Obviously I didn't post enough back of the envelope calculations estimating how truly massive a weapon like the Death Star's cooling system would have to be I thought I would return. That and discussions on facebook got into some rather intense levels of detail so I thought I might as well take advantage of the math I did and the excel file I created. Of the initial variable constraints I spent more time concerned with two of the more easily defined variables on the Death Star, the roughly known area of the exhaust port Luke shoots at and the smooth surface area of the original Death Star. (right now I'm going to deviate from my white board based art work as I am back in Alaska with my parents and use MS paint. That and I'm not exactly in the habit of traveling with a working surface (says the man who packed a desktop in his carry-on))
For later reference here is my size comparison between the exhaust port and that of a smoothed out original Death Star Surface. (Apologies for showing the original size but this was the only way to show that there was an actual pixel representing the exhaust port, for numerical reference the larger black box represents
1/20th the surface area of the Death Star)
To the newer calculations.
During my discussions with friends on facebook a friend of mine asked about another potential limiting variable, that of the temperature of the radiator being used to cool the Death Star. That provided some very interesting room to work with as to determining how big the cooling surface would have to be.
The equation for calculating the surface area is the same as used in the previous Death Star entry, I have simply rearranged which variables are bound and unbound.
For ease of calculation, and as the relative temperature difference makes it an effectively moot point anyways for how fuzzy the math here, I am assuming that the system is radiating into a perfectly cool ambient space existing at 0 degrees Kelvin. Additionally I am going to assume that the emissivity of the surface is perfect, Imperial engineers and scientists are so good that they have achieved a perfect black body material (well that and a few other rather impressive technological achievements.
Using the algebra on the equation to the right
we are left with our balanced solution.
the waste heat value is defined by the overall system efficiency, the Boltzmann constant is defined by wikipedia and other reputable sources as well ;)
ok now that we have our variables more or less defined, we can move onto tedious data tables. WaHooo!!!(let me know if you need any clarification I have gotten too used to speaking engineer)(also there may have been too many exclamation marks at the end of wahoo there, but I'm not going back on that one)
The Numbers of Importance
the first variables provided are here following links provided by @marylfl I will include the numbers associated with her tweet first
Assuming that the overall super laser system, from generators to powerlines, to firing mechanisms, are 99% efficient (thermally speaking). It is estimated that the Death Star must vent 1.17*10^25 watts continuously. To handle this tremendous quantity of waste thermal energy, Imperial Engineers and servants of the Sith Lords, have developed a perfect black body material that can operate at 20,000 degrees kelvin (these variables were suggested by a friend on facebook, as well as the reference to the Dark Side)
Using these tremendously advanced materials, Imperial Engineers still would find themselves requiring
5.29 nonilean square meters of radiator (5.29*10^30 m^2).
For personal editorial reasons I will add some data tables at the bottom of this entry at a later time.
Come back tomorrow for my estimates as to how Imperial Engineers might have solved their heat problem.
1/20th the surface area of the Death Star)
To the newer calculations.
During my discussions with friends on facebook a friend of mine asked about another potential limiting variable, that of the temperature of the radiator being used to cool the Death Star. That provided some very interesting room to work with as to determining how big the cooling surface would have to be.
The equation for calculating the surface area is the same as used in the previous Death Star entry, I have simply rearranged which variables are bound and unbound.
For ease of calculation, and as the relative temperature difference makes it an effectively moot point anyways for how fuzzy the math here, I am assuming that the system is radiating into a perfectly cool ambient space existing at 0 degrees Kelvin. Additionally I am going to assume that the emissivity of the surface is perfect, Imperial engineers and scientists are so good that they have achieved a perfect black body material (well that and a few other rather impressive technological achievements.
Using the algebra on the equation to the rightwe are left with our balanced solution.
the waste heat value is defined by the overall system efficiency, the Boltzmann constant is defined by wikipedia and other reputable sources as well ;)
ok now that we have our variables more or less defined, we can move onto tedious data tables. WaHooo!!!(let me know if you need any clarification I have gotten too used to speaking engineer)(also there may have been too many exclamation marks at the end of wahoo there, but I'm not going back on that one)
The Numbers of Importance
the first variables provided are here following links provided by @marylfl I will include the numbers associated with her tweet first
Assuming that the overall super laser system, from generators to powerlines, to firing mechanisms, are 99% efficient (thermally speaking). It is estimated that the Death Star must vent 1.17*10^25 watts continuously. To handle this tremendous quantity of waste thermal energy, Imperial Engineers and servants of the Sith Lords, have developed a perfect black body material that can operate at 20,000 degrees kelvin (these variables were suggested by a friend on facebook, as well as the reference to the Dark Side)
Using these tremendously advanced materials, Imperial Engineers still would find themselves requiring
5.29 nonilean square meters of radiator (5.29*10^30 m^2).
For personal editorial reasons I will add some data tables at the bottom of this entry at a later time.
Come back tomorrow for my estimates as to how Imperial Engineers might have solved their heat problem.
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