Rocket performance. Chapter Thrust

Transcription

1 Chapter 7 Rocket performance 7.1 Thrust Figure 7.1 shows a sketch of a rocket in a test stand. The rocket produces thrust, T,by epelling propellant mass from a thrust chamber with a nozzle. The test stand applies an opposite force on the rocket holding it at rest. The propellant (fuel+oidizer) mass flow rate is ṁ and the ambient pressure of the surrounding air is P 0. Figure 7.1: Rocket thrust schematic. Other quantities defined in Figure 7.1 are as follows. A s outside surface of the vehicle eposed to P 0 A c inside surface of the combustion chamber A e nozzle eit area ˆn outward unit normal P e area averaged eit gas pressure e area averaged eit gas density U e area averaged component of velocity at the nozzle eit (7.1) 7-1

2 CHAPTER 7. ROCKET PERFORMANCE 7- The vehicle is at rest and so the total force acting on it is zero. 0T + A s P I ˆndA + P I A c ˆndA +ṁu m (7.) The variable P is the gas pressure acting at any point on the surface of the rocket, is the viscous stress tensor, and ṁu m is the -momentum of the propellant injected into the thrust chamber. If the rocket were inactive so that there was no force on the restraint and the outside surface and thrust chamber were all at a pressure P 0 then 0 A s P 0 I ˆndA + P 0 I ˆndA A c. (7.3) In this situation the control volume contains fluid all at rest and 0 A c P 0 I ˆndA + P 0 I ˆndA A e. (7.4) The last relation can be written as 0 P 0 I ˆndA A c + P 0 A e. (7.5) Note that a unit normal vector that is consistent between the control volume and the outside surface of the vehicle points inward on A e. Thus equation (7.3) becomes 0 A s P 0 I ˆndA P 0 A e (7.6) and the original force balance (7.) can be written as 0T + P 0 A e + A c P I ˆndA +ṁu m. (7.7) Built into (7.7) is the assumption that when the engine is operating the eternal surface pressure and stress distribution is unchanged.

3 CHAPTER 7. ROCKET PERFORMANCE 7-3 A s P I ˆndA P 0 I ˆndA after engine turn on A s before engine turn on (7.8) In fact, the jet from the rocket mies with the surrounding air setting the air near the vehicle into motion leading to slight deviations in the pressure acting on the outside of the vehicle. For a rocket of reasonable size and thrust, this is a very small e ect. With the engine on, a balance of momentum over the control volume gives D Dt V Ū dv Ū V r ŪŪ + P I dv. (7.9) We are treating the case where the flow in the combustion chamber is stationary and the integral on the left hand side of (7.9) is zero. Our goal is to relate the thrust of the engine to flow conditions on A e and with this in mind we convert the right hand side to an integral over the surface of the control volume. 0 r ŪŪ + P I dv V ŪŪ + P I ˆndA + ŪŪ + P I ˆndA A c A e (7.10) Note that the unit normal that appears in the surface integrals in (7.10) is an inward pointing unit normal. On the surface A c, the velocity is zero by the no-slip condition (ecept over the injector holes) and on the surface A e we use area-averaged values of velocity, pressure, and density A c P I ˆndA +ṁu m + e U e A e + P e A e 0 (7.11) where the momentum of the propellant injected into the combustion chamber has been included. Small viscous normal forces on A e are neglected. Our force balance (7.7) now becomes Finally our rocket thrust formula is 0T + P 0 A e e U e A e + P e A e. (7.1)

4 CHAPTER 7. ROCKET PERFORMANCE 7-4 The propellant mass flow is T e U e A e +(P e P 0 ) A e. (7.13) and the rocket thrust formula is often written ṁ e U e A e (7.14) T ṁu e +(P e P 0 ) A e. (7.15) 7. Momentum balance in center-of-mass coordinates Let s look at the question of defining the thrust from a rather di erent point of view. Figure 7. depicts a rocket referenced to a system of center-of-mass coordinates. In the analysis to follow, gravity is taken to be zero. The e ects of gravitational acceleration will be taken into account later. Figure 7.: Center-of-mass description of the rocket and epelled propellant mass.

5 CHAPTER 7. ROCKET PERFORMANCE 7-5 For t<0 the rocket, its propellant and the surrounding atmosphere are all at rest. Sometime after ignition the rocket has translated to the right and the ehaust gases form a cloud o to the left of the center-of-mass. Because there is no eternal force on the system the center of mass remains at rest at the origin for all time. Conservation of momentum for the whole system of rocket vehicle, epelled combustion gases as well as the air set into motion by the drag forces on the rocket can be stated as D Dt V (t) Ū dv + M r (t) V r (t)! 0 (7.16) where M r (t) is the time-dependent rocket mass. At any given instant the rocket mass comprises all of the hardware and all of the propellant contained in the tanks, piping, pumps and combustion chamber up to the eit plane of the nozzle. The time-dependent rocket velocity is V r (t). We shall assume that all of the propellant mass contained in the rocket is moving at this velocity although there is always a small amount moving through the piping and combustion chamber at a slightly di erent velocity. The gas momentum is integrated over a control volume V(t) that completely encloses all of the moving gas outside the vehicle as shown in Figure 7.. Since there is a continuous flow of propellant mass into the combustion chamber and out of the rocket nozzle the volume required to contain the epelled gas must grow with time. This is depicted in Figure 7.. The control volume is cylindrical in shape. The left face A 1 moves to the left at a speed su cient to fully contain all the moving gas as well as any unsteady pressure disturbances generated by the rocket plume. The surface A 3 moves outward for the same reason. The upstream face A moves to the right at velocity V r (t) with the rocket. Finally the surface A s is attached to the rocket fuselage and outer nozzle surface. On the solid surface, the fluid velocity is equal to the rocket velocity V r (t) by the no slip condition. The last surface of the control volume is A e which coincides with the nozzle eit plane and translates to the right at the rocket velocity. The momentum equation integrated over V (t) is D Dt V (t) Ū dv A(t) Ū Ū Ū A + P I ˆndA. (7.17) The control volume is su ciently large so that the fluid velocity on A 1, A and A 3 is zero and the pressure is P 0. Therefore over most of the surface of the selected control volume no additional momentum is being enclosed as the surface moves outward. The pressure forces on A 1 and A nearly cancel ecept for a small deviation in pressure near the rocket nose in subsonic flight. For now we will cancel these forces but they will be included later when we develop an epression for the vehicle drag. The pressure forces on A 3 have no component in the direction. Thus the momentum balance (7.17) becomes

6 CHAPTER 7. ROCKET PERFORMANCE 7-6 D Dt V (t) Ū dv A s(t) A e(t) Ū Ū Ū A + P Ī ˆndA Ū Ū Ū A + P Ī ˆndA. (7.18) Now we can use an argument similar to that used in the previous section to relate the surface integral of the ambient pressure to an integral over the nozzle area. Recall 0 P 0 I ˆndA A s + P 0 A e. (7.19) The sign change in (7.19) compared to (7.6) comes from the change in the direction of the outward normal on A e compared to the control volume used in Section 7.1. Subtract (7.19) from (7.18) to get D Dt V (t) Ū dv A s(t) A e(t) Ū Ū Ū A +(P P 0 ) Ī ˆndA Ū Ū Ū A +(P P 0 ) I ˆndA. (7.0) On the no-slip surface of the rocket, the fluid velocity satisfies Ū (V r, 0, 0) and the control volume surface velocity is ŪA (V r, 0, 0). Therefore Now A s(t) Ū Ū Ū A ˆndA 0. (7.1)

7 CHAPTER 7. ROCKET PERFORMANCE 7-7 D Dt V (t) Ū dv A s(t) A e(t) ((P P 0 ) Ī ) ˆndA Ū Ū Ū A +(P P 0 ) I ˆndA. (7.) Near the rocket, the surrounding air is dragged along due to the no-slip condition and due to compressibility e ects that may generate shock waves as sketched in Figure 7.. Note that it is the deviation of the surface pressure from ambient, P P 0, that contributes to the change in air momentum due to the drag of the rocket. The combination of viscous skin friction drag, base pressure drag and wave drag are all accounted for by the integral over A s on the right-hand-side of (7.). Thus let D A s(t) (P P 0 ) I ˆndA. (7.3) Equation (7.) becomes D Dt V (t) Ū dv D A e(t) Ū Ū Ū A +(P P 0 ) I ˆndA. (7.4) Now consider the integral over A e on the right side of (7.4). All variables are area-averaged over A e.the-component of velocity of the gas passing through A e in the center-of-mass frame of reference is U V r + U e. (7.5) The nozzle ehaust velocity is the same velocity defined in section 7.1 (the velocity relative to the rocket) ecept that in this system of coordinates U e is negative. In this frame the speed of the surface A e is V r (t).

8 CHAPTER 7. ROCKET PERFORMANCE 7-8 A e(t) Ū Ū Ū A +(P P 0 ) I ˆndA (7.6) e A e (U e + V r )(U e + V r V r )+(P e P 0 ) A e. Now the momentum change of the epelled gas is D Dt V (t) Ū dv D ( e A e (U e + V r )(U e + V r V r )+(P e P 0 ) A e ). (7.7) Substitute (7.7) into (7.16) or D Dt (M r (t) V r (t)) + D ( e A e U e (U e + V r )+(P e P 0 ) A e ) 0 (7.8) M r (t) dv r (t) dt Note that + V r (t) dm r (t) dt + D ( e A e U e (U e + V r )+(P e P 0 ) A e )0. (7.9) dm r (t) dt e U e A e (7.30) and the second and fifth terms in (7.9) cancel. Remember that in the chosen set of coordinates, U e is negative and (7.30) is consistent with the fact that dm r /dt < 0. Finally our momentum balance in the center-of-mass system boils down to M r (t) dv r (t) dt e U e A e +(P e P 0 ) A e D. (7.31) In words, Equation (7.31) simply states Rocket mass Acceleration Thrust Drag. (7.3) The first term on the right-hand-side of (7.31) is the same thrust epression derived in the previous section.

9 CHAPTER 7. ROCKET PERFORMANCE E ective ehaust velocity The total mechanical impulse (total change of momentum) generated by an applied force, T,is I The total propellant mass epended is t 0 Tdt. (7.33) M p t 0 ṁdt. (7.34) The instantaneous change of momentum per unit ependiture of propellant mass defines the e ective ehaust velocity. C di dm p Ṫ m U e + A e ṁ (P e P 0 ) (7.35) This can be epressed in terms of the eit Mach number as follows C U e 1+ P ea e e U 1 e A e P 0 P e (7.36) or C U e 1+ 1 M e 1 P 0. (7.37) P e For a large area ratio ehaust with a large eit Mach number the pressure part of the thrust becomes a small fraction of the overall thrust. Let s estimate the theoretical maimum ehaust velocity that can be generated by a given set of propellants characterized by the heating value per unit propellant mass, q. Consider the simple model of a rocket thrust chamber shown in Figure 7.3. Between stations 1 and combustion takes place leading to a change in stagnation enthalpy of the propellant mass. h t h t1 + q h e + 1 U e (7.38)

10 CHAPTER 7. ROCKET PERFORMANCE 7-10 Figure 7.3: Thrust chamber propellant injection, heat release and nozzle epulsion. The last equality assumes adiabatic conditions between station and the nozzle eit. The quantity, h t1 is the stagnation enthalpy of the gases entering the combustion chamber and, in general, is much smaller than the heat added h t1 q. If the nozzle pressure ratio is very large P t /P 0 1, and the area ratio is large A e /A 1, then U e / h e and so we can define the theoretical maimum ehaust velocity as C ma p q. (7.39) The eit velocity is directly proportional to the amount of heat added through the combustion process. Making the approimation of constant specific heat and introducing the stagnation temperature, h t C p T t, the maimum velocity becomes, C ma p Cp T t r 1 RT t. (7.40) The gas constant in (7.40) is related to the universal gas constant by R R u M w (7.41) where M w is the mean molecular weight of the combustion gas. For this highly epanded rocket engine, the ehaust velocity is approimated by C ma s 1 Ru M w T t. (7.4) This last relation shows the key role of the molecular weight of the combustion gases on the ehaust velocity of the nozzle. The highest performing engines generally have the lightest

11 CHAPTER 7. ROCKET PERFORMANCE 7-11 weight ehaust gases. The most outstanding eample of this is the Space Shuttle Main Engine (SSME) which uses hydrogen and oygen, with water vapor as the main ehaust constituent. 7.4 C e ciency A very important characteristic velocity that is widely used in rocket motor testing is C defined by the mass balance ṁ P ta C. (7.43) The reason (7.43) is so useful is that it can be used to epress the combustion e ciency in the rocket chamber in terms of quantities that are relatively easy to measure: chamber pressure, propellant mass flow rate and nozzle throat area. These variables are much easier and less epensive to measure than the combustion chamber temperature and chemical composition. The C e ciency of a motor is defined as C Pt A ṁ measured P t A ṁ ideal. (7.44) The ideal value of C is determined using a thermochemical calculator such as CEA discussed in Chapter 9. Since the ideal calculation is assumed to take place at the same propellant mass flow rate and nozzle throat area, A,theC e ciency reduces to a simple comparison between the achieved chamber pressure and the chamber pressure that would be reached if there was complete miing and complete combustion under adiabatic conditions. C P t measured P tideal (7.45) 7.5 Specific impulse For historical reasons, the specific impulse has always been defined as the thrust per unit weight flow of propellant and so the gravitational acceleration at the surface of the Earth is always inserted. The specific impulse is defined as

12 CHAPTER 7. ROCKET PERFORMANCE 7-1 I sp T ṁg 0 C g 0 (7.46) even though the parameter g 0 9.8m/sec has no particular relevance to the problem. Generally one distinguishes between the sea level specific impulse of a vehicle where the ambient pressure detracts from the thrust and the ideal vacuum specific impulse where the ehaust is assumed to be epanded to the ehaust pressure P e with P 0 0. From the previous discussion of the theoretical maimum ehaust velocity, it is clear that the vacuum specific impulse at a very large area ratio e ectively characterizes a given propellant combination. Typical solid propellant systems have specific I spvac in the range sec. Liquid propellant systems using a hydrocarbon fuel with liquid oygen have I spvac around 360 seconds with hydrogen-oygen systems reaching 455 seconds. One needs to not take these I sp numbers too literally. Such values are often quoted for some typical real system such as the Space Shuttle Main Engine without stating the actual chamber pressure and area ratio and in some cases without identifying the system. The question of the area ratio corresponding to the ideal specific impulse is particularly important. For eample, for a hydrogen-oygen system at an area ratio of, say, 4, 000 the ideal I sp is over 500 seconds. An accurate specification of the specific impulse of a working system requires a knowledge of the chamber pressure, nozzle area ratio, combustion e ciency and nozzle e ciency. The chamber pressure is needed to determine the composition of the combustion chamber gas at the chamber temperature. This will become clear when we study the thermochemistry of gases in Chapter Chamber pressure The mass flow of propellant injected into the rocket engine and the amount of heat added between stations 1 and through combustion determine the engine chamber pressure. We can see this by considering the relationship between the mass eiting the nozzle and the stagnation conditions of the gas at station. In general, at any point in a channel flow of a compressible gas, the mass flow can be epressed as ṁ UA ( 1) P t A p RTt f (M) (7.47) where f (M) is the well known area-mach number relation

13 CHAPTER 7. ROCKET PERFORMANCE 7-13 f (M) A A ( 1) M +1. (7.48) 1+ 1 M ( 1) Equation (7.48) is plotted in Figure 7.4 for several values of C p /C v. Figure 7.4: Area-Mach number relation. The stagnation temperature at station is determined by the heat released through combustion. T t T t1 + q C p (7.49) To a first approimation, T t is nearly independent of chamber pressure, P t and is approimately known once the propellants are specified. Generally the chamber pressure is much larger than the ambient pressure P t /P 0 1 and so the nozzle throat is choked, M 1 and f (M ) 1. The chamber pressure is then determined by evaluating the mass flow at the nozzle throat +1 p +1 ( 1) RTt P t A ṁ (7.50) where adiabatic, isentropic conditions are assumed between station and the nozzle throat.

14 CHAPTER 7. ROCKET PERFORMANCE Combustion chamber stagnation pressure drop The stagnation pressure drop between stations 1 (near the injector) and station due to the heat addition is given by the conventional Rayleigh line relations. P t 1+ M1 P t1 1+ M 1+ 1 M! M 1. (7.51) The static pressure is P 1+ M1 P. (7.5) 1 1+ M At station 1, M 1 1, and we can approimate conditions at station in terms of just the Mach number at. P t 1 P t1 1+ M P 1 P 1 1+ M M (7.53) The Mach number at station is determined by the internal nozzle area ratio from to the throat. Assume isentropic, adiabatic, flow between A and A. A A ( 1) M M ( 1) (7.54) The relations (7.53) and (7.54) e ectively define a relationship between A /A, P t /P t1 and P /P 1, plotted in Figure 7.5 for several values of. The desire to keep stagnation pressure losses relatively small, while avoiding an ecessively large diameter combustion chamber, dictates the internal area ratio selected for the combustion chamber. It is clear from Figure 7.5 that an area ratio of about 3 is su cient to keep the stagnation pressure losses across the combustion chamber negligibly small. Practically all rocket thrust chambers have an area ratio of about 3 for this reason.

15 CHAPTER 7. ROCKET PERFORMANCE 7-15 Figure 7.5: Combustion chamber stagnation pressure loss. 7.8 The Tsiolkovsky rocket equation Consider the force balance on a rocket in flight shown in Figure 7.6. The variables identified in the figure are as follows. T vehicle thrust D vehicle aerodynamic drag V r vehicle velocity angle with respect to the horizontal ṁ nozzle mass flow M r vehicle mass g gravitational acceleration (7.55) The balance of forces along the direction of flight was derived earlier. Here we add the gravitational component of the force balance or M r dv r dt T M rgsin ( ) D (7.56)

16 CHAPTER 7. ROCKET PERFORMANCE 7-16 Figure 7.6: Rocket free body diagram. M r dv r dt C dm r dt M r gsin ( ) D (7.57) where C is the e ective ehaust velocity. Divide (7.57) through by M r. dv r dt C d (ln M r) dt gsin ( ) D M r (7.58) Let M ri initial mass at t 0 M rf final mass at t t b t b time of burnout. (7.59) Integrate (7.58) assuming constant C. The velocity change of the vehicle is where V r V rb V r0 V r ideal V r gravitational V r drag (7.60) V r gravitational V r drag tb 0 tb 0 D M r gsin ( )dt dt. (7.61)

17 CHAPTER 7. ROCKET PERFORMANCE 7-17 For a typical launch vehicle headed to orbit, aerodynamic drag losses are generally quite small on the order of 100 to 500 m/ sec. Gravitational losses are larger, generally ranging from 700 to 1500 m/ sec depending on the shape of the trajectory to orbit. By far the largest term is the equation for the ideal velocity increment Mri V r ideal C ln M rf (7.6) first derived in 1903 by the soviet rocket pioneer Konstantin Tsiolkovsky who is credited with developing much of the early theory of rocket flight. Equation (7.6) shows the dependence of the velocity achieved by a rocket on the e ective ehaust velocity (determined by the choice of propellants) and the initial to final mass ratio which is determined by what might be termed the structural e ciency of the vehicle and the density of the propellants. Notice the similarity of (7.6) to the Bruguet range equation discussed in chapter. In general, one seeks a very lightweight vehicle to carry high density propellants which after combustion produce very lightweight products. In practice these requirements conflict. Generally solid rockets use relatively dense, low energy propellants which do not produce very lightweight products of combustion. Whereas liquid rockets use more energetic propellants that produce light products but are not particularly dense. 7.9 Reaching orbit Orbital velocity at an altitude of 115 miles, which is about the lowest altitude where a stable orbit can be maintained, is approimately 7777 m/ sec. To reach this velocity from the Kennedy Space Center where the velocity due to the rotation of the Earth is approimately 47 m/ sec, assuming gravitational plus drag losses of 1700 m/ sec, requires an ideal velocity increment of 9050 m/ sec. A hydrogen-oygen system with an e ective average ehaust velocity (from sea- level to vacuum) of 4000 m/ sec would require M i /M f 9.7. This represents a very high level of structural e ciency and is the fundamental challenge being addressed by single-stage-to-orbit concepts. At the present time eisting launch vehicles require multiple stages to achieve orbit with a reasonable payload size. Strategies for reducing gravitational losses are mainly limited to optimizing the trajectory to orbit and epending the maimum amount of propellant as possible near the earth s surface (to avoid the work required to lift it to altitude). The latter strategy suggests that the most e cient way to orbit would be an artillery shell, however practical limitations prevent large acceleration loads on the payload. Most large launch vehicles are relatively delicate and require throttling back on thrust at low altitude to avoid large dynamic pressure loads on the vehicle.

18 CHAPTER 7. ROCKET PERFORMANCE 7-18 The drag losses can be minimized by designing a slender vehicle. follows This can be seen as V r drag tb 0 D M r tb dt 0 1 V r AC D M ri Mri M r dt A M ri tb 0 V r C D Mri M r dt (7.63) where A is the cross-sectional area of the vehicle. The integral on the right-hand-side is approimately independent of vehicle size and the initial mass of the vehicle is approimately proportional to the vehicle volume, M ri vehicle V vehicle. V rocket drag F rontalarea rocket tb density rocket V olume rocket 0 V rocket C D Mrocketi M rocket dt 1 Length rocket (7.64) The last result suggests that the vehicle should be long and thin, roughly like a pencil. Note that the drag losses go down as the mass goes up, and so the velocity loss due to drag tends to become smaller as the vehicle absolute size goes up. The length to diameter ratio of the vehicle does not come into the analysis directly but, in general, the drag coe cient, C d, decreases as the L/D goes up The thrust coe cient The thrust coe cient provides a useful dimensionless measure of engine thrust. C F T P t A ṁu e +(P e P 0 ) A e P t A Pe Ae P t A M e +1 P 0 P e (7.65) This rather complicated looking epression can be written in terms of the nozzle eit Mach number and pressure C F ( 1) M e +1 P 0 P e M e 1+ 1 M e 1 (7.66) where the nozzle flow has been assumed to be isentropic. For a rocket operating in a vacuum, with a very large epansion ratio M e! large, the thrust coe cient has an upper limit of

19 CHAPTER 7. ROCKET PERFORMANCE 7-19 C Fma (7.67) ( 1) The thrust coe cient is plotted in Figure 7.7 for several values of as a function of eit Mach number. Figure 7.7: Thrust coe cient versus Mach number. The thrust coe cient is also plotted in Figure 7.8 for several values of as a function of nozzle eit area ratio. Figure 7.8: Thrust coe cient versus area ratio. The thrust coe cient gives us a useful measure of the e ect of nozzle epansion on thrust. It is clear from Figure 7.7 that, in principle, epanding a gas with low would have the greatest benefit. However Figure 7.4 indicates that a large area ratio nozzle is required to

20 CHAPTER 7. ROCKET PERFORMANCE 7-0 reach the high eit Mach number required to obtain this benefit. We can see from Figure 7.7 and Figure 7.8 that fully epanding the flow, versus no epansion at all (a simple convergent nozzle), represents as much as a 50 % increase in the thrust generated by the nozzle. Generally, high temperature combustion gases have values of between 1. and 1.3 with the lower values characterizing high molecular weight products of combustion typical of solid rockets Problems Problem 1 - A monopropellant thruster using Argon gas at 100 psia and 1500 K ehausts through a large area ratio convergent-divergent nozzle to the vacuum of space. Determine the energy per unit mass of a parcel of gas at three locations: in the plenum, at the nozzle throat, and at the end of the epansion where the gas pressure approaches vacuum. What mechanism is responsible for the change of energy from one position to the net? How does your answer change if the gas is changed to Helium? Problem - The designer of a spacecraft maneuvering system needs to choose between Argon (atomic weight 40) and Helium (atomic weight 4) as propellants for a monopropellant thruster. The gas pressure and temperature in the propellant tank are N/m and 300 K respectively. The propellant tank volume is 1.0 m 3 and the empty mass of the vehicle is 10 kg. 1) Which propellant gas will give the largest velocity change to the vehicle? Estimate the vehicle velocity change for each gas? ) Suppose the vehicle mass is 1000 kg, which propellant would deliver the largest velocity change? Problem 3 - Consider two di erent systems used for space propulsion. System A uses propellants with an average density of gm/cm 3 and specific impulse of 00 seconds while system B uses propellants with an average density of 1 gm/cm 3 and specific impulse 300 seconds. The ideal velocity increment generated by either system is given by V I sp g 0 ln minitial m final (7.68) where g m/sec. Two missions are under consideration. 1) Mission I involves maneuvering of a large satellite where the satellite empty mass (mfinal ) is 000 kg and the required velocity increment is 100 m/sec.

21 CHAPTER 7. ROCKET PERFORMANCE 7-1 ) Mission II involves a deep space mission where the vehicle empty mass (m final ) is 00 kg and the required velocity increment is 6000 m/ sec. The design requirement in both cases is to keep the tank volume required for the propellant as small as possible. Which propellant choice is best for each mission? Problem 4 - Recently one of the popular toys being sold was called a stomp rocket. The launcher consists of a fleible plastic bladder connected to a 1.5 cm diameter rigid plastic tube. The rocket is a slightly larger diameter rigid plastic tube, closed at the top end, about 0 cm long. The rocket weighs about 10 gm. The rocket slips over the tube as shown in Figure 7.9. Figure 7.9: Stomp rocket toy. Jumping on the bladder pressurizes the air inside and launches the rocket to a height which the manufacturer claims can eceed 50 m. The area of the bladder in contact with the ground is approimately 100 cm. Use basic principles of mechanics to roughly estimate how much a child would have to weigh to be able to achieve this height. Problem 5 - Consider a class of monopropellant thrusters based on the use of the noble gases including Helium ( M w 4), Neon ( M w 0), Argon ( M w 40), Krypton, (M w 84) and Xenon (M w 131). Radon (M w ) is ecluded because of its radioactivity. The thruster is comprised of a tank that ehausts through a simple convergent nozzle to the vacuum of space. Onboard heaters are used to maintain the gas in the tank at a constant stagnation temperature T t as it is ehausted. 1) The thrust is often epressed in terms of an e ective ehaust velocity T ṁc. Show that the e ective ehaust velocity of this system can be epressed as ( C + 1) 1/ Ru T t. (7.69) M w

22 CHAPTER 7. ROCKET PERFORMANCE 7- ) The mass of propellant contained in the tank is M propellant P t initial V tan k M w R u T t. (7.70) The initial tank pressure is some rated value (a do-not-eceed pressure) independent of the type of gas used. The designer would like to choose the propellant gas so that the velocity increment produced by the propulsion system V is as large as possible for fied tank volume, initial pressure and gas temperature. The problem is to decide whether to choose a gas with low M w, thus achieving a high value of C but low propellant mass, or a gas with high M w reducing C but increasing propellant mass. By miing two or more gases, any mean atomic mass between 4 and 131 can be selected by the designer. Note that is the same regardless of what gas or miture of gases is used. Show that the maimum V occurs when the ratio M propellant /M structure is approimately 4 (actually 3.9). In other words, once the tank volume, pressure and temperature are determined and the vehicle empty mass is known, show that for maimum V the gas should be selected to have a mean atomic weight M w such that P tinitial V tan k M w R u T t M structure 3.9. (7.71) Problem 6 - The space shuttle main engine has a nozzle throat diameter of 10. in a nozzle area ratio of 77.5 and produces 418, 000 pounds of thrust at lift-o from Cape Canaveral. Determine the engine thrust when it reaches the vacuum of space.