Space_Dandy
Student
- Oct 29, 2023
- 1
Hey buds, I'm really struggling to figure this out. I'm trying to make my own home made fiberglass jacketed heat tape. I need it 8' long(a lil longer is ok, say like 3meters if you wanted to stick to a meter scale) And was planning on doing kanthal wrapped around a 8mm rod to coil(8' of the 8mm coil to be stuffed into 1/2" fiberglass jackets) I'm on 120v AC but have a plethora of DC voltage supplies too, including a 24v 14amp, stacks of PC powersupplies converted into powersupplies, and better or smaller. About what gauge and how much kanthal wire am I gonna need to make the coil?
I THINK from my understanding I'm gonna need about 28-32w/inch to make this?(The tape is gonna be made into a 1L heating mantle so it needs to be able to reach at least 700c)
I've been trying to use the guidance here:
And from kanthal.com
And chatgpt says
tho now that I'm copy pasting it I realize I asked the AI kinda weird.. so it's prob deff wrong?)
ME:
If I have a wire that is 1.8 Ohm/m how much wire would I need making a spiraled heating coil that would be 2.5m long and produce at least 5kW of heating
ChatGPT:
To calculate the length of wire needed to create a spiraled heating coil that is 2.5 meters long and produces at least 5 kW of heating, you can use the following steps:
First, determine the resistance of the wire required to produce 5 kW of heating. The formula for power (P) in a resistive element is:
P = V^2 / R
where P is the power in watts, V is the voltage, and R is the resistance.
Since you want to produce at least 5 kW (5000 watts) of heating, and you have a wire with a resistance of 1.8 Ohms per meter, you can rearrange the formula to solve for resistance:
R = V^2 / P
In this case, V is the voltage across the coil (which we'll assume to be constant), and P is 5000 watts.
R = V^2 / 5000
Now, let's calculate the resistance needed for 5 kW of heating. Assuming a constant voltage, you can calculate the resistance required:
R = V^2 / 5000
To achieve the desired heating power, you'll need a specific resistance value. However, for this calculation, we'll use a voltage of 120 volts (a common household voltage). Keep in mind that the actual voltage might be different depending on your specific application, so you should use the appropriate voltage value for your setup.
R = (120^2) / 5000 R = 14.4 Ohms
With the desired resistance of 14.4 Ohms, you can now calculate the length of wire required. You have a wire with a resistance of 1.8 Ohms per meter, so you can use the formula:
Length (in meters) = R / (Resistance per meter)
Length = 14.4 Ohms / 1.8 Ohms/meter Length = 8 meters
So, you would need approximately 8 meters of wire with a resistance of 1.8 Ohms per meter to create a spiraled heating coil that is 2.5 meters long and produces at least 5 kW of heating with a voltage of 120 volts.
But it seems to me the calculations in the guides demand I already know the answer to my question I'm trying to figure out? lol.. Maybe I'm just not getting it?
If someone could show their work with answer to I'd appreciate it, may help me understand better. TY so much ya'll. Feel dumb rn. I can do basic ohms law stuff but adding the coils into the mix my brain just won't do it lol.
I THINK from my understanding I'm gonna need about 28-32w/inch to make this?(The tape is gonna be made into a 1L heating mantle so it needs to be able to reach at least 700c)
I've been trying to use the guidance here:
And from kanthal.com
And chatgpt says
tho now that I'm copy pasting it I realize I asked the AI kinda weird.. so it's prob deff wrong?)ME:
If I have a wire that is 1.8 Ohm/m how much wire would I need making a spiraled heating coil that would be 2.5m long and produce at least 5kW of heating
ChatGPT:
To calculate the length of wire needed to create a spiraled heating coil that is 2.5 meters long and produces at least 5 kW of heating, you can use the following steps:
First, determine the resistance of the wire required to produce 5 kW of heating. The formula for power (P) in a resistive element is:
P = V^2 / R
where P is the power in watts, V is the voltage, and R is the resistance.
Since you want to produce at least 5 kW (5000 watts) of heating, and you have a wire with a resistance of 1.8 Ohms per meter, you can rearrange the formula to solve for resistance:
R = V^2 / P
In this case, V is the voltage across the coil (which we'll assume to be constant), and P is 5000 watts.
R = V^2 / 5000
Now, let's calculate the resistance needed for 5 kW of heating. Assuming a constant voltage, you can calculate the resistance required:
R = V^2 / 5000
To achieve the desired heating power, you'll need a specific resistance value. However, for this calculation, we'll use a voltage of 120 volts (a common household voltage). Keep in mind that the actual voltage might be different depending on your specific application, so you should use the appropriate voltage value for your setup.
R = (120^2) / 5000 R = 14.4 Ohms
With the desired resistance of 14.4 Ohms, you can now calculate the length of wire required. You have a wire with a resistance of 1.8 Ohms per meter, so you can use the formula:
Length (in meters) = R / (Resistance per meter)
Length = 14.4 Ohms / 1.8 Ohms/meter Length = 8 meters
So, you would need approximately 8 meters of wire with a resistance of 1.8 Ohms per meter to create a spiraled heating coil that is 2.5 meters long and produces at least 5 kW of heating with a voltage of 120 volts.
But it seems to me the calculations in the guides demand I already know the answer to my question I'm trying to figure out? lol.. Maybe I'm just not getting it?
If someone could show their work with answer to I'd appreciate it, may help me understand better. TY so much ya'll. Feel dumb rn. I can do basic ohms law stuff but adding the coils into the mix my brain just won't do it lol.