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=== Trivia === * Steam by itself '''m'''. In kilograms. It is very important to understand that amount of steam is '''mass''' (which is contrary to what it is called in game: steam volume). For example large turbine and small turbine can both contain the same amount of steam "volume" (say 500 kg) - but it will produce different steam density and different ''pressure.'' * Pressure '''P''' - this value is responsible for "production" of power and energy and for moving steam through the pipes. The more - the better. But things (pistons for now, but who knows what else they implement) may explode. Units are unknown. 1 Pa, I suppose (values are to high to be atmospheres) * Volume '''V''' - this is where steam is contained. Just good old cubic meters. For now any volume (boilers at most) may contain any amount of steam (reaching and exceeding sun core density) Steam is considered to be an ideal gas, for which this formula is true: <math>\frac{P*V}{T}=const.</math> Temperature '''T''' seems to be constant for any part of the steam system, and pressure is directly proportional to density, i.e. mass of steam contained in given volume '''V'''. It means that for FtD steam "physics" we can write down this equation as:<math display="inline">P*V=const.*m</math>. We can calculate that constant using pipes as we know their volume V from the item selection screen and pressure P and "volume" m are indicated when hovering over one of the pipe. To do so we use pipes with no piston attached to them and a pressure release vent to reach a stable system. We can easily find that this constant is equal to 1, so: <math> P * V = m </math>. Now using <math>PV=nRT</math>; the n (moles) can be replaced giving: <math>PV=\frac{M}{m_r}RT</math> we can rearrange this for density of the steam: <math>\rho=\frac{M}{V}=\frac{Pm_r}{RT}</math> and since we already know that <math>\frac{RT}{m_r}=1</math> we can then find that <math>P=\rho</math> So for steam in FTD the absolute pressure <math>P</math> equals the density <math>\rho</math>. Also we can work out the temperature of Neter using: <math>T=\frac{m_r}{R}</math> If we assume that the <math>m_r</math> of steam is 0.018 kgmol<sup>-1</sup> and that R is 8.314 Jmol<sup>-1</sup>K<sup>-1</sup> giving a temperature of '''~2.165x10<sup>-3</sup> K'''. Now knowing all of this we can use that to also find out the atmospheric pressure for Neter. For a sealed pipe that has a volume of <math>V</math> and a gauge pressure of <math>P</math> and the steam mass is <math>M</math> the density <math>\rho</math> is simply found by <math>\rho=\frac{M}{V}</math> and as we know the density equals the absolute pressure we can say <math>\frac{M}{V}=P+x</math> Where x is the atmospheric pressure. Now if we take some in game values to fill the previous equation: so for a single 1m sealed pipe with a volume of 0.2 m<sup>3</sup> and a pressure of 2260 Pa and a mass of 452 kg then the equation above can be solved for x. This gives an atmospheric pressure of '''zero'''. (This works for any pressure and volume and mass of steam) The pressure in the boilers always relates to the pressure in the pipe and the amount of steam produced:<math>P_{boiler}=P_{pipe}+Steam</math>. Where <math>P_{boiler/pipe}</math> is the pressure of the boiler and pipe respectively and <math>Steam</math> the amount of steam produced.
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