Calculating Resistivity across a mesh sheet

[AlexTheFirst]

Hi everyone!

I'm trying to figure out how to go about this in calculating resistivity of a titanium grade 1 mesh. From what I know the ρ for titanium is 4.3x10^-8 ohm x m. The mesh will be a 1x12inch sheet wrapped cylindrically end to end. Here are the stock values including a change in area. Les also say the temperature of the material is 20°C and grade 1 is equal to 99.9% pure so let's just round that to 100% and ignore impurities.

Area: 12x12inches
Wire diameter: 0.0020inches = 0.0508mm
Width Opening: 0.0080 inches = 0.2032mm
Wire Mesh (wires/in): 100 x 100
% open Area: 64%
ρ for titanium is 4.3x10^-8 ohm x m

I was thinking of using
R=(ρ)(Length)/(cross section area of wire)

im not an electrical engineer, so any guidance what's so ever would be a great help. Thank you everyone.

 

[AlexTheFirst]

also if you need any other values, please let me know and I'll try to get them to you.

 

[berkeman]

Hi everyone!

I'm trying to figure out how to go about this in calculating resistivity of a titanium grade 1 mesh. From what I know the ρ for titanium is 4.3x10^-8 ohm x m. The mesh will be a 1x12inch sheet wrapped cylindrically end to end. Here are the stock values including a change in area. Les also say the temperature of the material is 20°C and grade 1 is equal to 99.9% pure so let's just round that to 100% and ignore impurities.

Area: 12x12inches
Wire diameter: 0.0020inches = 0.0508mm
Width Opening: 0.0080 inches = 0.2032mm
Wire Mesh (wires/in): 100 x 100
% open Area: 64%
ρ for titanium is 4.3x10^-8 ohm x m

I was thinking of using
R=(ρ)(Length)/(cross section area of wire)

im not an electrical engineer, so any guidance what's so ever would be a great help. Thank you everyone.

Welcome to the PF.

Your geometry is not entirely clear to me. First you talk about a metal sheet, then you wrap it around a cylinder (?) and then there are some wires attached somehow? Can you post a sketch or some other way of illustrating what you are making? Also, how thick is the sheet? What are the wires made of?

 

[AlexTheFirst]

Its a mesh sheet made out of titanium. The goal is to wrap one end and connect it to the other to create a cylinder.

The thickness of the mesh is 0.0020 inches. I
Does that help?

 

[berkeman]

Oh, it's a mesh of bare wires, not a solid sheet. Got it. And the wires are made of titanium? How are they bonded at the crossings? What will be the resistance of the bonds?

Also, how will any voltages be coupled into this? Can you say what it will be used for? Which way will currents flow through this mesh?

 

[AlexTheFirst]

The mesh is made of titanium. Heres what it more or less should look like, albeit finer.

From what I know its a solid sheet that's been perforated to make it a mesh.
I can't really say what it'll be used for, but the current should be flowing upward.
As for coupling, I'm not entirely sure what you mean by that. Do you mean where the voltage energy will be going to once it passes through the wire mesh? then it will be wrapped around a porous wick.
Also thank you for being so patient.

 

[berkeman]

From what I know its a solid sheet that's been perforated to make it a mesh.

Ah, that makes it easier.

I can't really say what it'll be used for, but the current should be flowing upward.

If the current will be flowing vertically, then you can ignore the horizontal connections. Just combine all of the parallel vertical resistances.

As for coupling, I'm not entirely sure what you mean by that.

If you were using this mesh for E-field shielding, I was asking about how the E-field (or EM wave) would be coupling into it. But it sounds like it is not being used for EM shielding?

 

[Babadag]

In my opinion, if you'll supply voltage from top to bottom using tight fitted rings[see connection type A], I should agree with berkeman.

Then there are 1200 parallel columns [12/0.01] each one of 12 inches long and 0.002 inches diameter.

Each column resistance =length[m]*rotitanium[ohm.mm^2/m]/area[mm^2]

Rcolumn=12*25.4/1000*0.043/ 0.00202683=6.4665 ohm

Rtotal=Rcolumn/1200=6.4665/1200=0.00538871 ohm

However, if the outgoing wire is connected with a column only[see connection type B] the calculation is more complicated.