SPACE FLIGHT AT ISENTROPIC COST

SPACE FLIGHT AT ISENTROPIC COST (at minimal energy cost)

The Appolo program (Earth-Moon-Earth) cost 300,000 euro/kg sent on the moon. (1$=1.1 euro)
Dennis Tito paid his stay of 1 week on ISS 20M$, 22 Meuro for say 140kg, 160,000 euro/kg.
The current cost of placing in low orbit of the space shuttle is 18,000 euro the kilog .
The Japanese project Kankoh Maru (50 passengers in subspatial flight) would propose places with 100 000 euros for 100 kg, or 1000 euro/kg.
Thus the costs seem directed to lowering, but that remains expensive, how far can we lower the prices, and have I any chance to escape from earth one day?

Isentropic calculation:
Escape velocity from terrestrial gravity = 11.2 km/s either an energy of (11.2e3)²/2 = 6.27e7 J/kg = 17.4 kWh/kg or with 0.06 euro/kWh, 1.05 euro/kg
Escape velocity from lunar gravity = 1.6 km/s either an energy of (1.6e3)²/2 = 1.28e6 J/kg = 0.36 kWh/kg, or 0.021 euro/kg ( return ticket comparatively almost free ).
Speed of orbiting to an altitude of 100 km = root(9.81*6.478e6) = 7.972 km/s either an energy of (3.18e7+1e6) J/kg =9.11 kWh/kg or 0.55 euro/kg , half less.
Energy necessary for a subspatial flight: Mach 3 and 100 km of altitude (3*340)²/2+1e6=1.52e6 J/kg = 0.422 kWh/kg or 0.025 euro/kg , almost nothing.
This is well below the current costs and the prices actually charged (30 000 times less, at least).
Only the space elevator would make it possible to really approach some (we can expect its power efficiency with 80%).
But, in any event:

it will only make it possible to reach 2 orbits: the geostationary orbit and the infinite orbit. For any other orbit speed is not concordant with altitude and it is necessary to provide by way of engines a good fraction of the energy of the orbit to reach it.

this elevator will be inevitably anchored to the equator, which adds the cost of displacement at the equator (8 euro/kg).

its construction would require 24000 space shuttles for the transport of materials.

Thus let us forget the space elevator and concentrate ourselves on the propulsion by traditional engines.
The characteristics of various propellants are specific impulse I_sp, speed of ejection VE, end of adiabatic combustion temperature T_ad, enthalpy of reaction H_r, molar mass m.
We can calculate everything starting from the speed of ejection. Highest is for the reaction H₂ + 0.5 O₂ -> H₂O, 3690 m/s. It is the only (di)propergol which we will analyze.
For H₂O we have H_r = -241.8 kJ/mol=-1.34e7 J/kg = 3.73 kWh/kg. If we burn oxygen in crossed air we store only hydogene and obtain 1.206e8 J/kg = 33.5 kWh/kg.

The corridor of the continuous flight gives us a relation between speed of the spationef and altitude: 1 km/s by 10km altitude. From 80km altitude we are in satellite mode, below we can push on the air.
Therefore for 8 km/s and 80km of altitude, it is necessary to consider 3.73 kWh/kg for fuel. This corresponds at an average quadratic speed of 5.177 km/s, 40% more than the speed of ejection given above. Explanation: the value 3.690 km/s corresponds to the speed of ejection of a Laval conduit with expansion rate 100, engine which thus presents a specific loss of push of 29%. We will analyze later a model of engine able to recover most of the speed of ejection. For now let us call VE this speed of ejection, M the mass of the spationef without fuel, DM the ejecta mass throughput, VF the final speed. The question is to pass from 8 km/s to 11.2 km/s by limiting acceleration to 4g (that is to say 40 m/s²) for reasons of comfort.
Let U be the speed of the spationef at the moment T, F(t) the push, M(t) the instantaneous mass, we have F=DM.VE=M.dU/dt and dM/dt=-DM, therefore VE.dM/M=-dU, therefore M=M_11.2km/s = M_8km/s.e^-3200/VE. Then M_8km/s = M.e^3200/VE. If VE=5.177km/s, M_8km/s = 1.86M and if VE=3.69km/s, M_8km/s = 2.38M. But in fact, as soon as speed U is higher at the speed of ejection VE, it is disadvised to dilute ejecta with entering air, that would mean to provide them more kinetic energy (relative with the ground). The ideal would be to leave the almost motionless ejecta at a low altitude. But that supposes to modulate VE up to 11.2 km/s, impossible. Thus let us admit VE=constant from U=VE to U=VF=11.2 km/s.
M_VE = M.e ^{VF/VE - 1}. If VE=5.177km/s, M_VE = 3.20M and if VE=3.69km/s, M_VE = 7.65M. If we prolong this mode of propulsion up to the level of the ground, we obtain respectively M₀ = 8.92M and M₀ = 20.8M, that is to say energy consumption of 29.5 kWh/kg and 73.9 kWh/kg, isentropic efficiencies of 59% and 23.5%. If the first figure appears bearable, the second is aggravating. Thus let us analyze the atmospheric flight. Depending on speed U and altitude h, we can more or less press on the air and burn oxygen of it.
We chose an acceleration dU/dt=4G=40 m/s² and U=h/10, therefore dh/dt=400 m/s: impossible for the beginning, it is necessary to start by a phase of vertical acceleration dU/dt=4G-G=30m/s², U=30t, h=30t²/2=15t²=10U=300t, therefore t=300/15=20s, U=600m/s, h=6 km. After this phase, we can incline the space vessel. Let x be the horizontal distance traversed since the starting point, U²=(dx/dt)²+(dh/dt)²=(dx/dt)²+1.6e5 and dx/dt=h/10 and (d²x/dt²)²+(d²h/dt²-G)²=16G² thus when dh/dt=400 m/s, d²x/dt²=3.87G=38m/s², dx/dt=38(t-20), x=19(t-20)²m, h=(6000+400(t-20))=400(t-5)m and the trajectory is given.

Let S be the air inlet surface. The absorptive air flow is Dair=S.dx/dt.rho and even more at low speed (subsonic).
Variation of air density with altitude:
The curve to the left can be approximated by a line (to 10-15% near) passing by (0,1.2) and (100,1e-6).
Therefore rho(h)=exp((1-h/100)*ln(1.2) + (h/100)*ln(1e-6)) where h is in km.
rho(h)=1.2^(1-h/100) * 1e-6^(h/100) = 1.2*(1e-6/1.2) ^(h/100) = 1.2*e^(-0.14*h).
Then, because h=0.4*(t-5) and dx/dt=38(t-20)=38(2.5*h-15)=95*h-570, Dair=S*(95*h-570)*1.2*e^(-0.14*h)=S*(114*h-684)*e^(-0.14*h)
The maximum of Dair is obtained when Dair/dh=0, ie when e^(-0.14*h).(114-0.14*(114*h-684))=0=209.76-15.96*h so h=13.14km.
The total mass of air swallowed between 6km of altitude and the exit of the atmosphere is Mair=sum(S*114k*e^{(-0.14*(k+6))}) where k=h-6, Mair=sum(S*49.2*k*e ^(-0.14*k)), Mair=S*49.2*[-7.143*k-51.02]*e^(-0.14*k)=2510*S ie 2510 kg/m².
This is too weak an estimate. If we count since the ground the total mass of vertically crossed air, we obtain 101325/9.8=10300 kg/m², and the swallowed air mass is more significant than that, because of the slope of the trajectory and a additional speed of absorption.
Though it is, we can calculate the instantaneous power W required to have an acceleration of 4G: W=M*40*U=4000*M*h. In aerobic combustion and neglecting the kinetic and thermal losses, we have -dM/dt >= W/(33.5kWh/kg) =W/(1.206e8 J/kg)=3.317e-5*M*h=dM/dh*dh/dt=0.4*dM/dh. From where we draw 8.29e-5*h*dh=-dM/M=-d(ln(M)), then M=Msol*e^{(-4.14e-5*h²)}, where Msol is the mass on the ground (h=0), dM/dt=(-8.29e-5*h)*(dh/dt)*Msol*e^{(-4.14e-5*h²)} = - 3.317e-5*h*Msol*e^{(-4.14e-5*h²)}. It is the hydrogen flow in kg/s. The oxygen flow is 8 times higher, and the combustive air flow 38 times.

required and intake airflows vs altitude on the trajectory

There is thus Dair _comb = 1.26e-3*h*Msol*e^{(-4.14e-5*h²)}.
We can see that the combustive air flow becomes insufficient from 30 to 45 km, according to whether the S/Msol ratio varies from 1 to 6 m²/ton.
In addition, in combustion with the air, the maximum speed of ejection is 2.487km/s relative to the surrounding air, by neglecting the preliminary acceleration of hydrogen, 38 times lighter, and by supposing that the ramjet can compress the air entering isentropically, i.e. that the isentropic work of compression is equal to the kinetic energy: cp*(T1-T0)=U²/2 ie T1=T0+U²/2010. So for U=5km/s, T1=12400°C! As no material holds beyond 3700°C, that will be hard. In fact, as soon as U > Umax=2730m/s, it is not possible any more.
Incidentally, the ramjet, theoretical concept of which I did not see concretization (model making put aside) will not work beyond Mach 8. For 27 km of altitude and 2.7 km/s, it is necessary to close the input of the ramjet under penalty of destruction. It would however be interesting to continue to burn ambient air to avoid burning embarked oxygen already, can be as a practitioner a side air intake (kinetic filter). But let us leave that on side. From 2.7 km/s up, we use anaerobic combustion. There are then M_2.7km/s= M.e^8500/VE. If VE=5.177km/s, M_2.7km/s=5.17M and if VE=3.69km/s, M_2.7km/s = 10M.
But to get 2.7 km/s, we can use the ramjet. You see on the graphic to the left that the S/Msol ratio must be equal to 32/54=0.593~0.6 m²/ton.
In aerobic combustion and thrust 4G, we have 40M=S*Dair*(U2-U1), with U2=speed of ejection, U1=speed of input. At low speed, Dair=S*rho _i*a_i* .578, where rho_i is the impact density and a_i the speed of sound of impact. By using the relation of Barre de Saint-Venant: 1+ 0.2025*mach²=(T_I/T)=(P_i/P)^0.288=(rho_i/rho)^0.405=(a_i/a)² with mach=U/a, we have a_i=a*(1+0.2025*mach²)^0.5, rho_i=rho*(1+0.2025*mach²)^2.47, Dair=0.578*S*a*rho*(1+0.202*mach²)^2.97
this page is under construction, the continuation in a few days...

This picture indicates specific impulsions and static ejection velocities for various propulsion modes.
Specific Impulsion Is: time (in seconds) while a fuel mass can afford a thrust equal to its own weight. We aim to maximise it.
Ejection velocity Ve: gas relative speed (in fps -foot per second- : 1fps = 0.3048 m/s) downstream from jet.
We get a power function: Is = a.Ve^b involving (70fps,70000s) and (1800fps,4000s), so b=log(70000/4000)/log(70/1800)=-0.8815, et a=70000/(0.3048*70)^-0.8815=1.039e6 in metric system. Ejection velocity is necessariy greater than cruise velocity, otherwise thrust would be zero or less. How correct ejection velocity with cruise velocity?