That appears to potentially be a really good idea - but calculating the effective thermal resistance would be immensely difficult, whereas trying it in practice would be quite easy. I'll try neither at present (the latter is your assignment :-) ) but the following observations may help.
My first reaction was that the thermal transfer from case to spring (Rt_CS) would be very high, but you have added the (reasonable) assumption that enough thermal paste is used to make "pretty good" [tm] contact between case and spring.
Assuming that your spring area calculation is correct, that's the same area as plates of about
That's a useful piece of heatsink compared to a TO92 case.
You state datasheet figures of θJA = 160°C/W and θJC ~= 66 °C/w,
so θCA = 160 - 66 = 94°C/W
Even allowing for the poor thermal conductivity of the spring (steel?) I'd guesstimate a halving in effective θCA so you'd hope for a θJA of around 110°C/W.
That assumes θSA of about 50°C/W
For comparison, Aavid make several TO92 heatsinks of a "more usual" style.
The diaagrams and tables below were taken form page 68 of their excellent catalog - here. Effectiveness of course varies with size and area, but the tables suggest that they approach 10°C/W when used in a hurricane and more like 25°C/W as air speed approaches zero. Area of the smaller 575200 heatsink is somewhat under 400 mm^2 both sides. They have added some punch outs to improve airflow, the material is optimised for high thermal conductivity and there is a broad path for heat energy from TO92 case to the outer upper edges. So superior performance to your spring would be expected, and the original ~~~= 50°C/W is probably an OK starting guesstimate.
Note that the Aavid heatsink is shown in some places as 60 K/W. That's presumably a notional still air value. If the air is still there is no convective transfer, and air convection happens as temperature differential rises, but how well your spring does at encouraging this is TBD.
You could easily wind a "spring" using thick copper wire with a mandrel such that it would slip only a TO92 case and then be able to be lightly crimped in place with thermal paste to suit. If the spring was crimped against the TO92 case with eg pliers and then slightly more tightly just above the case to stop it sliding down onto the PCB then a reasonably stable, cheap and easily made heatsink could result. Or you could slide it right down to board level and solder the end to a pad.
Best of all would be to try it in real life.
Obtain a semi-sacrificial TO92 device - say a cheap bipolar transistor.
Use a consistent environment. Minimise artificial airflow and mount within an open box or whatever that roughly models intended use. Finger access required.
Connect say base to collector directly or via a resistor of choice and use a variable power supply to adjust dissipation until it was about too hot to hold onto for more than a few seconds. That's typically about 55°C. Use a thermometer that is suitable (eg IR non contact) if you have one, but fingers sufficeth.
Measure and save mW.
Now add spring, measuring temperature at TO92 case for similar result. Allow time to stabilise after adjusting power.
Measure and report back.
RELATED:
A zillion idea starters
Digikey AavidTO92 heatsink
Excellent Aavid heastink catalog
Aavid retail example - 36 C/W claimed
Aavid similar to above TO92 - 60 K/W claimed - Hmmm
Think as power supplys as a constant voltage, rather than the current they can provide. So a constant voltage supply will try to maintain the same voltage independent of the load you put on it (until, as you said, it blows up). So, a fan is designed to "pull" or "let pass" 1A for a given voltage while an air conditioning device is designed to "pull" 10A for the same given voltage. Thats why they pull different currents. And, while you can "force" more current with more voltage, some devices are smart enough that they will try to compensate for that using their regulators (switching or linear) by having their own constant voltage supplies on the inside, thus maintining about the same current consumption up to a given voltage. Normaly supplys fail not because they fail to push the current, but because they fail to provide the current that is being "pulled". If you have a constant current supply, when you try to "push" more current to a given resistive load, the voltage will rise.
About the phone, battery charging ICs will often have a limit to the current they can charge (as will the batteries). Often on cellphones that limit is close to 1A. Hence you can charge it faster on 1500mA. The 200mA rating is probably based on the USB standard max current, and is obviously easier for the manufacturer to supply you with the phone because its cheaper than a 1500mA supply.
p.s.: to better understand different current draws for same voltages: http://en.wikipedia.org/wiki/Ohm%27s_law also remember that not all loads are resistive(most arent)
Best Answer
If you currently require 28 volts and 2.4 amps, that's a power of:
$$ 28\:\mathrm V \cdot 2.4 \:\mathrm A = 67.2\:\mathrm W $$
67.2 watts. Unless you can somehow use the heat more efficiently, this is how much power you need, period. If you want less current you can do that by increasing the voltage such that the product of voltage and power equals 67.2 W. But a higher voltage battery, all else equal, will have a lower capacity, so you haven't actually increased battery life.
This is because while the battery geometry can be altered to have a higher voltage, unless you make it bigger, or use a different chemistry, the chemical energy in the battery has not changed.
The conversion of electrical energy to heat is nearly 100% efficient, so if you need to increase the battery life you'll have to look for efficiency improvements elsewhere. That means not heating things that aren't wax, such as by adding insulation or making the heating vessel as small as possible or using a more efficient battery.
There's a theoretical bound to the minimum amount of energy required to melt wax. Paraffin wax has a heat of fusion of about 200 joules per gram. A joule is one watt for one second. So if you need to melt 1000 grams of paraffin wax, you require at a minimum:
$$ 1000\:\mathrm g \cdot {200\:\mathrm J \over \mathrm g} = 200,000\:\mathrm J $$
Additionally the wax must be heated to its melting point. Paraffin wax has a specific heat capacity) in the neighborhood of 2.5 joules per gram kelvin. A change of one kelvin is the same as one degree Celsius. So say the 1000g of wax wax starts at 20°C and must be heated to the melting point of 37°C:
$$ 1000\:\mathrm g \cdot {2.5\:\mathrm J \over \mathrm {gK}} \cdot (37-20)\mathrm{^\circ C} = 85,000\:\mathrm J $$
This brings the total energy required to 285,000 joules. Given the 67 watt power of your current setup, that means it must run for a minimum of:
$$ {285,000\:\mathrm J} \cdot {\mathrm s \over 67.2\:\mathrm J} = 4241\:\mathrm s = 71\:\text{minutes} $$
71 minutes for each 1000 grams of paraffin melted.
In practice, additional energy is lost to heating other things like the air, the container, and the battery, so for best efficiency all you need to worry about is heating the wax, and only the wax.