# Finding current and temperature of platinum wire

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1. Dec 18, 2015

### 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 60oC. 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 1,768oC. 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?

2. Dec 18, 2015

### 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?

3. Dec 18, 2015

### Bystander

"Half mil?"

4. Dec 18, 2015

### 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.

Last edited by a moderator: May 7, 2017
5. Dec 18, 2015

### temaire

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 20oC.
$$\Delta R = \alpha R_s \Delta T$$
$$R - R_s = \alpha R_s \Delta T$$
$$R = R_s (1+\alpha \Delta T)$$
$$R_s = \frac{\rho L}{A}$$
$$R = \frac{\rho L}{A} (1+\alpha \Delta T)$$

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.
$$\Delta T = \left [1 - \frac{(3.93\times 10^{-3}) (1\times 10^{-3})^2 (1.06\times 10^{-7}) (600)}{\frac{\pi (1.27\times 10^{-4})^2}{4}^2 (\frac{21.37}{0.01^{3}}) (0.13)} \right ]^{-1} \left [ \frac{(1\times 10^{-3})^2 (1.06\times 10^{-7}) (600)}{\frac{\pi (1.27\times 10^{-4})^2}{4}^2 (\frac{21.37}{0.01^{3}}) (0.13)} \right ]$$

$$\Delta T = 0.14 K$$

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

Last edited: Dec 18, 2015
6. Dec 19, 2015

### jim hardy

wow thanks - make that 5 mil

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

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.

your result seems very reasonable.

I learned something just reading your wonderful presentation of the math

old jim

7. Dec 19, 2015

### 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.

8. Dec 19, 2015

### jim hardy

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.

9. Dec 19, 2015

### 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

Last edited: Dec 19, 2015
10. Dec 19, 2015

### dlgoff

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

11. Dec 20, 2015

### 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 ......

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

12. Dec 20, 2015