HW3_solution - ,WW,W HM Mg(saws/l#3 lei/m ELthsmfgzli The...

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Unformatted text preview: , ,,,,..,WW.,W HM Mg, (saws/l? #3 lei/m. ELthsmfgzli The exhaust velocity is Ve = g ISp = 4439 m/sec = 4.439 km/sec. Writing the rocket equation (7.9) as v - V ln (m ) ~ V ln ms + mp + m* * " e o/mf " e ms + m* we know everything except the payload mass m*. Put 8 == V*/Ve = 0.9664. Taking the exponential of both sides, the result is a linear equation for the payload mass, which can be solved to yield ms( 1 - e8) + m m*= GB - l P 1| 8525 kg when the calculation is carried out. For a round trip, the tug will carry payload on the outbound leg, but will return without payload. In order to return, we need 5 propellant mpr for the return leg sufficient for a velocity change of E 4.29 km/sec. The rocket equation becomes 5 pr m S m + m V* = Ve 1n ( ] for the return leg. This gives a propellant mass of m = ms(eB — 1) = 2117 kg to return the tug to low earth orbit. For the outward leg, then, this propellant must be considered to be payload. The available fuel for the outward trip is mpO = mp - mpr pr a 13883 kg. Recalling our earlier result ms( 1 ~ e3) + m o m* = .___._§._____~_.E.~ a 7225 kg e - 1 However, this value of m* includes the fuel for the return trip. The actual payload delivered to geosynchronous orbit is then i! m = 7225 kg ~ 2117 kg * 5108 kg This is about 2/3 of the mass delivered to geosynchronous orbit if the tug is expended. The economics of this situation must consider the COSt 0f the tug itself, and the cost of its fuel delivered to low earth orbit. The extra half of a shuttle payload required to deliver the fuel for a second trip must be considered against the cost of the tug itself delivered to the parking orbit. _wwfiw* MAE ('41. M», ewe/Z #3 @4065». 2/; W .6: N m _ 0k V* _ + Z Vek 1n m k=1 fk N mok = m* + mpk + msk + Z ( 53 + mpj ) j=k+1 N mfk:m*+msk+ Z ( sn+mpj) j=k+1 Also, note that spec1f1c impulse of stage k ls ISp k = Ve k / g. I I Taklng 1mpllclt partial derlvatlves w1th respect to ISp k and m*, I we have m ~ ~ 0k dV* ~ 0 - gC 1n { mfk } aIsp,k . m . f3 1 03 [ -- 2 J 3111* . m . . * is zero since we cannot 6m* ~ gc 1n ( m0k / mfk ) aIsp,k ““fi““““"“““““‘““"‘““***‘“*~*-- V O MM 14! ., {lemma/é #3 5040/2? ., 1 1 287 x 9.8 [ $135366” $616200 ] + 455 x 9.8 -l.__. - ..$1_._ z -2 622 X 104 846400 143300 ‘ meters/Sec/kg. The tradeoff ratios then become an; aI = 54.61 kg/sec sp,1 6m* 61 = 663.6 kg/sec sp,2 3/; , the solid rocket% boosters or the main engines, depends on economics and how feasible: the improvement is. The solids certainly are the least efficient parté of the system, but the main engines offer the greatest payback per: second of Isp‘ Eroblem 7-4. Starting at the top of the vehicle, and using Table 7~5,; the masses we need are mf2 = m* + m$2 = 291 kg m02 = mf2 + mp2 = 706 kg mf1 = m02 + msl = 819 kg m01 = mf1 + mp1 2 1986 kg Since we wish to consider two burn strategies, we will work out theg burnout speeds separately for each stage. The rocket law gives m v m V In ( 01 J = 2447.9 m/sec 1 e1 mfl v a V In 02 ] = 2449.3 m/sec 2 82 mf2 MM {4% [iammrg #3 2%Mém» The fact that this is an optimized rocket is shown by the fact that each stage shoulders an equal portion of the burnout speed. In the burn - burn - coast strategy, is vb0 = 4897.2 m/sec. unit mass the total burnout velocity Working out the conservation of energy per 1 2 .. 7 2 2 .. .5“ Vbo _ 1.199 x 10 m /sec — g Hmax gives a value of Hmax = 1223 km, indicating that we have exceeded theé bounds of the constant gravity assumption: this number is too large ai fraction of one earth radius! ' Using inverse square law gravity, energy is given by 1 2 u -————_. c- _ _ 2 2 2 vb0 "E; ~ 50.50 km /sec starting from the surface of the earth, r0 = 6378 km. 9 a O, the energy must equal ~u/r km, or Hmax = 1515 km. constant g assumption, real earth. Since at apogee: max’ giv1ng a value of rmax a 7893; This value is larger than that using the: since gravity drops off with altitude in the For the ,burn - coast - burn - coast method successive applications of energy conservation give, for the first burn —l— v2 - —E— = ~59.50 kmz/sec2 = ~ ~E~ 2 1 r r o 1 3 r1 = 6699.1km and, for the second burn —i~ v2 — ~E~ = ~56.500 king/sec2 = ~ ~E~ 2 2 r r l 2 a r2 = 7055 km This translates to a maximum altitude of Hmax = 677 km. This is much less than the burn ~ burn ~ coast method. Since the vehicle trades kinetic energy for altitude, doubling the burnout velocity results in‘ much more altitude than achieving the same burnout speed twice. MM (44 Ham/M #3 364/93», 17;“ Problem 7-5. We are asking what payload a certain configuration can; place in a low earth orbit. Using the rocket equation, the (required) burnout velocity is found from 2 mok v* = 31,000 ft/sec = Z vek ln [ } 4 x 81900 + 35000 + 703000 + m* 4 x 584600 + 35000 + 703000 + m* = 32.2 X 290 in + 32.2 x 455 ln [ 35000 + 703000 + m* 35000 + m* It is not necessary to be careful in the argument of the logarithms, and use pounds. The arguments are dimensionless, and m* will come out in kg. Performing the operations above not requiring m*, we have 3076400 + m* 31,000 = 9338 ln 1065600 + m* 738000 + m* + 14651 In ----—~—- ft/sec 35000 + m* (ii) Now, the payload mass is certainly much less than 3 x 106; kg. It is also probably negligible compared to l x 106 kg, and then? should probably be neglected compared to 700,000 kg as well, for the; sake of consistency. It is certainly not small compared to the dry; tank mass, 35000 kg, since this is the nominal payload of the usual? shuttle! Ignoring m* in these places, we have 31,000 ft/sec % 9900 ft/sec 738000 + 14651 ln 35000 + m* Solving for m* yields the value of m* x 139,800 kg : 307,000 lb or almost 5 normal shuttle payloads. This vehicle has a slightly larger payload than the Apollo Saturn 5 moon rocket. (The exact solution is m* = 167092 kg.) ...
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