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A Temperature of Plexiglass due to laser strike

  1. Nov 9, 2016 #1
    Hello,

    Any hep is appreciated. In the lab we want to use a 0.3mm sheet of plexi to reduce the intensity of the UV KrF laser.

    We want to determine the temperature of the plexi sheet after the laser beam strikes it. We have all the information about the laser...

    Is there a specific formula we can use to determine the temperature of our plexi sheet?

    It seems to be silly, but for some reason we cannot figure out what's missing in our heads...
    Please help!

    Thank you.
     
  2. jcsd
  3. Nov 9, 2016 #2

    davenn

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    well it would have been good to have given that .... particularly the power level
     
  4. Nov 9, 2016 #3
    KrF laser: wavelength (248 nm) - Power (25,000 MW) - Power Density (64,000 MW/m^2)
    *correction ... Plexiglas- PMMA: thickness (0.6 mm)
     
  5. Nov 9, 2016 #4

    Nidum

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    0,6 mm thickness Plexiglas seems slightly inadequate for your proposed use . Perhaps a few hundred metres of reinforced concrete or a small mountain might be more appropriate ?
     
  6. Nov 9, 2016 #5
    hahaa try telling that to a prof. who doesn't want to listen.

    how about for a pulsed laser of: Pulse Energy = 700 mJ & Average Power = 30 W.
     
  7. Nov 9, 2016 #6

    Andy Resnick

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    UV won't heat the plexiglass, it will ionize the molecules and over time, darken/yellow/craze the plastic. The required specification is optical density:

    http://www.lasersafetyindustries.com/Acrylic_Laser_Safety_Windows_s/553.htm
    http://www.phillips-safety.com/laser-safety/laser-windows/laser-acrylic-sheeting.html

    I agree with the above: 25000 MW of laser power is ummm... a lot. 700 mJ pulses are manageable. Read up on the ANSI Z136 standards for maximum exposure limits.
     
  8. Nov 9, 2016 #7
    Thank you! I'll read up on it tomorrow morning.

    Follow-up questions: 1. So basically a plexiglass sheet won't heat up, no matter what the power of laser is?

    2. The 500mJ energy is given by the PLD characterization booklet. They also say that the pulse rate is 20 ns. So doesn't that mean that power is 25,000 MW?!

    Sorry for the ridiculous questions, but yeah it's this bad. It doesn't make sense to me that the laser's power is that high in our PLD machine. Maybe there's some missing info in my mind.
     
  9. Nov 10, 2016 #8

    Andy Resnick

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    That's not true, either- it's a function of wavelength. Different processes happen upon absorption of UV versus IR- think about getting a sunburn as compared to burning yourself with a flame.

    2. The 500mJ energy is given by the PLD characterization booklet. They also say that the pulse rate is 20 ns. So doesn't that mean that power is 25,000 MW?! [/QUOTE]

    A pulse time (not rate) of 20 ns and pulse energy of 500 mJ gives a peak power of 25 MW (check your units). But that's not comparable to a 25 MW continuous output- to do that, you need the pulse repetition rate- say it's 10 Hz. Your time-averaged power output is then 5 W.

    Does this help?
     
  10. Nov 10, 2016 #9
    A pulse time (not rate) of 20 ns and pulse energy of 500 mJ gives a peak power of 25 MW (check your units). But that's not comparable to a 25 MW continuous output- to do that, you need the pulse repetition rate- say it's 10 Hz. Your time-averaged power output is then 5 W.

    Does this help?[/QUOTE]
    Yes that is extremely helpful. I've been working on this for 3months and the instructor fails to explain it as well as you do.

    I can't thank you enough.

    Also I was thinking of using this formula : heat flux = density.specific heat.(dT/dt)
    That way I can integrate over time to find the change in temperature of the sheet.

    Does that make sense?
     
  11. Nov 10, 2016 #10

    Andy Resnick

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    Not for this application, for the reasons I tried to explain. What do you propose to put in for the heat flux?
     
  12. Nov 10, 2016 #11
    I was thinking that the heat flux would be equal to the intensity of the laser multiplied by the exp(-ax). Where a is the absorption coefficient of plexi and x is the thickness.

    Thoughts?
     
  13. Nov 11, 2016 #12

    Andy Resnick

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    You seem bound and determined to perform the calculation, regardless of the applicability.
     
  14. Nov 11, 2016 #13

    Nidum

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    (1)
    Wouldn't it be better to use a commercially made laser beam attenuator of known characteristics ?
    or build a DIY version based on established principles ?

    (2)
    If you want to attenuate the laser beam for the duration of a long series of experiments you might be better off just getting a lower power laser .
     
  15. Nov 11, 2016 #14
    Well it really isn't up to me.
    Is there any way you can explain your point about the calculations not being applicable for what I have as characteristics for the laser?

    Again I appreciate your replies and helpful insights!

    Thank you
     
  16. Nov 11, 2016 #15
    It definitely would, but for all i know it would cost more than the prof is willing to put.

    Also, i think finding an attenuator with the required characteristics is a bit hard. I'll re-look into it for sure.
     
  17. Nov 11, 2016 #16

    Andy Resnick

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    As I have repeatedly stated, UV absorption does not heat the plexiglass, it degrades the plexiglass. Absorption of UV laser light by acrylic sheeting does not heat the acrylic sheeting, it degrades the sheeting.

    https://en.wikipedia.org/wiki/UV_degradation
     
  18. Nov 11, 2016 #17
    alright Andy, I got it. But my prof. seems to be certain it will heat up!
    I will definitely let you know what happens on monday.

    Thank you for the help
     
  19. Nov 12, 2016 #18
    The absorbed energy will definitely almost entirely go to heating the plexiglass. The mechanism that leads to chemical change occurs only occasionally and accumulates slowly over time.

    You will want to analyze the heating on two time scales: the heating caused by a single pulse and the heating caused by the average power. Generally the single pulse doesn't cause much heating on the short time scale, but 700mJ is a fair bit of energy, so you'll want to check both time scales.

    Unfortunately determining the heating is not trivial.

    You can make a reasonable estimate for the short time scale. 20ns is very short compared to heat flow. You can approximate that all 700mJ go to heat right where they are absorbed and calculate an upper bound on the fast temperature change before the heat diffuses away. For a quick estimate assume the intensity of the beam is uniform across the beam size. Take the area of the beam times the thickness of the sheet to give a volume of heating. Then calculate:

    Temperature rise = Pulse Energy*Absorption / [(Specific heat)*(density)*(Volume)]

    You can refine this by being more specific about where the heat is deposited. Break the volume down into little cubes and determine the temperature in the same way, but now using the amount of energy deposited in each cube. This will allow you to take into account the Beer's law absorption in the thickness dimension and the beam intensity profile in the other two dimensions.

    The steady state condition due to the average power load is much more complicated. Take the pulse energy times the rep rate of the laser to get the average power load. This goes into the small cubes with the same volume profile you used above. For steady state you have to determine the heat flow to the boundary conditions. What surfaces are being held at a fixed temperature because they are in good thermal contact with a heat sink? Other surfaces are in contact with the room air. Is there sufficient air flow that those surfaces can be treated as a single convective heat transfer constant to air at a fixed temperature? (Usually the case). Now you have to make cubes to represent the plexiglass all the way out to all the known boundary conditions. At each cube interface the heat flow across the boundary is the boundary area time a conduction constant. Internal to the plexiglass the conduction constant is the thermal conductivity. At the boundaries you will have conductivities to the heat baths. You have to find a temperature for each little cube such that the heat flow into and out of each cube is net zero.

    Computationally this can be done by the method of finite differences which is easy to understand and set up but slow to calculate, or the method of finite elements which is harder to understand but much quicker. If you choose to do this yourself in truth either will work fine. So what if it takes some minutes to calculate? You would probably learn more doing it with finite differences because it is closer to the actual physical process. You can also watch the temperature profile evolve.

    However, the smart money is on finding one of your mechanical engineering friends with access to a program which does Finite Element Analysis such as ANSYS handing them the heat profile and the boundary conditions and letting them get you the temperature profile.

    If you would like to do it yourself finite differences isn't difficult, it just requires a lot of book keeping because of the large number of elements. Start with all boxes at room temperature. For each box calculate the net heat flow across all 6 boundaries. Then for a small step in time dt take:

    dT = net heat flow*dt/(specific heat * density*element volume)

    Update the temperatures and iterate until the temperatures stop changing.
     
  20. Nov 12, 2016 #19

    Nidum

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    or just get 50p worth of Plexiglas and try it out .
     
  21. Nov 12, 2016 #20
    Yes! Actually I meant to say that. Much easier to just try it.
     
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