Enae 483/788D Lecture #03 (Rocket Performance) Problems

ENAE 483/788D LECTURE #03 (ROCKET PERFORMANCE) PROBLEMS – FALL, 2020

Blue Origin, owned and funded by Jeﬀ Bezos, is preparing for the inaugural launch of their New Glenn launch vehicle next year. When complete, it will be one of the most powerful launch vehicles in existence, capable of putting 45 mt1 into low Earth orbit or 13.6 mt into a geostationary transfer orbit. The masses and performance parameters of the two-stage New Glenn are given in the following table. For the following questions, assume we are talking about the version delivering payload into LEO, unless otherwise noted.

Stage empty mass (mt) propellant mass (mt) speciﬁc impulse (sec) nominal burn time (sec) First stage 110 1000 330 240 Second stage 12 120 440 500

(1) Calculate (a) Gross mass of the entire vehicle mo

mo = min,1 + mpr,1 + min,2 + mpr,2 + mpl = 110 + 1000 + 12 + 120 + 45 = 1287 MT

(b) Inert mass fraction for the ﬁrst stage δ1

mo,2 = min,2 + mpr,2 + mpl = 12 + 120 + 45 = 177 MT

min,1 110 δ1 = = = 0.0855 min,1 + mpr,1 + mo,2 110 + 1000 + 177

min,2 12 δ2 = = = 0.0678 mo,2 177

(d) Stage inert mass fraction for the ﬁrst stage 1

min,1 110 1 = = = 0.0991 min,1 + mpr,1 110 + 1000

(e) Stage inert mass fraction for the second stage 2

min,2 12 2 = = = 0.0909 min,2 + mpr,2 12 + 120

(f) Predicted value for 1 based on the heuristic equation from lecture 3 for storable (i.e., non-LH2) propellants From page 19 in the lecture slides, −0.275 storables = 1.6062 (Mstagehkgi) −0.275 1 = 1.6062 (110, 000 + 1, 000, 000) = 0.03485

1Throughout this class, “mt” (or occasionally “MT”) refers to “metric ton”, or 1000 kg 1 2 ENAE 483/788D LECTURE #03 (ROCKET PERFORMANCE) PROBLEMS – FALL, 2020

(g) Predicted value for 2 based on the heuristic equation from lecture 3 for cryogenic (LOX/LH2) propellants From page 19 in the lecture slides, −0.183 LOX/LH2 = 0.987 (Mstagehkgi)

−0.275 2 = 1.6062 (12, 000 + 120, 000) = 0.1141

(h) Velocity change provided by the ﬁrst stage ∆v1

mf,1 = min, 1 + mo,2 = 110 + 177 = 287 MT

mf,1 287 ∆v1 = −goIsp ln = −9.8(330) ln = 4853 m/sec mo 1287

(i) Velocity change provided by the second stage ∆v2

mf,2 = min, 2 + mpl = 12 + 45 = 57 MT

mf,2 57 ∆v2 = −goIsp ln = −9.8(440) ln = 4886 m/sec mo,2 177

(j) Total velocity change provided by the entire launch vehicle ∆vtotal

∆vtotal = ∆v1 + ∆v2 = 4853 + 4886 = 9739 m/sec

(k) Trade-oﬀ ratio for change in payload mass due to a change in ﬁrst stage inert mass ∂mpl ∂min,1

1 1 ∂m −gIsp,1 m − m pl = o,1 f,1 ∂m 1 1 1 1 in,1 gIsp,1 − + gIsp,2 − mo,1 mf,1 mo,2 mf,2

mo,1 = mo, and g appears in every term so we can divide that out (which we’ll do for the next steps) 1 1 ∂mpl −330 1287 − 287 = 1 1 1 1 = -0.1458 ∂min,1 330 1287 − 287 + 440 177 − 57 Note: this is a dimensionless number, so it is metric tons of payload lost per metric ton of inert mass, or kg of payload per kg of inert mass

(l) Trade-oﬀ ratio for change in payload mass due to a change in ﬁrst stage propellant mass ∂mpl ∂mpr,1

1 ∂m −Isp,1 m pl = o,1 ∂m 1 1 1 1 pr,1 Isp,1 − + Isp,2 − mo,1 mf,1 mo,2 mf,2

1 ∂mpl −330 1287 = 1 1 1 1 = 0.03803 ∂mpr,1 330 1287 − 187 + 440 177 − 57 ENAE 483/788D LECTURE #03 (ROCKET PERFORMANCE) PROBLEMS – FALL, 2020 3

(2) New Glenn, like the SpaceX Falcon 9, is designed to recover its first stage on a ship offshore from the launch site. Assume that, like Falcon 9, the New Glenn only uses 95% of its first stage propellants to launch the payload, reserving the remaining 5% for use in an entry burn and a landing burn.

(a) What is the revised ∆vtotal for the satellite launch for this case? 5% of the ﬁrst stage propellant (=50 MT) will be reserved for stage recovery

mf,1 287 + 50 ∆v1 = −goIsp ln = −9.8(330) ln = 4334 m/sec mo 1287

Second stage ∆v2 unchanged

∆vtotal = ∆v1 + ∆v2 = 4334 + 4886 = 9220 m/sec

(b) How much ∆v is available to the ﬁrst stage to be used for landing?

min,1 110 ∆vlanding = −goIsp ln = −9.8(330) ln = 1212 m/sec min,1 + mpr,landing 110 + 50

(c) Most of the commercial missions for New Glenn will be to deliver communications satellites to geostationary transfer orbit (GTO). Using the 13.6 mt mass of this payload, calculate the total mission ∆vtotal including reserving propellant to recover the ﬁrst stage.

mo = min,1 + mpr,1 + min,2 + mpr,2 + mpl = 110 + 1000 + 12 + 120 + 13.6 = 1255.6 MT

mo − mf,1 1255.6 − 950 ∆v1 = −goIsp,1 ln = −9.8(330) ln = 4570 m/sec mo 1255.6 min,2 + mpl 12 + 13.6 ∆v2 = −goIsp,2 ln = −9.8(440) ln = 7495 m/sec min,2 + mpr,2 + mpl 12 + 120 + 13.6

∆vtotal = ∆v1 + ∆v2 = 4570 + 7495 = 12,065 m/sec

(d) Regardless of the size of the launch vehicle, a potential customer will always come along wanting a larger payload. If they don’t try to recover the first stage and use all of the first stage propellants for launch, how much payload can they sent to GTO at the same ∆vtotal you calculated in the previous question? ∆v2 remains unchanged at 7495 m/sec; need to find the payload that produces the same 4570 m/sec for the first stage m −∆v1/goIsp −4570/9.8(330) f,1 r1 = e = e = 0.2434 = mo,1

min,1 + min,2 + mpr,2 + mpl 242 + mpl ∆v1 = −goIsp,1 ln = −3234 ln min,1 + mpr,1 + min,2 + mpr,2 + mpl 1242 + mpl

min,2 + mpl 12 + mpl ∆v2 = −goIsp,2 ln = −4312 ln min,2 + mpr,2 + mpl 132 + mpl

∆vtotal = ∆v1 + ∆v2 = 12, 065 m/sec

3 equations, 3 unknowns ⇒ solve numerically to get Mpl =17.8 MT 4 ENAE 483/788D LECTURE #03 (ROCKET PERFORMANCE) PROBLEMS – FALL, 2020

(e) It turns out that the customer wants even more payload to GTO, so Blue Origin decides to strap four GEM-63XL solid rocket boosters onto the Blue Origin first stage. Each rocket motor has a gross mass of 50.2 mt, with 46.3 mt of propellant. The specific impulse for this unit is 265 sec, and the burn time is 90 sec. Calculate the total ∆v for this configuration using the original payload of 13.6 mt to GTO, without first stage recovery. t − t 240 − 90 χ = b,c b,b = = 0.625 tb,c 240 ve,bm˙ b + ve,cm˙ c ve,bmpr,b + ve,c(1 − χ)mpr,c v¯e = = m˙ b +m ˙ c mpr,b + (1 − χ)mpr,c 9.8 × 265(4 × 46.3) + 3234(1 − 0.625)1000 = = 3023 m/sec 4 × 46.3 + (1 − 0.625)1000 mfinal min,b + min,c + χmpr,c + m0,2 ∆vo = −v¯e ln = −v¯e ln minitial min,b + mpr,b + min,c + mpr,c + m0,2 4 × (50.2 − 46.3) + 110 + 0.625(1000) + (12 + 120 + 13.6) = −3023 ln = 1468 m/sec 4 × 50.2 + 110 + 1000 + (12 + 120 + 13.6) min,c + m0,2 110 + (12 + 120 + 13.6) ∆V1 = −Ve,c ln = −3234 ln = 4000 m/sec min,c + χmpr,c + m0,2 110 + 0.625 × 1000 + (12 + 120 + 13.6) For this problem, ∆v2 is unchanged from (2)(c) and =7495 m/sec

∆vtot = ∆v0 + ∆v1 + ∆v2 = 12,964 m/sec

(f) How much payload could this conﬁguration carry to GTO at the same ∆v calculated in (2)(c)? Set up the equations from (2)(d) with mpl as a variable and solve numerically to ﬁnd the payload that gives you a ∆vtot=12,065 m/sec to get

∆v0 = 1459 m/sec; ∆v1 = 3940 m/sec; ∆v2 = 6666 m/sec; mpl = 20.5 MT