Sound intensity level of a thunder bolt

[Inertigratus]

Homework Statement

Find the sound intensity level of a thunder bolt, if a person is standing 200 m from the spot it hits. The bolt hits the ground vertically and the height of the bolt is 1500 m.
The bolt has an electrical current of 20 kA and a voltage of 30 MV.
It discharges thrice, each time lasts about 0.1 ms and the time between each discharge is about 30 ms.
A reasonable approximation is that the sound wave equals 1% of the thunder bolt's energy.

Homework Equations

P = UI
A = 4πrh (supposedly the sound wave from the bolt expands in the shape of a cylinder)
L = 10lg(I/I₀)
I = P/A

The Attempt at a Solution

At first it seemed simple, I thought I got the right answer but I've been trying with everything... still getting the wrong answer.
P = 6*10¹¹ * 0.01 = 6*10⁹
A = 2π*200*1500 ~ 6π*10⁵
I = (6*10⁹)/(6π*10⁵) = (1/π)*10⁴
Now I'm not entirely sure what to do.
I thought that maybe, intensity can be written as I = P/At.
Then I would just divide my intensity by the time the bolt lasts.
Which is (0.3+60)*10^(-3).
I also thought that maybe I could just multiply my intensity by three since it happens so fast anyway.
In either case I get the wrong answer after calculating the intensity level.

Any ideas?

 

[Inertigratus]

Actually, I think I did the area wrong... should be the area I wrote + 2*200^2 * pi to add the bottom and top circles. But it's still wrong, the answer is supposed to be 120 dB.

 

[Inertigratus]

Is this too "easy" for you guys? 50 views and no reply...

 

[Redbelly98]

Relax, it can take a while for somebody who both understands the problem and has the time to help to respond.

The Attempt at a Solution

At first it seemed simple, I thought I got the right answer but I've been trying with everything... still getting the wrong answer.
P = 6*10¹¹ * 0.01 = 6*10⁹
A = 2π*200*1500 ~ 6π*10⁵
I = (6*10⁹)/(6π*10⁵) = (1/π)*10⁴

Looks good so far. It would be nice to include units.

Now I'm not entirely sure what to do.
I thought that maybe, intensity can be written as I = P/At.

Well, you already wrote that I=P/A, which is correct.

Then I would just divide my intensity by the time the bolt lasts.
Which is (0.3+60)*10^(-3).
I also thought that maybe I could just multiply my intensity by three since it happens so fast anyway.
In either case I get the wrong answer after calculating the intensity level.

Any ideas?

Are you saying that (1/π)*10⁴ is wrong, even if you convert that to a decimal number with the proper units and number of significant figures? Because that answer looks right to me.

It seems to me that the time information and the fact that it is three bolts is extraneous info.

One final thought: is it possible that they want the answer expressed in decibels (dB)?

 

[Inertigratus]

Yes, I=P/A but I think I read somewhere that it can also be expressed as power per area and time unit. Maybe I'm just confusing it with the fact that P=W/t ?

Then I=W/At.

It's supposed to be in decibels, the level of sound intensity. The answer should be 120 dB or ~120 dB. I was thinking that I got the time wrong and that might be what is giving the wrong answer... or maybe the answer isn't supposed to be 120 dB?

P = [W], A = [m^2], I = [W/(m^2)]

And thanks for trying to help! :)

Edit:
I didn't include the final result in decibels because it depends on the intensity, if the intensity is wrong so is the final result.
P = 6*10^11 * 0.01 = 6*10^9 [W]
A = 2π*200*1500 + 2π*200^2 ~ 2.14*10^6 [m^2] (added the area of the top and bottom circle of the cylinder)
I = (6*10^9)/(3.4π*10^5) = (7/4π)*10^4 [W/(m^2)]
L ~ 167 dB which is way off.

 

[Redbelly98]

Yes, I=P/A but I think I read somewhere that it can also be expressed as power per area and time unit. Maybe I'm just confusing it with the fact that P=W/t ?

Perhaps. I is definitely P/A.

It's supposed to be in decibels, the level of sound intensity. The answer should be 120 dB or ~120 dB. I was thinking that I got the time wrong and that might be what is giving the wrong answer... or maybe the answer isn't supposed to be 120 dB?

What do you get for dB, given your I of (1/pi)*10⁴?

P = [W], A = [m^2], I = [W/(m^2)]

Yep, that's right.

And thanks for trying to help! :)

You're welcome!

 

[Inertigratus]

What do you get for dB, given your I of (1/pi)*10⁴?

~ 155 dB..

 

[Redbelly98]

I agree with your answer.

120 dB corresponds to an intensity of exactly 1 W/m². My best guess (and it's just a guess) is that the problem's answer-key author forgot to use the actual intensity when plugging numbers into the dB formula, and simply took 10 \log{\frac{1}{1*10^{-12}}}. Perhaps they even mistook the I in the numerator for the number 1.