Heating time of an aluminum plate

Question

I'm designing a circuit for my 3D printer's aluminium heating plate (200x200 mm) and I wanted to be sure before buying resistors. Also, I'm using a Sanguinololu 1.3a. I want it to heat at 120° fast.

So, I think ~150 Watts was enough.

I'll explain everything, people who'll have same problems may find solution here :).

Using 5 x 5ohm in parallel means I have a 1ohm load.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

$$ \frac{1}{R_{equivalent}}=\frac{1}{R_1}+\frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}+ \frac{1}{R_5} \\ \frac{1}{R_{equivalent}}=\frac{1}{5}+\frac{1}{5} + \frac{1}{5} + \frac{1}{5}+ \frac{1}{5} \\ \frac{1}{R_{equivalent}} = 1 \\ R_{equivalent} = 1 $$

I am using 12V to it, I am getting: $$ U = R_{equivalent}I \\ I = \frac{U}{R_{equivalent}} \\ I = \frac{12}{1} = 12 A $$

12A of current from the PSU. Which would give 12*12 = 144W, or 144/5 = 29W per resistor.

I wanted to use these: RESISTOR 50W 5R

Are my calculus ok ? If they are, do you know a way to find the time it'll take (in theory) to heat up it to 120 ° ?

To get an idea for the heating time, we need to know the mass of the plate. You wrote that it's 200x200 mm, but how thick is it? — Nick Alexeev, Nov 06 '13 at 23:58
Thanks a lot for your answers ! So, if I want to heat it faster, I can add more resistances in parallels. If I have 9 of 4.7 ohm and do the maths again, it can reach ~276W. With it, the plate can reach 110°C in 388 seconds, which is really good. I'm a bit lost with it. Sorry if I ask stupid questions, but I'm a bit afraid of burning the board. But, Is there a condition for a minimum intensity of the PSU ? Mine has 15A. As I said, I use a Sanguinololu 1.3a, what will happen if it doesn't ahve enough intensity ? Thanks — David Bensoussan, Nov 07 '13 at 16:02
What are the power and voltage capabilities of your power supply? With 15 A, you won't get more than 225 W into a 1 Ohm combination of resistors. But if your supply could do 15A/30V and 500 W, for example, you could make a 2 Ohm combination of resistors and get 450 W. Of course the closer you run to the limits of your supply capabilities, the more likely to run into reliability problems if you want to keep this running for days or weeks at a time. — The Photon, Nov 07 '13 at 17:37

Laszlo Valko · Answer 1 · 2013-11-07T00:04:48.820

Your calculations about the current and the power of your resistors are okay.

To know how much time it'll take to heat up your plate with those resistors to a certain temperature, you would need other data (most of which will not be available until you build this thing and start measurements). The time depends mostly on the initial temperature (the ambient temperatute), and on the ability of your plate to dissipate the heat (to transfer the energy to the surrounding air): the better it convects the heat, the longer it will take to heat up.

score 2 · Answer 2 · answered Nov 07 '13 at 00:08

You can get a lower limit on the time:

(0.902 J/g^oC)(20 x 20 x 1 cm³)(2.7 g/cm³)(100^oC) / (144 W) = 676 s

or about 11 minutes.

0.902 J/g^oC is the specific heat of aluminum, taken from the web.

20 x 20 x 1 cm is the volume of the plate (I assumed it's 1 cm thick)

2.7 g/cm³ is the density of aluminum, again from the web.

100^oC is what I assumed as the change in temperature you're looking for.

and 144 W is your input power.

As Laszlo points out though, at the same time you're heating the plate, it will also be cooling down due to convection and conduction, so this is only a lower limit on the heating time. The actual time could be quite a bit longer, expecially if the plate is in contact with any other thermally conductive materials, or has air blowing across it.

Heating time of an aluminum plate

2 Answers2