Article Hypersonic tube

Gas Tube Hypersonic Launcher*

Alexander Bolonkin

1310 Avenue R, #6-F, Brooklyn, NY 11229, USA

Tel/Fax 718-339-4563, aBolonkin@juno.com, http://Bolonkin.narod.ru

Abstract

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.

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* This Chapter as paper presented at the 38th AIAA Propulsion Conference, 7-10 July 2002, Indianapolis, USA (AIAA-2002-3927) and the World Space Congress, 10-19 Oct. 2002, Houston, USA (IAC-02-S.P.15). Detail material is published as A.A.Bolonkin, Hypersonic Gas-Rocket Launcher of High Capacity, JBIS, vol.57, No. 5/6, 2004, pp.162-172; Journal Actual Problems of Aviation and Aerospace Systems, Kazan, 8 (15), pp.45-58, 2003.

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];

Po - initial pressure of gas in tube.

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). Iraq made the most recent attempt to create a big cannon, which was destroyed by the UN.

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, Maryland. The objective of their paper [13] is to show that it is possible to build a device capable of accelerating small projectiles (weight range of 1 to 5 grams) to velocities in excess of 20,000 f/s (6 km/s) without subjecting the projectile to severe acceleration force. Their result is as follows.

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.

 

Description

 

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.

Methods of Estimation and Results of Computation

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 Po=1000 atm and the 40% gaseous oxygen we can reach the shot velocity about 10 km/sec for the tube length of 700 m (fig.4). This length is acceptable if we place the tube in water. We can essential decrease the tube length if we decrease the rocket air column, but the projectile speed is decreased. For example, if at 500 atm we decrease air column 15 to 10 m, the tube length decreases 650 to 430 m (fig.5), speed 7.2 to 6.45 km/sec (fig.3).

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 Anderson for computation of local air friction for a plate is given.

(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 7 to 10 km/sec).

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.

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