YAPQ(Yet Another Physics Question)

[cheapgoose]

Alright, so here's the deal, I'm lazzy, I'm suppose to find out how to do this, but I just worked from 5:30 am - 6:30pm, and my brain is dead.

I need to know if a certain pin with a certain diameter can handle the current going through it.

Of course I don't need someone to do the whole thing, just point me in the right direction. I have the max V and I, resistivity of the material, diameter and length.

what equation do I use? I know Ohm's law, but that won't tell me if the pin will work, will it? I guess I need to know how much heat is dissapated in the pin and how hot it'll get.

thanks

and the material is kovar, if anyone's interested.

 

[JohnCU]

how do you define how much it can take though? when it gets to what power?

 

[cheapgoose]

Originally posted by: JohnCU
how do you define how much it can take though? when it gets to what power?

to be honest, I don't know. we're currently using a .030" diameter pin, I need to know if a 0.025" diameter pin will work. I guess, I'm trying to find out if we pass say like 90watts, or 10 amp through it, will it melt or get extremely hot.

thanks for the interest.

 

[drinkmorejava]

Sssuming you have any safety margin on the .03, you should be fine with a .025. Is it going to hurt to just find out. If you know the current going through it, build a simple circuit with the pin and a resistor in series, hook it up to a power supply and see what happens.

 

[JohnCU]

well, Power =R*I^2, where R = length/(area*sigma) where sigma is the conductivity. what you need is the melting point of the material i would think.

power is in Watts which is J/s. Energy is in joules so we need an energy equation like Q = hf*m where hf is the heat of fusion for the metal, but then wouldn't you need the mass of the metal, which we could get from the density of the material. hmmm... im not sure i've never done this before.

density = g/cm^3 so mass = density*volume, where the volume of a pin is pi*r^2*d i think, can't remember. you need the density of the material and the heat of fusion, unless i am totally wrong and im tired as hell so i probably am.

 

[cheapgoose]

thanks guys. I'm sooo tired. Maybe I'll make the supplier do the work tomorrow. They should at least have it on some data sheet.

 

[Evadman]

Kovar from Carpenter? the stuff that seals with pyrex? I remember that stuff. Gimme sec I have the specs on it somewhere.

<edit>
Booyah:
Mass
kg/cu m - 8359
lb/cu in - .0302

Thermals:
Btu-in/ft²/hr/°F - 120
W/m-K - 17.3


Elec Conductivity
microhm-mm - 490
ohm-cir mil/ft - 294

melting point
2640 F, 1450 C

 

[JohnCU]

Originally posted by: Evadman
Kovar from Carpenter? the stuff that seals with pyrex? I remember that stuff. Gimme sec I have the specs on it somewhere.

thanks, im kind of interested in working it out now...

 

[JohnCU]

I found some more info on the net...

Here's my analysis:

Q = hf*m, where hf for Kovar = 267.8 J/g
l = length of pin, r = radius of pin
Imax = max current applied
R = resistance
P = R*I^2

Let's say you had 1 meter, so the resistance is 490x10^-6/mm ohms which makes R = 490x10^-3 ohms.

Now, the density is .302 lb/in^3 and converting this to metric and calculating the mass, I get 3.815 g. Since Q = hf*m => Q = 1021.66 J which means you'd need 1021.66 J of energy to melt the 39.4" (1 meter) and .03" diameter wire (i just used some random values).

J/s = R*I^2

solve for I = sqrt(J/R*s)

I get 456.6 amperes, but that's for one second. So, each second, you'd have 1021.66 J of energy in the wire, which would melt it.

 

[Evadman]

Check my thought process: (it has been a REALLY long time since I had to thibnk about this)

Wattage:
.025 pin = 0.635mm^2 area
Resistance: .000490 per mm^2
watts = amps^2 * ohms
watts = 10^2 * .000490
watts = 0.049 for a 1mm^2 pin size

0.049 * (1 +(1 - 0.635)) = .066885 watts

Dissipation:
17.3 W/m-K
17.3 * 0.001 = .0173 watts per mm/K
0.0173 * 0.635 = .0109 watts per K available
.066885 / .0109 = 6.1K

So at 10 amps the pin would rise 6.1 degrees K.

Do I have that right or did I mess up somewhere?

 

[JohnCU]

Originally posted by: Evadman
Check my thought process: (it has been a REALLY long time since I had to thibnk about this)

Wattage:
.025 pin = 0.635mm^2 area
Resistance: .000490 per mm^2
watts = amps^2 * ohms
watts = 10^2 * .000490
watts = 0.049 for a 1mm^2 pin size

0.049 * (1 +(1 - 0.635)) = .066885 watts

Dissipation:
17.3 W/m-K
17.3 * 0.001 = .0173 watts per mm/K
0.0173 * 0.635 = .0109 watts per K available
.066885 / .0109 = 6.1K

So at 10 amps the pin would rise 6.1 degrees K.

Do I have that right or did I mess up somewhere?

I think the area is off, he said the diameter of the pin is 0.025 inches, so 0.025 inches / 39.4 inches = 0.0125 meters divide by 2 to get the radius is .000317, square that times pi is .316 mm^2

and the resistance is a usually a function of length, but you have it as .000490 per mm^2, should be just mm...?

 

[Molondo]

well, Power =R*I^2, where R = length/(area*sigma) where sigma is the conductivity. what you need is the melting point of the material i would think.

R = Rha*lenght/Crosssection(area) (rha is hte Value of the ability of the material to conduct electricity, its a give value, look it up)

 

[Evadman]

Originally posted by: JohnCU
I think the area is off, he said the diameter of the pin is 0.025 inches, so 0.025 inches / 39.4 inches = 0.0125 meters divide by 2 to get the radius is .000317, square that times pi is .316 mm^2

You are absolutely correct. .635mm is the diameter not the area.
3.14159 * (.3175^2) = .3166mm^2

Wattage:
.025 pin = 0.3166mm^2 area
Resistance: .000490 per mm^2
watts = amps^2 * ohms
watts = 10^2 * .000490
watts = 0.049 for a 1mm^2 pin size

0.049 * (1 +(1 - 0.3166)) = .824866 watts

Dissipation:
17.3 W/m-K
17.3 * 0.001 = .0173 watts per mm/K
0.0173 * 0.3166= .00547 watts per K available
.824866 / .00547 = 150.7K

So at 10 amps the pin would rise 150.7 degrees K. Um. Owie. That's about 200 degrees C with ambient, 392 F. Little too hot there.

Is that better?

Originally posted by: JohnCU
and the resistance is a usually a function of length, but you have it as .000490 per mm^2, should be just mm...?

Well, I'm pretty sure it is actually surface area that changes the resistance (a stranded conductor can carry marginaly more amperage than a solid one), with length being more important to determining the resistance. But I could (and probably am) using it incorrectly. I'm trying to remember from classes I took about 8 years ago

 

[JohnCU]

Yeah, I think you got it. I need to read more about dissipation... good job.

Wait, K = C + 273.15, wouldn't it be 122.45 degrees Celsius per 10 amps?

oh you said with ambient, nevermind.

yeah, resistance is proportional to the length and inversely proportional to the area. but then i need to go re-read about the skin effect with AC frequency.

 

[Evadman]

Originally posted by: JohnCU
oh you said with ambient, nevermind.

Yea I was assuming about 50 C for a high ambiant. After all, if a simple pin is going to dissapate almost a watt, there is going to be some serious temps in there