Translating the trajectory

[iridiu]

I need your help to solve the following problem:
Let's suppose we have a pipeline as in the drawing attached and inside it circulates a fluid whose trajectory is changed. From this change of trajectory of the fluid, results a reaction force which is equal with the force exerted by the pipe walls on the forward path of the fluid, in opposite direction, while the velocity vector is changing.

 

When a fluid speeds up or slows down, inertial forces come into play. Such forces may be produced by either a change in the magnitude or the direction of the velocity since either change in this vector quantity produces acceleration.

The change in momentum of a mass is equal to the impulse given to it. (Newton's 2nd law of motion)

Impulse = Force x time

Momentum = mass x velocity

Change in momentum = Δmv

Newton's second law: Δmv = Ft

Δmv/t = F

Since Δv/t = acceleration "a", we get form of the law F=ma

The force is a vector quantity which must be in the direction of Δv. Every force has an equal and opposite reaction so there must be a force on the bend equal and opposite to the force on the fluid, but I do not know how to calculate this force.

 

 

[imatfaal]

First thoughts are that a fluid cannot circulate in that pipework as it does not form a circuit - if it did form a circuit then the forces would, off the top of my head, all cancel.

 

[Guesswork]If it is not a circuit then can you visualise it as a reservoir at the south end being denuded and the fluid moving to the northern reservoir. As you have no external force the centre of mass cannot move and no external torque means no change of ang mom of the system. The system will alter such that despite the transfer of fluid the CoM does not move nor change ang mom - that is the result of any net force on the pipes. [\Guesswork]

[studiot]

Fluid in a pipeline does indeed exert (side) thrust on any bend or change of direction.

This is why thrust blocks are placed at changes of direction along a pipeline, and why pilelines should be built straight except at defined changes of direction.

You are also correct in thinking that this force is calculated by using a momentum balance for the fluid.

 

Do you understand the principles of momentum balance in fluid dynamics?

[iridiu]

Do you understand the principles of momentum balance in fluid dynamics?

Most of my knowledge in fluid dynamics I have accumulated using google, now using this forum I hope to understand better from users who have advanced knowledge.

[studiot]

You have stated the acceleration version of Newton's law.

 

another version is that Force = rate of change of momentum.

 

This is the version used in the momentum balance or so called momentum equation in fluids.

 

It's too late to do this tonight, but tomorrow I will draw a diagram to show how this is applied to forces acting at bends in pipes.

[studiot]

The first thing to realise is that pipes have curved bends, not abrupt changes of direction as you have drawn.

This is because smooth profiles make for smooth flow and smooth flow changes.

 

It is not clear whether your pipebend is vertical or horizontal. Pipes do not normally twist in two directions at once.

So the analysis can usually be two dimesional as I have shown below.

 

So fig1 shows the flow entering horizontally along the x axis.

If the bend is in a horizontal plane the y axis is also horizontal and the weight of the fluid in the bend (W) is omitted from the following calculations.

If the bend is in a vertical plane then the y axis is vertical and W appears in the vertical summation.

The fluid passes through a bend in the pipe and is deflected through angle d in the positive y direction.

 

I have also shown the bend pipe diameter reducing as the most general case.

If the pipe has constant diameter then the velocity and area do not change.

 

We consider a segment of fluid between sections S₁-S₁ and S₂-S₂, which are edit :where the pipe is tangent to the entry and exit directions respectively.

 

For a constant volumetric flow rate, Q we have

[math]Q = {V_1}{A_1} = {V_2}{A_2} = const[/math]

 

and a mass flow rate of

[math]\frac{{dm}}{{dt}} = wQ = w{V_1}{A_1} = w{V_2}{A_2} = m[/math]
Which is also constant or fluid would accumulate in the pipe.

 

Where w is the fluid density.

 

If we consider the fluid inside the bend between sections S₁-S₁ and S₂-S₂ and draw a free body diagram we obtain fig2.

 

It is very important to realise that the following is not an equilibrium analysis, since the fluid is in motion.

 

Further the fluid in the bend segment is not a differential (small) element it is the whole amount.

 

So we have the basic equation that

 

Net force acting on the fluid = Sum of all forces acting = rate of change of momentum = mass flow rate x velocity change

 

Now both the forces and the momentum are vectors so may be resolved along the x and y axes as is done in the following calculations.

 

So R is the resultant pipe force on the fluid such that

 

[math]R = \sqrt {F_y^2 + F_x^2} [/math]
and

[math]\tan r = \frac{{{F_y}}}{{{F_z}}}[/math]

 

r is the angle the resultant makes with the x axis.

 

We calculate the forces on the fluid becasue this is easier than calculating directly the forces on the pipe.

Of course the force exerted by the fluid on the pipe is equal and opposite to R.

 

The other forces acting on the fluid are the weight if vertical (otherwise W=0 as previously noted) and the pressure forces S₁ and S₂ at each section.

 

[math]{S_1} = {P_1}{A_1}[/math]
and

[math]{S_2} = {P_2}{A_2}[/math]

Where P is the fluid pressure and A the section area at that section.

 

So in the x direction

 

[math]{S_1} - {S_2}cosd - {F_x} = wQ\left( {{V_1} - {V_2}\cos d} \right)[/math]

and in the y direction

[math]{F_y} - {S_2}\sin d - W = wQ\left( {{V_2}\sin d - 0} \right)[/math]

Where V the fluid velocity at the section.

These may easily be solved for Fx and F_y

 

This analysis is useful if we wish to provide a thrust block to receive -R as a direct force.

If, however, the pipe is fitted to brackets then the brackets resist the flow force in bending.

This can be seen as R applies an unbalanced counterclockwise moment to the fluid to turn it, so the fluid applies a corresponding clockwise moment to the bracket.

 

Does this help?

Edited March 5, 2014 by studiot

[Enthalpy]

The mass throughput multiplied by the (vector) speed change gives you the force at each angle.

 

The force is along the bisector, and if the angle of the speed change is a and the liquid's speed V, the speed change is 2*sin(a/2)*V.

[iridiu]

Thank you very much for your answers and I hope you will allow me to present another example related to trajectory translation.

 

In this example, reaction is represented by the force exerted by the pipe walls on the forward path of the fluid, and from there result a change of trajectory equivalent with the distance "d" between the point of entry of the fluid "1" and the output point of the fluid "2". From this change of trajectory of the fluid results a linear force (reaction force) which occurs simultaneously and opposite in direction, according to Newton's third law of motion, in the following conditions:

1. If the fluid goes out at the same velocity as it comes in, there should be no net force on the pipe.

2. Preserving the velocity vector, doesn't change the momentum of the fluid.

 

I did a practical experiment (rudimentary), I pushed water through a elastic spiral pipe with a diaphragm pump:

 

And the result:

https://www.youtube.com/watch?v=i0kb8dYL0lQ

 

The fluid comes back where it came from, but there is momentum exchange.
I think, with a system which generates custom compressibility and pressure wave propagation, there we can have short term instantaneous net forces on the pipe, but I don't have enough knowledge to make it possible.

[studiot]

 

In this example, reaction is represented by the force exerted by the pipe walls on the forward path of the fluid, and from there result a change of trajectory equivalent with the distance "d" between the point of entry of the fluid "1" and the output point of the fluid "2". From this change of trajectory of the fluid results a linear force (reaction force) which occurs simultaneously and opposite in direction, according to Newton's third law of motion, in the following conditions:

1. If the fluid goes out at the same velocity as it comes in, there should be no net force on the pipe.

2. Preserving the velocity vector, doesn't change the momentum of the fluid.

 

I did a practical experiment (rudimentary), I pushed water through a elastic spiral pipe with a diaphragm pump:

 

And the result:

https://www.youtube....h?v=i0kb8dYL0lQ

 

The fluid comes back where it came from, but there is momentum exchange.

I think, with a system which generates custom compressibility and pressure wave propagation, there we can have short term instantaneous net forces on the pipe, but I don't have enough knowledge to make it possible.

 

 

Agreed, but what was you question here?

 

Flexible small bore pipes and pulsed fluid action (unsteady flow) introduce many new variables.

[iridiu]

This system can be regarded as a possible thruster?
I know that currently, the idea is still in the incomplete stage, but look at it as an unconventional possibility.

[studiot]

 

This system can be regarded as a possible thruster?

 

 

You would need to elaborate on that and explain why it is superior (or at least no worse than) to more conventional arrangements.

[iridiu]

I think it can be superior to Ion Thruster with the same applications, by several aspects:

1. It can use only electricity (no fuel needed)

2. Does not expel reaction mass

3. It is not in contact with the environment

4. Reduced size and weight

 

"Therefore ion thrusters create very small levels of thrust(for example thrust engine Deep Space 1 approximately equals the weight of one sheet of paper^[1]) compared to conventional chemical rockets but achieve very high specific impulse, or propellant mass efficiencies, by accelerating their exhausts to very high speed." - Wikipedia - http://en.wikipedia.org/wiki/Ion_thruster

[imatfaal]

As I posted in my first response to you - if it is a circuit then you will get no net force. Think of it this way - put an imaginary box around the device and analyse the box and its contents as a system; what external forces act on the system? None. No external forces mean no change of momentum for the centre of mass of the system

[iridiu]

As I posted in my first response to you - if it is a circuit then you will get no net force. Think of it this way - put an imaginary box around the device and analyse the box and its contents as a system; what external forces act on the system? None. No external forces mean no change of momentum for the centre of mass of the system

 

Circuit is a system, even if it's not closed in a box and there is an external force: electricity.

[imatfaal]

Not external. And electricity is not a force.

 

But most importantly - Forces internal to a system cannot change the vector sum of the momenta of the parts of a system nor, in other words, the momentum of the centre of mass.

[iridiu]

Agree, electric current is not a force, it's energy. What I wanted to say is that between circuit and exterior is an exchange of energy, even if the circuit is closed in an imaginary box, it is not an isolated system, there is an interaction.

Let's suppose that in the circuit is no pump and the fluid is moved by electromagnetic fields (MHD), in this case we have direct interaction between electric current (elctromagnetic fields) and fluid. It is known that electromagnetic fields (waves) can carry energy and momentum.

[studiot]

Electric current is not energy either.

[imatfaal]

irridiu don't sweat the small stuff like how you are moving the liquid; consider this -

 

Forces internal to a system cannot change the vector sum of the momenta of the parts of a system nor, in other words, the momentum of the centre of mass.

 

Look at these ideas http://scienceworld.wolfram.com/physics/ConservationofMomentum.html

 

Your system has no external forces acting upon it - therefore it cannot under any circumstances change its overall momentum. Think about your other forms of propulsion and analyse them - you will be able to spot external forces to the smaller system or if you analyse as a whole system with no external forces you will note there is no change in momentum.

[iridiu]

Considering this article: http://www.popsci.com/article/technology/fuel-less-space-drive-may-actually-work-says-nasa and this two videos:

 

https://www.youtube.com/watch?v=i0kb8dYL0lQ

 

https://www.youtube.com/watch?v=bKDasXphJrE

 

In these rudimentary constructions I acquired a superior thrust force compared with space drive tested by NASA, using only electricity.