Finding current and temperature of platinum wire

[temaire]

I plan on using a 1 ft long, 0.005" diameter platinum wire as part of an electrode. The wire will be bare but will be inside a hollow plastic tube that has an inner diameter of 0.08". It needs to be able to carry a maximum of 1 mA for a length of 10 minutes, while not exceeding 60^oC. I'm not sure whether this wire meets this requirement, so I searched online and found the Preece equation (W.H. Preece, Royal Soc. Proc., London, 36, p464, 1884).

$$I_{fuse} = ad^{\frac{3}{2}}$$

where $a$ is the fusing constant, and is equal to 5172 according to the Standard Handbook for Electrical Engineers, 6th ed., Sec 15, p153, and $d$ is the diameter of the wire in inches.

Plugging these values into Preece's equation yields:

$$I_{fuse} = (5172)(0.005)^{\frac{3}{2}}$$
$$I_{fuse} = 1.83 Amps$$

So according to Preece's equation, the platinum wire can carry a maximum of 1.83 Amps before reaching its melting point of 1,768^oC. I'm not sure how accurate this answer is since it doesn't take into consideration the length of wire, elapsed time, or the increasing resistivity of the wire with increasing temperature.

Is there another way I can find out whether the wire can meet the requirement?

 

[jim hardy]

Here's another back of the envelope approach:

If i cipher correctly that half mil wire should be about 61.4 ohms per foot at room temperature
but I'm rusty - what do you get?


and that'll increase by 0.24 ohms per degc

1 milliamp through 61.4 ohms will dissipate only 61.4 microwatts

specific heat of platinum is 0.13 joule/gram.K
and density is 21.37 grams/cm³
http://hyperphysics.phy-astr.gsu.edu/hbase/pertab/pt.html
what's mass of your wire? How fast will 61.4 microwatts heat it?

 

[Bystander]

"Half mil?"

 

[dlgoff]

Here's another different approach. Hot-wire anemometers use convective heat transfer. The heat convection transfer equations there may be helpful. At least you'll need to consider convection with your device.

Consider[/PLAIN] a wire that's immersed in a fluid flow. Assume that the wire, heated by an electrical current input, is in thermal equilibrium with its environment. The electrical power input is equal to the power lost to convective heat transfer,

where I is the input current, Rw is the resistance of the wire, Tw and Tf are the temperatures of the wire and fluid respectively, Aw is the projected wire surface area, and h is the heat transfer coefficient of the wire.

 

[temaire]

Here's another back of the envelope approach:

If i cipher correctly that half mil wire should be about 61.4 ohms per foot at room temperature
but I'm rusty - what do you get?and that'll increase by 0.24 ohms per degc

1 milliamp through 61.4 ohms will dissipate only 61.4 microwatts

specific heat of platinum is 0.13 joule/gram.K
and density is 21.37 grams/cm³
http://hyperphysics.phy-astr.gsu.edu/hbase/pertab/pt.html
what's mass of your wire? How fast will 61.4 microwatts heat it?

I think I understand the approach you're suggesting, and I've tried calculating the change in temperature as follows:

1. I define the actual resistance based on the temperature coefficient of resistance and the standard resistance at 20^oC.
$$\Delta R = \alpha R_s \Delta T$$
$$R - R_s = \alpha R_s \Delta T$$
$$R = R_s (1+\alpha \Delta T)$$
$$R_s = \frac{\rho L}{A}$$
$$R = \frac{\rho L}{A} (1+\alpha \Delta T)$$

2. I introduce the electrical power equation.
$$P = I^2 R$$

3. I introduce the heat capacity equation and manipulate for change in temperature.
$$Q = m c_p \Delta T$$
$$\Delta T = \frac{P t}{m c_p}$$

4. I define equation for mass, in which $D$ is density.
$$m = DLA$$

5. The cross sectional area of the wire.
$$A = \frac{\pi d^2}{4}$$

6. I substitute equations 2 and 4 into 3.
$$\Delta T = \frac{I^2 Rt}{DLA c_p}$$

7. I substitute equation 1 into 6.
$$\Delta T = \frac{I^2 \rho L t (1+\alpha \Delta T)}{DLA^2 c_p}$$

8. The lengths cancel out and I substitute equation 5 into 7.
$$\Delta T = \frac{I^2 \rho t (1+\alpha \Delta T)}{D(\frac{\pi d^2}{4})^2 c_p}$$

9. I isolate for change in temperature.
$$\Delta T = \left [1 - \frac{\alpha I^2 \rho t}{\frac{\pi d^2}{4}^2 c_p D} \right ]^{-1} \left [ \frac{I^2 \rho t}{\frac{\pi d^2}{4}^2 c_p D} \right ]$$

10. I substitute values into equation 9. I obtained the resistivity and temperature coefficient, $\alpha$, from http://hyperphysics.phy-astr.gsu.edu/hbase/tables/rstiv.html. The specific heat capacity and density are from http://hyperphysics.phy-astr.gsu.edu/hbase/pertab/pt.html.
$$\Delta T = \left [1 - \frac{(3.93\times 10^{-3}) (1\times 10^{-3})^2 (1.06\times 10^{-7}) (600)}{\frac{\pi (1.27\times 10^{-4})^2}{4}^2 (\frac{21.37}{0.01^{3}}) (0.13)} \right ]^{-1} \left [ \frac{(1\times 10^{-3})^2 (1.06\times 10^{-7}) (600)}{\frac{\pi (1.27\times 10^{-4})^2}{4}^2 (\frac{21.37}{0.01^{3}}) (0.13)} \right ]$$

11. The answer.
$$\Delta T = 0.14 K$$

Therefore, the wire should be operating at about 20.14^oC at 1 mA. Is this correct?

 

[jim hardy]

"Half mil?"

wow thanks - make that 5 mil

i sure missed the decimal - time to clean my screen anf my bifocals

Therefore, the wire should be operating at about 20.14oC at 1 mA. Is this correct?

What an elegant post ! I sure wish i could master Latex (or even learn to crawl in it).

Your approach looks flawless.
At the moment I'm uncertain about denominators in 10:
area is meters^2 ?
density grams per meter^3 ?
specific heat J/gm.K ?
yields J/meter.K
ahhh numerator has meter in resistivity so legths cancel , as you said

i can't find anything amiss !

I'm old and work in circular mils
and learned in high school that 1 mil copper is 10.4 ohms per foot
platinum having resistivity 6.31X higher, 1 mil platinum would be 65.6 ohms per foot
5 mil would be 1/25th that , 2.62 ohms per foot
which at 1 milliamp is 2.62 microwatts per foot

gotta run now, when I've figured mass and rate of heat rise i'll be back

but it's intuitive that a milliamp won't heat 5 mil wire very much .
Industrial temperature detectors are wound with yet smaller platinum wire and at 1 milliamp the self heating i encountered was negligible. But we found a meter that applies 10 milliamps will heat them a degree.or so.
Here's some practical information from http://www.pyromation.com/Downloads/Doc/Training_RTD_Theory.pdf

your result seems very reasonable.

I learned something just reading your wonderful presentation of the math

old jim

 

[temaire]

Thanks for the feedback Jim, I really appreciate your help.

In step 10, the area is in m^2, the density is in gm/cm^3, but I divide it by a conversion factor to get gm/m^3, and the specific heat is in J/gm.K. All units cancel out except for K.

I'm from Canada, so we were taught everything in SI units. I've never heard of circular mils until now.

 

[jim hardy]

I've never heard of circular mils until now.

When you encounter wires larger than 4/0 they are designated by cross sectional area instead of gage .
500MCM is 500,000 circular mils , a short length of it makes a fearsome bludgeon.

 

[jim hardy]

My back of the envelope gave 0.0825 grams,
0.005² X pi/4 X12 = 2.36X10^-4cubic inches, multiplied by 2.54³ to get 0.00386 cm³
X 21.37 gm/cm³ = 0.0825 gm
X 0.13J/gmK = 0.0107J/K
and 61.4X10^-6 J/sec divided by 0.0107 J/K = .00572 degrees per second, X 600 seconds = 3.43 degrees

which seems too much. 100 ohm platinum RTD's don't change that much at a milliamp.

i must have made a mistake of arithmetic
will look again after dinner.

After Dinner
aha found my trouble a memory lapse power is 2.62 microwatts not 61.4...
i forgot about 5mil wire being 25 circular mils, used wrong power

2.62X10^-6J/sec divided by 0.0107 J/K is 2.45 X10^-4 K/sec,
X 600 seconds = 0.147 degrees

Close enough to yours for back of an envelope... and it agrees with what I've experienced with self heating in RTD's .

NIce Job, Mr Temaire !
old jim

 

[dlgoff]

a short length of it makes a fearsome bludgeon.

Such a waste of good copper. You should consider a piece of 1/2" water pipe.

 

[jim hardy]

MCM is a curious designation - why not KCM? Roman Numeral M instead of the familiar K ?
TIL Kilo is a French modification of Greek chilioi ...

Such a waste of good copper. You should consider a piece of 1/2" water pipe.

"But Judge - it was just a piece of wire, honest. "

 

[rbelli1]

From Latin:
https://en.wiktionary.org/wiki/mille
lots of Greek-Latin homophones that have totally different or opposite meanings.

BoB