Sum of torque

[EOW]

A closed container has a pipe enrolled on a grey disk. For enter the pipe inside the container I need to give the energy E1. All the device is at a linear velocity V, with V = constant. When the pipe move out the container I recover the energy E1. The container has 2 forces F2 and F4 on it, this forces give an energy E2 = (F2+F4)*V*t. All concentric forces on the pipe cancel themselves because the pipe has a mass and with rotation there is centrifugal forces. P2 > P1, and imagine the device with P1=0 in theory. There is no gravity.

[EOW]

In fact, all forces from pressure*surfaces cancel themselve => the container has no net force on it. The force come from the centrifugal forces only. The gas can't give a contrary force, I think the gas give a force that add the net force of the centrifugal forces.

[EOW]

The idea is to use the device I drawn in the message #40. But here I use pressure. No gravity. No modification of the volumes. I suppose no friction. It's a 2D but it's possible to imagine in 3D replace circles with spheres. Outside the pressure is constant at P (atmospheric pressure for example at 1 bar). A basic device is composed by an arm and a circle. The black arm can turn around the blue axis. The circle is fixed to the arm. The circle don't turn around itself. I drawn 2 basic devices but it possible to have N devices, in the third image I drawn 16 basic devices. Circles are smart, walls between circles can be removed : red in the drawing. I suppress wall like that a part of each circle receives a torque on it. Like all circles don't turn around the same center, the torque is not the same.

First drawing: the device turns from time=0 to time=2.
Second drawing: the device turns from time=0 to time=0.5

It's possible to look at torque on each arm, like the distance d1 is greater than d2, the torque is higher counterclockwise from time=0 to time=1.

At time=1, the pressure inside circles must changed. This don't need energy. It possible to think with a device with N basic devices (arm+circle) and X circles are time=0, X circles are at time=1, X circles are at time=2, etc. For change the pressure inside circle it's easy: take the pressure inside one and replace from another cirche that is in another time (or position). Like olumes are constant, change pressure don't need energy in theory, a little in practise.

The third drawing shows a device with 16 circles, 2 circles at each time, all 16 amrs turn at the same angular velocity. Two circles work together. At the exact position: time=1 time=4 time=6 and time=8 there is no work. At time=1it's necessary to put P/2 inside circles so change with the circles at time=4.

Inside circles the pressure is 2P from time=8 to time=1, from time=4 to time=6.
Inside circles the pressure is P/2 from time=1 to time=4, from time=6 to time=8.

[EOW]

First image: for look the sum of torque from time=0 to time=1.  There is no torque if radius are the same ! need one small circle and one bigger. Look at third image.

Second image:  look of the rotation of the gaskets. Even from time=0 to time=1, the device have a net torque on it and gaskets can't compensate it.

For change the pressure inside the circles it's possible :

a/ to exchange physically circles with a device with 16 circles (low angular velocity of the device)
b/ to use an external device, this device will need an energy but this energy will be transform in heating of gas

I explain with a gas but it's ok with a liquid too (less losses for change the pressure inside circles).

[EOW]

With gas: there is no torque if circles have the same diameter. The torque is not the same with 2 diameters but the angular velocity must be different for one circle. So the sum of energy is conserved.

Now, if I put liquid inside circles, with same diameter, centrifugal forces are not the same inside one circle to another because the centers of rotation are not the same. The torque must be different. Maybe the liquid move between 2 circles. The velocity in a fluid change like the speed of sound in the fluid, the pressure can't be faster than this. So if the device turn at high velocity, and like the red wall change all the time with the angular position, the pressure cannot be the same and the sum of torque is not 0.