# Electro Patenting Wire

joe shmo
Science 19 Mar '17 15:53
1. joe shmo
Strange Egg
19 Mar '17 15:531 edit
One of our process at the mill I currently work at is electro patenting. It is done in line ahead of a electro plating process. The wire comes in contact with a voltage source and ground as it is passing through the line causing it to heat up, which changes it microstructure etc.. I'd like to try to find the model for the temperature as a function of position between the electrical contacts. Neglecting (for now) heat transfer by radiation. I would like to try to take into account axial conduction, but because of my inexperience with using differentials, Im not sure I can do what I'm doing and am looking for some confirmation.

Start with a differential volume of the wire as it passes between contacts, and perform a power balance.

Cross Sectional Area = A
Length = dL

E'_stored = E'_in - E'_out (Eq1)

E'_stored = m' * c * dT

where,

m' = mass flow rate of steel wire
c = specific heat of steel
dT = differential change in temperature across the element

Next,

E'_in = V * I + k * A * dT/dL

where,

V = Voltage
I = Current
k = Thermal Conductivity of Steel
dT/dL = Thermal gradient across the element

Then,

E'_out = ( V - dV ) * I + dq_h
E'_out = ( V - dV ) * I + h* π * D * ( T - T_∞ ) * dL

where,
h = convection coefficient
T = Temperature as a function of position ( "L" )
D = diameter of wire
T_∞ = Ambient Temperature

Plugging all this back into Eq1, and simplifying:

m' * c * dT = I * dV - h* π * D * ( T - T_∞ ) * dL + k * A * dT/dL

dV is the voltage drop across the element

which is given by:

dV = I * dR = I * p * dL / A

where,

I = Current
p = coefficient of resistivity for steel ( assumed constant)
A = crossectional area of wire

Subbing that relationship in:

m' * c * dT = I² * p / A * dL - h* π * D * ( T - T_∞ ) * dL + k * A * dT/dL

Here is the part where I'm unconfident ( If i should be even less confident at this point let me know ). Divide through by the differential dL and re-arrange into standard form to give a second order linear ODE.

-k * A * d²T/dL² + m' * c * dT/dL + h* π * D * ( T - T_∞ ) = I² * p / A

Is the term for axial conduction I derrived here in bold legitimate?
2. 19 Mar '17 15:57
Originally posted by joe shmo
One of our process at the mill I currently work at is electro patenting.
Sorry I can't help with the math. Are you sure you spelled that right? I looked up 'electro patenting' and came up with nothing.
3. sonhouse
Fast and Curious
19 Mar '17 16:07
Originally posted by joe shmo
One of our process at the mill I currently work at is electro patenting. It is done in line ahead of a electro plating process. The wire comes in contact with a voltage source and ground as it is passing through the line causing it to heat up, which changes it microstructure etc.. I'd like to try to find the model for the temperature as a function of pos ...[text shortened]... T - T_∞ ) = I² * p / A

Is the term for axial conduction I derrived here in bold legitimate?
Pardon me for sticking in side issues but I have some experience with heat control and one thing for sure, the only time you can ignore heat loss through atmosphere is if there IS no atmosphere, like in a vacuum. In our sputtering and ion milling machines, the substrates gets hot and stays that way since radiation is a poor heat conductor compared to atmospheric heat loss.

We have a pallet coming out of the vacuum system heated to a couple hundred degrees C and sometime there is a delay before we extract the pallet and we find even hours later the pallet is still pretty hot. As compared to exchange to atmosphere and the temperature goes down to handling temps within 5 minutes or so.

Just saying, I think it will be wrong to ignore atmospheric heat loss.

BTW, that is exactly how a TC gauge works, Thermocouple gauge for vacuum. Atmospheric heat loss is so severe that the TC gauge is only accurate to about 1 torr, 1000 millitorr, after that the heat loss is so severe there is no more reading available to the TC. It is reasonably accurate from a few millitorr to 1 torr though because there is so much less air in that level of vacuum.
The TC gauge works like this: There are two very thin wires crossed like an X and one wire goes to a current source with a very well calibrated current source similar to your wire. That current is constant so there is a constant heat flow from the wire.

The other wire is a TC gauge, two wire device. It sits in the center of the heated wire.

It's job is simply to measure the temperature of the heated wire which will go up and down depending on the ambient vacuum in the system. The more air in the system, the lower the temperature reading and the opposite for less air, TC gauge reads hotter. So that is correlated with a vacuum reading, X current out of the TC gauge = Y vacuum reading.

This is a well known way to read vacuum pressures in the 1 to 1000 millitorr range and other techniques are available for ranges outside that level but the gist of this whole thing is the readings are totally unreliable for pressures even at 2 torr, a pretty good vacuum, about the level of the air on Mars.

So just trying to get across your expected temperatures will be pretty far off because of atmospheric heat loss.
4. joe shmo
Strange Egg
19 Mar '17 16:24
Sorry I can't help with the math. Are you sure you spelled that right? I looked up 'electro patenting' and came up with nothing.
It might be a bit obscure.

Our predecesor ( Bethlehem Steel Corp ) holds the US patents on the processes that we use in our mill.
5. joe shmo
Strange Egg
19 Mar '17 16:331 edit
Originally posted by sonhouse
Pardon me for sticking in side issues but I have some experience with heat control and one thing for sure, the only time you can ignore heat loss through atmosphere is if there IS no atmosphere, like in a vacuum. In our sputtering and ion milling machines, the substrates gets hot and stays that way since radiation is a poor heat conductor compared to atmosp ...[text shortened]... o get across your expected temperatures will be pretty far off because of atmospheric heat loss.
"Pardon me for sticking in side issues but I have some experience with heat control and one thing for sure, the only time you can ignore heat loss through atmosphere is if there IS no atmosphere, like in a vacuum. In our sputtering and ion milling machines, the substrates gets hot and stays that way since radiation is a poor heat conductor compared to atmospheric heat loss."

I'm not ignoring convective heat loss (atmosperic as you call it). I'm ignoring heat loss from radiation ( even though I shouldn't, because the wires are glowing red hot and radiation is most likely significant at these temps ). The problem with adding it is it makes the model differential eqaution non-linear, which is a class of equations I know almost nothing about finding a solution for.
6. 19 Mar '17 17:39
Originally posted by joe shmo
It might be a bit obscure.

Our predecesor ( Bethlehem Steel Corp ) holds the US patents on the processes that we use in our mill.
Thanks. Interesting.
7. sonhouse
Fast and Curious
19 Mar '17 20:54
Originally posted by joe shmo
"Pardon me for sticking in side issues but I have some experience with heat control and one thing for sure, the only time you can ignore heat loss through atmosphere is if there IS no atmosphere, like in a vacuum. In our sputtering and ion milling machines, the substrates gets hot and stays that way since radiation is a poor heat conductor compared to atmos ...[text shortened]... on non-linear, which is a class of equations I know almost nothing about finding a solution for.
Ok, sorry for butting in. Maybe you could introduce a figure for radiative loss by measuring it and adding a factor to the final equations.

I see the patents for that process goes back to 1918! 100 years and the one from Bethlehem Steel to 1955. Been around a while. When did they start using the term "patenting'?
8. DeepThought
19 Mar '17 21:541 edit
Originally posted by joe shmo
One of our process at the mill I currently work at is electro patenting. It is done in line ahead of a electro plating process. The wire comes in contact with a voltage source and ground as it is passing through the line causing it to heat up, which changes it microstructure etc.. I'd like to try to find the model for the temperature as a function of pos ...[text shortened]... T - T_∞ ) = I² * p / A

Is the term for axial conduction I derrived here in bold legitimate?
I think the equation should be first order. You can't divide dT/dL by dL to get d^2T/dL^2. From memory alone the equation ought to be mcdT not mcdT/dL. So I think you've divided by dT too many times. I'll have a think about it, and see if I can sort out the radiative term as well as it's relevant to a problem I'm interested in anyway. Are you assuming that the heating is uniform throughout the wire and that current flows uniformly in an effectively two dimensional slice of the wire?
9. Soothfast
0,1,1,2,3,5,8,13,21,
19 Mar '17 22:06
Originally posted by joe shmo
It might be a bit obscure.

Our predecesor ( Bethlehem Steel Corp ) holds the US patents on the processes that we use in our mill.
So...that link is to a U.S. patent...for a process of annealing and quenching steel wires...that is itself called "patenting"...

A cute trick for keeping a patent quasi-secret: makes it harder to search for in a database.
10. joe shmo
Strange Egg
19 Mar '17 22:47
Originally posted by Soothfast
So...that link is to a U.S. patent...for a process of annealing and quenching steel wires...that is itself called "patenting"...

A cute trick for keeping a patent quasi-secret: makes it harder to search for in a database.
Hmm... I'm not sure on the origins of the term, but your scenario sounds plausible!
11. joe shmo
Strange Egg
19 Mar '17 23:12
Originally posted by DeepThought
I think the equation should be first order. You can't divide dT/dL by dL to get d^2T/dL^2. From memory alone the equation ought to be mcdT not mcdT/dL. So I think you've divided by dT too many times. I'll have a think about it, and see if I can sort out the radiative term as well as it's relevant to a problem I'm interested in anyway. Are you assumi ...[text shortened]... t the wire and that current flows uniformly in an effectively two dimensional slice of the wire?
"You can't divide dT/dL by dL to get d^2T/dL^2."

I had a hunch that most likely was incorrect.

I was talking about a power balance ( not energy ) but the original term for stored power was

m' * c * dT
m' = mass flowrate of steel wire ( the element is traveling between points of electrical contact at a constant speed "v" )

"I'll have a think about it, and see if I can sort out the radiative term as well as it's relevant to a problem I'm interested in anyway"

I believe you would tack on the term "- dq_rad = -π*D*ε*σ*(T^4 - T_surr^4)*dL" to the RHS to account for radiation heat transfer.

ε = emissivity
σ = Stefan Boltzmann Constant
T_surr = Temperature of Surroundings

"Are you assuming that the heating is uniform throughout the wire"
Yes uniform throughought the crossection (the only temperature gradient is over length)

"current flows uniformly in an effectively two dimensional slice of the wire"
Yes, assuming current per unit volume is constant.

just as a marker, If I ignore axial conduction ( and radiation ) I'm fairly certian the model will be given by:

m' * c * dT = I² * p / A * dL - h* π * D * ( T - T_∞ ) * dL

m' * c * dT/dL + h* π * D * ( T - T_∞ ) = I² * p / A ( First Order - Linear)
12. joe shmo
Strange Egg
20 Mar '17 20:452 edits
I think I've figured out the proper way to handle the conduction term in the model. If I'm correct about this, I believe that my original term was correct...However, as you pointed out it was not derived correctly. Even crazy squirrels get a nut every now and then!

The power entering the element via conduction is given by

E '_in = q

E '_out = ( q + (dq/dL)*dL)

q = -k* A * dT/dL

E '_in - E '_out = -(dq/dL)*dL = k*A*d²T/dL² * dL

The model ( ignoring radiation) is then,

m' * c * dT = I² * p / A * dL - h* π * D * ( T - T_∞ ) * dL + k * A * d²T/dL² * dL

Then divide through by dL

m' * c * dT/dL = I² * p / A - h* π * D * ( T - T_∞ ) + k * A * d²T/dL²