# Trying to heat a wire Capacitor

1. Feb 25, 2007

### foahtein

Hello all, I am trying to heat a wire (nitinol, shape memory alloy, or nickel titanium).

Well my problem here is that i need to heat the wire with a DC current from the wall outlet, usually generate 110 volts but using a AC to DC converter to generate approximently 20 volts, and need a certain capacitor to store the electric charge that upon release will instantaneously heat that wire to approx. 90 degress celcius.

Givens:
wire diameter is .085 inches or .00212 meter
wire length is 7 inches or .1778 meter
measured with multimeter resistance of wire is .5 ohms
calculated the current to be 30 amps
calculated the voltage with ohms law to be 15 volts, but due to losses of heat add an additional 5 volts(approx. 1/3 of the total volts)

now i used the Heat Transfer thermal equilibrium equation: to calculate current

density(roe) * specific heat(Cp) * volume(V) * change in tem.(dT)/change in time(dt) =
resistance(R) * current^2(i^2) - convection ht. transfer coeff. (k)* area (A)(Tem max - Tem room).

used Ohms law : to find voltage
V = IR

I'm looking at Q = C * V which is the charge = capacitance * voltages
and i know that Power = IV and I = change in Q/t and Power = Work / time
Using those equation i need a 2 farad capacitor to actuate the wire within 1 sec and at a temperature of 90 degrees Celcius.
Using the same equation and inputting it for 5 sec yields a larger capacitor, which doens't make since.

here is some properties of nitinol
http://www.tiniaerospace.com/sma.html [Broken]

Last edited by a moderator: May 2, 2017
2. Feb 25, 2007

### waht

Resistance of Wire = 0.5 Ohms

Current to heat Wire to 90 deg = 30 Amps

Right?

We can now calculate the required power

$$P = I^2 R = 450 Watts$$

That means it will take 450 Joules of energy every second to maintain the 90 degree temperature.

Another formula for energy stored in a capacitor

$$E = 1/2 CV^2$$

C is capacitance, and V voltage. So you have to decide for how long you want to heat the wire. Calculate the total energy required from the first formula, and then solve for C or V, you can play around with those.

Hope that helps.

Last edited: Feb 25, 2007
3. Feb 25, 2007

### Integral

Staff Emeritus
Caution!
It is not that easy to get a wire temperature from your given data. While it is easy to know how much energy you put into the system your losses are little better then a guess.

Your best bet would be to control the environment of your wire as closely as you can. Then run experiments to determine the temperature resulting from a given current.

4. Feb 25, 2007

### light_bulb

this is what graphs were made for, you have a changing resistance as the temp increases.

5. Feb 25, 2007

### foahtein

well the wire temperature is the Austenite peak from the Differential Scanning Calorimeter (DSC) test. my Teaching Assistant said that the wire will contract at approximately 70-90 degrees celcius. so that is how i found the temperature needed for the actuation in a control environment. but our group will probably run several tests to determine the speed and electrical current needed to actuate the wire.

But for the capacitor...if i need to send a current say in less then 1 msec. what type of of capacitor is needed.

Using I = change in Q/t (use Qo as 0) and Q = C V my capacitance yields 001875 farads.

so if my calculations were correct, if i want to heat the wire at a quicker time (1 milliseconds) then i need a smaller capacitor???

Thanks for the previous posts, its of great help to me.

6. Feb 25, 2007

### waht

When discharging capacitors, peak current is rushed immediately after closing the circuit, then it dies down exponentially.

It will be difficult to have a controlled discharge without other electronic magic, but you can calculate the ballpark timing range.

The half life of a capacitor is

$$t = RC$$

That is the time the voltage at the cap will diminish by half.

Give you an example, if C = 1 Farad, and R = 0.5 ohm then t = 0.5 seconds. That means, 0.5 seconds after the cap is unloading into the wire, its initial voltage has dropped by half. Another 0.5 seconds later it would drop by another half and so on. This is similar to radio-active decay concept.

So you could use this to tweak the required timing. The first half life will be the most important since most current will be released.

Note, we are still assuming your heater wire is acting linearly. I suspect it doesn't which could farther throw off this model.

Good luck.

Last edited: Feb 25, 2007