. Peraire 16.07 Dynamics Fall 2004 Ⅴ ersion1.1 Lecture D27-Variable Mass Systems: The Rocket Equation In this lecture, we consider the problem of a body in which the mass of the body changes during the motion that is, m is a function of t, i.e. m(t). Although there are many cases for which this particular model is applicable, one of obvious importance to us are rockets. We shall see that a significant fraction of the mass of a rocket is the fuel, which is expelled during fight at a high velocity and thus, provides the propulsive rce for the ro We shall start by considering a body with velocity v and external forces F, gaining mass at a rate i= dm/dt Let us look at the process of gaining a small amount of mass dm. Let u' be the velocity of dm before it is captured by m, and let f represent the average value of the impulsive forces that dm exerts on m during he short interval dt, in which the capturing takes place. By Newtons third law, dm will experience a force f, exerted by m, over the same dt We can now examine the capture process from the point of view of dm and equate the impulse,-fdt,to e change In nne fat= dm(u+ Here, v+dv is the velocity of m(and dm) after impact. Analogously, from the point of view of m, we write Fdt+fdt=m(u +du)-mu= mdv The term dm dv in equation 1 is a higher order term and will disappear when we take limits. The impulse due to the contact force can be eliminated by combining equations l and 2 Fdt-dm(a-v)=mda
J. Peraire 16.07 Dynamics Fall 2004 Version 1.1 Lecture D27 - Variable Mass Systems: The Rocket Equation In this lecture, we consider the problem of a body in which the mass of the body changes during the motion, that is, m is a function of t, i.e. m(t). Although there are many cases for which this particular model is applicable, one of obvious importance to us are rockets. We shall see that a significant fraction of the mass of a rocket is the fuel, which is expelled during flight at a high velocity and thus, provides the propulsive force for the rocket. We shall start by considering a body with velocity v and external forces F, gaining mass at a rate ˙m = dm/dt. Let us look at the process of gaining a small amount of mass dm. Let v ′ be the velocity of dm before it is captured by m, and let f represent the average value of the impulsive forces that dm exerts on m during the short interval dt, in which the capturing takes place. By Newton’s third law, dm will experience a force −f, exerted by m, over the same dt. We can now examine the capture process from the point of view of dm and equate the impulse, −fdt, to the change in linear momentum of dm, −fdt = dm(v + dv − v ′ ). (1) Here, v + dv is the velocity of m (and dm) after impact. Analogously, from the point of view of m, we write Fdt + fdt = m(v + dv) − mv = mdv. (2) The term dm dv in equation 1 is a higher order term and will disappear when we take limits. The impulse due to the contact force can be eliminated by combining equations 1 and 2, Fdt − dm(v − v ′ ) = mdv, or, dividing through by dt, m dv dt = F − (v − v ′ ) dm dt = F + (v ′ − v) dm dt . (3) 1
Here, v'-v is the velocity of dm relative to m. This expression is valid when dm/dt>0(mass gain)and when dm/dt<o(mass loss). The term(u'-v)dm/dt is an additional force on m which is due to the gain d(mu F+ which shows that, for systems involving variable mass, the usual expression stating conservation of linear momentum,d(mu)/dt= F, is only applicable when the initial(final) velocity of the captured(expelled) mass. 0. Is zero The Rocket Equation We consider a rocket of mass m, moving at velocity v and subject to external forces F(typically gravity and drag). The rocket mass changes at a rate rn= dm/dt, with a velocity c relative to the rocket. We shall assume that the magnitude of c is constant F m(t) The velocity of the gas observed from a stationary frame will be v'=U+c. Thus F+ The term T= cdm/dt is called the thrust of the rocket and can be interpreted as an additional force on the rocket due to the gas expulsion Equation 4 is a vector equation which can be projected along the direction of v(tangent to the path). Thus, mdf=Ft-cdt=fitt, (5) where Ft is the tangential component of F, v and c are the magnitudes of v and c respectively, and we have assumed that c is parallel and has opposite direction to u. The magnitude of the thrust is T=-crm. Not that for a rocket in will be negative(mass is lost). If the force Ft is known, this equation can be integrated in time to yield an expression for the velocity as a function of time Let us consider some simple cases If gravity and drag effects are neglected we have du
Here, v ′ − v is the velocity of dm relative to m. This expression is valid when dm/dt > 0 (mass gain) and when dm/dt < 0 (mass loss). The term (v ′ − v)dm/dt is an additional force on m which is due to the gain (or loss) of mass. Equation 3 can also be written as d(mv) dt = F + v ′ dm dt , which shows that, for systems involving variable mass, the usual expression stating conservation of linear momentum, d(mv)/dt = F, is only applicable when the initial (final) velocity of the captured (expelled) mass, v ′ , is zero. The Rocket Equation We consider a rocket of mass m, moving at velocity v and subject to external forces F (typically gravity and drag). The rocket mass changes at a rate ˙m = dm/dt, with a velocity c relative to the rocket. We shall assume that the magnitude of c is constant. The velocity of the gas observed from a stationary frame will be v ′ = v + c. Thus, m dv dt = F + c dm dt . (4) The term T = c dm/dt is called the thrust of the rocket and can be interpreted as an additional force on the rocket due to the gas expulsion. Equation 4 is a vector equation which can be projected along the direction of v (tangent to the path). Thus, m dv dt = Ft − c dm dt = Ft + T, (5) where Ft is the tangential component of F, v and c are the magnitudes of v and c respectively, and we have assumed that c is parallel and has opposite direction to v. The magnitude of the thrust is T = −cm˙ . Note that for a rocket ˙m will be negative (mass is lost). If the force Ft is known, this equation can be integrated in time to yield an expression for the velocity as a function of time. Let us consider some simple cases: 1) Ft = 0 If gravity and drag effects are neglected we have, m dv dt = −c dm dt 2
or, integrating between an initial time to, and a final time t AU=U-U0=-c(In m-In mo)=-cIn Alternatively, this expression can be cast as the well known rocket equation m no e which gives the mass of the rocket at a time t, as a function of the mass initial mass mo, Au, and c. The mass of the propellant propellant, Is given by From the above equations, we see that for a given Aw and mo, increasing c increases m(payload) and decreases propellant. Unfortunately, we can only choose c as high as the current technology will allow Note that if ri is constant, we can write m(t)=mo +rit, and substitute into equation 6 to obtain an expression for v as a function of t, U=0-cln(1+ Recall that according to the convention used, i is negative as the mass decreases with time 2)Ft=-mg A constant gravitational field acting in the opposite direction to the velocity vector can be easily corporated. In this case, equation 5 become dt ng Thich can be integrated to give U=vo-cIn--gt= vo -cIn - no 3)Ft f--mg-D The effect of the drag force, D, is harder to quantify. It turns out that for many important applications drag effects are very small. The drag force is characterized in terms of a drag coefficient, CD. Thus, where A is the cross-sectional area of the rocket. The air density changes with altitude z, and may be approximated by
or, integrating between an initial time t0, and a final time t, ∆v = v − v0 = −c(ln m − ln m0) = −c ln m m0 . (6) Alternatively, this expression can be cast as the well known rocket equation, m = m0 e −∆v/c , (7) which gives the mass of the rocket at a time t, as a function of the mass initial mass m0, ∆v, and c. The mass of the propellant, mpropellant, is given by, mpropellant = m0 − m = m0(1 − e −∆v/c). From the above equations, we see that for a given ∆v and m0, increasing c increases m (payload) and decreases mpropellant. Unfortunately, we can only choose c as high as the current technology will allow. Note that if ˙m is constant, we can write m(t) = m0 + ˙mt, and substitute into equation 6 to obtain an expression for v as a function of t, v = v0 − c ln(1 + m˙ m0 t) . Recall that according to the convention used, ˙m is negative as the mass decreases with time. 2) Ft = −mg A constant gravitational field acting in the opposite direction to the velocity vector can be easily incorporated. In this case, equation 5 becomes, m dv dt = −mg − c dm dt , which can be integrated to give v = v0 − c ln m m0 − gt = v0 − c ln m m0 − g m − m0 m˙ . 3) Ft = −mg − D The effect of the drag force, D, is harder to quantify. It turns out that for many important applications drag effects are very small. The drag force is characterized in terms of a drag coefficient, CD. Thus, D = 1 2 ρv2ACD , where A is the cross-sectional area of the rocket. The air density changes with altitude z, and may be approximated by ρ = ρ0e −z/H , 3
where h a 8000 m is the so-called"scale height" of the atmosphere, and po is the air density at sea level. It turns out that the differential equation that results for the velocity cannot be integrated explicitly and, in prac eeds to be int Numerically. It ng to note. however. that the effect of drag losses is usually quite small, and it is often reasonable to ignore it in a first calculation In order to see the importance of D versus the the effect of gravity, we can estimate the value of the ratio D/mg. At conditions typical for maximum drag, p a 0. 25 kg/m and v=700 m/s. Considering a rocket of 12, 000 kg with a cross section of A= 1 m2 and Cp=0.2, we have pACD20.25×1×0.2×7002 2×12.000×9.8 0.021, which indicates that the drag force is only about 2% of the gra force Example Single vs. Two Stage Rockets [1] To achieve an orbital speed of v= 7600 m/s, we require an ideal Au of about 9000 m/s where the extra velocity is needed to overcome gravity and drag. Chemical rockets produce exhausts jets at velocities of A 2500-4500 m/s. Using the higher c, if we wish to place a payload in orbit with a single stage rocket re have a mass ratio(mass at burn out, mf, divided by initial mass, mo) of 9000/4500 =e-2=0.135 The mass mf must include all components of the rocket infrastructure, including the engine, empty tank guidance equipment, etc, as well as the payload. The mass of the propellant will be mo-mf =(1 0.135 )mo=0.865mo. If we assume that the tank plus engine are the main contributors to the weight of the rocket and weigh 10% of the propellant, 0.087mo, we have maod=(0.135-0.087)mo=0.048mo Clearly, there is not much margin here, and, in fact, single-stage-to-orbit vehicles have yet to be engineered ccessfully. The alternative solution is to subdivide the rocket, so that empty tanks are dropped when they are no longer needed Two stages Consider now a two stage vehicle. We shall again assume that each empty tank plus its engine weighs 10% of the propellant it carries. The required Av is now subdivided into two Aus of 4500 m/s each First stage If mo is the initial mass and mi is the mass after burn out, we have m1=c-45004500m0e-1mo=0.368m0
where H ≈ 8000 m is the so-called “scale height” of the atmosphere, and ρ0 is the air density at sea level. It turns out that the differential equation that results for the velocity cannot be integrated explicitly and, in practice, needs to be integrated numerically. It is interesting to note, however, that the effect of drag losses is usually quite small, and it is often reasonable to ignore it in a first calculation. In order to see the importance of D versus the the effect of gravity, we can estimate the value of the ratio D/mg. At conditions typical for maximum drag, ρ ≈ 0.25 kg/m3 and v = 700 m/s. Considering a rocket of 12, 000 kg with a cross section of A = 1 m2 and CD = 0.2, we have, ρACDv 2 2mg = 0.25 × 1 × 0.2 × 7002 2 × 12, 000 × 9.8 = 0.021 , which indicates that the drag force is only about 2% of the gravity force. Example Single vs. Two Stage Rockets [1] Single stage To achieve an orbital speed of v = 7600 m/s, we require an ideal ∆v of about 9000 m/s where the extra velocity is needed to overcome gravity and drag. Chemical rockets produce exhausts jets at velocities of c ≈ 2500 − 4500 m/s. Using the higher c, if we wish to place a payload in orbit with a single stage rocket, we have a mass ratio (mass at burn out, mf , divided by initial mass, m0) of mf m0 = e −9000/4500 = e −2 = 0.135 . The mass mf must include all components of the rocket infrastructure, including the engine, empty tank, guidance equipment, etc., as well as the payload. The mass of the propellant will be m0 − mf = (1 − 0.135)m0 = 0.865m0. If we assume that the tank plus engine are the main contributors to the weight of the rocket and weigh 10% of the propellant, 0.087m0, we have, m 1stage payload = (0.135 − 0.087)m0 = 0.048m0 . Clearly, there is not much margin here, and, in fact, single-stage-to-orbit vehicles have yet to be engineered successfully. The alternative solution is to subdivide the rocket, so that empty tanks are dropped when they are no longer needed. Two stages Consider now a two stage vehicle. We shall again assume that each empty tank plus its engine weighs 10% of the propellant it carries. The required ∆v is now subdivided into two ∆v’s of 4500 m/s each. First stage If m0 is the initial mass and m1 is the mass after burn out, we have, m1 = e −4500/4500m0 = e −1m0 = 0.368m0 4
The fuel burnt will be mo -m1=(1-0.368)mo=0.632mo, and the weight of the tank and engine to be opped will be 0.0632mo, leaving an initial mass for the second stage, m2, of m2=0.368m0-0.063m0=0.295m0 After burn out The fuel burnt in this stage will be m2-mf =(0.295-0.109)mo=0.186mo and the weight of the tank plus engine will be 0.019mo, leaving for the payload dng=(0.109-0.019)m0=0.090m which is still very small but about twice the size of the payload obtained for the single stage rocket Note Specific Impulse The specific impulse Isp, with units of seconds, is often used in practice to characterize the performance of a rocket engine. The definition of the specific impulse is the magnitude of the thrust divided by the propellant weight flow rate T g 9 Typical values of Isp are around 300 s for solid propellants and up to 500 s for higher energy fuels ADDITIONAL READING J. L Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition References [1 M. Martinez-Sanchez, Unified Engineering Notes, Course 95-96
The fuel burnt will be m0 − m1 = (1 − 0.368)m0 = 0.632m0, and the weight of the tank and engine to be dropped will be 0.0632m0, leaving an initial mass for the second stage, m2, of m2 = 0.368m0 − 0.063m0 = 0.295m0 . Second stage After burn out, mf = e −1m2 = e −1 0.295m0 = 0.109m0. The fuel burnt in this stage will be m2 − mf = (0.295 − 0.109)m0 = 0.186m0 and the weight of the tank plus engine will be 0.019m0, leaving for the payload m 2stage payload = (0.109 − 0.019)m0 = 0.090m0, which is still very small but about twice the size of the payload obtained for the single stage rocket. Note Specific Impulse The specific impulse Isp, with units of seconds, is often used in practice to characterize the performance of a rocket engine. The definition of the specific impulse is the magnitude of the thrust divided by the propellant weight flow rate, Isp = T mg˙ = c g . Typical values of Isp are around 300 s for solid propellants and up to 500 s for higher energy fuels. ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 4/6, 4/7 References [1] M. Martinez-Sanchez, Unified Engineering Notes, Course 95-96. 5