lightning heat transfer calculation 4

[purpletoes]

I am stumped trying to find the thermal rise on a brass strip during a lightning strike. The expected energy at .000070s is 271.1 Joules.

I tried using the heat transfer equn for conduction but I'm getting a really high delta T. Can anyone please help, am I using the wrong equation, or am I missing a step? Thanks very much in advance.

 

[IRstuff]

Energy of the strike is irrelevant. What is the current going through the strip? What is the resistance?

TTFN

FAQ731-376

 

[Zesti]

Volts = Joule / (Ampere x seconds)

We have the values for time and the Joules involved.

Now for the Amperes and Volts.

I guess if you have found the values for time and charge then you should also be able to find how many volts lightning gives.

You then have the Amps going through the wire.

Resistance of the copper wire should not be a problem.

IRstuff: How to go from here? Does I^2 x R x t apply?

 

[purpletoes]

I guess that's where I went wrong, because I just divided the energy over time, which gave me a 3.87 MW.

How would I go about calculating the rate of heat flow or Q?
Is it just as simple as dividing the energy/Amp*sec? I feel like there's a step I'm missing. Thanks

 

[ginsoakedboy]

I would follow these steps:

You already have the power.
Estimate or use potential difference (voltage) and find current (P=VxI).
Find resistance R from geometry and material properties.
Use I^2xR to find the heat generated, Q.
Find deltaT using specific heat and mass of the conductor.

 

[Zesti]

Look here:

http://en.wikipedia.org/wiki/Joule_heating

I think this is what is going on in the copper strip when the current from the lightning runs through it.

Do you have the voltage of the lightning and did you calculate the current as I described?

You should be able to calculate the total electrical resistance of the copper strip. Then current^2 x Resistance gives the amount of energy heating up the copper. With the total mass of copper involved and the specific heat of copper you can then calculate the temperature rise.

But, I have never heard of copper lightning conductors melting and these have a diameter of less than 10mm.
Why are you expecting problems with your copper strip??

You calculated 3.87MW correctly but then looking at conduction is not the right phenomenon. It's an electrical current running through a resistor that is causing the temperature rise of the copper strip. So I think that is where you went wrong.

 

[Zesti]

htlyst beat me to it..

 

[purpletoes]

So what equation should be used to calculate the thermal rise?

This is actually for a brass strip that will be installed in a baseplate. Thanks very much for all the replies

 

[ginsoakedboy]

Q=m x c x deltaT

 

[purpletoes]

I did that but somehow I think my thermal rise is pretty low, considering that a 3.87 MW was dissipated. The thermal rise that I got is only 21 C. Does anyone have a good explanation on why this is so.

Thermal Mass is 12.575 and energy is 271 J

I got the energy from I^2*R*T
Thermal Mass = (density*vol)*specific heat

 

[prex]

What is important for your problem is the energy dissipated in the conductor, power is nonsense here, as it is not continuous, it lasts just a few microseconds.
Your calculated temp.rise is simply the demonstration of what Zesti pointed out: lightning conductors are normally just 10 mm dia. or less.

prex

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[Compositepro]

Lightning is more analogous to an arc welder than a resistance heater. Most of the heat is in the arc and transfered to the metal rather than generated in the metal.

 

[ginsoakedboy]

I think Compositepro makes sense. Electrical resistive heating is indeed small.

With that in mind, you would have to look at radiative heat transfer as well. And, note that the arc does not transmit 100% of its energy to the conductor.

Unless you have already factored the following, your energy will be lost in many different ways:

Sound
Heating up of air
Ionization of air
Radiative losses to the surroundings
Heating up of substrate

Not to mention the energy absorbed by the brass strip.