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Introduction to Astronautics Sissejuhatus kosmonautikasse Vladislav Pustõnski 2009 Tallinn University of Technology.

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Presentation on theme: "Introduction to Astronautics Sissejuhatus kosmonautikasse Vladislav Pustõnski 2009 Tallinn University of Technology."— Presentation transcript:

1 Introduction to Astronautics Sissejuhatus kosmonautikasse Vladislav Pustõnski 2009 Tallinn University of Technology

2 2 Basics of the rocket propulsion Because of the law of momentum conservation, the only way a rocket may change its velocity in space is momentum exchange with its parts, i.e. the rocket should throw away some parts of itself in one direction to move the payload in the opposite direction. Exhaust gases play role of parts thrown away; they are accelerated by the rocket engine and ejected with a high speed in the direction opposite to the direction of the motion of the rocket. Let us derive the equation of the rocket motion. We may assume that the exhaust gases are ejected with a speed unaltered relative to the rocket’s body. This condition is generally met except for some cases which will be considered later (for example, motion in the atmosphere). Let the V e be effective exhaust velocity – the speed (relative to the rocket) the exhaust gases have after leaving the nozzle. In fact, it is not the real speed of gas, since gases at different zones of the nozzle move with slightly different velocities and also since external pressure (if present) leads to the difference between the true velocity of the gas and its effective velocity, as we shall see further. So, V e should be regarded as an effective value of the gas exhaust velocity. When a mass portion of dM is ejected, it carries away the momentum V e dM. The velocity V of the rocket will change by dV, so if the mass of the probe is M, we get from the law of momentum conservation Ideal rocket equation This expression should be integrated with the additional condition V = 0  M = M 0 : at the start, when the rocket is motionless, it has an initial mass M 0. The result of the integration will be as follows:

3 3 or, alternatively, This is the so-called ideal rocket equation, or Tsiolkovsky rocket equation. It is the principle formula for the rocket motion, the foundation stone of the rocketry. Using the ideal rocket equation, we may compute the characteristic velocity of the rocket with the effective exhaust velocity V e and mass ratio R = M 0 /M (M is the mass of the rocket when its velocity gain totals V. It is seen that this velocity gain does not depend on the force the exhaust gases exert on the rocket (i.e. on the thrust) and thus on its acceleration. That means that two rockets with identical effective exhaust velocities and R will have identical characteristic velocities, even if one of them possesses more powerful engine than another one. If V e is not constant, the general idea remains the same: we may divide the trajectory into pieces so that at each piece V e be constant, when calculate velocity gains for each piece (using the R valid at this piece) and finally sum up all the gains to get the final velocity. The very first look to the ideal rocket equation leads us to the conclusion that we need very high R to get a high final velocity. For most liquid propellants (LOX /kerosene, hypergolic propellants), V e  3200 m/s (may be some higher in upper stages). To put a rocket into a 200 km circular orbit, a characteristic velocity of about 9600 m/s is needed, that is ~ 3V e. To accelerate a rocket to 3V e, the R should have the value of exp(3)  20, and if we need a characteristic velocity of 5V e (for a mission to return of a lunar soil sample, for example), the R of ~ 150 is indispensable. However that means that the final mass of the rocket should constitute only a very small fraction of its initial mass: a huge rocket may accelerate only a tiny part of itself to a velocity of several units of V e. The largest fraction of the initial mass M 0 should be the propellant. Actually, if we use a one stage rocket, we cannot provide high R at all. The R of 150

4 4 means that if we need 150 tons of the lift-off mass to accelerate 1 ton to ~5V e. However, this 1 ton should include not only the payload, but also the mass of the rocket body with all its systems: tanks, engines, guidance system etc. But it is clear that it is impossible even to build a 1-ton tank that could hold 150 tons of the propellant, never mind the engines that could lift that mass from the ground. This leads us to the need to use multistage rockets, which will be discussed further. Lets us introduce some basic concepts regarding to the mass of a rocket. This mass may be divided into 3 parts: 1) the payload mass M d, i.e. the mass of the spacecraft; 2) the structural mass M s, i.e. the mass of the construction of rocket stage, including tanks, engines etc.; 3) the propellant mass M p, i.e. the mass of the expendable solids, liquids or gases which are used as working mass to accelerate the rocket. From these masses, several other concepts are deduced: the empty mass M e = M d + M s i.e. the mass of the rocket after depletion of the propellant; the full mass M 0 = M e + M p i.e. the total mass of the fuelled rocket stage. Some important coefficients are based on these masses: – payload ratio, the fraction that the payload makes up in the mass of the fuelled stage without the payload. High values mean that large payloads may be lofted with a moderate rocket stage; – structure coefficient, small values mean that a moderate rocket stage is capable to hold much propellant. Smaller values are highly desirable, since a small structural mass means that more payload can be taken for the same R. From this formula it is seen the importance of decreasing . It is seen also that it is impossible to use high values of, since they would mean a useless R ~ 1. R may be deduced from these coefficients as

5 5 Multistage rockets The mass ratio of a rocket cannot exceed the value of 1/ , even if there is no payload onboard. But there are technical limits which do not allow to infinitely minimize  : It is impossible to allocate many tons of propellant in a very lightweight tank, it is also impossible to attach very light engines since the rocket should have enough thrust to be lofted from the ground. Every gram of additional structural mass takes one gram from the mass of the payload of this stage, since the structure of the stage is accelerated to the same velocity as the payload. For the V-2 ballistic missile   0.3, that is a huge value. Now launch vehicles possess much smaller . For example, the 1 st stage of the Saturn-V moon rocket had   0.06. The record was the first Atlas main stage with   0.02. Generally,  strongly depends on the density of the propellant: more dense propellants need smaller tanks. This is why low density is one of the highest disadvantages of the LOX/LH2 propellant: possessing high specific impulse (see further), this combination has very low density, only ~0.28 kg/m 3, and needs large tanks. However, solid fuel motors need thick casings, so in spite of the high density of the propellant, their structure coefficients are large as well. For example, for the Shuttle SRB   0.14. In the case of the Atlas, it had very light balloon tanks, i.e. with no internal framework inside; the tanks did not collapse due to the constant pressure inside (so the tanks were inflated like balloons). – (propellant) mass fraction, determines the portion that the propellant makes up in the total initial mass. The following relation with R is valid: MF = 1-1/R.

6 6 The value of   0.06 is close to the feasible minimum. That would mean the maximum R  17, and from the ideal rocket equation we get the maximum velocity that can be reached, it is ~2.8V e. For the LOX/kerosene propellant with V e  3400 m/s it would mean the maximum final velocity of only 9500 m/s, and that is without any payload onboard. This velocity is close to that needed to launch a satellite to LEO (launch losses included). So, using a one-stage rocket hardly will we be able to put anything useful to LEO: we may hope only put the structure of the rocket itself, with no payload at all or a minimum payload. Never mind higher orbits or interplanetary probes: they would be technically unfeasible. There is probably only one example in the history of Astronautics when a one-stage rocket was used for orbital launches, although it was not a pure one-stage rocket. The Atlas launch vehicle (launch mass ~120 tons) based on the Atlas ICBM had 3 engines at the launch, but two engines (with the total mass of ~3 tons, which made up about one half of the mass of the stage) were discarded in-flight. It was done not only to lighten the structure, but also to decrease g-loads on the rocket, since its mass significantly decreases by the end of the flight. Discharge of the engines made it possible to deliver a payload of ~3 tons to LEO. So, it is the mass of the rocket structure what impede to the rocket to achieve higher final velocities. Although it is not possible to reduce the lift-off mass of the structure (below the level the state of technology permits), it is possible – and necessary – to reduce it in-flight. We may jettison parts that have become unusable and not to carry them along, wasting propellant to their acceleration. In this way we reduce the useless weight accelerated to the final velocity (in general, to high velocities) and thus we may replace it by a useful payload.

7 7 What are the parts that may be discarded in-flight? First of all, these are some tanks. During the flight the amount of the propellant reduces and tanks became nearly empty. So the propellant for the first part of the flight may be held in separate tanks which may be discarded when the propellant inside is depleted. Since tanks make up most of the mass of a rocket, in this way we will get rid of a significant portion of the useless mass. For instance, the Soviet lunar probes E-8 (Luna-16, Luna-17/Lunokhod and others) discarded some tanks at lunar landing. Another thing that may be discarded is some engines (that is the case of the Atlas). Their discharge is useful also because during the flight the mass of the rocket decreases, and if the thrust remains constant, the acceleration increases. So the construction of the rocket should be stronger and thus heavier to bear the increased g-loads. It is much simpler to jettison some engines (or discard all of them and replace them by engines with smaller thrust) when to create deeply throttable engines. In addition, engines for the second part of the flight work mostly in vacuum and may be designed for more effective work than the engines working in the atmosphere (to have higher expansion ratio of the nozzle). Summing up, the rocket may be divided into stages (two or more) which work in sequence or simultaneously and are discarded after the propellant inside their tanks is depleted, leaving the remaining rocket on a sub-orbital trajectory. Each stage may have its own tanks and engines. Sometimes parts smaller than an entire stage are discarded, for instance, stand-alone engines (like on the Atlas) or tanks (like on the Space Shuttle). Only the upper stage makes it to orbit together with the payload. Sometimes, if the payload have its own engine and propellant (being intrinsically an independent rocket stage), the the upper stage may even leave the payload on a sub-orbital trajectory, and the final orbital injection is performed by the

8 8 engines of the payload. This is done to additionally increase the mass of the payload and to avoid pollution of the space by debris (what is the upper stage after it has done its work and jettisones). The rocket composed of several stages is called multistage rocket. Let us analyze an example of how useful a multistage rocket may be. Assume a one-stage rocket of total mass 100 tons, let it payload be 2 tons, and the exhaust velocity V e = 3200 m/s. Let the structure coefficient be  = 0.08. So, the mass of the rocket without payload is 100 - 2 = 98 tons, from which 98 · 0.08 = 7.8 tons is the structure and 98 - 7.8 = 90.2 tons is the fuel. The mass ratio of this rocket is R  100/(7.8+2) = 10.2, so the final velocity is V  3200 · ln(10.2)  7400 m/s. So, such rocket will never make it to orbit, since a characteristic velocity of ~ 9600 m/s is necessary to launch a satellite to LEO. Let us divide the rocket into 2 stages working in sequence, the second having the total mass (with the payload) of M 02 = 30 tons and the first having the total mass of M 01 = 70 tons. Assume the structural coefficients of the stages is the same,  1 =  2 = 0.08, and V e1 = 3200 m/s, V e2 = 3400 m/s (the exhaust velocity of the upper stage may be higher due to the larger nozzle extension). We will get for the first stage: M 1d = 30 tons, M 1s = 70 · 0.08 = 5.6 tons, R 1  100/(30 + 5.6) = 2.8, V 1  3200 · ln(2.8)  3300 m/s. For the second stage: M 2d = 2 tons, M 2s = (30-2) · 0.08 = 2.2 tons, R 2  30/(2 + 2.2) = 7.1, V 2  3400 · ln(7.1)  6700 m/s. The total final velocity of the rocket will be V = V 1 + V 2  10 000 m/s, so the payload will be delivered to the orbit (and not very low). Extra stage gave the velocity gain of more than 2.5 km/s. The total mass ratio decreases to 100/(2+2.2)  23.8. If we divide the rocket into 3 stages of masses 10, 20 and 70 tons and repeat the analogical calculation (assuming V e1 = 3200 m/s, V e2 = V e3 = 3400 m/s ), we will get the final velocity of V  10 900 m/s. So, additional ~900 m/s have been gained. The total mass ratio is 35.7. It may seem that increasing the number of the stages, it is possible to get higher and higher

9 9 final velocities. But it is not so due to technical limitations. The size of a stage cannot be smaller than some limit. The mass of the engines correlates with their thrust, and if we have small stages with tiny and very light engines, their thrust will be small, launch accelerations will be small and gravity losses will increase very much. The mass of other systems has also some limit: these are the guidance system, the steering hardware, electrical equipment, hydraulics etc. There are also the problems of reliability and of cost. Each stage is actually is a standalone rocket, and although smaller stages are cheaper than larger ones, two small stages are generally more expensive than one stage of the same total mass, since the number of systems (engines, steering devices etc.) in two stages is larges. Also, the reliability of the rocket drops with raise of the number of its systems, so the rocket with 4 stages is generally less reliable than that with 2 stages. Finally, the total R is limited by the mass of the payload M d (for a given launch mass M 0 : R < M 0 /M d ), the velocity of the rocket will be limited in any case (in our example by ~ 3300 · ln(100/2)  12 300 m/s); so each additional stage will give smaller and smaller gain, until this gain becomes minuscule. The table below illustrates the masses which a rocket with the launch mass M 0 = 100 tons and identical V e1 = 3200 m/s and   0.06 for all stages may deliver to LEO with V char  9600 m/s (actually the number in the second row divided by M 0 is the payload coefficient for the rocket as whole). Number of stages n 123456 Payload, tons- It is seen that the 4 th and higher stages give nearly no gain in the payload mass. Actually, since their size cannot be very small, they will only add hardware and so the payload mass may even decrease, never mind reliability and the raise of costs. So, generally an optimal number

10 10 of stages exists. This optimum depends on many parameters: on the required characteristic velocity, on the exhaust velocity, on the cost and reliability of the systems. If we assume the structure ratios are equal for all stages as well as the exhaust velocities, the following approximation may be used in a wide range of  (0 <  < 0.4) : V char /V e Optimum number n 1234567 We see that while V char 4.6 km/s and are performed in two stages (Luna-16). The Mars escape velocity is ~ 5 km/s, so obviously two stages are needed to start from the Mars to the Earth (additional delta-v is necessary to compensate losses and perform corrections). Even a launch to the Martian orbit from the surface may be better to perform in 2 stages, if we consider the launch losses. LOX/LH2 propellant has V e  4300 m/s, making possible to fly missions with V char  6.0 km/s in one stage. This number should be rounded; rounding down is often preferred to increase reliability and to decrease costs. The optimum number n for different limiting characteristic velocities V char of the mission is tabulated below (   0.06, all data are approximate).

11 11 A launch to LEO needs V char  9.6 km/s, so if LOX/kerosene or hyperbolic propellants are used in all stages, the optimum number of stages is 2 or 3. If LOX/LH2 is used at least in one of the stages, the optimum number is 2. This is the reason why many launch vehicles have 2 or 3 stages. If the launch vehicle is used for more energetic missions, like delivering satellites to GTO or sending space probes to planets, the optimum number increases to 3 or even 4 stages. For example, Russian Soyuz and Proton launch vehicles fly in 3-stage variants to LEO but the 4 th stage is added for delivering satellites to higher orbits (Soyuz changes its name to Molniya, and the name of the upper stage is added to the name of Proton). If solid propellant is used (with V e  2400 m/s) in all stages, the optimum number of stages is 4. That is the case of the first Japanese launch vehicles Lambda-4S and Mu-3 (actually these rockets included also boosters that may be considered the 5 th stage; they worked in parallel with the 1 st stage for a very brief time and then were jettisoned). However, since rocket optimization does not always mean the highest payload coefficient (so the optimization criterion presented here is insufficient, see further), the actual number of stages may differ from what follows from the table. This is also true for the case of the parallel staging, the case when one of the stages has much smaller V char than the others or/and than structure coefficients are very different from 0.06. Actually, in the case of parallel staging, when two stages work simultaneously, it is sometimes problematic to define the number of stages in this way, specially if V char of some stages is much smaller than that of the others. Such stages are called strap-on boosters and sometimes they are not regarded as an authentic stage; designations like “stage 0” or “2.5-stage rocket” can be met. For example, Delta II, Atlas V and other US rockets are frequently launched with varying number of solid rocket boosters working in parallel with their first stages. These boosters have high thrust and help the heavy rocket to start, and they are quickly discarded. Usually they are not regarded as

12 12 a stage. The same situation is with the N2O4/UDMH liquid propellant boosters of CZ-3B/3C Chinese launch vehicle, although the working time of these boosters is not much shorter than that of the first stage. Sometimes strap-on boosters do not separate simultaneously but fall apart in groups, as it happens with Delta II in its versions with 9 boosters. The Space Shuttle SRB are named stage 0 because of their short working time, although they constitute more than one half of the lift-off mass. However, liquid propellant boosters of the Soyuz launch vehicle are counted as the separate stage, since they burn for a quite long time. Ideal velocity of a multistage rocket Ideal velocity V ideal of a rocket is the velocity to which can accelerate its payload in empty space. To compute ideal velocity of a multistage rocket where stages work in sequence (serial staging) the ideal velocity is simply the sum of characteristic velocities of all its stages: V char,i is the characteristic velocity of the stage #i. If stages work in parallel, calculations are more complicated. Some of the parallel stages are jettisoned earlier than others (for example, strap-on boosters) and to compute the mass of the rocket at the moment of their separation it is necessary to compute also the amount of the propellant consumed by the stages which continue in-flight and add this amount to the propellant consumed by the stages to discard. So the rate of propellant consumption is required for such calculation. The situation is simpler if the stages working in parallel have identical V e ; in this case we may use the ideal rocket equation with the corresponding mass ratio. However, if V e are not identical, calculations become even more complicated, since different stages working in parallel have different characteristic velocities, and we should use a value that is some mean of these velocities in the ideal rocket equation. For the case of

13 13 If a rocket has more than one stage, a question about distribution of characteristic velocities among the stages arises. It can be shown that if the exhaust velocities V e are the same for all stages, the optimum distribution occurs when payload ratios of the stages are equal. If structure coefficients  are equal as well, this optimum means that mass ratios R of the stages are are also equal. In the more general case, if structure coefficients are not equal, the fulfillment of the following condition optimizes the mass of a multistage rocket, providing minimum lift-off weight for a given mass of the payload: Optimization of multistage rockets. Rocket families i, j – stage numbers. So, the payload ratios should be proportional to the structure coefficients. If the exhaust velocities are also different, the expressions for the optimization criterion are more complicated. Actually, the most important parameter that should be optimized for each rocket is the cost. In the previous analysis it was assumed that from two rockets delivering the same payload to the orbit the lighter one is the cheapest. However, it should not necessarily be true. A heavier, but simpler rocket (or a more reliable one, since it has less chances to cause a mission loss) may be cheaper than a lighter, but more complex or less reliable rocket. The same is true for a rocket built from standardized components: it is cheaper than a stand-alone launch vehicle. V e1 & V e2 are the effective exhaust velocities of the 1 st and 2 nd stages,  = (0  1) is the ratio of the propellant burning intensity by the first stage to the overall propellant burning intensity. Analogical formula may be obtained for arbitrary number of stages. two stages it may be shown that the analog of the ideal rocket equation is as follows:

14 14 Standard components enable to reduce production costs of the unit. Thus at the present time many launch vehicles are built from standard components: core stages, upper stages, engines, strap-on boosters etc. So a rocket family forms, its members have many identical blocks. Individual members are not optimized by the weight of the payload, and even the cost of an individual launch vehicle may not be the optimal. But the production costs of the whole family are optimized due to the large number of produced items, and the overall reliability also increases since each component flies frequently and initial defects quickly become apparent and are removed. The number of stages and their individual characteristic velocities in a representative of such rocket family are defined by the structure of the family. Let us study different versions of the Delta IV launch vehicle as an example. The basic version is Delta IV Medium, it consists of a Common Buster Core (CBC) as the first stage (LOX/LH2 propellant is used); the second stage also burns LOX/LH2 propellant, actually it is a Delta III second stage with stretched tanks. This version delivers ~4.2 tons of payload to GTO. Two GEM 40 solid rocket strap-on boosters may be added to increase the GTO payload to ~5.8 tons; this version is Delta IV Medium+ (4,2). The upper stage may be replaced by a larger variant with a larger interstage and firing (5 m instead of 4 m in diameter). This version, Delta IV Medium+ (5,2), may deliver larger payloads with masses up to 4.6 tons to GTO. Two additional GEM 40 boosters may be added, so the version Delta IV Medium+ (5,4) with 4 strap-on boosters delivers ~6.6 tons to GTO. Additional boosters increase the total thrust at the lift-off, thus decreasing gravity losses. The largest member of the family, Delta IV Heavy, consists of one central CBC and two additional strap-on CBCs working in parallel with the central core, they are separated earlier than the central core. This version delivers more than 13 tons to GTO. So, the smallest and the largest members in the family differ more than 3 times in their payload capacity. Various fairings are available for different

15 15 Thrust of the rocket engine. Specific impulse Assume a static firing of a rocket engine. Let the engine consume propellant with a constant mass flow rate of q = dm/dt, and the mean (constant) velocity of exhaust gases is V mean. A mass portion dm leaving the engine carries away the momentum of dp = dm · V mean. So, the reaction force which exerts on the engine is F react = dp/dt = qV mean. There is another force that exerts on the engine, it is related to the pressure. Let the ambient pressure be p, the static pressure of the exhaust gases at the nozzle exit be p a and the nozzle exit section be S a. The static pressure of the exhaust gases exerts the force of p a S a on the engine, this force pushes in the same direction as the reaction force. The ambient pressure exerts the force pS a in the opposite direction. So, the total force related to the pressure is F press = S a (p a - p), it acts in the same direction as the reaction force. Thus, the total force which is applied to the rocket engine (named atmospheric thrust or simply thrust) is members. The family may be expanded by adding more boosters to the Medium versions, although it may need some time to develop since loads that the booster experiences change. It is seen that this force is larger when the ambient pressure is smaller (contrary to a popular idea that the ambient pressure “helps” the rocket to fly). So, as a launch vehicle raises from the ground and the ambient atmospheric pressure drops with the height, the thrust of its engines increases. In vacuum, This is the vacuum thrust. In most cases the reaction component of the thrust is much higher than the pressure component. The difference

16 16 between the thrust of an engine at the sea level and in vacuum depends on the nozzle exit section, nozzle expansion ratio and the propellant; typical values are 10% - 15%. The pressure component is typically included into the concept of the effective exhaust velocity V e, which is defined as T/q: Historically, the efficiency of a rocket engine was measured by the thrust which the engine produces per weight flow rate – the amount of propellant consumed per unit time. This amount was calculated in weight units (the normal weight is meant, i.e. the mass m has the normal weight of W = mg 0, with g 0 = 9.80665 m/s 2 – the standard gravity). This efficiency parameter is called specific impulse I sp. By definition, So, V e depends on the ambient pressure, increasing from the minimum value at the lift-off to the maximum value in vacuum. By definition, the thrust of the engine is T = qV e. Thus, the specific impulse is directly proportional to the effective exhaust velocity V e, the standard gravity g 0 being the proportionality factor: V e = I sp /g 0. It is obvious that the units for the specific impulse are [(m/s)/(m/s 2 )] = [s]. If I sp and the thrust T given in weight units are known, the mass flow rate may be found as q = (T/g 0 )/I sp. (Sometimes the specific impulse is defined otherwise, namely as the engine thrust per mass flow rate. The specific impulse by this definition is larger than the common specific impulse by the factor of g 0 and is measured in m/s. It is clear that defined so, the specific impulse is identical with the effective exhaust velocity V e ). I sp (in seconds) is a very common characteristic of rocket engines, and in aerospace engineering literature it is usually met instead of V e. So, the ideal rocket equation may be

17 17 written using I sp : Specific impulse highly depends on the propellant. Typical vacuum values for chemical propellants in use (close to achievable maxima) are illustrated in the following table. Sea level values are typically 10% - 20% lower, depending on the propellant (LOX/LH2 tend to have larger losses in the atmosphere). In the same table mean densities of the propellants are given (may vary depending on the oxidizer to fuel ratio). The two propellants in the end of the table PropellantI sp, s density, kg/m 3 Solid2701350 LOX/Kerosene3501020 N 2 O 4 /UDMH (hypergolic) 3301180 LOX/LH2450280 LOX/LCH4370420 LF2/LH2470460 (in italics) have never been used: LF2 is very reactive and toxic and thus not safe, LCH4 is better than kerosene, but is cryogenic and have lower density. I sp of other types of engines may be much higher. For example, a nuclear engine may have I sp > 1000 s (such engines have not been used up to now since they are not very safe), electric propulsion (ion thrusters) demonstrate I sp > 5000 s (but they have very low thrust and work in vacuum only). These issues will be discussed in the further lectures. As it is seen, the importance of the specific impulse is the same as that of the effective exhaust velocity. Generally, every second of the specific impulse is valuable. If the I sp of an LOX/kerosene engine is increased by, say, 5 s, the thrust will raise by 1.5 % and the payload

18 18 mass may increase by several %. To understand what factors define exhaust velocities of chemical propellants, let us recall the expression for molecule velocities of heated gas (mean velocity, root mean square velocity etc): In this expression T is the gas temperature,  is the mean molar weight, const is a constant depending on what mean velocity is calculated. Molecules of the heated gas inside the thrust chamber have some mean velocities and leave the nozzle with the speed which is directly related to this mean. The value of the corresponding factor is related to the properties of the gas (mainly to the adiabatic exponent ), on the pressure inside the chamber, on the nozzle extension and on the ambient pressure. But in most cases this dependence is of the second order of magnitude, and the general dependence on the temperature and the molar weight keeps. So, to increase I sp, it is necessary to increase the temperature inside the thrust chamber and to decrease the molar weight. However, the temperature may be raised only till the limits set by the cooling issues: if the temperature is too high, it may become impossible to cool the walls of the chamber and the nozzle. The temperature is also limited by the chemical energy of the propellant per unit mass. The inverse dependence on  makes it necessary to decrease the molar weight of the exhaust gases. Another reason to decrease  is that the more precise expression V e includes dependence on degrees of freedom of the molecules: more complex molecules keep energy in the internal degrees of freedom which are useless for their velocity. That is the reason why often the oxidizer to fuel ratio in the thrust chamber is not stoichiometric, i.e. there is usually an excess of fuel (fuel-rich mixture) and not all molecules of the fuel participate in chemical reactions. Although this leads to incomplete burning of the

19 19 components, the resulting  of the exhaust is lower than it would be in presence of stoichiometry. For instance, LOX/LH2 is usually burnt with the mixture ratio of about 1:5.5, and the exhaust contains 7 molecules of H 2 per 4 molecules of H2O: so its mean molar weight is ~8 g/mol (instead of 18 g/mol for H 2 O exhaust, if all the molecules reacted). The specific impulse, being one of the most important efficiency factors, is not the only such factor. Some other factors should also be taken into account, and the density of the propellant is one of them. For example, LOX/LH2 is one of the best propellants due to its high I sp. But the density of this propellant is extremely low (because LH2 has very low density), and this means that huge tanks are needed to held a certain amount of the propellant. But larger tanks mean large structure mass, so the structure coefficient will be large. Specially large would be the tanks of the first stage, since its mass totals about 70% of the lift-off mass or more. This is the reason why this propellant is generally used on the upper stages, where the mass of the propellant is not very high (Delta IV is the recent exception, it burns LOX/LH2 in the first stage as well). Another example is LOX/LCH4, that might be an alternative to LOX/kerosene, since its I sp is markedly higher. But its density is much smaller, so it would need larger tanks and thus the efficiency of higher I sp would be partially or totally equilibrated by these disadvantages. In addition, CH 4 is cryogenic and thus it is less convenient to treat than kerosene. Other factors are also important. For example, solid rocket motors are generally much cheaper and reliable than liquid propellant engines (they contain much less parts, much less moving parts and nearly no technically complex devices like turbopumps, sophisticated

20 20 hydraulics etc.). It is also much simpler to scale them up than liquid propellant engines (for example, the thrust of the Space Shuttle SRB is nearly as twice as high than the thrust of F-1 engine, one of the 5 powerplants of the Saturn-V rocket, but is was much simpler to develop by scaling up Titan strap-on boosters). This is the reason why solid motors are widely used as strap-on boosters: they provide high thrust at the lift-off and help the launch vehicle to overcome the gravity force (many rockets, like Space Shuttle or Arian, even would not start without these boosters due to insufficient thrust-to-weight ratio, see further). Their low I sp is not a great issue on the first stage since it is a “cheap” increase of mass. But in the upper stages solid propellants are rarely used due to underperformance of such motors and thus the necessity to use more massive solid engines: the increased mass of the upper stages would make it necessary to significantly increase the mass of the first stages and thus of the whole rocket. The exception are whole-solid light rockets which are sometimes built thanks to their simplistic design; the examples are the Israeli Shavit rockets and the first Japanese rockets the Lambda-4S and the Mu series (now Japan uses the LOX/LH2-based H-II rocket with solid rocket boosters). Sometimes solid propellants are used on kick stages of launch vehicles which mostly fly without them, but need an additional stage to send a satellite or a spaceprobe to an exceptionally energetic trajectory. The example is the Star 48 kick motor used to send the New Horizons space probe to the Pluto atop the Atlas V launch vehicle and the Star 27 kick motor used on the Titan-Centaurs for sending the Voyagers to the giant planets. A liquid- propellant stage would not be so small and would not have enough thrust, so a solid motor is more appropriate in such occasions.

21 21 Thrust. Thrust-to-weight ratio. G-force The ideal rocket equation does not include thrust of the engine. That means that two rockets with identical mass ratios and specific impulses would achieve identical velocities in vacuum independently of their thrust (since their characteristic velocities are equal). But in actual space maneuvers thrust does matter. We have already seen that transfer maneuvers (like the Hohmann transfer) are performed most effectively if the burn times are as short as possible, so that the propellant could be consumed exactly in the place where the efficiency is provided (for example, near the planet, if the Oberth’s effect does matter). There are also other situations when the thrust is very important. One of such cases is a lift-off from the surface of a planet. To keep the rocket aloft, the thrust should be larger than the weight of the rocket: otherwise the thrust will not overcome the gravity force and the rocket will not start. The difference should not be small as well. If it is small, the rocket’s acceleration will be very small, and the propellant will be expended in vain (these are so-called gravity losses which will be discussed later). So it is better to have a higher thrust at the lift-off in order to accelerate the rocket quicker and not to spend much propellant to overcome the gravity force. The ratio of the thrust (expressed in weight units) and the lift-off weight of the launch vehicle is called thrust-to-weight ratio (T/W). However, it is wrong to think that there is sense to increase the T/W ratio to very high values. If it is high, the velocity of the rocket raises too quickly, and soon the aerodynamic forces (proportional to the velocity square) become very large. This leads not only to

22 22 increment of the aerodynamic loads to the construction (mainly to the fairing), but these forces also decelerate the rocket (these are so-called drag losses which will be also discussed later). So a compromise between the gravity losses and the drag losses is needed. Sometimes launch vehicles even throttle their engines back after launch, when the aerodynamic forces raise to significant values. This permits to reduce drag losses. When the rocket raises higher and the density of the ambient air drops, the engines are throttled up again. The example is the Space Shuttle: its main engines are throttled back to 65% near the 26 th second after the lift-off and are throttled up to the full thrust near the 60 th second, when the area of maximum aerodynamic pressure is passed. There is another reason why extreme T/W ratios should be avoided. In flight, while the propellant burns and leaves the rocket, the mass of the rocket decreases. Since the thrust remains more or less constant (it slightly raises as the rocket leaves the dense atmospheric layers), the total acceleration of the rocket increases. This means the raise of the g-force – the force that the rocket experiences due to acceleration. G-force is expressed in units of the normal weight as a being the net acceleration of the rocket, g being the gravitational acceleration at the place where the rocket is, g 0 = 9.80665 m/s 2 being the standard acceleration. In the absence of other forces except for the gravity and the thrust, the g-force is equal to the ratio T/m, T being the thrust and m being the instantaneous mass of the rocket. By the end of the time of work of the stage, the mass of the rocket decreases several times (since the propellant of a stage is ordinarily several times heavier that the empty stage and the

23 23 next stages with the payload). So the increasing g-force may destroy the launch vehicle. Thus the construction should be stronger and heavier to bear additional g-loads. An excessive structure weight decreases the performance. In the case of launch vehicles for manned spacecraft, excessive acceleration is of special consideration since the human body cannot withstand high g-forces. Sometimes it is even makes sense to decrease the thrust at the end of the working time of stage. However, it is usually difficult and impractical to build a deeply throttled engine. But if there are several motors on the stage, one or more of them may be cut off earlier than the rest of the motors. This will lower the thrust and the g-force will grow slower. The Saturn-V is an example of such solution. It had 5 F-1 engines on the first stage, one central and 4 outboard. The central engine cut-off occurred around the 135 th second, while the outboard engines continued working for about 25 seconds more. G-force is the second reason why very high T/W ratios are generally not used. Common values for the T/W ratios are 1.3-1.8 at the lift-off. It may be demonstrated that these are optimum values. But there are also few exceptions. One of them are launch vehicles built as a result of ICBM conversion. ICBMs should start quickly to overcome the ballistic missile defense, so launch vehicles on their base often have high T/W ratio. Dnepr (converted from R- 36M2 ICBM) is an example. Another exception are many full solid rockets. Their motor casings are strong enough to withstand high g-loads and aerodynamic loads, and high drag may not be a great issue if quick start permits to decrease gravity losses significantly. The Japanese Mu-3 series had an extreme T/W ratios of ~3.5, the Lambda-4S even more (about 6). In space T/W ratio is generally limited by the g-force and such technical issue as the size

24 24 the engine. On one hand, it is useful to have higher thrust to perform maneuvers quickly, on the other hand the thrust of the engine is proportional to its mass, so to provide higher thrust, it is necessary to carry a heavy engine. A compromise is needed. Thus, the size of the engine for a spacecraft is generally defined by the maneuvers the spacecraft should perform. If rapid and energetic maneuvers are needed, the spacecraft is usually provided with a large engine with high thrust. If such maneuvers should be many, a liquid-propellant engine is generally chosen since solid rocket engines are quite difficult to restart, they usually (but not always!) burn until depletion of the propellant. For a single maneuver, a solid engine may be used due to its simplicity, reliability and high thrust; that was the case of the Mercury spacecraft retrorockets and the Surveyor lunar probes. If the rapidity of the maneuver is not the issue (for example, station keeping or orbital corrections during an interplanetary transfer), low thrust engines may be preferred. Electric propulsion is often chosen for such purposes now. Having very low T/W ratio, it possesses an extraordinary performance because of its very high I sp. A soft-landing to the Moon (or a planet where the thrust is used to cancel the velocity) is a special issue. The orbital velocity should be cancelled not very far from the surface of the planet (to provide efficiency according to the Oberth’s effect). Thus, the maneuver should be performed quickly, so that the spacecraft does not collide with the planet. That means that the engine should have high thrust (and the T/W ratio should be ~1 for the case of the Moon). But when the probe is near the surface, it should maneuver carefully and decrease its velocity slowly, so that the velocity changes were not very quick and the probe could gently land. However, the thrust of a powerful engine cannot be controlled with sufficient accuracy, and small control interventions would lead to large changes of the acceleration. The situation is

25 25 similar to the task to stop a quickly moving car near a wall: it is impossible to stop it by an instantaneous braking in the vicinity of the wall, one should gently decrease the speed and to come closer at a low velocity. If, for example, a lunar probe has a mass of 4 tons on the lunar orbit, it will have a mass of ~2.2 tons at landing. If its engine has a nominal thrust of 4 tons, it will accelerate the probe to ~4·9.81/2.2  18 m/s 2 near the lunar surface. However, the lunar gravity acceleration is only ~1.6 m/s 2 near the surface, thus the acceleration of the probe will be more than 10 times greater than the lunar acceleration. To be capable to move and maneuver slowly near the lunar surface, the probe should have a thrust of 4·1.6/18  0.36 tones, i.e. much less than its thrust during the braking. However, it is a difficult task to build an engine throttable to that extent (the rare exception is the engine of the Descent Stage of the Apollo Lunar Module, which had the nominal thrust of ~4.8 tons and was throttable back to 10% of the nominal). Thus, two separate engines may be used for landing. The first, high- thrust engine cancels the orbital velocity and is cut of, and then the second, low-thrust engine is ignited to perform the soft landing. Such method was applied on the Surveyors which cancelled the orbital velocity by a solid rocket motor. The motor casing was then was jettisoned, and the further landing was performed with the aid of 3 liquid-propellant verniers. The same method was used on the Luna-16 and its successors: the main liquid-propellant engine was cut off at the height of ~20 m and the landing was performed with the aid of low- thrust nozzles.

26 26 End of the Lecture 8

27 Atlas launch vehicle Launch of Atlas with a Mercury spacecraft (By source)source Launch of Atlas-Able (By source)source

28 Delta IV launch vehicles Delta II, Delta III and Delta IV (By source)source Delta IV Heavy on the launch pad (By source) source

29 p – ambient pressure; p a – pressure at the nozzle exit; S a – nozzle exit section (By V.I.Feodosjev, Osnovy tehniki raketnogo poleta.) Thrust related to pressure

30 Saturn-V Saturn-V at the launch pad. Note that the second stage (LOX/LH2) has the same diameter as the first stage (LOX/RP-1) and not much shorter than the first stage. However, it is ~4 times lighter than the first stage. (By source)source

31 Lambda-4S and Shavit Japanese Lambda-4S full-solid rocket on its ramp. (By source)source Launch of Israeli full- solid rocket Shavit. (By source)source

32 Star 48 solid motor Star 48 solid motor used as a kick stage to launch the New Horizons probe to the Pluto. (By source)source

33 G-force for Saturn-V G-force and acceleration of the Apollo 8 Saturn V launch vehicle during the orbital insertion 1 – first stage ignition, lift-off; 2 – first stage central engine cut- off; 3 – first stage outboard engines cut-off; 4 – second stage ignition; 5 – oxidizer to fuel ratio change; 6 – second stage engines cut-off; 7 – third stage ignition; 8 – third stage cut-off. (By source)source

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