Method to calculate work while moving two walls

[Mary2100]

Hi guys,

It is not some homework, so I think it is better to post here. I would like to know if my method is correct for the work of the walls x and y.

I suppose no mass, no friction and the gas keeps its temperature constant.

An external device modify the shape of a triangle but it keeps constant the area. The external device turns the walls x clockwise around the dot O and the wall y counterclockwise around the dot O. The wall x rotates at the same angle as the wall y. At the same time, the external device moves the wall z up. Inside the triangle there is a gas and the pressure of the gas is higher than outside. There are gaskets between x/z and y/z: the gas doesn't escape.

I noted:

At start, a=b=√2/2 m and c=1 m
Difference of pressure P=1 Pa
Area of the triangle = 1*1/2 = 0.5 m² = constant
Calculation for an angle of 1e-1 rad
Depth of the device = 1 mEnergy from x and y :

The area is constant so a*b=0.5, so a=0.5/b
Pythagoras's theorem: c²=a²+b²
b=c*cos(α):

c²=0.25/(c²*cos²(α))+c²*cos²(α)
<=> (c²-c²*cos²(α))*(c²*cos²(α))=0.25
<=> c⁴*(1-cos²(α))cos²(α)=0.25
<=> c=(0.25/((1-cos²(α))*cos²(α)))^0.25

There are two walls but I need to divide by two (the moment), so the work is:

$\int_{\frac{pi}{4}}^{\frac{pi}{4}+0.1} (\frac{0.25}{(cos^2(x)*(1-cos^2(x)))^{0.25}} dx$

Is it ok for you ?

 

[CWatters]

Not sure I follow that but I have some comments...

An external device modify the shape of a triangle but it keeps constant the area.

So no change in volume or pressure? Work is being done against what exactly?

I can see that the centre of mass of the gas moving vertically so I suppose it's KE (and possibly PE) is changing which may require work but otherwise I don't see it.

 

[Mary2100]

So no change in volume or pressure?

No, the volume is constant and the pressure too.

Work is being done against what exactly?

The external device needs to give an energy to move the z wall.

I can see that the centre of mass of the gas moving vertically so I suppose it's KE (and possibly PE) is changing which may require work but otherwise I don't see it.

I supposed there is no mass, even the gas, I would like to simplify the calculations.

I drawn the triangle at start and at the end (pink):

 

[CWatters]

The external device needs to give an energy to move the z wall.

Not that I can see. Apart from some effects due to turbulence the z wall is not moving relative to the gas because the volume is constant.

Work = pressure * change in volume.

No change in pressure or volume means no work.

Consider a bicycle pump mounted on a bike. The handle of the pump might well be moving at 30mph due to the bikes motion but so is the rest of the pump. The volume in the pump isn't changing so it takes no work to move the handle of the pump at 30mph like this. Your situation appears identical.

 

[Mary2100]

"No change in pressure or volume means no work."

I agree, the external device recovers an energy from the walls x and y but it needs to give the same energy to move the wall z. There is a difference of pressure, inside the triangle there is the pressure 2P and outside there is P, so there are forces on the walls x, y and z. I'm interesting of the energy recovered from the walls x and y.

The length of three sides of the triangle is changing.

 

[CWatters]

I don't see how that changes anything.

The energy flowing through the system from z to x and y is independent of anything going on in the system. The more you push in the more you get out but the amount you can push in is entirely up to the external system. You can't calculate it any way I can think of.

 

[Nidum]

I'm not convinced that the constant volume stipulation is even achievable with that geometry .

Personally I don't think that the problem as given is worth pursuing further .

 

[CWatters]

It's not achievable indefinitely but see no reason why it can't be constant over a range.

 

[Mary2100]

" I'm not convinced that the constant volume stipulation is even achievable with that geometry ."

The geometry can respect the volume constant with that:

The area is constant so a*b=0.5, so a=0.5/b
Pythagoras's theorem: c²=a²+b² (a, b, c are not x, y, z)
b=c*cos(α):

The length of the z wall is : 2*0.5/(sqrt(2)/2-x)

"You can't calculate it any way I can think of." Three walls are independent, I can rotate x and y and move z.

 

[Khashishi]

If the area isn't changing (hard to see how that is possible), then there is no work done.
The differential work is given by
$dU = PdV$
if all other things are constant. if dV is also 0, then no work.

 

[Mary2100]

"If the area isn't changing (hard to see how that is possible), then there is no work done." I agree, but I want the work from x and y not x +y +z.

Yes, the area doesn't change because I rotate clockwise the wall x, I rotate counterclockwise the wall y and I move up the wall z, the external device moves the wall z to keep constant the area, the device is under control.

 

[CWatters]

For small displacements of wall z the length of the wall can be considered constant. The volume swept out by wall z can be calculated by multiplying it's length by displacement. A notional Energy Input can be calculated using pressure * swept volume.

You can do the same for walls x and y giving a notional energy output.

Since the swept volumes are the same by design you just end up proving energy in equals energy out for any small displacement.

The fact that the shape of the swept volumes is different (rectangular for z vs triangular for x an y) makes no diffeence

 

[Khashishi]

"If the area isn't changing (hard to see how that is possible), then there is no work done." I agree, but I want the work from x and y not x +y +z.

x, y, and z are pneumatically connected. The work is not well defined. Basically, if you have two sides pushing on opposite sides of a massless wall, all the work is external to the wall, and depends on the pushers.

 

[Mary2100]

"Basically, if you have two sides pushing on opposite sides of a massless wall, all the work is external to the wall, and depends on the pushers." I don't understand, the walls don't pushing themselves. The difference of pressure (inside/outside) give a force on the walls. I rotate independently the wall x, independently the y, independently the wall z. But sure, the external device controls the angle of rotation and the translation to keep constant the volume.

"The fact that the shape of the swept volumes is different (rectangular for z vs triangular for x an y) makes no diffeence" I agree.

I would like to know if my calculations are correct for the walls x and y. I don't know what you called small displacements but I would like to calculate the work for an angle of 0.1 rd so the length d= 6.77 cm

 

[Khashishi]

Consider a pneumatic lift with incompressible fluid. When I push one piston down, the other one goes up. Now, if I put a car on top of one piston, it takes more work for me to push down on the other piston than without a car. The details of the pneumatic lift are irrelevant to the calculation.

 

[Mary2100]

"Consider a pneumatic lift with incompressible fluid. When I push one piston down, the other one goes up. Now, if I put a car on top of one piston, it takes more work for me to push down on the other piston than without a car. The details of the pneumatic lift are irrelevant to the calculation."

I can't understand why I couldn't calculate the work of a wall in rotation. If you're right, I can't calculate the work from a pneumatic actuator ? With an actuator, I can suppose the pressure constant (it is easier to calculate), the cylinder gives an energy : dP*S*x, with S the surface, dP the difference of pressure, and x the length.

 

[Khashishi]

Maybe we need to know more about the actual machine so we can help you ask the right question. Right now it sounds a little contrived so we can't come up with reasonable assumptions.

 

[CWatters]

The walls x an y must be connected to wall z at the corners. Indeed something must provide a force between z and xy that counters the pressure difference or the seals will open or the system explode. When calculating the force on wall z you have to subtract these forces. You are left with a net zero force on wall z. Walls x an y pull upwards on z with a force exactly equal to the pressure differential. The result is it takes no force to move wall z.

See diagram.

 

[Mary2100]

"or the system explode." An external device controls the walls. The walls x, y and z are independent. That external device needs to give an energy to move in translation the z wall and recover the same energy from x and y. I would like to know if my calculations are correct for the walls x and y because it is not only a question of physics it is a question about maths, I would like to know if my integrals are correct.

 

[sophiecentaur]

I'm not convinced that the constant volume stipulation is even achievable with that geometry .

The shape of the container is not relevant here - the prism shape just gets in the way of understanding, I think.
You could get exactly the same situation in principle by having a cylinder withe pistons at each end. As piston A moves left, so does piston B (moved by the same actuator). Clearly, nothing will happen to the energy situation except some increase in total KE. Doing it infinitely slowly will eliminate even that. If you move the cylinder, instead, the internal situation is identical and KE change is zero. A case of 0=0, I think.

 

[CWatters]

I would like to know if my calculations are correct for the walls x and y because it is not only a question of physics it is a question about maths,

Your calculations are not correct. Your physics is wrong so your maths is irrelevant. We've been trying to explain that to you for some time now.

"or the system explode." An external device controls the walls. The walls x, y and z are independent.

They cannot be "independent". The pressure inside creates a force that pushes wall z and xy apart. The "external device" must counteract that to stop the system exploding. It cannot do that without also applying a force between z and xy. I refer you to Newtons third law.

If you disagree then please provide a diagram showing how the external device works. How is your "external device" different to the clamp or spring I proposed?

 

[Nidum]

sophiecentaur :

Certainly true but I don't think op is interested in any explanations other than ones that would confirm her fixed ideas .

CWatters and Khashishi have already tried very hard to explain the physics of this problem .

 

[Mary2100]

"We've been trying to explain that to you for some time now." Sorry if I need time for that.

"If you disagree then please provide a diagram showing how the external device works. How is your "external device" different to the clamp or spring I proposed?" I don't know what you called a diagram but for example, I have 2 motors on the dot O, each motor controls the rotation of one wall x or y. A stepper motor controls the translation of the z wall. I can measure with a wattmeter the power of each motor and sum to have the energy, no ? Sure, I know exactly the efficiency of each motor.

"Certainly true": it is 100% true.

 

[CWatters]

OK so wall z is controlled by a stepper motor. What is the body of the motor mounted on? Newton's third law requires that the body of the motor pushes against something or the output shaft cannot push against wall z. For example a car jack cannot lift a car without pushing against the ground.

 

[Mary2100]

"What is the body of the motor mounted on?" the motor of the z wall is fixed on the ground, the motor for the z wall is a linear motor or use a gear (the z wall moves in translation).

 

[CWatters]

OK so both point O and the motor are fixed to the ground. They are not independent they both "brace" themselves against the ground. eg The pressure inside is ultimately resisted by the ground.

When calculating the power needed to move wall z this additional force must be taken into account. However because the ground force and pressure force are equal and opposite the net force is zero.

 

[Mary2100]

When calculating the power needed to move wall z this additional force must be taken into account. However because the ground force and pressure force are equal and opposite the net force is zero.

I don't understand. There are 3 motors, each motor don't consume any energy ?

 

[sophiecentaur]

I am totally confused as to what the OP is trying to achieve here. Thinking up a wacky experimental setup is the easiest thing in the world but any added complexity will produce errors if you want good results. If the OP wants to verify the Gas Laws then she should be using (even in a thought experiment) the simplest set up. Why do you think all the textbooks use simple cylinders and pistons for all the basic stuff? Extras are only bolted onto an experiment when it's desired to eliminate some inconvenient extra factor. What is the point of using such a set up as the prism shape that's been described? You can make a list of some of the possible shortfalls of such a needlessly complicated set but you are bound to miss some out. So we're asked to consider and even quantify a set of losses of Energy due to the inadequacies of the experiment. To what end?

 

[CWatters]

I don't understand.

I refer you to my post #18.

To calculate the power required to move any of the walls you need to consider all of the forces acting on the wall NOT just the pressure difference.

You have been ignoring the force that keeps the system together when it's not moving. I have explained this in post #18 and again just above. This force is provided by the clamp, spring or motor and is equal and opposite to the pressure force. So the net force is zero.

Therefore the power required to move the wall is zero.

Whatever you use (motors, clamps or springs) to stop the walls moving apart against the pressure difference consumes no power.

 

[CWatters]

I don't understand. There are 3 motors, each motor don't consume any energy ?

Why would they consume any energy?

Work = force * displacement

When everything is stationary there is no displacement going on so no work is being done.

When things are moving the net force is zero (see above) so again no work is done.

I'm assuming there is no friction and the motors are "ideal".

 

[Mary2100]

Why would they consume any energy?

Work = force * displacement

When everything is stationary there is no displacement going on so no work is being done.

When things are moving the net force is zero (see above) so again no work is done.

I'm assuming there is no friction and the motors are "ideal".

Nothing move ? Look at that image:

At start, the position is 1, at final it is 2. The motor 1 take the black wall. The motor 2 take the pink wall. The motor 3 take the green wall. The black arm turns, no ? the green wall moves, no ? for me the motors 1 & 2 recover an energy. The motor 3 needs the energy recovered by the motors 1 & 2.

 

[sophiecentaur]

@Mary:
This argument seems to be along the same lines as the arguments people try to use to prove that they have invented a Perpetual Motion Machine. You have made a highly complicated experiment and done some maths which seems to produce a strange result.
1. No net work is done
2. Some net work is done
?
It's all lost in the fog of your over complicated model. Do you think that, somehow, a complicated enough model will produce a result that proves the gas laws are wrong? Personally, I would always look for a (possibly subtle) error in the model if it yields a result that goes against conventional Science.

 

[CWatters]

Nothing move ? Look at that image:

At start, the position is 1, at final it is 2. The motor 1 take the black wall. The motor 2 take the pink wall. The motor 3 take the green wall. The black arm turns, no ? the green wall moves, no ? for me the motors 1 & 2 recover an energy. The motor 3 needs the energy recovered by the motors 1 & 2.

Yawn. Go back and read my post again. I described two situations... one while it's stationary and one while it's moving.

I think I'm just about done with this thread. All the information you need to understand the problem has been well covered.

I'll just repeat.. In in order to work out the energy required to move any of the walls you need to consider all the forces acting on the wall. You haven't done that.

 

[berkeman]

This thread will be closed unless the next post by the OP is extremely lucid and helps us to clear up your confusion. Fair warning.