- #1
Saladsamurai
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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?
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?