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一个气球从空间?

需要多长时间为一个商店买橡胶气球填满空间与空气从空间外大气和地面联系?它会流行吗?

添加评论后:理想情况下,你想要一个气球装满14 PSI下降之前,它保持它的形状一直到地面。但作为一个正常的气球有它自身的问题。这只会取一个PSI的环境空气填充一个派对气球在空间和任何超过它会流行。我不需要在考虑温度如何影响橡胶。阳光的一面会远高于熔点的橡胶。地球的阴影冻结一边橡胶将冻结变得脆弱和打破。

室外温度将下降的压力通过大气中流行气球吗?

回答

你不能离开“温度”外,温度是关键因素知道气球会流行。让我们设置一些假设的问题所以我们可以计算:1。气球内的气体的行为根据[理想气体定律][1]:PV = nRT 2美元。气球可以容纳30毫米汞柱的压差(= 4000 Pa)(实验)[2]。3所示。一个气球的最大体积是0.015米^ 3美元作为建议[这][3]。4所示。气球开始了它的最大大小。这个我们可以计算有多少摩尔的气体在太空气球膨胀时能在给定的温度。的温度我假设这将是介于的温度(宇宙微波背景辐射)[4](如果在树荫下)~ 2.7 K (-270°C),和黑体的温度被太阳地球轨道附近,也就是大约278 K(约5°C)。 This assumes that the balloon is small enough so that heat can be transferred efficiently by diffusion from one side to the other (the calculation of this value can be seen in the code at the end and is based on the equation [here][5]). With this range of temperatures we can find the volume of the balloon on the surface assuming a final pressure of 1 atm (101325 Pa) and a temperature of 15 °C. We get the following relationship: [![enter image description here][6]][6] In red are the final volumes for which the balloon would explode and in blue when it wouldn't. The critical temperature in between is -262 °C (~11 K). Therefore, **the balloon would not pop as long as its initial temperature is higher than -262 °C**. This also assume that the hottest point in the trajectory is the surface. This means that we are assuming that the balloon will pass through the [thermosphere][7] fast enough to do not reach thermal equilibrium with it (it can be extremely hot). This could be a reasonable assumption given the very low density of that layer. Of course this also assume the rubber of the balloon do not change its properties with temperature. So, only once you specify the height at which you release the balloon and its initial temperature, then you can start calculating fall time. And that will be more difficult, because even if you assume the balloon is at free fall terminal velocity the whole time, this velocity will change along the fall line due to changes air density and balloon size. Air density and balloon size will affect air resistance and, therefore, terminal free fall speed. You will need to do an analytical or numerical integration of the fall to get the total time. ---------- Matlab code used to generate the above figure R=8.3145;%Ideal gas constant [J/mol K] Lsol=3.828e26;%Solar luminosity [W] Dsol=149597870700;%Distance earth-sun [m] sigma=5.670373e-8;%Stefan–Boltzmann constant [W m^-2 K^-4] alpha=0;%albedo epsilon=1;%emissivity % Temperature of a black body (rotating or small enough to transfer heat efficiently from the illuminated side to the dark side) Tbb=((1/4)*((Lsol*(1-alpha))/(4*pi*epsilon*sigma*(Dsol^2))))^(1/4); % Temperature range T=linspace(2.7,Tbb,1000);%[K] Vmax=0.015;% Exploding volume [m³] Pspace=4000;%Maximum pressure hold by the balloon [Pa] n=Pspace*Vmax./(R*T);% number of moles that the balloon would hold Patm=101325;% Atmospheric pressure [Pa] Tsurface=15+273.15; % temperature at the surface [K] % Volume once back on earth surface Vfin=n*R*Tsurface./Patm; explodeIdx=Vfin>Vmax; Tlim=find(explodeIdx,1,'last'); figure('Color','w'); hold on box on plot(Vfin(~explodeIdx)*1000,T(~explodeIdx)-273.15,'b','LineWidth',2); plot(Vfin(explodeIdx)*1000,T(explodeIdx)-273.15,'r','LineWidth',2); plot([0 max(Vfin)]*1000,[1 1]*T(Tlim)-273.15,':r','LineWidth',2) ylim([0 max(T)]-273.15); xlabel('Final volume [liters]') ylabel('Initial temperature [°C]') [1]: https://en.wikipedia.org/wiki/Ideal_gas_law [2]: https://www.youtube.com/watch?v=fwh-i0WB_bQ [3]: https://answers.yahoo.com/question/index?qid=20090704001445AA47gCq [4]: https://en.wikipedia.org/wiki/Cosmic_microwave_background [5]: https://en.wikipedia.org/wiki/Effective_temperature#Surface_temperature_of_a_planet [6]: https://i.stack.imgur.com/pkE7k.png [7]: https://en.wikipedia.org/wiki/Thermosphere

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\ begingroup美元顺便说一下,如果你有一个气球承载1 atm (psi) 14日(类似轮胎),这将是一样的,如果空气原本很冷,你可以得到大量的空气里,当温度升高,它就会爆炸,即使从1开始自动取款机。问题可以简化如果气球内的空气总是在相同的温度。 \ endgroup美元
- - - - - -卡米洛·Rada
2019年2月17日在3

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