Calculating heat loss and conduction through a wire

In summary, the project involves using two copper wires to melt solder paste by passing current from a 9V battery through them. The current needed is being calculated using the resistivity of the paste and Ohm's law, but the internal resistance of the battery also needs to be taken into account. It is recommended to use a 12V car battery instead of a 9V battery, as the latter may not be able to deliver enough current and there is a risk of safety hazards.
  • #1
electricnov
3
0
I'm creating a project where I am attempting to melt solder paste using the current running through two copper wires, in a kind of lap joint. I'm trying to figure out how much current would actually be going through the wires if I use a 9V battery. I'm not sure where to take into account heat losses and the conduction of heat through the wire.
I'm new to circuits and all things electrical, so any help would be great! thanks!
 
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  • #3
Thank you for the post. I understand that part. My question is about the current. The resistivity of solder paste is very low and so the current I calculate is very high (over 1000 A). This isn't actually a realistic number right? Is there some kind of loss in practice?
 
  • #4
The highest resistance in your circuit will be the internal resistance of the battery. Small 9V batteries won't deliver more than a few amps of current and only for a short time.

On the other hand a 12V car battery could deliver 500A or more, but DON'T TRY THAT EXPERMENT. You run a serious risk of getting burned, and also the battery may release explosive gas which could be ignited by a spark when you try to disconnect the circuit. (You willl quite likely wreck the car battery as well, but that's a minor problem compared with the safety issues).
 
  • #5
The 9V battery short circuit current is close to 5A at the beginning.
The 9V battery internal resistance is between 1.7Ω to 2.8Ω.
 
  • #6
So, in calculating the current, I take into account the resistance of the battery? and not the solder paste or the wire? Is that just because it is negligible in comparison with the battery?
 
  • #7
electricnov said:
So, in calculating the current, I take into account the resistance of the battery? and not the solder paste or the wire? Is that just because it is negligible in comparison with the battery?
Yes, the battery internal resistance is the main factor limiting the short circuit current.
At least for 9V battery.
 

Related to Calculating heat loss and conduction through a wire

1. How do you calculate heat loss through a wire?

The formula for calculating heat loss through a wire is Q = k*A*(T1-T2)/L, where Q is the heat loss, k is the thermal conductivity of the wire, A is the cross-sectional area of the wire, T1 and T2 are the temperatures at each end of the wire, and L is the length of the wire.

2. What factors affect heat loss in a wire?

There are several factors that can affect heat loss in a wire, including the thermal conductivity of the wire material, the cross-sectional area of the wire, the temperature difference between the two ends of the wire, and the length of the wire.

3. How does conduction affect heat loss in a wire?

Conduction is the transfer of heat through a material, and it plays a major role in heat loss through a wire. The thermal conductivity of the wire determines how easily heat can be transferred, and the temperature difference between the two ends of the wire drives the heat flow.

4. Can you reduce heat loss in a wire?

Yes, there are several ways to reduce heat loss in a wire. One way is to use a material with a higher thermal conductivity, such as copper or silver. Another way is to increase the cross-sectional area of the wire, which will decrease the resistance and thus reduce heat loss.

5. How does insulation affect heat loss through a wire?

Insulation can greatly reduce heat loss through a wire. By adding a layer of insulation around the wire, the thermal conductivity is decreased, which reduces the heat flow through the wire. This is why wires are often covered in insulation, especially in electrical systems.

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