pressure increases to 105 bar, then energy is conserved.

 

(If you redo your sketch/spreadsheet with the pipe being 100m long, you will more clearly see what I refer to)

Edited by jamese, 21 October 2015 - 05:54 AM.

[samayaraj]

You can change the pipe length to 100m in the spread sheet and see the midpoint pressure. It will not be same.

Again im saying you are not getting any free energy. You are rotating the pipe right? That adds the head. You can do a simple experiment. Take a 10m length pipe. Fill it with water in horizontal position and see the pressure at both ends. It will read atm pressure. Next you made the pipe vertical with water, the top end pressure will be atm and bottom pressure will be atm pr + rho x g x h. Do you agree this? Try to make a energy balance calc.

[jamese]

According to you, after the pipe is rotated we have more pressure in the pipe than we had in horizontal situation.

Since the top end pressure is the same, and everything from this point downwards is then increased.

This equals more energy.

Edited by jamese, 21 October 2015 - 07:52 AM.

[samayaraj]

Yes. The energy of vertical pipe is more than horizontal pipe. This is because you are doing work to make the pipe vertical from horizontal.

[jamese]

There is no work done to rotate the pipe, since the pipe rotates at the mid point the pipe is mass balanced.

[samayaraj]

If the pipe is mass balanced it would remain horizontal. Then how it will rotate about the mid point?

[jamese]

The work required to raise it would only be a function of the inefficiency of the bearing it rotates on, which is external to the closed pipe system.

 

(whilst one end of the pipe raises, the other end drops)

(top end requires work to be raised, lower end does work whilst being lowered = no net work)

Edited by jamese, 21 October 2015 - 12:24 PM.

[samayaraj]

Your concept is wrong. Its the TORQUE you are giving to rotate the pipe. The top end tries to resist and bottom end aids the pipe to rotate in anticlockwise about its centre. If both equals it won't rotate. For rotation you are giving a work i.e. Torque.

[jamese]

OK agreed on the torque.

 

But putting it another way, in the absence of resisting forces, ie drag and bearing friction (both which are external to the pipe);

 

If tiny force is instantaneously applied to the pipe (and then removed) the pipe will then start turning.

 

In the absence of any more force being applied the pipe will continue turning with no additional force applied.

In your scenario the total energy in the pipe varies dependent on its position?, why?

Edited by jamese, 23 October 2015 - 03:48 AM.

[samayaraj]

Yes. The pressure of the system will vary as the height of pipe . In your case, the pipe is keep rotating about a center and hence the pressure will vary as the orientation of pipe varies.

 

You are giving a force to bring the pipe from horizontal to vertical position. Let the force you are applying at the intial (i.e. when pipe is horizontal) be Fh and force at the final (i.e. pipe at vertical) be Fv.

 

Case 1: Horizontal to vertical

 

Force required Fh will be maximum at horizontal & Fv required will be zero at the vertical position. When the pipe is rotating to vertical position, the force Fh applied is used to increases the pressure by increasing the height of pipe where as applied force decreases gradually as the elevation of pipe increases and when it reached vertical position, it will become zero.

 

Case 2: Vertical to horizontal

 

Due to inertia, the pipe continue to rotate and come back to horizontal position. The force Fh recovered back from the system when it attains horizontal position and pressure in pipe reduced to initial value.

 

Again it will go to vertical position, pressure will rise and its a cyclic process. 

 

So, the total energy of system remains same. When the pressure increases to maximum, the force required will be zero and when the pressure comes to initial value, force required is maximum.

 

This is how the energy is conserved in your hypothetical case.

Edited by samayaraj, 25 October 2015 - 11:08 AM.

[shantanuk100]

Dear Jamese,

Good Day.

 

Please see the spreadsheet in which I have drawn a diagram to illustrate my explanation and confirm if the diagram is in line with your expectations. It is way more easier visualizing it when we draw the diagram.

This answer is with respect to the initial spring shaped pipe you were talking about.

 

1. We do not change the System Volume nor are we compressing the spring boundaries.

The structure of the spring and thus the system volume does not change. So the pressure at the top cannot

change unless you are accounting for realistic spring buckling (See Spreadsheet) in which case the pressure at the top changes just a wee bit due to pressure increase in fluid which is a consequence of decrease in overall system volume because of buckling. This buckling occurs because no Spring can be compressed without Buckling as it is impossible to do so. The buckling is what stores the elastic energy in the spring.

 

2. The pressure at the bottom changes due to the reduction in height of fluid above the datum and if you account for the Spring buckling as in the above case in order to be realistic, then that too. Because the pressure increase due to buckling is equally distributed at the top and the bottom.

 

3. As for the conclusion from your PhD friend, the pressure at the top of the spring only depends on the fluid above that cross section (assumed here as atmospheric air), and the spring buckling, so it cannot change in this case. 

 

Please see the drawing in the Spreadsheet.

 

Regards,

Shantanu

Attached Files

•  SPRING PIPE.xlsx   36.28KB   20 downloads

Edited by shantanuk100, 18 November 2015 - 01:00 AM.