Torque on a square screw full with water

[Gh778]

If the thread starts turning, The pressure balance will change

I don't understand this point. The thread is always the same, the water up and down is always here. Can you explain ? If the thread starts turning, it has always water up and down.

The phase angles are eventually irrelevant because the torques are canceled at point

Have you see the pressure on the drawing, it decrease like the angle increase and this prevent to cancel last torque. Example, we have 10 at start, after we have -9, after +8 after -7 etc. The sum can't be to 0 I think.

 

[Hassan2]

Can you please explain how the films are kept there? Each film has two sides, one side is in touch with the thread, how about the other side? in touch with a stationary part?

 

[Gh778]

The film is film of water I supposed ? If yes, you're right the film of water has one side on the square thread and one side on the fixed thread (up and down). For me a torque is on the square thread and another torque on fixed threads. But like the square thread turn only, this give energy I suppose.

 

[Hassan2]

How can such a thread rotate? The fixed thread has locked it. As soon as it starts to move it is stopped by the upper thread.

 

[Gh778]

Not at all, the square thread is a thread building from a circular thread:

1/ I draw a circular thread
2/ I substract a rectangle cuboid
3/ I take intersect with a bigger rectangle cuboid

When I say square thread, it's at the top view, but in fact at origin it's a circular thread which can turn in a circular thread (sure with the good pitch). I only deleted parts for have a thread like I study. And all square thread I drawing in this topic come from circular shape and this method.

I put an example of equation for the torque (it's not a real case only an example for see the next torque can't cancel the last)

I hope you can understand ;)

 

[Hassan2]

Sorry, I had no problem with the shape of the thread but I was assuming the system to be a screw pump. In a screw pump, the thread turns without axial displacement. By turning, it displaces the fluid in the space between the thread, so no fixed thread can be there.

In case of a screw with degree of freedom in axial direction, We can have fixed thread then. Something became clear now. The upward force ( buoyancy farce) is converted to torque through the reaction of the fixed thread on the turning thread. It's also compatible with common sense too. For example a screw made of a low density material tends to pop up when placed in water. If its threads face an obstacle with low friction, it starts turning. Perhaps this is what you were trying to prove all along. If so, congratulations! But the conservation of energy is still held. Energy was spent to push the screw inside water.

 

[Gh778]

Sorry, I had no problem with the shape of the thread but I was assuming the system to be a screw pump. In a screw pump, the thread turns without axial displacement. By turning, it displaces the fluid in the space between the thread, so no fixed thread can be there.

Yes, here the grey threads don't turn, in contrary the square thread turn and so move up or down.

Perhaps this is what you were trying to prove all along.

no, no, no ! I'm looking for the truth, I know it's not possible, I just want to see where the another torque is but I don't know where it is.

The upward force ( buoyancy farce) is converted to torque through the reaction of the fixed thread on the turning thread.

don't forget:

1/ the up surface = down surface
2/ water at up is not connected to down surface, so the up force = down force
3/ In standar system we must give energy for put air in water for example, here the energy can be near 0 so I'm sure there is another torque somewhere

Energy was spent to push the screw inside water.

I know I have changed the system, sorry for that, at first we put thread in water (thread that can be of density like water), but in the second system I put all the system in air, I put only film water at up and down surfaces. In one case like another I don't need energy (in theory). In the first case I takeoff very few water in second I put very few water, this need very low energy compared to the angle phase of torque.

Maybe you can see where is the problem, thanks for your patience ;)

 

[Hassan2]

If instead of water, we had pressurized air in the gaps, both pressures the same, would your calculation still yield a net torque ?
Also in one of the figures I see a plot. What is the horizontal axis?

 

[Gh778]

If instead of water, we had pressurized air in the gaps, both pressures the same, would your calculation still yield a net torque ?

No torque because pressure don't decrease with altitude. I'm agree with that. It's the angle phase of the radius which give problem for me.

The plot (the integration ?) it's an example for see if the pressure decrease with altitude (100-x) the angle phase of the radius (sin(x)-sin(x-3.14)) the sum of the product is not 0. If you do the calcul with 100 not (100-x) you'll obtain 0 or near (it depend where you stop the integration).

 

[Hassan2]

Are you sure that sin(x-3.14) is correct. I guess it should be sin(3.14-x)!

 

[Gh778]

sin(3.14-x)=sin(x) I think so if you look at the phase angle you can see the courbes have pi phase. Sure if you integrate (sin(x)-sin(x)) * (100 -x) this give 0, ok. But the radius is dephasing.

 

[Hassan2]

Yes they are equal!. But for a constant pressure, ( 100 ), the torque should become zero without integration. The two sines must cancel out. I don't know how the sines came into your equation, but check it once again.

 

[Gh778]

See the drawing, sine is for simulate up radius and down radius. I have put sine for example of a phase angle, it's not the true mean radius by the real pressure. It's only for show the next torque can't cancel the last (it's logical but with an equation we can see if it's true or not). Maybe the curves are not exactly sine but there is a phase angle. What I'm sure is the radius change and the pressure in the same time and the radius of down surface is dephasing with up surface. I would like to have the true equations of the surface for do calculations but it's difficult I need time for that. I'm happy for your help :)

 

[Gh778]

I have think about the equation of the radius. it's R/cos(x) from 0 to π/4 and √2*R*cos(x) from π/4 to π/2 (repeat if for 2π), sure it depend where you're starting the pitch. It's a function of sin (or cos) and it's like that I can draw the radius like the drawing show above. For the integration, I can put the cos or sine I think. Sure, I need the complete calculations but it's logical, no ? I don't know where is another torque or maybe something else is wrong.

Have a good day

Edit: I have choose (sin(x)-sin(x-3.14))*(100-x) it's the worst case, if the phase angle is lower 1 rd for example, we have (sin(x)-sin(x-1))*(100-x), the result is 45 but not to 0.

 

[Hassan2]

I think you are wasting your energy and time by integrating and also considering different surfaces and radii.

Cut the thread with planes perpendicular to the axis. Think of the in-plane component of the force which causes toque. The torque is canceled by an opposing torque due to the same film of water. To make it simple, assume the thread to be very thick. I think the opposing torque must be found there not in the lower film. Can you make a cut of the thread and film by your software and show it in 2-dimensional space?

 

[Gh778]

I join two different square threads with a bottom view (it's like a cut view because the thread is cut perpendiculary to the axis). I can't exactly show the film with my software, but why it's important (don't forget the film can be like 1µm of thickness) ? Here we can see r2>r1, don't forget the slope is always the same all along the thread because it's a circular thread at start. Ask me if you want more views of the thread. I put 2 examples with differents phase angle.

 

[Hassan2]

Thanks for the figures. We should not ignore the side force, both inside and outside. There must be a pressure on both sides, other wise the water gets out. The side pressure cause forces. If the inner surface was circular/cylindrical, would cause no torque, For a square threads they cause torque. Do you agree with me?

For simplicity, I have assumed a part of the thread spanning 90 degrees only. Also the lower surface is assumed to be flat, a plane parallel to xy plane ( the thickness of the thread increasing by angle), so it causes no torque. Both upper and lower surfaces are under equal constant pressure ( pressurized gas). As we know, the net torque must be zero. According to your calculation, There is a large torque because the lower one causes no torque to cancel it out. Now think about side forces and their contribution to the torque.

 

[Gh778]

But the thread is not full in water. It's the second case I study, I put only film of water only to up and down surface, not external/internal surfaces. There is only 2 surfaces in contact with water up and down only.

 

[Hassan2]

1. In the first case if study, when you removed some part of the thread to make it square you should not ignore the force due to those removed parts.

2. In the second case, the water can't remain there without some side pressure. What contains the water on from the sides? When you assume water ( without flowing) there, You are assuming a side pressure too.

3. Seems you had no problem with the circular thread at the beginning. But after you cut it by the cuboid, You found a contradiction. When we remove, we should consider the forces due to the removed parts.

 

[Gh778]

2. In the second case, the water can't remain there without some side pressure. What contains the water on from the sides? When you assume water ( without flowing) there, You are assuming a side pressure too.

Gaskets remain water. Torque from gaskets ? So small if the film is 1µm or less I think we can consider like 0.

3. Seems you had no problem with the circular thread at the beginning. But after you cut it by the cuboid, You found a contradiction. When we remove, we should consider the forces due to the removed parts.

Yes, I'm finding the solution of the problem;

 

[Hassan2]

Gaskets with their weight can't contain the water. There must be some side pressure. In case of circular threads, the external side is contained by the pipe wall, or if it is in touch with water , by water pressure.

If you focus on these points, you would find the problem faster than I can do. I'm waiting for the result!

 

[Gh778]

You forget the square thread is in sandwich between 2 fixed circular threads. The gasket is 1µm of thickness only. One side of water is on square thread, one side of water is on fixed thread. The gasket is fixed on the square thread. The gasket is a line of 1µm of thickness which is all around the square thread.

 

[Hassan2]

The gaskets can be treated as a part of the fixed thread then. Give a 1 cm thickness for water film and analyze it. If you think the water wight makes thinks more difficult, fill the 1 cm gap with gas. Doesn't your calculation give a net torque again, if the total angle is not multiple of 360 ? There is a problem here too.

 

[Gh778]

I think the problem is enough difficult like that. I don't want to put a gasket with 1cm of thickness, this is more complicated, my gasket is very small. It's important to see where the torque is canceled. There is only 2 surface with water, no weight, the gasket can be very small. I'm looking for where is the another torque

 

[Hassan2]

Sorry, I recommended to ignore the gasket. But increase the water film thickness.
by the way, jf you cut the turning thread only, and the fixed thread is still circular, what fills the space between the two fixed threads?

And if you cut both, what fills the space inside and also outside?

 

[Gh778]

If I cut fixed thread, there is a big problem for retain water, it's not this case I study.

I cut only square thread. The fixed threads are circular. See message #10 for look the yellow square thread (but be carreful the message is for first method not second, look the drawing). The water is between square thread and fixed thread, this water follow square thread.

 

[Hassan2]

The removed regions are filled with air then. Again what prevents water from ejecting to the air from the sides? gaskets? gaskets are on the top and down. Gasket fringes? What forces push the gasket fringes against the water? Water pressure is small? The torque is small too.

 

[Gh778]

The pressure is the height of the thread, if the thread is 10 m we have at bottom 1 bar.

Sure gaskets receive pressure from water but the force = pressure * surface, like surface can be near 0 the force is near 0.

Imagine the gasket like a cord with a radius of 0.5µm for example, put this cord between square thread and fixed thread at up and down surfaces. Sure there is friction in practise but we can imagine no friction in theory. I would like to know where is the torque which cancel the torque give from phase angle of the radius.

I add drawing with gasket and another with gasket and water (blue color)

 

[Hassan2]

I think I have found the problem but I am not sure how to explain it the best. When the waters are static (the bottom of the film is closed), as it is in your case, at any point on the surfaces, the force is normal to the surface but the surface can not move in that direction. So the force can not do any work for us. Question:　What are the opposing forces or torques? 　Answer: The combination of forces and reaction forces ( or the due torques) which keep the thread fixed in that direction. So, even without the lower water, the net torque is still zero. Maybe you can explain this better.

 

[Gh778]

First, for the torque I think it is like:

dM=r*P*de*dl*sin(θ)

With r= radius,
P=pressure at the point
de=small proof
dl=small lenght
θ=slope
dS=small surface=de*dl

With this we can see the Moment is not the same at each point (if the torque was the same in all points in the thread we will have not angle phase torque)

Second, I understand what you say, it's a good idea, ok all displacements perpendicular to the force don't give energy (or take) in the contrary if the thread has a torque it turn and give energy. Don't forget there is no up or down force. Never I see a case in physics where we have a torque and where the force is perpendicular to the movement (or when a case exist the system can't turn, here the system can turn and it is always in the same state). If another torque don't exist it's very strange (for me !).

regards