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Calculating Planck's integral for finite range of wavelength

  1. Apr 22, 2017 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I am asked to calculate how much of the total radiated power of a light bulb at temperature ##T=2300##K is contained within ##400##nm and ##750##nm. I am also given the average emissivity of tungsten ##\epsilon_\text{ave}=0.288## and the emissivity within the range of wavelength ##\epsilon_\text{visible} \approx 0.45##.

    2. Relevant equations

    Stefan-Boltmann law: ##R=\epsilon(\lambda) \sigma T^4##
    Planck's law of radiation: ##\frac{dR}{d\lambda} = \frac{2\pi h c^2}{\lambda^5} \frac{\epsilon(\lambda)}{\exp (\frac{hc}{\lambda k T}) -1}##

    3. The attempt at a solution

    First I calculated the total power per unit area ##R_\text{ave}## using Stefan-Boltzmann law:

    ##R_\text{ave} = \epsilon_\text{ave} \sigma T^4 = 457##kW##\cdot##m##^{-2}##

    I spent some time trying to think of how to estimate the part of the power contained within the range of wavelength. The only solution I found so far is to use the Planck's law of radiation and integrate from ##\lambda_1 = 400##nm to ##\lambda_2 = 750##nm, but that isn't easy:

    ##dR_\text{visible} = \frac{2\pi h c^2}{\lambda^5} \frac{\epsilon_\text{visible}}{\exp (\frac{hc}{\lambda k T}) -1} d\lambda##
    ##\implies R_\text{visible} = \frac{2 \pi (kT)^4}{h^3 c^2} \epsilon_\text{visible} \int_{u_1}^{u_2} \frac{u^3}{\exp(u)-1} du##

    Here I have used substitution with ##u=\frac{hc}{\lambda kT}## like one would normally do with the integral from ##0## to ##\infty##. Then with ##\frac{1}{\exp(u)-1} = \sum_{n=1}^{\infty} e^{-nu}##:

    ##R_\text{visible} = \frac{2 \pi (kT)^4}{h^3 c^2} \epsilon_\text{visible} \sum_{n=1}^{\infty} \int_{u_1}^{u_2} u^3 e^{-nu} du##
    ##= \frac{2 \pi (kT)^4}{h^3 c^2} \epsilon_\text{visible} \sum_{n=1}^{\infty} \bigg( (\frac{u_1^3}{n} + \frac{u_1^2}{n^2} + \frac{6 u_1}{n^3} +\frac{6}{n^4}) e^{-n u_1} - (\frac{u_2^3}{n} + \frac{u_2^2}{n^2} + \frac{6 u_2}{n^3} + \frac{6}{n^4}) e^{-n u_2} \bigg)##

    after performing integration by parts several times and inserting the limits of integration (pfiuu). I calculated the first 5 terms of the sum using Matlab, and this gives me the result

    ##R_\text{visible} = 0.011## W##\cdot##m##^{-2}##

    which I find way too low to be true. Did I make a mistake somewhere? Some websites seem to confirm my calculation of the integral (for example here: http://www.spectralcalc.com/blackbody/inband_radiance.html), but maybe there is another way to calculate/approximate this calculation? I think I cannot use Rayleigh-Jeans approximation, since the peak of radiation is at intensity ##\lambda_\text{max} = 1.26 \mu##m, which means that the Rayleigh-Jeans approximation goes towards ##\infty## on the left of the peak, which is not the case with Planck.

    Any advice here? I think I might have gone too far with the integral, when maybe there is another path to the solution.

    Thank you very much in advance.


    Julien.
     
  2. jcsd
  3. Apr 22, 2017 #2
    Actually I made a typing mistake in Matlab. The result is now ##R_\text{visible}= -18.493##kW##\cdot##m##^{-2}## which is better than before, except for the minus sign. Maybe I inverted my limits of integration by mistake somewhere.
     
  4. Apr 22, 2017 #3
    I found the mistake, it was in my first substitution: either I forgot a ##-## or to invert the limits of integration. The result is now ##R_\text{visible} = 18.493##kW##\cdot##m##^{-2}## which represents about ##4{\%}## of the total electromagnetic power. I think it is a reasonable figure, what do you guys think?

    Still I would be curious to know if there is another method of estimating this value. Maybe something with Wien displacement law?

    Thanks a lot in advance.


    Julien.
     
  5. Apr 22, 2017 #4

    vela

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    Since you're using Matlab, you could just have it evaluate the integral numerically. I got 21.9 kW/m2.

    You could try estimating the integral in different ways, but I think you always have to integrate ##dR/d\lambda## in some way.
     
  6. Apr 23, 2017 #5
    @vela Thanks for your answer. Indeed I didn't know how to numerically integrate using Matlab until now. Thanks a lot, I got now the same result as you!


    Julien.
     
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