How to calculate the thrust

[Ashiq_A]

please help me to solve this problem
The air platform is used to support an object on a thin layer of air to
reduce the friction when the object is moving. A flat plate (1m2) is placed
on the platform, the thickness of the air cushion is 1mm, and the gas
viscosity coefcient μ=2.18×10-5n·s/m2. What is the value of thrust
required to make the plate move in parallel with 1m/s.

 

[MacGyverS2000]

Show us your work, we'll help aim you in the right direction... but we don't directly solve your homework problems for you.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[waross]

You have a couple of factors missing.
What is the mass of the moving parts.
What rate of acceleration do you intend to use.
In a word:
Inertia.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Bill, I think steady state is assumed so no inertia needed. Likewise air cushion thickness given, so the need to find the vertical load disappears in this simplified problem.

op,

Find a definition of viscosity.

Does that definition relate to a velocity gradient? (rate of change of velocity with respect to distance in the perpendicular direction)

Can you find a velocity gradient in your problem?

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(2B)+(2B)' ?

 

[itsmoked]

Hmmm. So let me see if I have this correct.

The weight will fall out of the viscosity and the fact that it's mushed down to 1mm thick.

I thought that if you had something on a cushion of air you essentially have no friction. I guess not for this problem.

The question is a steady-state question so for this question there is friction which is somehow gotten from the area, weight, and desired speed.

That correct?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[electricpete]

> I thought that if you had something on a cushion of air you essentially have no friction.

They give a non-zero viscosity for the air/gas, so the friction won't be zero (although you're right it's probably very very low).

In a fluid film bearing, the relative movement of two surfaces with a viscous fluid (oil) between them creates friction in the process.
That's a similar situation to the moving air cushion, except the air/gas plays the role of the oil (*)

(*) in reality the dynamics of an air cushion are more complicated than a simple plain sliding bearing ... it's more like a rotating bearing with injected lift oil because of the air injected under pressure into the clearance which complicates the velocity profile in the clearance. But from the given information it seems pretty clear they intend to use a very simple model where the horizontal gas velocity Vx varies linearly across the depth of the air film in the y direction but is otherwise uniform across the 1m^2 area (no velocity variation in the z and x directions). The fluid shear dVx/dy of the viscous fluid creates friction.

> The weight will fall out of the viscosity and the fact that it's mushed down to 1mm thick.

Yes, the weight will not be part of the problem if that other info is specified. I don't think you can compute a weight from the limited info provided and the simple model suggested by the question.

> The question is a steady-state question so for this question there is friction which is somehow gotten from the area, weight, and desired speed.

Yes, fluid friction will be computed from film thickness, viscosity, area and speed. Weight not required if the other stuff is known.

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(2B)+(2B)' ?

 

[waross]

Does the viscosity vary with the density of the air, as influenced by the applied air pressure?
If so, the viscosity will vary from the center of the plate to the edges as the pressure drops.
The weight will determine the average air pressure required to lift the plate 1mm.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[GregLocock]

I can't help wondering if there might be an equation relating the velocity gradient, the shearing force per unit area, and the viscosity. Perhaps a textbook might provide an answer.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[electricpete]

Bill I think in the real world you'd be correct. But this is the student / textbook world. You solve from the information given. Viscosity is given but no viscosity-pressure coefficient (if pressure was important why wouldn't they tell you the pressure that their viscosity value corresponded to, and why give viscosity but not pressure coefficient if both are needed). Even if you searched to find such coefficient, there's still a ton of info missing to do anything with it and you're probably end up needing numerical computer solution involving the Navier Stokes equations. I don't think that's what the question had in mind for student purposes.

I'm with Greg on this one.

Meanwhile op logged in once to post and hasn't been back...


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(2B)+(2B)' ?