Let's consider two windings containing thin diameter wire(0.3 mm) and thick diameter wire(4 mm). These windings operate under the following conditions.
1- The same power is applied.
2- They have the same NI (Ampere.turn).
The question is: Which diameter wire is more advantageous in terms of the temperature issues that occur during operation? Which wire should I use in this dc electromagnet design, the thinner one(0.3 mm) or the thicker one(4 mm)?
Heat may conduct outwards and in the axial direction quicker if the conductors are thicker, and the diameter and length of the coil will be smaller, all because there will be less electrical insulation inside the winding, which tends also to be a bit thermally insulating. Another consideration is what DC power supply do you want to use? 16 A at 0.9 mV or 0.9 mA at 16 V, for example. The supply current needed is proportional to the area of the wire (or wires if you use several in parallel).
We have a 48 volt, 50 A 2000W DC power supply. The only thing we don't know is whether we should choose thin wire or thick wire for the same NI and power. Which one is more advantageous in terms of temperature? Since we will be applying high power, temperature issues may cause trouble.
You need about 1 ohm. That is about 720 m of 4 mm diameter copper, or 4 m of 0.3 mm copper. 4 metres with 2 kW is 500W/m or 5 W/cm which is a lot of heat in a thin wire. It would vaporise. You will need to use some of it in parallel, so it is longer for the same resistance. In that case it is better to use thicker wire because it has less insulation to add weight and to slow down heat transfer.
720 m of 4 mm diameter wire occupies about 9 litres and weighs about 82 kg. You should be able to get lots of ampere turns.
Thank you very much for your valuable comments. But I need to explain the situation to you more specifically.
I need a coil with 11250 NI (amper.turn). According to my calculations, 710W power is required for wire with a diameter of 0.41 mm and 715W power is required for wire with 1.02 mm diameter. In addition, for these coils , 4123 meters of 0.41 mm wire and 480 meters of 1.02 mm diameter wire are required. The total resistance of these coils is 542 ohms for the 0.41 mm wire and 12 ohms for the 1.02 mm wire. Here, I calculated it assuming that I could pass 1.15 Ampere current through 0.41 mm wire and 8.4 A current through 1.02 mm wire (I found these current values from technical documents on internet). From these current and length values, I obtain NI = 11250 from both coils. Of course, since I could get 48 V from my power supply, I determined that I had to divide the 0.41 mm winding into 13 parallel branches, and the 1.02 mm winding into 2 branches. The temperature problems here are unknown to me, which one will be more advantageous, the wire with a diameter of 0.41 mm or the wire with a diameter of 1.02 mm? At the same time, in my calculations taking into account the insulators of these two different windings, the winding widths of both windings are approximately equal to 15 mm (when I wind them on the same core).
As you are putting in the same amount of heat into the coil in roughly the same places in both cases, the main problem is how quickly it can come out. In that case the ticker wire should be significantly better because there will be a significant temperature drop across each layer of insulation and each contact between the layers where they touch or don't quite touch.
I presume you will be taking heat away from the ends of the coil and from the outer diameter. For both these direction the layers of insulation are across the direction you want the heat to flow, and more layers of insulation and more gaps between wires will result in a higher temperature in the middle of the coil, compared to the outside, to get the same heat out at the edges. For that reason using thicker wires should mean that the temperature in the centre of the coil is lower.
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