Extraction (rise) of hydrogen from the atmosphere of Uranus

Imagine a spacecraft (SC), the front part of which consists of a central cone of the fairing and an annular gas inlet along its edges, while the ratio of the area of ​​the base of the cone and the annular gas inlet is selected so as to ensure minimal heating of hydrogen, which makes up the bulk of the gas entering the gas inlet, when the spacecraft moves through the atmosphere of the planet. The ideal situation would be a complete rejection of the cone of the fairing, but this element hides the mechanisms and devices of the spacecraft as well as the tank of commercial hydrogen, therefore, if possible, it should be as small as possible but cannot have a zero area.





Let's divide the incoming stream of hydrogen into two, the mass ratio between which we will establish later. Let us subject the first stream to a sharp significant compression by narrowing the channel through which it flows and, as a consequence, to a significant increase in the temperature of the stream. At the same time, it will cool the first stream at the expense of the second. When a certain pressure of the first flow is reached, we remove it from the heat exchange path and subject it to a sharp expansion, which leads to its condensation. The result of this process is the liquefaction of atmospheric hydrogen delivered on board the spacecraft, which is sent to the tank of commercial hydrogen.





The second stream of hydrogen heated by the first is directed into a direct-flow solid-phase nuclear engine where we heat it up to a temperature above 3000



and eject it from the back side of the spacecraft through a nozzle with a specific impulse I_{SP}=9.0/



to compensate for the atmospheric resistance and increase the spacecraft mass due to commercial hydrogen.





An interesting feature of such a motion is that the Tsiolkovsky formula does not apply to it, because during a given motion the spacecraft speed remains constant and only its mass changes.





Let us determine what should be the mass ratio between the two streams of hydrogen, neglecting various losses associated with imperfect design.





V1=15.061/



, V_E=2.590/



, V_{atm}



:





V_ {atm} = V_1-V_E = 15.061-2.590 = 12.471km / s

m_1



:





m_1 = \ frac {V_ {atm}} {I_ {SP}} = \ frac {12.471} {9.000} = 1.385 (6)

2.385(6)



1.385(6)



, 1.0



.





.





V_H=5.933/



, V_{UE}



, :





V_ {2-1} = (\ sqrt {2} -1) V_1 = (\ sqrt {2} -1) 15.061 = 6.238km / s V_ {UE} = \ sqrt {V_H ^ 2 + V_ {2-1} ^ 2} = \ sqrt {6.238 ^ 2 + 5.933 ^ 2} = 8.609km / s

m_2



:





m_2 = e ^ {V_ {UE} / I_ {SP}} - 1 = e ^ {8.609 / 9.000} -1 = 1.6027

m_3



, :





m_3 = (m_1 + 1) (m_2 + 1) -1 = (1.385 (6) +1) (1.6027 + 1) -1 = 5.209

1.0



5.209



, .





32,2 . , , .






( ) , , , .





, , .





 +6.0/



  , ( ), . . - .





The second stage accelerates the Oxygen Collector filled with commercial hydrogen up to speed +2.6/



, also in the pericenter of the planet, but in one stage. After reaching the required speed, the second stage separates from the Oxygen Collector and immediately begins a braking maneuver to return to the highly elliptical orbit of the planet, where, like the first stage, it performs an aerial braking maneuver in the planet's atmosphere.





Thus, the 32.2



 Oxygen Collector filled with commercial hydrogen is sent on a  summer flight, and spacecraft designed to operate in the atmosphere of Uranus do not leave the vicinity of the planet.








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