Pedal Control | Amp Input + Deep Dive Into HI and TC
Posted: Tue Aug 30, 2016 8:52 pm
Pedal Control Relative to Amp Input + Deep Dive Into the Dark Side of Heat Input and Thermal Conductivity
The idea here isn't necessarily an idea, it's more of a theory of theories in my way of notation for sharing. The initial focus is on how the Amperage range is affected by the travel of pedal. One day I might include the 22k vs. 47k Ohm differences... Later it goes into the physics and sciences relative to welding, as I try to theorize how it all "clicks" together. Read on if you dare and call me crazy if you'd like!
My Basic Everlast 22k Ohm pedal has 10 clicks for full pedal travel. The total travel is 25 degrees. The first half of the pedal is easy to push but after 50% push the tension in the spring is significantly increasing giving a slightly harder press and stronger return i.e. less control.
To puddle my metal I have to have Y number of Amps at X work piece temperature in N seconds. These numbers are determined by what you know and how fast you can go.
By taking the total of 10 clicks and using only half which is 5 for my easy ride, and using the upper half of the 6-10 for starting the puddle, I can then do some tests and math to find a pretty good set of values to use with the pedal and it can also help me figure out just how many Amps I've been using.
For my aluminum I was running 150 Amps at the dial and with a part at 110-110F I was able to use the 10 clicks to start my puddle, then feathering around 6-8 for about the first 1/2 in, and then towards 2-5 for the rest and I'd generally get a decent bead that was shiny with a good travel speed I am comfortable with.
With 150A as the high and 0 as the low with the 10 points of travel in the pedal we can infer the increments are 15A per click so I'm starting out with 150A and tapering off between 90-120A as I begin travel and pulse the pedal, then continue after that 1/2in at about 30-75A.
Without clicks, we can infer every degree equals 6 Amps since 150A / 25 degrees = 6A per Degree.
If it was 200 Amps I'd have 20 Amps per click or 8 Amps per degree.
You can take this information to make some affirmations about where you're at or where you need to go. There is more to it but I think this is good for those ball park figures.
Now I'm about to get deep in an area I'm researching but don't fully understand.... This is not ME telling YOU how it is, these are my notes regarding all my research thus far. I'm wondering if I'm missing something or if this is what I should be thinking about as I delve into the science/physics of welding.
So on to the theoretical notes:
I would imagine we should take this further and consider the solidus and liquidus states of the materials. I've listed some examples below:
Aluminum
Solidus 582 °C 1080 °F AA; Typical
Liquidus 652 °C 1205 °F AA; Typical
11% difference between solidus and liquidus
125F Degrees difference between solidus and liquidus
plain carbon steel
Soliuds 1,130 °C (2,070 °F)
Liquidus 1,492 °C (2,718 °F)
24% difference between solidus and liquidus
648F Degrees difference between solidus and liquidus
Stainless steel 304L
Solidus 1400 °C 2550 °F
Liquidus 1450 °C 2640 °F
4% difference between solidus and liquidus
90F Degrees difference between solidus and liquidus
What this would tell me is that between my lowest and highest points (once things are going) of my pedal press should be relative to the difference between solidus and liquidus so if the difference is 10% then the difference between the high and low pedal press should be around 10% of the necessary Amps for the metal to remain between solidus and liquidus.
The question that would drive your range is whether you'd want to keep in the liquidus state or go to or below solidus to freeze before your next pulse since this would set your range. I'd imagine the solidus/liquidus difference percentage should be equal or be close to the difference percentage you've come up with for your pedal travel vs. Amps needed to melt relative to how much melt you want.
You might be able to find this range easier by finding what the highest Amp you can run and not melt the metal and the lowest Amp you can run to begin melting the metal. Sound easy enough but this is where thermal conductivity ensures you will "never" have a constant Amp value.
Let's really fall at terminal velocity down the rabbit hole....
Now I do not understand a direct way to write out how thermal conductivity plays it's part but Jody did give the Heat Input equation that helps determine some of this in one of his videos, just can't remember which one... I'm posting all these thoughts and the rest of this information to see if perhaps some science expert can make some of this simpler, tell me I'm completely wrong, and/or tell me I'm missing something...It's also for me to reference as time goes on to see if I can't make more sense of it and this wasn't easily put together.
Heat Input equation:
Heat Input(KiliJoules per mm or inch (kJ/mm or kJ/in)) = (Voltage X Amperage X Time) / (DistanceTraveled in/mm X 1000)
1 Joule = 4.186 Calories
1 Joule = .000239 Kilocalorie
1 Joule = .0239 Calorie
1 Joule = 0.000526565076466 Celcius Heat Units
1 Calorie = 4.1868 Watts Per Second
1 Amp = 12 Watts / 12 Volts
1 Amp = 9 Watts / 9 Volts
1 Amp = 9999 Watts / 9999 Volts
Watts = Amps * Volts
1000 Watts = 100 Amps * 9 Volts
1 Calorie is required to increase the temperature of one gram of water by 1C.
1 Celcius heat unit is required to raise the temperature of 1Lbs of water by 1C.
TIG perfect arc is ~9 Volts
Thermal Conductivity of Materials
http://www.engineeringtoolbox.com/therm ... d_429.html
- k - W/(m K)
Material 25C 125C 225C
Aluminum 205 215 250
Carbon Steel 54 51 47
Stainless Steel 16 17 19
Water 0.58
Conductive Heat Transfer through an Aluminum Pot Wall with thickness 2 mm - temperature difference 80C
Thermal conductivity for aluminum is 215 W/(m K) (from the table above). Conductive heat transfer per unit area can be calculated as
q / A = (215 W/(m K)) (80C) / (2 10^3 m)
= 8600000 (W/m^2)
= 8600 (kW/m^2)
Conductive Heat Transfer through a Stainless Steel Pot Wall with thickness 2 mm - temperature difference 80C
Thermal conductivity for stainless steel is 17 W/(m K) (from the table above). Conductive heat transfer per unit area can be calculated as
q / A = (17 W/(m K)) (80C) / (2 10^3 m)
= 680000 (W/m^2)
= 680 (kW/m^2)
http://www.engineeringtoolbox.com/condu ... d_428.html
Conduction as heat transfer takes place if there is a temperature gradient in a solid or stationary fluid medium.
With conduction energy transfers from more energetic to less energetic molecules when neighboring molecules collide. Heat flows in direction of decreasing temperatures since higher temperatures are assCiated with higher molecular energy.
Conductive heat transfer can be expressed with "Fourier's Law":
q = k A dT / s (1)
where
q = heat transfer (W, J/s, Btu/hr)
A = heat transfer area (m2, ft2)
k# = thermal conductivity of material (W/m K or W/m C, Btu/(hr F ft2/ft))
dT = temperature gradient - difference - in the material (K or C, F)
s# = material thickness (m, ft)
Example - Heat Transfer by Conduction
A plane wall is constructed of solid iron with thermal conductivity 70 W/mC. Thickness of the wall is 50 mm and surface length and width is 1 m by 1 m. The temperature is 150C on one side of the surface and 80C on the other.
The conductive heat transfer through the wall can be calculated
q = (70 W/m C) (1 m) (1 m) ((150C) - (80C)) / (0.05 m)
= 98000 (W)
= 98 (kW)
Heat Transfer through Plane Walls or Layers in Series
Heat conducted through several walls or layers in thermal contact can be expressed as
q = (T1 - Tn) / ((s1 / k1 A) + (s2 / k2 A) + ... + (sn / kn A)) (2)
where
T1 = temperature inside surface (K or C, F)
Tn = temperature outside surface (K or C, F)
Example - Heat Transfer through a Furnace Wall
A furnace wall of 1 m^2 consist of 1.2 cm thick stainless steel inner layer covered with 5 cm outside insulation layer of insulation board. The inside surface temperature of the steel is 800 K and the outside surface temperature of the insulation board is 350 K. The thermal conductivity of the stainless steel is 19 W/(m K) and the thermal conductivity of the insulation board is 0.7 W/(m K).
The conductive heat transport through the layered wall can be calculated as
q = ((800 K) - (350 K)) / (((0.012 m) / (19 W/(m K)) (1 m^2)) + ((0.05 m) / (0.7 W/(m K)) (1 m^2)))
= 6245 (W)
Thermal Conductivity Units
Btu/(h ft^2 oF/ft)
Btu/(h ft^2 oF/in)
Btu/(s ft^2 oF/ft)
MW/(m^2 K/m)
kW/(m^2 K/m)
W/(m^2 K/m)
W/(m^2 K/cm)
W/(cm^2 C/cm)
W/(in^2 oF/in)
kJ/(h m^2 K/m)
J/(s m^2 C/m)
kcal/(h m^2 C/m)
cal/(s cm^2 C/cm)
1 W/(m K) = 1 W/(m C) = 0.85984 kcal/(h m C) = 0.5779 Btu/(ft h F) = 0.048 Btu/(in h F)
A fairly complete list of metals and their thermal conductive properties:
http://www.engineeringtoolbox.com/therm ... d_858.html
Another site with a nifty nomograph calculator that explains heat input in great detail for various steels although it fCuses on stick/MIG. The principles should be the same:
http://www.arcraftplasma.com/welding/we ... errous.htm
For thermal conductivity which includes equations for geometry relative to heat convection:
http://www.roymech.co.uk/Related/Thermo ... nsfer.html
Lastly, here's a site that explains all of this in about the most simplest terms and puts a lot of the thoughts here together fairly well with useful equations and charts.
Lesson 1:
http://www.physicsclassroom.com/class/t ... t-Transfer
Lesson 2:
http://www.physicsclassroom.com/Class/t ... u18l2a.cfm
So to put everything together Thermal Conductivity per material based on existing temperature and thickness, and the desired expectation of being solidus vs. liquidus determine the amount of Heat Input and travel speed necessary to perform your welding job. To understand the amount of wattage, which will determine Amps when using the 9V "perfect arc length for TIG" it seems best to use Joules converted to wattage, or perhaps using Caloric energy by using an equation that considers thermal conductivity of water in relation to the metal to be welded since water is the basis of Caloric energy measurements.
Understanding these requirements would, I believe, set the precedence for the maximum/minimum Amperage desired for any welding pulse/non-pulse parameters. When using a Foot Pedal, one would also have to calculate the variance between the maximum/minimum Amperages and the desired degrees of travel the user intends to use/has available on their foot pedal.
How the hell do you take all these equations and make one complete equation to determine all of this? Have fun!
The idea here isn't necessarily an idea, it's more of a theory of theories in my way of notation for sharing. The initial focus is on how the Amperage range is affected by the travel of pedal. One day I might include the 22k vs. 47k Ohm differences... Later it goes into the physics and sciences relative to welding, as I try to theorize how it all "clicks" together. Read on if you dare and call me crazy if you'd like!
My Basic Everlast 22k Ohm pedal has 10 clicks for full pedal travel. The total travel is 25 degrees. The first half of the pedal is easy to push but after 50% push the tension in the spring is significantly increasing giving a slightly harder press and stronger return i.e. less control.
To puddle my metal I have to have Y number of Amps at X work piece temperature in N seconds. These numbers are determined by what you know and how fast you can go.
By taking the total of 10 clicks and using only half which is 5 for my easy ride, and using the upper half of the 6-10 for starting the puddle, I can then do some tests and math to find a pretty good set of values to use with the pedal and it can also help me figure out just how many Amps I've been using.
For my aluminum I was running 150 Amps at the dial and with a part at 110-110F I was able to use the 10 clicks to start my puddle, then feathering around 6-8 for about the first 1/2 in, and then towards 2-5 for the rest and I'd generally get a decent bead that was shiny with a good travel speed I am comfortable with.
With 150A as the high and 0 as the low with the 10 points of travel in the pedal we can infer the increments are 15A per click so I'm starting out with 150A and tapering off between 90-120A as I begin travel and pulse the pedal, then continue after that 1/2in at about 30-75A.
Without clicks, we can infer every degree equals 6 Amps since 150A / 25 degrees = 6A per Degree.
If it was 200 Amps I'd have 20 Amps per click or 8 Amps per degree.
You can take this information to make some affirmations about where you're at or where you need to go. There is more to it but I think this is good for those ball park figures.
Now I'm about to get deep in an area I'm researching but don't fully understand.... This is not ME telling YOU how it is, these are my notes regarding all my research thus far. I'm wondering if I'm missing something or if this is what I should be thinking about as I delve into the science/physics of welding.
So on to the theoretical notes:
I would imagine we should take this further and consider the solidus and liquidus states of the materials. I've listed some examples below:
Aluminum
Solidus 582 °C 1080 °F AA; Typical
Liquidus 652 °C 1205 °F AA; Typical
11% difference between solidus and liquidus
125F Degrees difference between solidus and liquidus
plain carbon steel
Soliuds 1,130 °C (2,070 °F)
Liquidus 1,492 °C (2,718 °F)
24% difference between solidus and liquidus
648F Degrees difference between solidus and liquidus
Stainless steel 304L
Solidus 1400 °C 2550 °F
Liquidus 1450 °C 2640 °F
4% difference between solidus and liquidus
90F Degrees difference between solidus and liquidus
What this would tell me is that between my lowest and highest points (once things are going) of my pedal press should be relative to the difference between solidus and liquidus so if the difference is 10% then the difference between the high and low pedal press should be around 10% of the necessary Amps for the metal to remain between solidus and liquidus.
The question that would drive your range is whether you'd want to keep in the liquidus state or go to or below solidus to freeze before your next pulse since this would set your range. I'd imagine the solidus/liquidus difference percentage should be equal or be close to the difference percentage you've come up with for your pedal travel vs. Amps needed to melt relative to how much melt you want.
You might be able to find this range easier by finding what the highest Amp you can run and not melt the metal and the lowest Amp you can run to begin melting the metal. Sound easy enough but this is where thermal conductivity ensures you will "never" have a constant Amp value.
Let's really fall at terminal velocity down the rabbit hole....
Now I do not understand a direct way to write out how thermal conductivity plays it's part but Jody did give the Heat Input equation that helps determine some of this in one of his videos, just can't remember which one... I'm posting all these thoughts and the rest of this information to see if perhaps some science expert can make some of this simpler, tell me I'm completely wrong, and/or tell me I'm missing something...It's also for me to reference as time goes on to see if I can't make more sense of it and this wasn't easily put together.
Heat Input equation:
Heat Input(KiliJoules per mm or inch (kJ/mm or kJ/in)) = (Voltage X Amperage X Time) / (DistanceTraveled in/mm X 1000)
1 Joule = 4.186 Calories
1 Joule = .000239 Kilocalorie
1 Joule = .0239 Calorie
1 Joule = 0.000526565076466 Celcius Heat Units
1 Calorie = 4.1868 Watts Per Second
1 Amp = 12 Watts / 12 Volts
1 Amp = 9 Watts / 9 Volts
1 Amp = 9999 Watts / 9999 Volts
Watts = Amps * Volts
1000 Watts = 100 Amps * 9 Volts
1 Calorie is required to increase the temperature of one gram of water by 1C.
1 Celcius heat unit is required to raise the temperature of 1Lbs of water by 1C.
TIG perfect arc is ~9 Volts
Thermal Conductivity of Materials
http://www.engineeringtoolbox.com/therm ... d_429.html
- k - W/(m K)
Material 25C 125C 225C
Aluminum 205 215 250
Carbon Steel 54 51 47
Stainless Steel 16 17 19
Water 0.58
Conductive Heat Transfer through an Aluminum Pot Wall with thickness 2 mm - temperature difference 80C
Thermal conductivity for aluminum is 215 W/(m K) (from the table above). Conductive heat transfer per unit area can be calculated as
q / A = (215 W/(m K)) (80C) / (2 10^3 m)
= 8600000 (W/m^2)
= 8600 (kW/m^2)
Conductive Heat Transfer through a Stainless Steel Pot Wall with thickness 2 mm - temperature difference 80C
Thermal conductivity for stainless steel is 17 W/(m K) (from the table above). Conductive heat transfer per unit area can be calculated as
q / A = (17 W/(m K)) (80C) / (2 10^3 m)
= 680000 (W/m^2)
= 680 (kW/m^2)
http://www.engineeringtoolbox.com/condu ... d_428.html
Conduction as heat transfer takes place if there is a temperature gradient in a solid or stationary fluid medium.
With conduction energy transfers from more energetic to less energetic molecules when neighboring molecules collide. Heat flows in direction of decreasing temperatures since higher temperatures are assCiated with higher molecular energy.
Conductive heat transfer can be expressed with "Fourier's Law":
q = k A dT / s (1)
where
q = heat transfer (W, J/s, Btu/hr)
A = heat transfer area (m2, ft2)
k# = thermal conductivity of material (W/m K or W/m C, Btu/(hr F ft2/ft))
dT = temperature gradient - difference - in the material (K or C, F)
s# = material thickness (m, ft)
Example - Heat Transfer by Conduction
A plane wall is constructed of solid iron with thermal conductivity 70 W/mC. Thickness of the wall is 50 mm and surface length and width is 1 m by 1 m. The temperature is 150C on one side of the surface and 80C on the other.
The conductive heat transfer through the wall can be calculated
q = (70 W/m C) (1 m) (1 m) ((150C) - (80C)) / (0.05 m)
= 98000 (W)
= 98 (kW)
Heat Transfer through Plane Walls or Layers in Series
Heat conducted through several walls or layers in thermal contact can be expressed as
q = (T1 - Tn) / ((s1 / k1 A) + (s2 / k2 A) + ... + (sn / kn A)) (2)
where
T1 = temperature inside surface (K or C, F)
Tn = temperature outside surface (K or C, F)
Example - Heat Transfer through a Furnace Wall
A furnace wall of 1 m^2 consist of 1.2 cm thick stainless steel inner layer covered with 5 cm outside insulation layer of insulation board. The inside surface temperature of the steel is 800 K and the outside surface temperature of the insulation board is 350 K. The thermal conductivity of the stainless steel is 19 W/(m K) and the thermal conductivity of the insulation board is 0.7 W/(m K).
The conductive heat transport through the layered wall can be calculated as
q = ((800 K) - (350 K)) / (((0.012 m) / (19 W/(m K)) (1 m^2)) + ((0.05 m) / (0.7 W/(m K)) (1 m^2)))
= 6245 (W)
Thermal Conductivity Units
Btu/(h ft^2 oF/ft)
Btu/(h ft^2 oF/in)
Btu/(s ft^2 oF/ft)
MW/(m^2 K/m)
kW/(m^2 K/m)
W/(m^2 K/m)
W/(m^2 K/cm)
W/(cm^2 C/cm)
W/(in^2 oF/in)
kJ/(h m^2 K/m)
J/(s m^2 C/m)
kcal/(h m^2 C/m)
cal/(s cm^2 C/cm)
1 W/(m K) = 1 W/(m C) = 0.85984 kcal/(h m C) = 0.5779 Btu/(ft h F) = 0.048 Btu/(in h F)
A fairly complete list of metals and their thermal conductive properties:
http://www.engineeringtoolbox.com/therm ... d_858.html
Another site with a nifty nomograph calculator that explains heat input in great detail for various steels although it fCuses on stick/MIG. The principles should be the same:
http://www.arcraftplasma.com/welding/we ... errous.htm
For thermal conductivity which includes equations for geometry relative to heat convection:
http://www.roymech.co.uk/Related/Thermo ... nsfer.html
Lastly, here's a site that explains all of this in about the most simplest terms and puts a lot of the thoughts here together fairly well with useful equations and charts.
Lesson 1:
http://www.physicsclassroom.com/class/t ... t-Transfer
Lesson 2:
http://www.physicsclassroom.com/Class/t ... u18l2a.cfm
So to put everything together Thermal Conductivity per material based on existing temperature and thickness, and the desired expectation of being solidus vs. liquidus determine the amount of Heat Input and travel speed necessary to perform your welding job. To understand the amount of wattage, which will determine Amps when using the 9V "perfect arc length for TIG" it seems best to use Joules converted to wattage, or perhaps using Caloric energy by using an equation that considers thermal conductivity of water in relation to the metal to be welded since water is the basis of Caloric energy measurements.
Understanding these requirements would, I believe, set the precedence for the maximum/minimum Amperage desired for any welding pulse/non-pulse parameters. When using a Foot Pedal, one would also have to calculate the variance between the maximum/minimum Amperages and the desired degrees of travel the user intends to use/has available on their foot pedal.
How the hell do you take all these equations and make one complete equation to determine all of this? Have fun!