Black- Adiabatic.
This is the temperature gradient which equals [(gamma - 1)/(gamma)] * [(t/P)] * [(g*rho)]
Red- Actual
This was calculated using the 'old' method. This was adding and taking away 1 from each point to get x1 and x2. Their corresponding temperatures were found (y1 and y2). The slope of these were calculated:
(y2-y1)/(x2-x1)
Blue- Actual
This was calculated from the 'new' method:
For every point along the x-axis, its preceding and following points were taken to be x1 and x2 respectively.
The y-values for each of these (y1 and y2) were found, and the slopes were calculated as above.
It can be seen that the two methods for the actual temperature gradients are pretty much equivalent.
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Adiabatic Temperature Loss of Cell
I have managed to calculate the temperature change of a cell as it travels a distance.
The plot below is of the temperature loss of a cell of initial T=3E6 Kelvin, and r_initial = 0.7R.
When factored in, this does indeed slow the cell's acceleration down, so that a cell of 2.5E6 Kelvin at r_initial = 0.7, it takes 2133.9 seconds to rise. This is still quite fast.
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Adiabatic Temperature Loss of Cell
I have managed to calculate the temperature change of a cell as it travels a distance.
The plot below is of the temperature loss of a cell of initial T=3E6 Kelvin, and r_initial = 0.7R.
Hmm... I am convinced that the dT/dR graph should not be noisy. The T graph is so smooth: http://1.bp.blogspot.com/-U7PXmpgroEk/UJE5f5ONJkI/AAAAAAAAADU/38uAikl8TyM/s1600/sunstats1.png
ReplyDeleteThat the derivative should also come out smooth... Try increasing the spacing of the points you are using to calculate the slope until it becomes smooth.