Article Hypersonic tube
Gas Tube Hypersonic Launcher*
1310 Avenue R, #6-F,
Tel/Fax 718-339-4563, aBolonkin@juno.com, http://Bolonkin.narod.ru
The present articlebes
a hypersonic gas rocket, which uses tube walls as moving compressed air
container. Suggested burn programs (fuel injections) enable to use of the
internal tube components as a rocket. A long tube (up to 0.4-0.8 km) provides
mobility and serves to aim in water. Relatively inexpensive oxidizer and fuel
are used (compressed air or gaseous oxygen and kerosene). When a projectile
crosses the Earths atmosphere with an angle more then 150, loss of
speed and the weight of required thermal protection system are small. The
research shows that the launcher can give a projectile a speed of up to 5-8
km/sec. The proposed launcher can deliver up to 85,000 tons of payloads to
space annually at a cost of one to two dollars per pound of payload. The
Launcher can also deliver about 500 tons of mail or express parcels per day
over continental distances and one be used as an energy station and
accumulator. During war, this launch system could deliver military munitions to
targets thousands to tens thousands of kilometers away from the launch site.
------------------------
* This Chapter as paper
presented at the 38th AIAA Propulsion Conference, 7-10 July 2002,
Keywords: Hypersonic
launcher, space launcher, gas rocket, Bolonkin Launcher.
Main Nomenclature (metric system):
Cp
- heat capability at constant pressure Cp=1.115
[kJ/kg.Ko] for air, Cp=1.069 for oxygen.
Cv - heat capability at constant volume;
Gp - mass of the projectile (payload)[kg];
k= Cp/Cv - coefficient, k=1.34 for air; k=1.2 for
gas of liquid
L - length of launch tube [m];
Lr - length of rocket air column [m];
M - mass of rocket [N];
Mk
- rocket mass in end of acceleration [N];
Mo
- rocket mass in beginning of acceleration [N];
V - speed of rocket [m/s];
Vo - speed of the rocket after initial
acceleration [m/s];
Vp - projectile speed after shot on a
sea level [m/s];
DV - increment of rocket velocity [m/s];
w - speed of gas after nozzle [m/s];
P - pressure of gas after nozzle [atm];
Q - amount of heat [Wt];
R - gas low constant, R=287
for air, R=325 for rocket gas rocket.
t - time [sec];
T - absolute temperature, OK.
q - trajectory
angle to a horizon [degree].
b - fuel
rate [kg/sec]; m=Mk/Mo - relative mass of rocket.
1. Introduction
Unfortunately, most specialists, when they see projectiles in tubes,
think of a gun (cannon). They know quite well that a gun cannot give a
projectile speed of more than 1-2 km/sec [1]-[5]. They also know that a gas gun
using hydrogen can give a speed of up to 3-4 km/sec, which however is not
enough for space flight.
The
high-speed ground based gun is very large, complex and expensive, and can shoot
only from one point of the Earth or Space [7]-[12]. Therefore they do not want
to read or hear about a new revolutionary method.
As is
shown in [12] the offered installation is a gas rocket in a tube. The
installation looks like the solid traveling charge gun [13]-[14]. However, the
offered tube rocket uses air and kerosene as fuel (not powder as in the
traveling charge gun).
Short history of the high speed gun. The first
high-speed gun was the Krupp gun in World War 1.
After WW2 research of a big high-speed gun was made in project HARP [11].
The
cannon had the caliber of 16.4 in (417 mm), the barrel length 45-126 calipers
(18.8-52.5 m), and the weight of the gunpowder 780 lbs (351 kg). Maximum
pressure reached 340 MPa (3400 atm).
The projectile weighting 38 lbs (17 kg) had speeds up to 6000 ft/s - 6135 ft/s (2 km/sec).
A gun
with a solid traveling charge was studied by the Ballistic Research Laboratory
[13]-[14]. They used solid gunpowder as rocket fuel and a small bore (5/8)(9.5 mm). The traveling charge gun was initiated at the
Ballistic Research Laboratories, Aberdeen Proving Group,
In
theory there is no upper limit to the velocity, that
can be attained. All that is required is more propellant and a longer barrel.
However the shot time is about 4 ms and the requested high burning rate of 2000
in/sec is three orders of magnitude greater than the
burning rate of a solid gun propellants. This problem could not be solved by
using a porous (grains) propellant (gunpowder). It was noted that could be
possible for the propellant to detonate. Some experiments with ball powder in a
long tube (analogous to a long grain of porous propellant) have indicated that
rapid burning of the propellant may change over a detonation.
The
suggested method and installation uses conventional compressed air (as
oxidizer) and conventional kerosene (as fuel). The barrel is long (up to 0.4 -
1 km) and the time of the shot is long (up 0.5 - 1.2 sec). The injection fuel
program keeps a constant pressure in the rocket (constant thrust) and avoids
detonation. Other differences and advantages are shown later.
Fig.1a
shows a design of the tube of the suggested Hypersonic
gas-rocket system. The system is comprised of the following: tube, piston with
fuel tank and payload, nozzle connected to piston and valves.
The
tube rocket engine can be without a special nozzle (Fig.1b). In this case, fuel
efficiency of the gas rocket engine will decrease but its construction becomes
simpler.
The
tube may be placed into a frame (Fig.1c). The frame is placed into water and
connected to a ship for mobility and aiming.
The
launch sequence is as follows:
The moveable piston with the fuel tank (containing liquid fuel), and
payload are loaded into the tube. The piston is held in place by the fasteners
or closed valve 17 (Fig.1). The direction and angle of the launch tube is set.
Fig.1. a - Space Launcher with the gas rocket and a rocket
nozzle in tube. The system is comprised of the following: 1 - tube, ,
2 - payload (projectile), 3 - fuel tank, 4 -piston , 5 - fuel pipeline, 6-
nozzle connected to piston, 7 rocket air column, 8 - combustion chamber, 9 -
injectors of a combustion chamber, 12
tube frame, 14 - additional injectors, 15 - lower tube injectors.2 - 9, 16 - air pipeline, 17 - lower valve, 18 -
upper valve, 19 - top valve, 20 - air lock, 21- gas pipe, 22 electric
engines.
b - Space Launcher
with the gas rocket without the rocket nozzle . c
Launcher in frame.
Valve 19 (Fig. 1) is closed and a vacuum (about
0.005 atm) is created in the launch tube volume above
the payload/piston to reduce the drag imparted to the payload/piston as it
moves along the launch tube. The tube, of a length of 630 m and a diameter of
10 m, contains 61 tons of air at atmospheric pressure. If this air is not
deleted, the payload must be decreased by the same value. If air pressure is
decreased down to 0.005 atm, the parasitic air mass
is decreased to 300 kg. This is acceptable parasite load.
Valve 17 is closed and an
oxidizer (air, oxygen, or their mixture) is pumped into the volume below the
payload/piston.
Liquid fuel (benzene, kerosene) is injected
into the volume below the nozzle 6 through the launch tube injectors (item 15,
Fig. 1) and ignited. The valve 17 (Fig.1) is opened. The hot-combustion gas
expands and pushes the payload/piston system along the launch tube together
with the air column (item 7) between piston and nozzle.
When the piston reaches the maximum gun speed (about 1 km/sec), the
compressed air column begins to work as a rocket engine using one of the
special injection fuel programs (see [12]).
As the payload/piston approaches the end of the launch tube, valve 19 is
opened and the airlock (item 20) begins to operate. After the payload/piston has left the launch
tube, valve 18 closes the end of the launch tube and re-directs the
hot-combustion gases down the bypass tube (item 21) to various turbo-machines
preparing compressed air for next shot and electricity for customers.
If a high launch frequency is required, then
the internal tube water injectors are used to quickly cool the launch tube.
After the payload/piston system leaves the
launch tube, the payload (projectile) separates from the piston and the empty
fuel tank. The payload continues to fly along a ballistic trajectory. At
apogee, the payload may use a small rocket engine to reach orbit or fly to any
point on the Earth.
The method by which the fuel is injected and
ignited within the launch tube is critical to high-speed (hypersonic)
acceleration of the payload. The author has developed the five fuel injection programmes for the launch system [12].
In these programs the thrust (force) are
constant at all times. It means that pressure and all parameters in the rocket
engine are constant. Part of the programs has two steps. In the first step the
fuel is injected into compressed air at the lower part of the tube to support a
constant pressure and provide the initial acceleration of the rocket (together
with air column Lr)
to the velocity Vo. In the
second step the rocket engine begins to trust and support the constant pressure
and temperature into the rocket combustion chamber. The result is that the
trust force of the gas rocket engine remains constant. In the reference article
the author considered only a simplified model (Fig.1b) when a rocket nozzle is
absent.
1)
Estimation of missile velocity
Take the well-known rocket equation
.
(1)
and its solution
,
(2)
where DV =
increment of speed [m/s], m = Mk/Mo = relative mass of rocket; Mk = rocket mass at the end of acceleration [kg]; Mo = rocket mass at the
beginning of acceleration [kg]; b = fuel (gas consumption) rate
[kg/sec](constant); w = speed of gas
at nozzle exit [m/sec](constant), P =
force of pressure at nozzle exit [N](constant).
It is
well-known from the theory of nozzle expansion the following relations
P =104gPnSn
; b = rcwcSc ;
where sub-index c notes parameters
in a critical (most narrow area) nozzle area; sub-index o notes
the gas parameters into tube rocket before combustion; sub-index n
notes parameters in the nozzle exit, S
is the cross-section area of the tube [m2], T is the gas temperature in combustion chamber [oK],
p is the gas pressure in the rocket
[kg/cm2], k, R are gas
constants, g=9.81 m/sec is gravity
In
this article the computations are given only for the simplest case: rocket
without special nozzle (the tube is used as a nozzle of the constant cross
area), where: S=Sn=Sc , w = wc
, Pn=Pc.
Let us substitute the above expressions into
(2):
(3)
where k=1,4, R=287
for air and k=1.2, R=325 for rocket gas, Vo
is the initial speed [m/sec]. Substituting k=1.4,
R=287 for air into above expressions, we get the following equations
,
wc =18.13(T)0.5, (wc +
P/b) = 31.68(T)0.5,
rc = 0.634ro ,
pc= 0.565po , Tc=0.833T . (4)
If we substitute k=1.2, R=325 for rocket gas
, (5)
The
conditional temperature T in the
combustion chamber without molecule dissociation and with constant coefficient
of the heat capacity at a constant pressure Cp,
may be calculated by equation (for kerosene)
, T =
2880 + DT,
(6)
where n =
percent of an oxygen in the air (n =
21% - 100%). Q=0.067e=2.95.103 is the amount of heat [joules] when 0.067kg
kerosene was fully combusted in 1 kg of air used 21% of oxygen, e=44.106 [J/kg] is heat capacity of kerosene, 1.115=Cp
is heat capacity of gas at constant pressure.
The
computation gives: n =21%
(conventional atmosphere air), T =27680
K; n = 40%, T =47500 K, n =60%, T=66000 K; n =100%,
T @ 98500 K.
In
this linear model the top temperatures is higher than the temperatures of
molecular dissociation. However the high pressure resists the dissociation. For
example, the dissociation of water vapor at atmosphere pressure starts at temperature
2500oK and finishes 4500oK. If the pressure is 100 atm, the water vapor dissociation begins at 3500oK
and finishes at 6500oK. We use the pressure from 300 to 1200 atm (and more) where dissociation is not so much. We can
also use it because in the rocket nozzle, when the gas expands and the
temperature decreases, the molecules recombine and the energy of the
dissociation returns (comes back) to gas. We have the same situation in
conventional internal combustion engines, rocket engines, and guns.
The
initial velocity of the tube rocket is taken 1 km/sec. It requires
approximately 1/3 additional length of the compressed air rocket column Lr=30m.
Fig.2
shows an estimate of the hypersonic velocity V via a given initial (under piston) pressure po and ratio n of oxygen in
air [Eq.(5)], with cargo
(piston plus payload) weight Mk =20
tons (44,000 lb), the tube diameter is 10 m (15 ft)(S=78.5 m2) and rocket air column length Lr=15m.
As one see, the speed of the projectile reaches 7.2 km/sec when P =500 atm for
conventional air (oxygen 21%) and reaches 8.2 km/sec for 30% gaseous oxygen.
For P=800 atm the speed is V=7.5 and 8.8 km/sec, respectively. In the conventional tank gun
the pressure reaches up to 4500 atm and the
temperature up to 45000C.
The
high speed of the projectile is not a surprise because we can get a very high
mass ratio Mo/Mk=1/m in this launcher. For example, this
ratio reaches 30 for P=500 atm, while it equals maxim 8-10 for one stage conventional
rocket. Specific impulse equals 181 sec for air and 280 sec for 60% oxygen.
Increasing diameter and rocket air column increase the charge, and projectile
speed. Decreasing air column 15 to 10 m decreases the projectile speed 7.2 to
6.5 km/sec (fig.3).
The
rocket nozzle decreases the required fuel (increases efficiency), but this
nozzle requires more tube length. Unfortunately, the brevity of this paper does
not allow the presentation of the many computations (see the note at the end of
the paper).
Fig.2. Projectile speed for load 20 tons, tube
diameter 10 m, rocket air column 15 m and percentage oxygen in air 21 - 60%.
Fig.3. Projectile speed for load 20 tons, tube
diameter 10 m, rocket air column 10 m and percentage oxygen in air 21 60%.
2) Estimation of tube length.
The
acceleration requires two assessments: acceleration of load together with
rocket air column from 0 to speed Vo=1 km/sec (the tube is used as a gun) and the
rocket acceleration of the load from Vo
to exit V by the rocket air column 7.
a) Length for the initial acceleration
from 0 through Vo=1 km/s
(tube is used as a gun) (unmarked equations show the process of getting the
estimation):
L1 = Vo2/2a ; t1
= Vo/a ; a = F/Mo
; F = 104gpoS
; Mo = Mk+Mf+Ma , Ma
= 1.225.Sp0Lr/q ,
(7)
where po = pressure of gas in tube [kg/cm2],
Lr = length of rocket air column [m], a = acceleration [m/s2], F = moving force [N], Ma = mass
of air column [kg], 1.225 [kg/m3] = air density in pressure 1 atm, Mf =
mass of fuel [kg], Mk -
mass of a load (piston + payload)[kg], Mo
= full mass of rocket [kg], q is the
correction of air compressibility for Po>250
atm (for Po<250
atm q=1), t1 = acceleration time [sec] of distance L1 [m] when installation work as gun.
b) The length of acceleration as a
rocket after substituting (3) into (4) and with the numerical data (k, R for air) is:
(8)
where
t2 = (Ma + Mf)/b ; b = rcwS ; m = Mk/Mo ; , Here t2
= time of acceleration in distance L2
, rc =
gas density in nozzle, M0=Mk+Ma+Mf = initial
mass of rocket, Mk = mass
of shoot [load (projectile or payload) + piston], Mk =20000
kg, Ma.=
mass of air in the rocket, Mf = mass of fuel in the
rocket [see (7)].
c) The full length of the launch tube
and the acceleration times are:
L = L1 + L2 + L3 , t = t1 + t2 , (9)
where L3=1.3Lr - length air column plus air column for initial acceleration (30% from Lr).
Fig.4
represents the barrel length required for the speeds of Fig.2. At pressure 500 atm, the tube length equals L=650 m for conventional atmosphere air (V=7.2 km/sec) and L=1400
m for 39% additional oxygen (total oxygen is 60%)(V=10.9 km/s). This is no surprise,
because we use the SAME pressure in the shot for air and oxygen. If the tube length is less then the
requested length for a given air pressure, the use of oxygen (for a given
program and same pressure) does not give an increase in speed but only
increases the load (two and more times).
Fig.4. Required tube length for load 20 tons, tube
diameter 10 m, rocket air column 15 m and percentage oxygen in air 21 60%.
The
oxygen allows the combustion time to increase (length of tube) and the exit
projectile speed to increase. For pressure
Fig.5.
Required tube length for load 20 tons, tube diameter 10 m, rocket air column 10
m and percentage oxygen in air 21 60%.
3) Loss of the velocity in atmosphere.
The
loss of velocity of the projectile in air is small, if the projectile is
properly shaped and its trajectory more then 15 degrees to horizon. For an
exponential density of the atmosphere, a constant speed of sound (a=300 m/s), we receive the following
equation for an estimation of a decrement of the loss speed (unmarked equations
show the process of computation):
MpVp2/2 =A , D=2a2r0aVpSp/b ,
A/Mp = 2a2roaVpSp/Mpsinq , , (10)
where: Vp = initial projectile speed after shot [m/s], Sp = specific projectile area
(m2), a = relative thickness (0.1), q = angle trajectory to horizon, r = r(h) density
of atmosphere (h -altitude), ro=1.225 kg/m3 - atmosphere density in sea
level, Mp = mass of
projectile ( kg), A = work of passing
[J], D = wave drag of projectile (90% of total drag). The drag of the projectile depends linearly
on speed for large Mach numbers.
It is
shown, that the loss of energy in an exponential Earth atmosphere equals the
loss of energy in an atmosphere with constant density ro=1.225
kg/m3 and altitude h=8900
m.
The
results of the computation for different Mp/Sp
[ton/m2] are shown on Fig.6. The loss does not depend on speed
because than the more speed then less the time of braking. It is about
30-100 m/sec.
Fig.6. Loss velocity (m/s) of projectile
in atmosphere versus shot angle for specific load 2, 4, 6, 8 ton/m2.
Some
people think that passing through atmosphere gives a big heating of projectile.
They know that space ship Shuttle need in good heating protection when one
re-entry to Earth atmosphere. However, the Shuttle must decrease the speed V1=8 km/sec to V2=0.3 km/sec. That for mass m=20 tons equals energy /2= 6.4.1011
J. All this energy is converted in a heat. In our case the speed decreases V1=8 km/sec to V2=7.9 km/sec. That for mass m=20 tons equals energy /2= 0.16.1011
J. It is 40 times less.
The
same problem exists in electromagnetic launcher. Research in [17] p.395 show
that a sharp nose five ton projectile needs only 0.9 kg a lithium refrigerant
for protection projectile head when one across atmosphere with angle 150.
4. Load delivered to orbit
Load delivered to orbit can be computed by
following equations
(11 )
where J - ratio speed V to circular speed Vc in
given point [m/sec], R0 Earth radius [m], p parameter of elliptic orbit, e eccentricity of elliptic orbit, ra radius apogee [m], H
altitude [m], Va
speed at apogee [m/sec], q - trajectory angle to horizon, Vd
requested additional speed at apogee [m/sec], Vc,H requested
circular speed [m/sec], Mk/M0
ratio final (useful) and initial mass, w
speed of rocket exhaust gas [m/sec].
Result
of computation for rocket engine with impulse w =2000 m/sec is presented
in figs. 7-8.
Example of using. We have tube diameter 10 m2, load
20 tons and the tube pressure 500 atm. From fig.2 and air we find for air that the
shot speed is 7.2 km/sec, from fig.4 the requested tube length is 650 m. From
fig.6 the loss of velocity in atmosphere for shot angle 180 and M/A =4 ton/m2 is 45 m/sec.
From fig.7 we find apogee of orbit altitude H=1000
km for the shot angle 180 and speed 7155 m/sec.
Assume, that in 20 tons of load a piston weight equals 5
tons and 15 tons is the orbit load. From fig. 8 for V=7.155 km/sec and the shot angle 180 we find that
useful part of orbit load is 47% or about 7 tons. The eigth
tons is rocket fuel for increasing apogee speed to circular speed at given
altitude.
Fig.7. Apogee altitude versus shot
angle for initial projectile speed.
Fig.8. Relative useful mass of projectile for
initial speed 5.5 8 km/s and specific impulse 2000 m/sec.
5. Production cost
of delivery.
The
production cost of delivery is the most important feature of the suggested
launcher [12],[22]. In [12,22]
Fig.18 shows computations of delivery cost versus annual payload in thousand
tons and an initial cost of the installation.
The following data were used: life time of the installation 20 years,
cost of fuel (kerosene) is $0.25 per liter, requested time for preparing of a
shot is 0.5 hours, 10 men with average salary $20 per hour, payload of one
launch is 5 tons (11,000 lb) (25% from 20 tons of piston-projectile system).
The load (20 tons) is: payload - 5 tons, a solid engine for getting additional
speed of 3 km/s at apogee - 10 tons, piston - 5 tons. Maximum capability is 86,400 tons per year.
The
closed computations are presented in fig.9.
Fig. 9. Delivery cost of payload versus annual payload for
installation cost 200 1000 millions $, life time 20 years, annual maintained
10 millions $/year, kerocine cost 0.25 $/kg.
One can
see, the production delivery cost may be about 1-2 $/kg if the payload is
approximately 50-80 thousand tons per year. About 70% of this cost is fuel cost
(kerosene or gasoline).
6. Excess of energy
Hypersonic Launcher uses a very high compression ratio,
It means that its efficiency coefficient is very high about 85-95%. Such
efficiency do not have any a head engine. Part of this
energy is spent for a load acceleration. Part is loss
for non-useful gas extension. However, the Hypersonic Launcher has a huge
excess of shot energy over a gas compression energy.
If the Launcher has a top valve, this energy can be used for production of
compression air, electricity, oxygen, etc.
Let us
to estimate this energy. After shot the valve closes the tube. The pressure in
tube and possible work of one kg gas can be computed as (fig.10):
(12 )
where: T
is temperatures in marked points, K0,
T1=2880C; DT
increasing temperature from fuel: it equals 24800 C for air, 46600C
for adding 20% oxygen,
66140C for adding 40% oxygen; P
pressure in market (sup index) points, atm; W work on market length [J]; WT isotherm work, WA adiabatic work; R= 287 thermodynamic coefficient; k=1.4 thermodynamic coefficient; L tube length [m]; L3 tube length where a
pressure can be constant [m]; N
number of cooling radiators in the compression process.
Fig.10. Thermodynamic diagram of launcher. 1- point of atmospheric pressure; 2
point of common air columns volume and tube pressure; 3 point after fuel
combustion; 4 point of tube volume; 5 point after gas exhaust.
Fig.11. Adiabatic expansion (after shot) and
compression work of one kg gas for cooling radiators N=2 8, charge length 20 m, tube length 650 m.
Result of computation is presented in fig.11.
One kg tube gas for N=4, p=500
atm has the excess of energy about 2535 kJ. Our
Launcher for p=500 atm has about M=rV/q=1.225.500.78.5.20/1.72=559
tons compressed air. It means, after the short and compressing air for next
shot we have Ea= 2535.559=1417065 MJ = 1.4.1012
J of the excess energy.
The
requested fuel for shot is Mf =0.067.559=37.45
tons. It has energy E=Mfe=37.45.103.44.106=
1.65.1012 J. The
energy of projectile is
Ep= mV2/2=20000.72002/2=0.144.1012
. The excess energy (which
can be used) is Ea=1.4.1012
J . The total efficiency (without loss compression gas to atmosphere) is (Ep+Ea)/E=(1.4+0.144)/1.65=0.93.
It is more then any current heat machine because we use a very high compression
ratio p=500 atm. The exhaust gas has a temperature
T=9800K (p=500 atm0. Utilization of this high level head can give
additional energy.
7. Launcher as
accumulator of energy
There
is a big problem in electric energy industry. The power stations have high
efficiency when they work in constant regime. It means, they produce excess
energy in night and lack energy in day. For balancing energy the electric
industry build special hydro-accumulator station. These stations pump water to
a compensation (balancing) reservoir when the excess energy and produce
additional energy when it is lack.
We have
a gigantic reservoir for a high compression air. When the launcher does not
have work, one can be used as accumulator of energy. The Launcher gets energy
from electric stations, pumps a compression air to the tube. When there is lack
of energy, the Launcher produces energy from the compression air and send to customer. If it is use by a sea water the cooling
air in during of compression and heating of air by a sea water in during extension, the
efficiency of this method may be high (80-90%), especially if we will use the
sea depth cool (T1=70
C) water for cooling and warm (T2=20-250C)
surface water for heating. The equation for heating is below
(13 ),
where W1
= work compression, W2 =
work extension, p2=tube
pressure, h = efficiency coefficient, N = number cooling radiators.
Result
of computation is presented in fig. 12. At 500 atm
one kg air gives 580 kJ when it is isotherm expending. In tube V=SL=78.5.650=51025
m3 under pressure p=500 atm we have
M=rpV=1.225.500.5.1.104/q=
30.6.106 kg compressed air. They contain E=580.30.6.106= 17.748.109
kJ of energy. The Installation can work
as 100 MW power electric station in during 17.748.109
/3600/105= 49.3 hours and h=0.88
if N=8, T1=T2.
Fig.12. Compression and expansion work versus gas
pressure (atm) of one kg gas for cooling radiators N=2 8.
8. Gun Recoil
The
recoil of the Launcher can be computed by theoretical mechanics equation
m1V1=m2V2 , (14
)
where m1,
m2 mass and V1
, V2 speed of projectile and installation
respectively.
For example, if the projectile mass equals 20 tons and speed is 8
km/sec; the water inside of installation frame 15x15x650 m3 is 14.6.104
tons, the installation speed after short will be V2=20.8000/14.6.104=1.1
m/sec.
9. Tube curvature
A tube straightness is check up by laser beam and is
supports by engines located along the tube. The maximum force from the tube
curvature can be calculated by equation
(15 )
where n
centrifugal overload [g]; V
projectile speed [m/sec]; g=9.81 m/sec
Earth gravity; R radius of
curvature [m]; L length of tube
[m]; h deviation of tube from
straight line [m].
For
example: if tube middle has deviation 0.1 m, length 1000 m and projectile speed
in the tube middle V=5000 m/sec, the centrifugal overload will be n=2g.
It is very small force in comparison with acceleration along tube equals about
1000g.
10.
Wall thickness
The tube wall thickness can be estimate the
following way. Take the tube length L=1
cm. The tensile force equals F =PDL,
where P is pressure, atm [kg/cm2], D is a tube internal diameter [cm], F tensile force [kg]. The cross-section area [cm2] of
the tube walls is s=F/s,
where s is admissible tensile stress [kg/cm2].
The wall thickness equals s/2.
For example, if the pressure is P=500
atm, the tube diameter is D=10 m =1000 cm, L=1cm,
then the force is F=PDL= 500 tons. If
the tube is made from a composite material having maximum s =500 kg/mm2 coefficient safety 5, and admissible tensile
stress s=100 kg/mm2, then the
wall thickness will be 25 cm.
If the tube is made from a steel
having maximum s=250 kg/mm2 , coefficient
safety 5, and admissible tensile stress s=50 kg/mm2 , then the wall thickness
will be 50 cm for tube diameter 10 m.
11.
Gas Friction
Let us to estimate a gas friction about wall. We use the method
described in [5].
Below, the equation from
(T*/T)
=1 + 0.032M2 + 0.58(Tw/T -1) ; m*=1.458x10-6T*1.5/(T*+110.4);
r* = rT/T*; Re*=r*Vx/m* ; Cf,l =
0.664/(Re*)0.5 ; Cf,t=0.0592/(Re*)0.2 ;
DL = 0.5Cf.lr*V2S ; DT = 0.5Cf.tr*V2S . xLr/3 . (16)
Where: T*, Re*, r*, m* are
reference (evaluated) temperature, Reynolds Number, air density, and air
viscosity respectively. M is Max
Number, V is speed, x is length of plate (distance from the beginning of the cable), T is flow temperature, Tw is
body temperature, Cf.l
is a local skin friction coefficient for laminar flow, Cf,t is a local skin friction coefficient for turbulent
flow. S is area of skin [m2]
of both plate sides, it means for the cable we must take 0.5S; D
is air drag (friction) [N]. Our gas column is moved with acceleration. It means
(from aerodynamic), that we have the laminar layer.
The result of computation of laminar friction are presented in
fig.13. The average friction is about 0.4 ton/m2 . The friction area decreases with increasing
speed. If average area is S=pDLr/2=3.14x10x15/2=235.5 m2
, the average wall drag is 94 tons. It is small value with comparison
the drive force 20.104.0.5=105 tons.
Fig.13.
Wall laminar drag of 1 m2 versus speed for gas pressure 200 600
atm.
Advantages of the offered
space launcher.
One
advantage of this launch system over opposed fixed gun systems is that of
placing the launch system in water. The installation becomes mobile and can be
aimed (by adjusting the azimuth) to any point in space or on the Earth. The
launch tube may be integrated into a watertight enclosed frame (Fig. 1c). Water
can then be forced into and out of the frame to control the position of the
launch tube like a submarine. Electric motors may be attached to the frame for
maneuverability and to maintain the frame in a desired location. The launch
tube is fixed inside the frame by control cables. The azimuth of the launch
tube, hence the aim, is obtained by sinking one end of the launch tube/frame.
The launch tube is brought into the horizontal position to transport the system
to another location or to conduct maintenance on the system.
Additional advantages, each of which is an important innovation,
that this launch system has over a conventional fixed gun system are
listed as follows:
1)
In a conventional gun system, powdered fuel is ignited all at once to move
the projectile. The velocity of the projectile cannot exceed the velocity of
the combustion gases, which is usually the speed of sound (at most 1 to 2
km/sec). In the proposed launch system, the mass of gas in the launch tube is
pushed along with the payload/piston, which later works as a rocket engine.
This rocket engine is different from conventional rocket engines since the
proposed engine:
a) Does not
have to move heavy tanks of compressed oxidizers;
b) Is not
limited by the speed of sound;
c) Is allowed to reach a very high ratio of fuel
mass (compressed air + fuel) to payload mass (up to 30 and more);
d) Allows the
payload to reach hypersonic speeds (up
2)
The vacuum (less 0.01 atm) above the payload/piston
eliminates a significant amount of air resistance, which could prevent the
payload from reaching high velocities when the launch tube is very long.
3) The launch tube can be long
(0.4 to 2 km), which is not conducive in ordinary gun systems. In addition, by placing the launch system in
water, the payload can be aimed to any point in space or on the Earth. A fixed
conventional gun system cannot be moved or aimed. The use of water also
allows the launch system to be hidden below the surface and in the event of an
accident, the high pressures under the water can help contain the explosive
materials and avoid injuring people.
4)
Conventional fuels and oxidizers (such as air or gaseous oxygen) can be used
with the proposed launch system. In regular rockets for example, gaseous oxygen
can not be used since the fuel tank weights would be too high and liquid oxygen
requires special equipment for storage and transportation.
The main advantage of the proposed launch system is a very low cost for
payload delivery into space and over long distances. Expensive fuels, complex
system control, expensive rockets, computers, and complex devices are not
required. The cost of payload delivery to space would drop by a factor a
thousand. In addition, large payloads
could be launched into space (on the order of thousands of tons a year) using a
single launch system. This launch system is simple and does not require
high-technology equipment. The launch system could easily be developed by any
non-industrialized country and the cost of this launch system is ten times
lower than that of contemporary rocket systems.
Computations show that if the launch tube is designed to a diameter of 2
- 7 m, a length of 1 - 4 km, and gas pressure of 200 - 800 atmospheres, then 3
20 tons of payload per launch could be deliver to
Earth orbit. If the launch frequency is 30 minutes, then 15 - 45 thousand tons
of payloads could be delivered to space per year at production costs of 2 - 10
dollars per kg.
During peace time, this launch system can be also used for the delivery
of mail or express parcels over long distances (for example, from one continent
to another) and give a profit of 5-10 millions dollars per day. The
Installation can be also used as energy storage. However, during war, this
launch system could deliver munitions to targets thousands to tens of thousands
of kilometers away. It is known, that 90% of loads delivered to orbit are items
that can endure high acceleration.
In this article I wanted to show the feasibility of the suggested
method. The parameters of the launcher
are not optimized. Selection of the tube diameter, rocket nozzle, and length of
rocket air column could possible increase the projectile speed, weight, and
decrease the tube length and fuel consumption. The hot gas after the shot may be
used for electricity production or as energy accumulator.
The author has more detailed research of this
concept and innovations which solve problems that could appear in development
and design of the suggested Launcher. The author is prepared to submit his research and to discuss the
problems with serious organizations wanting to research and develop this
project.
Discussion
The computations in this paper are not exact but only estimations to
show that the suggested simple gas tube rocket can reach a hypersonic speed 4-8
km/sec. Hypersonic speeds depend only on the right gas tube rocket
design and the right fuel injection program.
The estimates are with the conventional technical accuracy of 5-10%
except may be temperatures (and projectile speed) with high level of oxygen
(more 40%) when temperature exceeds the conventional combustion chamber
temperature 4000oK and the gas dissociation decreases the
temperature. However, the velocity of 5.5 7 km/sec (Fig.2), which is reached
before this temperature, is hypersonic. The more exact computation is complex
and need in finance support. There are some factors which allow an increasing
of the temperature in the combustion chamber and utilizing it in the tube
nozzle: (1) The heat is not spent on evaporation and heating of the liquid
oxygen from 90oK to 288oK (we use the gaseous oxygen);
(2) We use a very high pressure, which decreases the temperature dissociation
by approximately 500oK per every 100 atm pressure. (3) With the tube
nozzle the recombination time is high enough (0.5-1 sec). In conventional short
rocket nozzle the gas outflows in vacuum space and recombination time is not
enough (0.001 sec). The ionization for the given temperature is small.
There are also good margins for increasing the efficiency of the gas
tube rocket. Some metals (Be, Li, B, Mg, Al, Si) have
heat capability of 1.5 2 more then kerosene. However their application is
limited because in a high extension nozzle used in the conventional rocket
these metals condense from gas to liquid and decrease efficiency of a nozzle.
In our case we have a nozzle that has a small extension (or the rocket without
nozzle, as in this article). It means we can use a liquid fuel with a high
concentration of the metal powder and achieve the high efficiency (projectile
speed).
The
author has many the detailed computations of different gas tube rocket designs
and programs for the fuel injections. Organizations interest in these projects
can address to the author.
Patent
applications: 09/013,008 of 01/216/98; 09/344,235; 10/051,013.
References
1.
Gun
propulsion technology, 1998, 568 p., AIAA.
2.
Development
in high-speed-vehicle propulsion systems, AIAA, 1996.
3.
12th
Symposium Propulsion System, AIAA,1995.
4.
Space
Technology & Application. Internation. Forum,1996-1997,
5.
6.
Henderson
B.W., Livermore Proposes Light Gas Gun For Launch of
Small Payloads, Aviation Week & Space Technology/July 23, 1990, pgs.98-99.
7.
8. Wolkomir, R., Shooting right for the stars
with one gargantuan gas gun. Smithsontan, Vol.6, No.10; p. 84-91.
January, 1996.
9.
Tuler, V.M., "Gun barrel" launching,
Space/Aeronautics, February 1959, p.52-54.
10.
Seigel,
11.
Bull,
G.V., Murphy, C.H.,
12.
Bolonkin
A.A., Hypersonic Launcher, Actual Problem
of Aviation and Aerospace, KAI, No.1, 2003.
13.
Baer,
P.G., The Traveling Charge Gun as a Hypervelocity Launching Device, Ballistic
Research Laboratories, Maryland, 1960.
14.
Vest,
D.C., An Experimental Traveling Charge Gun, Ballistic Research Laboratory
report No. 773 (1951).
15.
Palmer,
M.R., "A revolution in Access to Space through Spinoffs
of SDI Technology", IEEE Transactions on Magnetics,
Vol.27, No.l, January 1991, p. 11-20.
16.
Palmer,
M.R., "Economics and Technology Issues for Gun Launch to Space",
Space Technology, 1996, part 3, p.697-702.
17.
Palmer,
M.R., and Dabiri A.E., Electromagnetic Space Launch,
IEEE Transactions on Magnetics, Vol.,No.1, January
1989, pp.393-399.
18.
Bolonkin
A.A., Hypersonic
Gas-Rocket Launcher of High Capacity, IBIS, Vol. 57, No. 5/6, 2003, pp. 162-172.
19.
Bolonkin A.A. ,
Hypersonic Launch System of Capasity up 500 tons per
day and Delivery Cost $1 per Lb, IAC-02-S.P.15, 53rd International Asronautical Congress. The World Space
Congress-2002, 10-19 Oct. 2002, Houston. Texas, USA.