Compute the average effective exhaust velocity for the propulsion system as a whole and from this also the average specific impulse.

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1 2 2.1 Ariane 5 Thrust Initially at lift-off of an ARIANE 5 launch vehicle two P230 solid propellant boosters plus the main Vulcain engine are ignited. The two different components of the launch vehicle have the following characteristics: Effective exhaust velocities: v,v ulcain = 3285 m/s v,p 230 = 2355 m/s Mass flows: ṁ V ulcain = 255 kg/s ṁ P 230,booster = 1835 kg/s Compute the average effective exhaust velocity for the propulsion system as a whole and from this also the average specific impulse. Now we have parallel staging. The two boosters and the main engine are running simultaneously. Our total thrust can easily be combined adding up the three thrust sources: F,tot = k ṁ p,i v,i F,tot = ṁ V ulcain v,v ulcain + 2 ṁ P 230 v,p 230 = 9, 480, 525 N For the average I SP we add the three mass flows to find the total mass flow ṁ tot = ṁ V ulcain + 2 ṁ P 230 = 3925 kg/s With this mass flow we can compute the average v and I SP v,avg = F,tot ṁ tot = m/s I SP,avg = s 1 Tutor:

2 2.2 Multiple Stage LEO Rocket Calculate the gain in total v of using the three-staged rocket (parameters below) to a simple single stage to orbit (SSTO) rocket (using a standard structural factor and the maximal v of the three-stager). The payload mass that has to be put into orbit is 800 kg. m s,1 = 5 t m p,1 = 40 t m s,2 = 1 t m p,2 = 10 t m s,3 = 0.2 t m p,3 = 1 t v,1 = 2300 m/s v,2 = 2500 m/s v,3 = 2900 m/s To compute the total v of the rocket we have to find the individual ε i and λ i for each stage (1,2,3). To do this we are writing down the correct masses following the definitions in the lecture. The total launch mass m 0,1 was calculated by adding up all partial masses. 1 st stage: m 0,1 = 58 t m f,1 = m 0,1 m p,1 = 18 t 2 nd stage: m L,1 = m 0,1 m p,1 m s,1 = 13 t def = m 0,2 m 0,2 = 13 t m f,2 = m 0,2 m p,2 = 3 t 3 rd stage: m L,2 = m 0,2 m p,2 m s,2 = 2 t def = m 0,3 m 0,3 = 2 t m f,3 = m 0,3 m p,3 = 1 t m L,3 = m 0,3 m p,3 m s,3 = 0.8 t With these masses we can compute the structural factor and the payload factor of each stage 1 st stage: ε 1 = m s,1 m 0,1 m L,1 = Tutor:

3 2 nd stage: λ 1 = m L,1 m 0,1 m L,1 = ε 2 = m s,2 m 0,2 m L,2 = rd stage: λ 2 = m L,2 m 0,2 m L,2 = ε 3 = λ 3 = m s,3 m 0,3 m L,3 = 0.16 m L,3 m 0,3 m L,3 = To verify these results we compare the total payload factor λ using two different equations. The first equation gives a total payload factor of λ = m L,3 m 0,1 = 0.8 t 58 t = The second way to compute this value is to use the values of the three different stages λ = 3 λ i 1 + λ i = = Now we can use the propulsion capability equation for multiple staged rockets that is given in the formulary. This equation sums up all three v s. v 3 Stager = λ1 v 3 Stager = v,1 ln + v,2 ln ε 1 + λ 1 ( 1 + λi v,i ln ε i + λ i ( 1 + λ2 ε 2 + λ 2 ) ) + v,3 ln v 3 Stager = 2701 m/s m/s m/s = 8393 m/s 1 + λ3 ε 3 + λ 3 No as we know the v maximum for the three stage rocket we have to compute the SSTO value. For this case we assume a standard structural factor of 10% and the maximal v of the three-stager. We have to use the same liftoff mass of 58 t and the same payload mass of 800 kg. This leads to the same total payload factor λ = λ = m L,3 = 0.8 t m 0,1 58 t = λ v SST O = v,max ln = 2900 m/s ln = 6344 m/s ε + λ So we can find an increase in v of v 3 Stager v SST O = 8393 m/s 6344 m/s = 2595 m/s 3 Tutor:

4 2.3 Optimal Stage Number The following data is known for a launch vehicle (Scout): Thrust: 1.Stage: kn 2.Stage: kn 3.Stage: 97.4 kn 4.Stage: 25.8 kn Lauch mass: 17, 500 kg Acceleration at launch: 2.71g 0 Specific Impulse: ground: 240 s vacuum: 264 s Payload(556 km - Orbit): 115 kg Velocity demand( v): 9.0 km/s Under the assumption of an identical structural factor for all stages, calculate the structural factor for the rocket. Is the choice of 4 stages optimal for this structural factor? We are looking at a tandem staged rocket with 4 stages. Each stage has a different thrust but the same structural factor ε. In the first step we consider the total payload factor λ to be optimized. With the given data we can compute λ with the given formula: λ = m L,n 115 kg = m 0,1 17, 500 kg = Here n is the number of stages and n = 4. From the formulary we can use the second definition of λ which is as follows λ = n λ i 1 + λ i This product can be transformed in an easier form only if all payload factors λ i are equal (as stated in the text above). Then you can simplify as follows λ = n n λ i λ = 1 + λ i 1 + λ 4 Tutor:

5 From now on we only write λ to save some writing. We can now compute λ λ = λ1/n = λ 1/n To compute the structural factor ε we use the Ziolkovsky equations for staged rockets which is v = n 1 + λi v,i ln ε i + λ i Here again we can use the fact that all structural and payload factors are equal throughout the 4 stages. This leads to This can be transformed to yield ε v = n 1 + λi 1 + λ v,i ln = n v ln ε i + λ i ε + λ v ε = (1 + λ) e n v λ The only variable missing is the v which we only know on the ground (240 s) and in vacuum (264 s). We assume that the rocket flies 1/3 of the time in the atmosphere and 2/3 of the time in vacuum. This leads to a average value of v ( 1 v,avg = 3 v,ground + 2 ) 3 v,vacuum = This leads to an structural factor of ε = Now we want to check if the choice of 4 stages is an optimal choice for this rocket and the given v. The first possibility is the rule of thumb equation stated in the lecture n 1.09 v v 3.91 So we are probably in the right magnitude when we are using 4 stages. This easy to use formula is derived from this equation n opt v ( 1 + ε e 1 ) 4.11 v 2 We used our calculated ε = and we see again that 4 stages is the closest value to the computed optimum. The last way to calculate the optimum would be to compute the optimal total payload factor first using the equation (derived in the lecture) λ opt = 1 ( 1 e ε ) = e 1 2 This values can be inserted in the Ziolkovsky equation which has be transformed to yield the optimal values of stages Four stages should be the optimum for this rocket. v n opt = = λopt v ln ε+λ opt 5 Tutor:

6 2.4 Sounding Rocket For a two-stage sounding rocket there are two strategies for ascent available: a) After burn-out of the first stage the rocket proceed until the velocity is zero before the second stage is ignited. b) Immediately after burn-out of the first stage the second stage is ignited. Which strategy gives the highest altitude for the two-stage sounding rocket? We are comparing the two different ascent strategies a) and b): Which strategy brings our payload into the higher orbit? First we define the three masses of our rocket: total mass of the rocket m 1 mass without the 1 st stage m 2 mass without 1 st and 2 nd stage m L (payload mass) Then we define three heights as follows: Final height of the rocket H Height through 1 st stage h 1 Height through 2 nd stage h 2 6 Tutor:

7 We apply the principle of energy conservation to the complete system. We assume that the burn phase is comparable short to the flight time and can be neglected. So directly after the rocket launch the rocket has burned the fuel of the first stage it has no potential energy but the velocity v 1 and the kinetic energy of E kin = 1 2 m 1 v 2 1 We wait until the velocity of the rocket becomes the zero at the height h 1. Now the rocket has no kinetic energy but the potential energy: Due to the energy conservation we can write: E pot = m 1 g h 1 The same equations can be stated for the second stage m 1 g h 1 = 1 2 m 1 v 2 1 m L g h 2 = 1 2 m L v 2 2 In case a) the total height H can be calculated through adding the height h 1 to h 2. Simply spoken we are consecutively launching two single-stage rockets. The first rocket starts on the ground and puts the second stage and the payload in to a given height h 1. When the second stage s velocity reaches zero we start the second single-stage rocket (= 2 nd stage) which puts the payload in the final height H. The leads to H = h 1 + h 2 = 1 2g (v2 1 + v 2 2) For the scenario b) we again assume that the burn time of both stages is very small compared to the complete flight and that both stages are burned without a pause between the burn phases. So after the initial burning of both stages the rocket has the kinetic energy: E kin = 1 2 m L (v 1 + v 2 ) 2 The rocket reaches its maximal height at H and has the final potential energy If you combine the two equations you get E pot = m L g H H = 1 2g (v 1 + v 2 ) 2 = 1 2g (v2 1 + v v 1 v }{{} 2 ) So the second scenario reaches a greater height due to the last term. So the conclusion is that cruising phases during the ascent have to be avoided. 7 Tutor: