You must also consider what is your heat flux from the plate to surroundings. Is your plate insulated (i.e. q_loss =0 W/m2) or is there convection and radiation heat transfer. Basically you have to solve heat transfer problem with known boundary conditions. E.g. if your plate is insulated, you may heat it with very weak heating element, but if it is souuounded by cold flowing air, the power required will be greater. I recommend you find the book "Heat transfer: fundamentals and applications" by Cengel. And learn from there. There are many examples.

Hi,

here is a pragmatic approach, backed up by experience:

A 3D-Printer's heat bed with 200 mm x 200 mm consist of a 3mm aluminum plate, heated by PCB-wires.

It is typically driven by 12V consuming something like 8-10A which leads to 120W. This will not get you 150°C but something like 70 to 80°C at 25°C ambient temperature.

Math:

with about 120W/400cm², a temperature swing of 50K can be achieved.

You want 125 K increase considering 25°C ambient +125°C.

You'd need roughly 2.5 times the power resulting in

120*2.5W/400cm² and scale it with the smaller area as you only want to heat up 150cm².

About 112.5W by pure math.

Adding a little margin, I assume 150W will do fine.

Cheers - Martin

Time will also be a factor when you try to choose which heating element , 175W will bring the increase in the above mentioned stainless steel 150°C (i.e. from 25°C to 175°C) in round about 10 Minutes. If you have a 1.7KW heat source it might take less than 1 minute. Please select the device accordingly to your specification and availability. The increase of temperature is the consequence of increase in the internal energy or the work , since the work is 0, internal energy has to be increased (considering boundary to be insulated). You required round about 100000W , but as I told you its time dependent. You can instantaneously increase or increase with time. Please choose accordingly to your requirement



Ajay Konakanchi which formula did you use in your calculations. did you consider power loss due to radiation and conduction

I used First law of Thermodynamics, and Closed System. I added 15% losses due to Convective heat transfer losses and Radiation losses which might be close to realistic situation. If you require exact Losses and values it would be better to simulate the system. Since I have limited information on the scenario , I would stick to the additional power than require , very similar to Martin Schulz solution which is close to realistic approach

Martin Schulz Did you consider time in your calculation

Karol Pietrak thanks for your answer

Dear Polycarp Onyebuchi,

in my printer, going from room temperature to 80°C takes just a few minutes.

As you did not mention in your question that speed is essential, the assumption was, that you need to reach the 150°C and be able to keep it stable with a minimum power demand.

Of course, if you need to speed things up, higher power is a necessity as Ajay Konakanchi mentioned.

One approach would be to consider the thermal capacity of the heat plate, calculate the energy needed to change the temperature and divide it by the time you want to spend.

Considering iron as an example as you not mentioned the alloy your steel is made of, the thermal capacity acc. to datasheets is ~450J/(K kg), density is 7.87g/cm³.

Math again:

Your steel plate is 15 x 10 x 0.1 cm³ =15cm³ with a mass of ~0.120kg

Energy to increase its temperature by 125K is W1=450J/(K kg)*125K*0.12kg=6750J or 6750Ws - with no losses considered.

Without losses, heating this up with 150W takes about 45s, with losses considered it will be a little longer - maybe it doubles to 90s.

So of course, if you want to heat it up in a matter of single digit seconds, your heat source needs to provide kW instead.

Hope it helps - Martin