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Homework Help: I still have trouble with Dummy Variables

  1. Jul 12, 2010 #1
    I am reading a paper and I am coming across some usage of a dummy variable and it is becoming increasingly clear that I never really understood how to use these things. I will demonstrate once usage that I do understand and one that I do not. Let me also introduce the basic physics behind the equation since it might be of some help.

    We have a closed chamber in the form of a sphere. It is filled with a fuel and oxidizer. A spark ignites the gaseous mixture at the exact center of the chamber. We assume that the resultant flame propagates spherically outwards until it hits the wall of the chamber. We also assume that the flame is is of negligible thickness. If we "freeze" the flame at some instant in time, it will divide the chamber into 2 parts: An "inner" sphere of burned gas and an "outer" sphere of unburned gas.

    We will denote x to be the mass fraction of burned gas: x=mb/M, where M is the total mass M = mb + mu. Also, v is called the specific volume of the burned or unburned gas. It is defined as the volume per unit mass (inverse of density).

    Here is the first equation that I DO understand (I think):

    If we denote V as the total volume of gas (which is conserved) we have

    [itex]V/M = \int_0^xv_b\,dx' +\int_x^1v_u\,dx' \qquad(1)[/tex]

    So in the above equation (1), x' is the dummy variable. For some value of x we integrate the first term over x' from 0 to x and the second term over x' from x to 1. I get it.

    Here is the second equation I am a little lost on:

    The average temperature of the burned gas is given as

    [tex]\bar{T}_b = \frac{1}{x}\int_0^xT_b(x',x)\,dx' \qquad(2)[/tex]

    Now in (2), I have conviced myself that this is similar to (1) in that we "fix" some value of x, but then I do not understand why there is an x' inside the argument of the function? It makes sense in the dx' since we are integrating over in interval and at each infinitesimal sub-interval we are "substituting in" for dx'. But in Tb(x', x) it looks like it takes x' as an argument.


    ~Any comments or thoughts on this?
     
  2. jcsd
  3. Jul 12, 2010 #2

    Office_Shredder

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    Tb has two inputs, x and x'. One is being integrated over, one is not. When calculating the integral x will just be treated as a constant. For example, if Tb=[tex]x'e^{x}[/tex]

    Then we have [tex]\int_0^x x'e^{x} dx' = e^{x} \int_0^x x'dx'[/tex] because x is treated as a constant inside the integral. In general of course it won't be so easy to get rid of the x's in the integral
     
  4. Jul 12, 2010 #3

    Office_Shredder

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    Tb has two inputs, x and x'. One is being integrated over, one is not. When calculating the integral x will just be treated as a constant. For example, if Tb=[tex]x'e^{x}[/tex]

    Then we have [tex]\int_0^x x'e^{x} dx' = e^{x} \int_0^x x'dx'[/tex] because x is treated as a constant inside the integral. In general of course it won't be so easy to get rid of the x's in the integral
     
  5. Jul 12, 2010 #4
    I think that makes sense to me. I guess it was the fact that the actual function T(x',x) was not given. Though now that I think of it, if we put integration aside and speak only of the function Tb=[itex]x'e^{x}[/itex], as you stated, then what does the 'dummy' variable x' mean without the context of an integration? I am trying to keep in mind the physical meaning of the function. Tb is the temperature of the burned gas when some fraction x of the gas has burned. But what does the x' represent now? Perhaps it is meaningless without an integration? That is, perhaps the function Tb itself has an integral? Perhaps something to the effect of Tb(x',x) = [itex]\int_{x_o}^x f(x')dx'[/itex].

    Just an idea. Thoughts?
     
  6. Jul 12, 2010 #5

    Office_Shredder

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    Your proposed function doesn't actually depend on x' as a function, so it's unlikely. x' is a dummy variable in the integral, but should not be a dummy variable in Tb itself, unless you have something weird going on
     
  7. Jul 12, 2010 #6
    Meh. I will never understand this. If I may back up a little: We were initially given that:

    [tex]\bar{T}_b = \frac{1}{x}\int_0^xT_b(x',x)\,dx' \qquad(2)[/tex]

    Which says to take the function [itex]T_b(x',x)[/itex] and integrate it over x'. So you are telling me that you think that in this case, x' is NOT a dummy variable?
     
  8. Jul 13, 2010 #7

    Office_Shredder

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    x' is a dummy variable in the integral. But just looking at the function

    Tb(x,x')

    It's a function that depends on two variables, x and x'. It's like if you integrate sin(v), v becomes a dummy variable, but just given the function sin(v) v is not a dummy variable
     
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