THE UNIVERSITY OF MICHIGAN INDUSTRY PROGRAM OF THE COLLEGE OF ENGINEERING DESIGN STUDY OF A LIQUID AND A SOLID ROCKET PROPELLANT SYSTEM William Whieher Theodore Petersen January, 1960 IP-411

PREFACE There are few places in the literature wherein the rocket propulsion system and vehicle performance have been treated together. In addition, a design study of two propulsion systems - one based on a liquid propellant and the other upon a solid propellant is quite interesting for comparison purposes. The two design studies presented here were carried out by the authors as part of the requirements for the Rocket Propulsion course in the Department of Aeronautical and Astronautical Engineering. In the interest of analytical facility, simplified assumptions were made for both preliminary designs. These assumptions, however, would not seriously affect the results. This report was reproduced in the interest of supplying the Industry Program members with a concise design study of two systems based on the same design objectives and employing the same techniques of construction. It also clears up some of the arguments concerning the relative advantages and disadvantages of solid and liquid rocket motor systems. ii

TABLE OF CONTENTS LIQUID ROCKET DESIGN William Whicher Page NOMENCLATURE.....................**..*...*........................ v VEHICLE SPECIFICATIONS.............................. 1 Additional information............................. 2 Determine chamber pressure as a function of time.... 5 Calculate Helium Data * *.............................. 55 Calculate amount of Figerglas needed...3.....7... SOLID ROCKET DESIGN Theodore Petersen NOMENCLATURE.......................................... 55 DESIGN SPECIFICATIONS............................ 56 Propellant.....*................*............. 56 Specifications................... ~~~..a..,......... 56 Weight Estimate................................... 57 NOZZLE SIZING.......................*..*.o..*................. 58 GRAIN SIZING.....*.....*..........*....*..*.................. 63 DERIVATION OF EXPRESSIONS RELATING PACKING FRACTION AND BURNING PERIMETER TO CONFIGURATION DIMENTIONS.6.....3............ 63 DISCUSSION OF CHOICE OF GRAIN CONFIGURATION................... 65 SELECTION OF CORRECT GRAIN CONFIGURATION........ o............ 66 GRAIN CHARACTERISTICS..............................*......... 67 DISCUSSION OF INITIAL CONFIGURATION DESIGN PLOT............... 69 DETERMINATION OF BURNING PERIMETER AS A FUNCTION OF TIME...... 71 iii

TABLE OF CONTENTS (CONT 'D) Page DETERMINATION OF PROPELLANT CROSS-SECTIONAL AREA AND WEIGHT AS A FUNCTION OF TIME.................... 80 THRUST FOR VARYING Pc AND PA o......e... o.. o..o...........oo 81 PERFORMANCE EQUATIONS........ o.. 0......0.0...0.000 85 COMPUTATION OF MACH NUMBER AND DYNAMIC HEAD....o.9 o...o.. oo 92 DRAG CALCULATION............00.oo.a...o.o.o0o.o 96 ACCELERATION COMPUTATION.............................. 98 DISCUSSION OF STORAGE TERM 0.................0.0 90000... 102 DETERMINATION OF VELOCITY AT END OF GRAIN.. 0009.0900.0.000. 105 COMPARISON WITH LIQUID PROPELLANT ROCKET DESIGN** *.o..o.o. 106 SUMMARY................ 00 a 0 000000. a 0 0.000. a a0. o 107

LIQUID ROCKET DESIGN William Whicher

NOMENCLATURE sp gr - specific gravity psi - pounds per square inch Ae - exit area of rocket nozzle q. - dynamic pressure PC - rocket chamber pressure F - thrust (also T is used for thrust) X - rocket nozzle correction factor to account for divergence of flow at rocket exit Ve - gas velocity at exit of rocket nozzle Ri - mass flow Tc - exit temperature of rocket gas Pe - exit pressure of rocket gas AT - rocket nozzle throat area P2/P1 - pressure ratio across oblique shock in exit of rocket nozzle 9 cone angle of rocket nozzle at exit Ml - Mach number before oblique shock in exit of rocket nozzle +1 r y (7 2) 2(7 - 1) t - time T - thrust D drag on rocket M - mass of rocket at time, t (also M - Mach nanmber) h - altitude of the rocket g - gravitational constant 0 w - weight flow v

V - velocity of the rocket L - length Co - drag coefficient Re - Reynold's number S - area vi

Vehicle Specifications Problem Number 3 Given the following information, design a rocket. Mass ratio = 20/1 (includes payload) Payload wt. = 10% dry wt. of bird. Fineness ratio = L _ l5 D 1 iake bird out of Fiberglas. Sp.gr. Fiberglas = 1.80 Working stress = a = 80000 psi Use a monopropellant sp.gr. = 1.00 400 lbs. thrust Motor - lb. engine wt. Use a gas pressurization system. Ae in a spherical container. PHe = 3500 psia initially. Use a heater charge in Helium-tank L so that Helium always expands isothermally. 25% pressure drop across plumbing. 25% pressure drop across injector face. Lift off wt. = 25,000 lbs. Also make plots of: a) head suppression vs. time. b) chamber pressure vs. time. c) drag vs. time. d) acceleration vs. time. e) altitude vs. time. f) Mach number vs. time (only until q drops off appreciably) g) 9 vs. time (q p 2)..~~~~~ h _I

-2 - Additional information Use an engine with same geometry as in Problem 2 except chop off nozzle so that flow never separates. When designing tankage only consider hoop tension. Assume that a regulator valve is in the Helium line that controls the pressure on top of the propellant at a constant value until the helium runs out. Assume a head suppression valve is installed in the propellant lines so as to regulate the pressure in the chamber at a constant 150 psia until the helium runs out. Size rocket Mass ratio = 20/1 Wt. fuel at lift off = (0.95)(25000) = 23750 lbs. Wt. fuel Tank volume — Wt. fuel density fuel 23750 62.4 = 381 ft3 Calculate tank dimensions V = L215 L = 15D 4 D 15D3 V = D3 = (381)(4) 324 15ic D = 3.18 ft. L = 47.7 ft. Use L/D = 3 for nose cone LTotal 47.7 + (3) (35.l18) = 57.24 ft. * Refers to design of a rocket motor chamber and nozzle previously carried out

Calculate Engine Characteristics Characteristics for Pc = const. = 150 psia Calculate Fo 0 FO = X m Ve + PeAe from Problem 2 1/2 Fo = 577 MeTe / + PeAe for Ae- = 3.011; pe- 0.06526; - = 0.6345 AT Pc T Me = 2.40 (from gas tables) so F = (577)(2.40)(3172)l/2 + (144)(9.789)(2.68)(3.011) = 78000 + 11350 = 89350 lb. The thrust at any altitude is given by F = F - AePa Verify that nozzle does not separate P2 = 1 + 71Q = 1 + (1.2)(2.4)2(o.1745) Pi (M2-1)1// [(2.4)2-1]1/2 = 1 + 1.205 =1 + 0.552 (4.76)1/2 P2 - (150)(0.06526)(1.552) - 15.19 psia. This is above atmospheric pressure of 14.7 psia so separation does not occur. Calculate thrust after 45.0 seconds Pc from graph F = 6oo Pc

-4 -TABLE I Altitude Atmospheric Pressure AePa F ft lb/ft2 lb lb 0 2116 17060 72290 4000 1828 14730 74620 8000 1572 12680 76670 12000 1346 10830 78520 16000 1147 9240 80110 20000 972.5 7840 81510 2,4000 820.2 6610 82740 28000 687.8 5550 83800 32000 573.3 4670 84680 36000 474.7 3825 85525 40000 391.7 3156 86194 44000 323.2 2604 86746 48000 266.7 2148 87202 55000 190.5 1535 87815 60000 149.8 1207 88143 65000 117.8 950 88400 70000 93-51 753 88597 75000 73.70 594 88756 80000 58.01 467 88883

-5 - t = 50.0 F = (600)(130.6) = 78360 lbs. t = 55.0 sec. F = (600) (118.4) = 71040 lbs. t = 60.0 F = (600)(105.0) = 63000 lbs. t = 65.0 F = (600)(89.0) = 53400 lbs. t = 67.26 - tBO F = (600)(83.3) = 49980 lbs. Determine chamber pressure as a function of time A head suppression valve is installed in the propellant lines to keep the pressure in the chamber constant at 150 psia. However due to the limitations on the amount of helium that can be carried the chamber pressure decreases during the last part of burning. At some time the head suppression term goes to zero and the chamber pressure is determined solely by the helium pressure, the hydraulic head, and the line plus injector impedance. After the critical time (the time at which the chamber pressure starts to fall) the chamber pressure can be determined by an equilibrium equation relating the line impedance, the helium pressure and the hydraulic head. PHe + H - Z Pc ~

The chamber pressure was constant up to 45 seconds. Therefore it is convenient to take our time origin at 45 seconds. Doing this we can rewrite the equation t PiVi (h45 -f hdt)Pp F P(t)....to... + z Pc(t) V45 + Vdt W45 o i Where Pi = initial helium pressure Vi = initial helium volume V45 = helium volume at 45 seconds 0 V= time rate of change of helium volume h45 = level of the propellant at 45 seconds 0 h = time rate of change of propellant level pp = density of propellant W45 = weight of rocket at 45 seconds 0 C weight flow F thrust Pc(t) = chamber pressure Z = ratio of tank pressure to chamber pressure, previously calculated as 16/9 e The chamber pressure and the chamber temperature are related through the equilibrium constant Ko. However since we do not know the chemical composition of the monopropellant we are forced to assume that the reaction products and therefore the chamber temperature are constant. As a consequence of this assumption we note that the mass flow is sensitive only to changes in chamber pressure.

o r Pc At m = - ac 0 m = -i p p c ci By using this relation we can solve for V, h, and (n as functions of Pc alone. At 45 seconds we are at about 100,000 ft. Therefore we can ignore drag, and also assume that we are exhausting into a vacuum. The last assumption allows us to solve for F as a function of PC alone. F = X m Ve + Pe Ae 0 = Pc Ve + PC Ae Pc45 c = [k m45 Ve + -Ae] Pc Pc45c x = 0.9924 0 m^5 Pc45 Ve 11.9 0.07933 -50 =- Me 7 R Te Tc Tc = 6600 ft/sec F = 599.9 Pc 0 O = g m = (32.2)(0.07933) Pc = 2.554 Pc V = =_ 2.554 Pc = 0.04094 Pc pp 62.4 0 h = 0 V = o.o4094 Pc = 0005155 P 7.94 inside tank area Substituting these values into the original equation we obtain (3500)(16.83) 2935.4 + 0.04094 otPc(t)dt = Pc(t) t + (13,15 - 0.005155 J Pc(t)dt) (62.4) (599.9 Pc(t) ) (7765 - 2.554 J PC(t)dt) (144)

-8 - The initial values at t=O (45 seconds) have been included in the above equation. This equation gives Pc as a function of time. PC = Pc(t) We note that this equation is a non-linear integral equation. An. analytical solution is not possible. However if we make certain assumptions we can obtain an approximate solution to this equation. First from the first mean value theorem of integral calculus we note t f Pc(t)dt = PC(O)t where Pc () is some average value in the interval. Now unless we know t we cannot locate P(~) in the interval. However for calculation purposes we will assume that ~ lies in the middle of the interval. This is true if the interval 0 to t is small. For our purposes we will use 2.5 second intervals. The equation to be solved is now PiVi + [ho - (0.005155) Pc(~)t] 259.96 Pc(~)] 16 p (e) Vo + 0.04094 Pc({)t Wo - 2.554 Pc()t 9 c there V0, ho, and Wo are the values of the helium volume, the propellant level and the rocket weight at the end of the previous interval. The rest of the terms are as noted above. The above equation is cubic in Pc(0) and can be solved most easily by iteration. The final calculation for each interval is presented below. 58905 + (13.15 - 0.1289 Pc)(260)(Pc) 16 Pc 2935.4 + 0.10235 Pc 7765 - 6.385 Pc 9

-9 - 45 0 < t < 47.5 try Pc = 142 58905 + [(13.15)-(1.830)](26o)(142) _ 16 p 293.4 + 14.53 7765 - 906 9 c 58905 + (11.32)(260)(142) 16 p 307.93 6859 9 c P = 9 (191.2 + 60.9) = 9 (252.1) = 141.9 c -16 16. Use P. = 142.0 47.5 < t < 50.0 try Pc = 134.5 58905 + (11.32 - 1.732)(260)(134.5) = 16 p 307.93 + 13.78 6859 - 858 9 c 16 p = 58905 + (9.588)(260)(134.5) 9 c 321.71 6001 p = 9 c 16 (183.0 + 55-9) = 9 (238.9) = 134.5 16 Use P. = 134.5 50.0 < t < 52.5 try Pc = 127.5 58905 321.71 + 13.07 (9.588 - 1.642)(260)(127.5) _ 16 6001 - 8149 c 16 p 58905 + (7.946)(260)(127.5) 9 c 334.78 5187 P = 9 c 16 (175.9 + 50.7) = 9 (226.6) 16 = 127.5 Use P = 127.5

-10 - 52.5 < t < 55.0 try Pc = 121.0 58905 334.78 + 12.39 + (7.946 - 1560)(260)(121) = 16 5187 - 775 9 Pc 58905 + (63.586)(26o)(121) = 16 po 347.17 44 14 9 P = 96 (169.8 + 45.4) 9= (215.2) i 6 i 6 = 121.0 Use PC = 121.0 Psia c 55.0 < t < 57.5 16 pc 9 58905 + (63.86 - 1.479)(26o)(114.8) 4414 - 732 347.17 + 11075 = 58905 + 358.92 (4.907)(260)(114.8) 3682 P =9 (164o + 39.8) (208) 16 = i'g ' " 6;' (2o3.8) Pc = 114.8 Use Pc = 114.8 57.5 < t < 6oo try Pc = 108.0 16 9 58905 + (4,907 - 1.391)(260)(108) 3682 - 689 358.92 + 11.o6 _ 58905 + 569.98 (3.516) (26o)(o18) 2995 PC 9 (159.1 + 33.0) 9 (1924) = 16 i6 = 108,0 Use Pc = 108.0

-11 - 60.0 < t < 62.5 try Pc = 101.0 16 p g c = 58905 369.99 + 10.35 + (3.516 - 1.291)(260)(101) 2993 - 645 58905 (2.225) (260)(101) 580.5 33 2548 P = 9 (154.8 + 24.9) = 9 (179.7) 16 16 = 101.0 Use Pc = 101.0 62.5 < t < 65.0 try 93.0 psi 16 p 9 = 58905 380.5533 + 9.53 + (2.225 - 1.198)(260)(93) 2348 - 593 - 58905 + (1.027) (26o) (93) 389.86 1755 P 9 - (151.0 + 14 1) = 9 (165.1) =935.0 psia Use 93 To determine burn out time we note that the helium volume at burn out is given by VHe = Vol propellant tank + initial vol He. = 397.83 ft3 From this we can determine the chamber pressure at burn out. p (3500) (16.83) PcBO 397.83 = 83.287 x 83.3 psia

-12 - Burnout will occur before 67.5 seconds. We can find the burnout time if the average chamber pressure during the last period is known. Estimate this from the average pressure for the last two times c65 = 93.0 (101 - 93) = 89.o The average pressure over the interval 65 - -tBO is _cav 83.3 + 89.0 PCav =- - = 86.15 psia Now since the helium volume must be 397.83 ft3 at burnout VHe65 + (0.04094) Pc(~) tb 397.83 389.86 + (o.o4o94)(86.15) t = 397.83 7.97 - (86.15)(0.04094) = 2.26 tBO = 65.0 + 2.26 = 67.26 seconds after lift off.

-13 -TABLE II Time Chamber Pressure Weight w Sec psia lb lb/sec (over interval) 45.0 46.25 47.50 48.75 50.00 51.25 52.50 53.75 55.00 56.25 57.50 58.75 60o.00 61.25 62.50 63.75 65.00 67.26 150.0 142.0 7765 362.2 6859 134.5 343.2 6001 127.5 325.6 5187 121.0 309.4 114.8 293.0 3682 108.0 275.6 2993 101.0 258,0 2348 93.0 89.0 83.3 237,2 223~8 1755 1250

-14 - D Performance Calculations Using Newtons second law MdV T - D - Mg dt Now if we consider thrust and drag constant over the time interval and express M as M(t) we obtain. dV T - D dt Mo - mt g(T D) Wo - Wt ~Mg Y~V t g(T-D)dt t f dV = ( T- - gdt Vo o Wo - t 0 Vo T - D W o - gt V - dt T-D Wo dt dh = Vodt + - J- n W -~ o- fgtdt ho 0o m 0 Wo0-t o after consulting an integral table we find 0 h = ho + Vot gt2 + ToD[t W~- t Qn.] 2 m L W0-ct Note that I have integrated from 0 to t. This is equivalent to considering only the time intervals with initial conditions applying from the end of the previous time interval. 0 < t < 5 F - D= 72290 lbo

-15 - t = 5 V = 72290 (in 25000 - in 23085) - 161 11.9 = 6070 [10.12663 - 10.04693] - 161 = 6070 (0.07970) - 161 = 484 - 161 = 323 ft/sec. h = 6070 [5 - (23085)(0~,797) ] - (25)(16.1) 383 Since the [ ] involve the difference of numbers very close together, use log tables instead of a slide rule so as to improve accuracy. log 235085 = 4.56335 log 0.07970 = 8.90146 - 10 15.26479 - 10 log 585 = 2.58520 10.68159 - 10 = log 4.8049 h = 6070(5 - 4.8049) - 403 = 1185 - 403 = 780 ft Re = kL L = 47.7 + (35.18) (3) = 47.7 + 9.54 = 57.24 Re = (0.002310)(323)(57.24) (p and p. from NACA TN 3182) 35.699 x 10-7 = 105.5 x 106 M = 525 = 0.29 (a at various altitudes from NACA TN 3182) 1114

-16 - From graph CD = 0.18 D = CDq S (S based on frontal area) q = p V2 (2.310 x 10-3)(323)2 2 2 = 120.3 lb/ft2 Since the estimation of CD from the graph is not precise use S based on the inside area of tank. S = (1.59)2 E = 7.94 D = (0.18) (120.53) (7.94) = 172.1 lbs. 5 < t< 10 Thrust at 780 ft = T - D = 72700 72700 lb - 172 = 72528 lb t = 10 V = 323 + 72528 [,n 23085 11.9 - in 21170] - 161 V = 323 + 6100 (10.04693 = 162 + 6100 (0.08659) = 162 + 528 = 690 ft/sec - 9.96034) - 161 h = 780 + (323)(5) - 403 + 6o00 [5 - (21170) (0.8659)] ~ ( 0. 0 8659)3 4.32572 8.93747 - 10 15.26519 - 10 2.58520 10.67999 - 10 5.0000 -> 4.7862

-17 - h = 780 + 1212 + 6l00 (0.2138) = 1992 + 1302 = 32 94 ft Re = (0.002158)(690)(57.24) 3.669 x 10-7 = 232 x 106 M - 6.90 = 0.123 1107 - CD = 0.17 q = (0.002158)(690)2 = 51 lbft2 2 D = (0.17) (514) (7.94) = 695 lbs 10 < t < 15 T - D = 74200 - 695 = 73505 lbs t = 15 V = 690 - 161 + 73505 n 11.9 21170 - en 19255] = 529 + 6170 [9.96034 - 9.86552] = 529 + 6170 [0.09482] = 529 + 584 = 1113 ft/sec h = 3294 + (690)(5) - 403 + 6170 [5 - (19255) 383-~ (0.09182)] 4.28454 8.97690 - 10 153.26144 - 10 2.58320 10.67624 - 10 5.0000 -> 4.7671 h = 6341 + (6170)(0.2329) h = 6341 + 1437 = 7778 ft

-18 - Re = (0.001881)(1113)(57.24) 3.561 x 10-7 = 336 x 1o6 M = 1113 1 011 1100 CD - 0.295 = (0.001881)(1113 5)2 1164 lbs/ft2 2 D = (o.295)(1164) (7.94) = 2726 lbs 15 < t < 20 T - D = 76600 - 2726 = 73874 lb t = 20 V - 1113 + 73874 [Qin 19255 - 2n 17340] - 161 11.9 - 952 + 6205 [9.86552 - 9.76077] = 952 + (6205) (0o10475) = 952 + 651 - 1603 ft/sec h 7778 + 5565 - 403 + 6205[5 (- 7340)(0.10475)] 383 4.23905 9.02015 - 10 15.25920 2.o 58520 10.67600 - 10 - 4.7424 h 12940 + (6205)(0.2576) = 12940 + 1597 1= 4537 ft Re = (1.520 x 10-3)(1603)(57.24) 35.441 x 10-7 = 405 x 106

-19 - M = 1603 = 1.512 1060 ---- CD = 0.275 (1.520 x 10-3)(1603)2 q = 2 D - (0.275)(1953)(7.94) = 4260 lb 20 < t < 25 T - D = 79500 - 4260 - 75240 lb t = 25 = 1953 lb/ft2 V = 1603 - 161 + 75240 11.9 L[n 17340 - 9n 15425] V =- 1442 + 6320 (9.76077 - 9.64374) = 1442 + (6320)(0.11703) 1442 + 740 = 2182 ft/sec h = 14537 + (1603)(5) - 403 + 620 [5 (15425) (0.11703) 383 4.18823 9.06830 - 10 13.25653 - 10 2.58320 10.67333 - 10 -, 4.7133 h = 22149 + (6320)(0.2867) = 22149 + 1812 = 23961 ft Re ( (1.103 x 10-53) (2182) (57.24) 3.234 x 10-7 = 426 x 106 M 218 = 2.12 1021

-20 - CD = 0.245 = (1.103 x 10-l (2182)2 2 D = (0.245)(2620)(7.94) = 5100 lb 25 < t < 50 T - D = 82700 - 5100 = 77600 t = 30 = 2620 lb/ft2 V = 2182 - 161 77600 11.9 [in 15425 - in 13510] = 2021 + 6520 (9.64374 - 9.51118) = 2021 + (6520)(0.13256) = 2021 + 864 = 2885 ft/sec h = 23961 + (2182)(5) - 403 + 6520 [ 5 - (13510)(013256) ] 383 4.13066 9.12241 - 10 153.253507 - 10 2. 58520 10.66987 - 10 -> 4.6760 h = 34468 + (6520)(0.324) = 34468 + 2110 = 36578 ft (6.92 x 10-4) (2885) (57.24) 2.96 x 10-7 = 386 x 106 M = 2885 = 2.98 968 CD = 0.215 q= (6.92 x 10-4)(2885)2 2 = 2880 lb/ft2

-21 - D = (0.215)(2880) (7.94) = 4910 lb 30 < t < 35 T - D = 85600 - 4910 = 80690 t = 35 80690 V = 2885 - 161 + 11.9 [Ln 13510 - In 11595] = 2724 + 6790 (9.51118 - 9.35832) = 2724 + (6790)(0.15286) = 2724 + 1038 = 3762 ft/sec h = 56578 + (2885) (5) - 403 + 6790 [5 - (11595) (0.15286)] 383 4.06426 9.18429 - 10 13.24855 - 10 2.58320 10.56535 - 10.- 4.6276 h = 50600 + (6790)(0.3724) = 50600 + 2524 = 535124 ft Re = (3.118 x 1o-4) (3762)(5724) 2.96 x 10-7 = 226.5 x 106 M = 3762 = 3.88 968 CD = 0.19 (3.118 x 10-4)(3762)2 = 2205 lb/ft2 D = (0.19) (2205)(7.94) = 3320 lb 35 < t < 40 T - D = 87700 - 3320 = 84380

-22 - t = 40 84580 V = 3762 - 161 + 19 [+n 11595 - in 9680] = 3601 + 7100 [9.35833 - 9.17782] = 3601 + (7100)(0.18051) = 3601 + 1281 = 4882 ft/sec h = 535124 + (3762)(5) - 403 + 7100 [5 - (9680)(0.18051)] 383 3.98588 9.25650 - 10 153.242358 - 10 2.58520 10.65918 - 10 4.5622 h = 71531 + (7100)(o.4378) = 71551 + 3104 = 74635 ft Re - (1.115 x 10-4) (4882) (57.24) 2.96 x 10-7 = 105 x 106 4882 M.....503 971 CD = 0.17 (1.115 x 10-4) (4882)2 2 D = (0.17) (1328) (7.94) = 1790 lb 4o < t < 45 T - D = 88700 - 1790 = 86910 lb = 1328 lb/ft2

-23 - t = 45 86910 V = 4882 - 161 + 11.9 [in 9680 - in 7765] = 4721 + 7300 (9.17781 - 8.95738) = 4721 + 161o = 6331 ft/ sec h = 74635 + (4882)(5) - 403 + 7300 [5 - (7765)(0.22043) 3.890141 9.543271 - 10 153.233412 - 10 2.583199 10.650213 - 10 - 4.4690 h = 98642 + (7300)(0.5310) = 98642 + 3880 = 102,522 ft Re = (2.96 x 10-5) (6331)(57.24) 2.96 x 10-7 = 36. 2 x 106 M - 697331 = 6.52 971 CD = o.18 (2.96 x 10-5)(6331)2 2 D = (0.18)(595)(7.94) = 848 lb = 595 lb/ft2 An expression for altitude as a function of time can be obtained analytically if an approximate expression is used for chamber pressure as a function of time. However, the altitude integral becomes rather messy, so it was decided to continue the numerical integration using 2.5 second intervals and the average mass flow over each interval.

-24 - t = 47.5 V = 6331 - 80 + 7565 In 6765 = 6251 + 933 = 7184 ft/sec h = 102522 - 100 + (6331) (2.5) + 7565 [2.5 - 659 n 77 5 = 118249 + 7565 [2.5000 - 2.5348] = 118249 + (7565)(0.152) = 118249 + 1150 = 119399 ft t = 50.0 V = 7184 - 80 + 7565 n 6859 = 7104 + 1013 = 8117 ft/sec h = 119399 + (7184)(2.5) - 100 + 7565 [2.5 - 64.2 n 659 ] = 137259 + (7565)(2.500 - 2.342) = 137259 + (7565)(0.158) = 137259 + 1194 = 184535 ft t = 52.5 V = 8117 - 80 + 7565 in 6oo = 8057 + 1104 = 9141 ft/sec h = 1384535 - 100 + (8117)(2.5) + 7565 [2.5 525.6 on 5187] = 158645 + (7565)(2.500 - 2.324) = 158645 + (7565)(0.176) = 158645 + 1352 = 159977 ft

-25 - t = 55.0 V = 9141 - 80 + 7565 Yn 5187 4414 = 9061 + 1221 - 10282 ft/sec h = 159977 - 100 + (9141)(2.5) + 7565 [2.5 - 441 n 5187] = 182729 + (7565) (2.500 - 2.5304) = 182729 + (7565)(0.196) = 182729 +, 1483 = 184212 ft t = 57.5 V = 10282 - 80 + 7565 Yn 4 -= 10202 + 1375 = 11577 ft/sec h = 184212 + (2.5)(10282) - 100 + 7565 [2.5 - 3682 9 444] = 209817 + 7565 (2.500 - 2.282) = 209817 + (7565)(0.218) = 209817 + 1650 = 211467 ft t = 60.0 V = 11577 + 7565 tn 3682 - 80 2995 = 11497 + 1570 = 13067 ft/sec h = 211467 - 100 + (11577)(2.5) + 7565 [2.500 - 2995 682] = 240309 + (7565)(2.500 - 2250) = 240309 + (7656)(0.250) = 240309 + 1890 = 242199 ft

-26 - t = 62.5 V = 13067 - 80 + [in 2993] 7565 2348 = 12987 + (7565)(2432) = 14829 ft/sec h = 242199 - 100 + (13067)(2.5) + 7565 [2.5 - 25348 n 299] = 274766 + 7565 (2.500 - 2.216) = 274766 + (7565)(0.284) = 274766 + 2150 = 276916 ft t = 65.0 v = 14829 - 80 + 7565 in 2348 1755 = 14749 + 2204 = 16953 ft/sec h = 276916 - 100 + (2.5) (14829) + 7565 [2.5 - 755 n 2348 237.2 1755 = 313888 + 7565 (2.500 - 2.158) = 313888 + (7565)(0.542) = 315888 + 2590 = 316478 ft t = 67.26 = tBO V = 16953 - (32.2)(2.26) + 7565 in 1755 1250 = 16953 - 73 + 2565 = 19445 ft/sec h = 316478 - (161)(2.26)2 + (16953)(2.26) + 7565 [2.26 1250 1755] = 354710 + 7565 [2.26 - 1.895] = 354710 + (7565)(0.365) = 554710 + 2690 = 357400 ft

-27 - If missle flies in a constant gravitational field then Md2h dt2 or d2h — h = - g h = -gt + Clt + C2 2 at t = 0 o h = 19445 h = 357400 h =-gt + 19445t + 357400 2 h = - gt + 19445 = 0 tsummit = 19445 = 604. 5 sec 32.2 hsummit = - (162)(604.5)2 + (19445) (6o4.5) + 357400 = - 5875000 + 11740000 + 357400 = 5865000 + 357400 = 6222400 ft = 1180 miles Calculate hydraulic head H = hydraulic head = (L - ht)pp(a + 1) 0 V _ 6.15 IR2 7t(1.59)2 = 0.775 ft/sec 0< t< 5 (47.7)(62.4)(2.89) 144 = 59.7 psi

-28 - 5 < t < 10 H (47.7-3.875)(62.4)(3.i4) 144 _ (43.825) (62.4) (3.14) 144 = 60.4 psi 10 < t < 15 H= (43.825-3.875) (62.4) (3.47) 144 (39.950) (62.4) (3.47) 144 = 60.0 psi 15 < t < 20 (39.950-3.875) (62.4) (3.83) 144 (36.075) (62.4) (3.83) 144 = 59.8 psi 20 < t < 25 H (36.o75-3.875)(62.4)(4.34) 144 (32.2) (62.4) (4.34) 144 = 60.5 psi

-29 - 25 < t < 30 H 50 < t < 55 H 55 < t < 40 H = (32.300-5.875)(62.4)(5.03) 144 (28.325) (62.4) (5.03) 144 = 61.6 psi (28.325-3.875)(62.4)(5.98) 144 (24.45) (62.4) (5.98) 144 = 63.3 psi (24.450-3.875) (62.4)(7.29) 144 (20.575) (62.4)(7.29) 144 = 65.o psi 40 < t < 45 (20.575-53.875) (62.4) (8.99) 144 (16.7) (62.4) (8.99) 144 65.1 psi

-30 - 45 = t ~H =(13-15) (62.4) (11.48) 144 = 65.4 psi For t > 45 sec the hydraulic head was a necessary part of calculating the chamber pressure curve. The values listed below are from those calculations time, sec 46.25 48.75 51.25 53.75 56.25 58.75 61.25 53.75 67.26 Hydraulic head, psi 60.9 55.9 50.7 45.4 39.8 3355.0 24.9 14.1 0 Calculate accelerations F-D-W = - a g a = [ - 1l]g t = 0 to 5 72290 (25000 - l)g = (2.89 - 1)g = 1.89 g

-31 - t = 5 to 10 a = (72508 - 1)g = (3.14 - l)g = 2.14 g t = 10 to 15 a = (73505 -)g 21170 = (3.47 - 1)g = 2.47 g t = 15 to 20 a = (75874 - 1)g 19255 = (3.83 - l)g = 2.83 g t = 20 to 25 175240 a = (i7540 - 1)g = (4.34 - 1)g = 3.34 g t = 25 to 30 a = (77600 - )g 15425 = (5.03 - 1)g = 4.03 g

-32 - t = 50 to 55 80690 a = (13510 - ))g = (5.98 - l)g = 4.98 g t = 55 to 4o 8438o a = (84580 - 1)g 11595 = (7.22 - l)g = 6.22 g t =40 to 45 f86910 a = (9680 - l)g = (8.99 - l)g = 7.99 g t = 45 a- (89100 )g 7765 = (11.48 -:)g = 10.48 g t = 50 a = (78560 - 1)g 6001 = (13.05 -:)g = 12.05 g

-33 - t = 55 (71o4o 1) a 4414 - )g = (16.1 - 1)g = 15.1 g t = 60 65000 - )g a = (2993 l)g = (21.05 - l)g = 20.05 g t = 65.0 a = (53400 - l)g 1755 = (30.4 - 1)g = 29.4 g t = 67.26 a 49980 l)g a = 1250 -)g = (40.o - 1)g = 39.0 g Calculate Helium Data Let the inside diameter of the helium tank equal the inside diameter of the propellant tank. Volume = 4 ir3 - (4t)(159) = 16.83 ft3 3 3 Use ideal gas law for helium calculations PV = mRT (use standard temperature)

-34 - m = (3500)(144)(16.83) (1545)(518) 4 = 42.4 lbs He Find the required He pressure at the top of the propellant 25% drop across the injector face X - 0.25X = 150 X 4- 150 = 200 psia 3 25% drop across the plumbing X - 0.25X = 200 X = 4 200 = 266.6 psia 3 hydraulic head just before lift off = (477) (62.4) 144 = 20.6 psi throttle the helium so that its pressure at the top of the propellant is 246 psia. Find the time at which the pressure of the helium throughout the system is 246 psia. O 0 ft3 _ mg V = _. sec pp = (11.9)(32.2) 62.4 = 6.15 ft3/sec PlVl = mRT = P2V2 V2 = (3500)(16.83) - 239.6 ft3 Vol of He in tank = 239.6 - 16.8 (assume volume of helium pipes negligible) = 222.8 222.8 = 36.3 sec tl = 6.15

-35 - Calculate BTU's necessary to make process isothermal Starting with the first law Sq = dU + PdV = mCvdT + PdV since we assume helium is an ideal gas Then dT = But 0 so bq = PdV p = RT V so V2 = - f RT dV (1544)(518) 397.83 = (4)(778) ~n 16.83 = 814.5 BTU/lb He = 34,450 BTU Total Calculate head AP = suppression 16 head suppression = PHe + H - PC 9c 0 < t < 5 AP = 246 + 59.7 - 266.7 = 39 psi 5 < t < 10 AP = 246 + 6o.4 - 266.7 = 39.1 psi 10 < t < 15 AP = 246 + 6o.o - 266.7 = 39.3 psi

-36 - 15 < t < 20 AP = 246 + 59.8 - 266.7 = 59.1 psi 20 < t < 25 AP = 246 + 6o.5 - 266.7 = 39.8 psi 25 < t < 30 AP = 246 + 61.6 - 266.7 = 40.9 psi 30 < t < 35 AP = 246 + 63.3 - 266.7 = 42.6 psi t = 35 - 40 AP = 246 + 65.0 - 266.7 = 44.3 psi At t = 36.3 the pressure throughout the helium system is 246 psia. For each time interval (t > 36.3) the helium pressure must be calculated by P1V1 - P2V2 P " Pi Vi PHe - Vi + Vt Where Pi = initial He pressure Vi = initial He volume 0 V = time rate of change of He volume t = time

-37 - t = 40 p _ (3500)(16.83) PHe 16.83 + (6.15)(4o) (5500)(16.83) 262.8 = 233 psia AP = 233 + 65.1 - 266.7 = 31.4 psi Find time this is: that chamber pressure begins to decrease. The condition for PHe + PH - 16 Pc= P = 0 9 (3500)(16.853) + 65.1 266.7 16.83 + 6.15t 58900 = (16.83 + 6.15t)(201.6) = 3390 + 1240t = 55510 124o = 44.8 seconds Since this is very close to 45 seconds consider for calculations that the chamber pressure starts to decrease at 45 seconds. Calculate amount of Fiberglas needed From the head suppression graph head suppression is 44.6 psi. Therefore, bottom of the tank is 266.6 + 44.6 psia Maximum pressure = 311.2 psia it is seen that the maximum the maximum pressure at the

-38 - F/2 F F/2 F = (511.2)(l44)(5.l8)(l) = 142,300 lb/ft Working Stress = 80,000 lb/in2 Neglect atmospheric pressure in calculating wall thickness. This gives an additional safety factor. t 142,300 t = (2) (80,000) (12) =.0741 inch 2 2 r VT = r 2h + 2 ir AV = 2jtrhLr + 2xcr2Ar or Vol Fiberglas = (itDh + 2itr2)Ar = it(Dh + 2r2)t [(35.18)(47.7) + (2)(1.59)2] 0'~74 12 (151.7 + 5.o05)(0.0741):c 12

-39 - (156.75) (0.0741)i 12 = 3.04 ft3 Helium Tank t F = (3500) (144) (1.59)2t = 4,000,000 lb A = tr2 dA = 2trdr A = DAr = = Dt = (4,oo000,000) (12) (144)(80,000)(3.18)t = (348) (12) = 0.417 inches AV = 4ir2t

(4i)(1.59)2(o.417) 12 = 1.106 ft5 Total volume of fiberglas = 3.04 + 1.106 = 4.146 Wt. fiberglas = (4.146)(1.8)(62.4) = 465.0 lbs. Wt. engine = 89,350 400 = 223.37 lb Total Wt. used Wt. Helium = 42.4 Wt. fiberglas = 465.0 Wt. engine = 223.4 Total Wt. used = 730.8 lb. Wt. left over for payload, plumbing, telemetry, etc. = 519.2 lb Payload from problem statement = 125.0 lb. Wt. left over for plumbing, telemetry, etc. = 394.2 lb

-41 - The following performance curves and data sheet present a summary of the calculations for the liquid rocket motor system.

-42 - 70 60 0) t0 w CX - 0, 30 20 10 THRUST, LBS x I0 Figure 1. Rocket Motor Thrust as a Function of Altitude.

-43 - 3000 NA I^ 2000._1 P 03 03.-J C,) w Gr. 0 Z i a). < I 1000 0 0 5 10 15 20 25 30 TIME, SECONDS Figure 2. Dynamic Pressure Versus Altitude.

6001 5000 4000 40 0 3000 CD 2000 1000 15 20 25 30 35 40 45 50 TIME, SECONDS Figure 3. Drag Versus Time

-45 - 7 6 5 w 4 CD 2 _ ~0 10 20 30 40 50 60 TIME, SECONDS Figure 4. Mach Number as a Function of Time.

16C 15C 14C 13C 12C lk 0 o U) (n cr. an I 0 I. IIC IOC 9C 8C ) I --- —'a —...... I________I 7C 6C i 50 40 30 I I I -------- calculations. I i Approximation ---------- I I -0 —0- Connects calculated pressure points. I I used in performance I 1 Feroge chamber 20 10 0 L 0 10 20 30 40 50 60 70 TIME, SECONDS Figure 5. Rocket Chamber Pressure as a Function of Time.

co 60000 Cn30000 - _ _ ____ 0 10 20 30 40 50 60 70 TIME, SECONDS Figure 6. Rocket Motor Thrust Versus Time Until Burnout. 30000 ----------------------------------------------- - -— ~~~~~~~~~~~~~~~~~~~~ 0 10 20 30 40 50 60 70~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ i~ ~ TM,iEOD Figure 6. Rocket Motor Thrust Versus Time Until Burnout.~~~~~~~

x N~ 0) (.) LL z 0 or wJ ti 0 W I i~ 70 TIME, SECONDS Figure 7. Vehicle Acceleration Versus Time.

-49 - 70 60. 50 10 2 - 0 I0 0 10 20 30 40 50 60 70 TIME, SECONDS O~~~~~~TM EOD Figure 8. Hydraulic Head Versus Time.

-50 - 50 4! c 4< 3( 0n Z 21 0 ~n a- 2< Ca O3 < 1 I 5 5 5 - -- 5 I( ( w v0 10 20 30 TIME, SECONDS 40 50 Figure 9. Pressure Drop Across the Head Suppression Valve as a Function of Time.

-51 - 20000 15000 d w 0 0 -J _,1 I> 5000 50 60 70 TIME, SECONDS Figure 10. Vehicle Velocity Versus Time During Powered Phase.

-52 - 400000 300000 - 200000 W I00000 100000 I I 2 I I ILIA I I -. n L 0 10 20 30 40 TIME, SECONDS 50 60 70 Figure 11. Altitude Versus Time During Powered Phase.

-53 - DATA SHEET Time Thrust Dynamic Drag Pressure Sec lb lb/ft2 lb 0- 72290 0 0 0+ 72290 0 0 5 72700 120.3 172.1 10 74200 514.0 695 15 76600 1164 2726 20 79500 1953 4260 25 82700 2620 5100 30 85600 2880 4910 35 87700 2205 3320 40 88700 1328 1790 45 89100 595 848 46.25 47.50 48.75 50.00 78360 51.25 52.50 53.75 55.00 71040 56.25 57.50 58.75 60.00 63000 61.25 62.50 63.75 65.00 53400 67.26 49980 Acceleration Head Chamber Hydraulic Velocity Altitude Mach Suppression Pressure Head Number ft/sec2 Psi Psi Psi ft/sec ft 0 150 20.6 1.89g 39.0 150 59.7 0 0 0 2.14g 39.7 150 60.4 323 780 0.29 2.47g 39.3 150 60.0 690 3294 0.623 2.83g 39.1 150 59.8 1113 7778 1.011 3.34g 39.8 150 60.5 1603 14537 1.512 4.03g 40.9 150 61.6 2182 23961 2.12 4.98g 42.6 150 63.3 2885 36578 2.98 6.22g 44.3 150 65.0 3762 53124 3.88 7.99g 31.4 150 65.1 4882 74635 5.03 10.48g 0 150 65.4 6331 102522 6.52 142.0 60.9 7184 119399 134.5 55.9 12.05g 8117 138453 127.5 50.7 9141 159977 121.0 45.4 15.10g 10282 184212 114.8 39.8 11577 211467 108.0 33.0 20.05g 13067 242199 101.0 24.9 29.40g 39. OOg 93.0 89.o 83.3 14829 276916 14.1 16953 316478 0 19445 357400 Summit Altitude is 1800 Miles.

SOLID ROCKET DESIGN Theodore Petersen

0 m 7-Ex M T c P a 7 A t Ex F n r q NOMENCLATURE mass flow - velocity at exit of rocket nozzle - Mach number - temperature - denotes chamber conditions - pressure - speed of sound - ratio of specific heats - area denotes throat conditions - denotes the conditions at the exit of the rocket nozzle - thrust - density - denotes exponent in burning rate low - burning rate - dynamic head r R 7 + 1 1 /^2 \ 2(y - 1) - ga72 7t + 1) - gas constant -55 -

-56 - DESIGN SPECIFICATIONS Assume propellant chamber with internal dimensions as shown: f 3' I 40 ' Propellant: NH4NO3/C2H40/CATALYST Properties: Adiabatic Flame Temperature Average Molecular Weight 7 = 27000~F = 22 lb./mol. = 1.26 Typical Sea Level Isp Characteristic Velocity r @ Pc = 1000 PSI 70 ~F r Exponent - n Specific Weight Lower Combustion Limit Pressure Limit 195 Sec. 4000 ft./sec. 0.1 in./sec. 0.4 0~056 lb./in. 100 PSI 3000 PSI Specifications: General: Fill up to Base of Nozzle Consider Star Grain Burning time 20 -> 60 seconds Payload = 125 lb. "Fiberglas" Casing - 80,000 PSI Design Stress Protective Heat Material -. 050 ino of Gunk at same Sp. Gr. as "Fiberglas." Motor Weight: 70 lbo Thrust Lbo motor wt, Leave out Volume Increase with Respect to Time but Comment on its Effect, Estimate Velocity at Back End of Grain. Size Nozzle so PX = 1/2 PATM at Sea Level so Don't Need to Worry About Separation, Maximum Pressure Fluctuation = 50% Junk Weight = 150 lb. Lift Off at 4go

-57 - Weight Estimate: Total Cross-Sect. Area of Propellant Assume a loading fraction e = Cross-Sectional Area of Motor Cross-Sectional Area of Motor =.80 (Preliminary Assumption) Volume of propellant charge: (.80)) = 226 Ft,5 Weight of charge: (226)(.056)(1728) = 21,850 lb. Weight of fiberglas casing: 36 ino x 1000 PSI = 36,000 lb. working stress = 80,000 PSI, So the thickness is: 225 in. thick 80,000x 2 The volume of fibyerglas for the walls of the chamber is then given by () (3)(40) (.225)(1/12) = 7.07 fto3 Estimate the same thickness of fiberglas for the top of the chamber and roughly assume 1 ft. of lap joint: l 4+t (5) 1 = 3t(1 + ) Volume = 53Tt(1-75)(.225)(1/12) = 3509 ft3,5 Total volume = 7 379 ft,5 Weight = 7,379 x 62~4 x 1.8 = 828 lb. Weight of insulating material: Volume of insulating material: (At)(5)(40)(.050)(1/12) + (i)(9/4)(.050)(l/l2) = 1,6 fto3 Weight = (lo6) x (62.4) x (1.8) = 1.79o8 lb. Payload weight = 125 lb. Junk weight = 150 lbo Total weight - Motor weight = 23,133 lb. Assume motor weight = 1400 lb. Total Weight = 24,533 lb. For 4g lift-off thrust - thrust = 98,132 lbo - which corresponds to an engine weight of 1400 lb.

-58 - NOZZLE SIZING It was specified to size the nozzle so that PEX = 1/2 PATMO This specification, along with a knowledge of the chamber properties and the desired thrust at sea level, enables the nozzle to be sized using isentropic relations and the basic thrust equation; the relations to be used are: (1) F = m VEX + (PEX - PATM)EX (Assuming Ideal Nozzle) (2) m = -A ~a~~ ~yl () T 2 (4) Pc _ (l + Z M2)7/P 2 - A _1 1+ -1 M2 2(y A...Y- 2 (6) VE = MEX aEX = MEX JaRTEX Now, substituting (2) into (1): (7) F - Zf~MtAtPt *v ( P)A (7) F VEX + (PEX PATM)AEX at and substituting in (6) and cancelling constant terms: (8) F = yMtAtPtMEx(TEx/Tt)2 + EX TM) X and, from the isentropic relationships: Hence, 1Txl/2 - + Y/Y-' pE l1/z7' (9) t /- Y' )2 yPC Also, 2,7/7-1 (10) Pt = 7+) P and, Pc = (1 y- 2 )7/7Y PEX '' MX Y- [ (- F ) X = {1/-1 111/2 (11) {c / -lI

-59 - Now, divide (8) by At: F (12) -t PtMEX t (TEX/Tt) + (PEX - PATM)AE/At and the terms on the right of (12) can be evaluated with the known conditions: Pc = 1000 PSI Tc = 3160~R y = 1.26 PEX = 7.35 PSI From (10): Pt = (0.5532)(1000) = 553.2 PSI From (11): 2064 1/2]} MEX = 7. 69 [ (l3 6.1) }1 / I] = {7.69 [2.76-1]}1/2 = 3.678 From (9): From (5): (EX)1/2 r ( 85 6.1031 = (1.13). (.00735) =.64 AEX - At.678 (.885)(2.758) 3.678 L J = 13505 Then, from (12) 98,152 At 98,132 = (1.26)(553.2)(3.678)(.64) - (7535)(1305) = (1638 + 96) At A = ~ ~ ~ S q ~ ~ I n. ~ Ata =-63'w.h6 Sq. In. and now the mass flow can be computed: o MA P m = at (1.26)(63.6)(553 2) [(1.26)(32.2)(1544) (3160) 1 /2 22 1.13 = 15.7 slugs/sec,

-6o Figure 1. Wagon Wheel Grain Configuration.

-61 - I I / / / / / I / DASHED LINE REPRESENTS |I/ BURNING SURFACE AFTER I / PROPELLANT HAS BURNED I / DISTANCE X. X I - Figure 2.

-62 - CENTER LINE BETWEEN SPOKES I W / V D 2 / / (2 - w ) U. 0 m -4 L Figure 3.

GRAIN SIZING Now, in general, the basic differential equation governing continuity for solid propellant rockets is: Acr Pp - d (pcVc) + 7MtAtPt Propellant Mass Burned - Storage Term -- — Mass Flow through Nozzle and assuming that the storage term is negligible, this equation becomes: Acr Pp = Mt at and from the propellant properties: pp = (.056)(1728) = 96.8 lb./ft 3 n 4 r = a P =a( a( 15.9) =.1 ino/sec. =.00833 ft./sec. c a =.1/15o9 =.00629 so, A = (15.7) (32.2) 626 ft.2 c = (.oo0833)(96.8) 626 ft and, as the chamber is 40 ft. long the burning perimeter must be 15,66 ft. DERIVATION OF EXPRESSIONS RELATING PACKING FRACTION AND BURNING PERIMETER TO CONFIGURATION DIMENSIONS Burning Perimeter: The burning perimeter may be found by considering the half spoke shown in the following sketch and multiplying the result by 10o The perimeter is made up of three portions: The arc subtended by the angle 9, the line of the spoke inclined at the angle P to the

-64 - center-line and the vertical line of length mo The burning perimeter and loading fraction will be analytically determined, The burning perimeter is initially given by the following relationship: B.P.i = 10(Sl + S2 + S3) where: S = (D W) 1 2 s = L- L - ZW- W= ( cos G) Z sinp cos (. - 9) sin S = m= --- ) sin: Now, the cross-sectional area of the propellant can be found. This will be done by subtracting the pie-shaped segments from the total area - then adding back in the triangles of base m and altitude L sin3 - and finally subtracting out the parallelograms of base m and altitude Z. Hence the cross-sectional area of the propellant is given by: D W = D2 5(D 2 W - - n- cos 10 mZ 1 4 2(- sinp cos ( -9) Now, after the propellant has burned a distance x the burning perimeter is: ~BP x 10 (S1 + S2 + S3 + S4) where: S1 = (D - [W-x]) 9 S2 = x (- + ) ( W sin- cos 9) Z 3 ns (a - 9) sinp S4 = m x s ins

-65 - and the cross-sectional area of the propellant after it has burned a distance x can be found as: Ax = Ai - 5 {(2 - W + x) - (2 - W) 9 _ 5 x2 {2 - Q + A} -10 ix - - -10 sin -W cos - Z_ x 10 mx - - -- 10 ~......... tan: 2 L cos (- - i) sinp tanp - x tan ~ DISCUSSION OF CHOICE OF GRAIN CONFIGURATION A sketch of the grain configuration chosen is shown on Page 60. Preliminary calculations were carried out for several star configurations but these all showed that it would be extremely difficult to meet the 4g lift-off specification for any decent loading factor without a highly regressive burning configuration resulting. It was therefore decided not to consider a star (internal) grain configuration any further. A cylindrical grain was briefly considered as it has the advantage of a uniform burning surface but it was immediately obvious that it would not be possible to meet the 4g lift-off requirement with any sort of loading fraction that was acceptable. Also, as it was specified in class not to use this configuration, its consideration was dropped, It was then decided to try the wagon wheel configuration, and a five spoke wagon wheel, as shown in the sketch on Page60, was decided upon as it has only slightly progressive burning as can be seen by the pressure trace on Page 61. An analytical procedure was then determined to size the configuration for a 4g lift-off and an analytic method of determining the burning surface as a function of x, the distance burned, was derived.

-66 - The actual design would differ slightly from the sketch on Page 60 in that the acute angles of the open or port area should be replaced by small fillets to prevent cracking during storing and handling. Also, although theoretically none are needed, there should be some sort of an inhibitor or structural member (i.e., a fine screen) placed along the center line between the spokes to guard against chunks of the propellant being carried downstream to the nozzle as a result of uneven burning -- i.e., see sketch: laig^ -- Unburned propellant Uneven burning in this region Inhibitor or structural would cause chunk of propellant restraining member. to break off except for restraining action of inhibitoro The analytic solution of the optimum grain configuration follows: SELECTION OF CORRECT GRAIN CONFIGURATION It is first necessary to know the relationship between burning perimeter and loading fraction e for a 4g lift-offo This relationship may be plotted by preceding in a manner similar to that used in the preliminary calculations from the calculations on Page62, F *2 At = 542 ino and from the calculations on Page63: 0 cA^ = m~ = o0064 F ft. when F is in units of rpp pounds

-67 - and to. determine the trust necessary for a given loading factor it cah be seen from the calculations on Pages 57and58that: Total weight = T.W. = 1283 + (27350)e + motor weight Motor weight = F/70; F = 4 x T.W. and the values of thrust and loading factor that satisfy these relationships are given below along with the corresponding values of Ac and burning perimeter: for e =.60: Total weight = 18753 lb. Thrust = 75,012 lb. Ac = 480 ft2 B. P. = 12 ft. for C =.70 Total weight = 21648 lb. Thrust = 86,600 lb. Ac = 554 ft.2 B. P. = 13.85 ft. for e =.75 Total weight = 23103 lb. thrust = 92,412 lb. Ac = 591 ft.2 B. P. = 14.79 ft. for e =.80 Total weight = 24533 lb. Thrust = 98,132 lb. Ac = 627 ft.2 B. P. = 15,67 fto for e =.85 Total weight = 25983 lb. Thrust = 103932 lb. Ac = 665 ft.2 B. P. = 16.63 ft. GRAIN CHARACTERISTICS Consider a grain with a four inch web and five spokes (the five spoke configuration was chosen as it gives the most neutral progressivity ratio). Then: W = 4"; 2 = 18; 3 = 36~ =.628 rad.; sink =.588; 9 = 56~ - sin-1 1 = 36.0~ - 16.6~ = 19.4~ =.3385 rad. cos 9 =.943; 'cos @-Q)= cos 16.6~ =.959

then, S1 S2 s3 = (14)(.553385) (14 - 6.42).959 = 6.8o in. = 4.74 in. - 1.7z = 7.91 - 1.7z Therefore, 10 B. Pi = 1- (19.45 - 1.7z) ft. and the cross-sectional area is: Ai = -0O8 - 5 x 196 x.3385 + 5 = 1018- 332 + 158 - 68z = so the loading factor is given by: 844-68z Ei = Then, for z = 0, B.P.i = 16.20 ft., for z = 1, B.P.i = 14.8 ft., for z = 2, B.P.i = 135.537 ft., W= 39 = 18; = 3 = 18; = 6~; = 56 sin-1 =.53475 rad. cos 9 =.940; cos (P-$) =.961 then, S1 = (14.1)(.3475) = 4.90 in. S2 = (14.1 -6.24) 1.7z = 8.961 S3 = 6.64 in. x 6.8 x 7.91 x.588 - 10 x 6.8 z 844 - 68 z Ei Ei ci 14.1 =.829 =.761 =.695 = 36~ - 16.05~ = 19.950.18 - 1.7z Therefore, B.P.i = 10 (19.72 - 1.7z) and the cross-sectional area is: Ai = 1018 - 5 x 198.8 x,3475 + 5 x 6.64 x 8.18 x.588 - 10 x 6,64z = 1018 - 346 + 159.5 - 66.4z = 832 - 66.4z

so, the loading factor is given by: ei = 832 - 66o4z 1018 Then, for z for z for z w = 595? D cos 9 then, S1 S2 S3 = 0 BoP.i = 16.42 ft., ci =.817 = 1 BP.i = 15 ft., ci = ~752 = 2 B.P.oi = 13.6 ft., i=.687 = 8, = 3P6, 9 = 36~ 6~ sin-1 f = 3 56~ - 1631~ = 19.690 14.05 =.942, cos (P-) =.960 3435 rad, = (14.o05)(.3435) = 4.83 in. (14.05 - 632) 1.7z = 806 - 17z.960 = 6.71 in. Therefore, B.P. = 12 (19.6 - 1.7z) and the cross-sectional area is: A. = 1018 - 5 x 197.53 x.3435 + 5 x 6.71 x 8.06 x.588- 10 x 6.71z = 1018 - 339 + 159 - 67,lz = 838 - 67.lz so the loading factor is given by: 838 -67.lz i 1018 Then, for for for z =0 z =1 z =2 B.Pi = 16,32 fto, BoPoi = 14,91 fto, BoP.i = 1355 ft,, ci 1. 1 = 823 =.757.691 DISCUSSION OF INITIAL CONFIGURATION DESIGN PLOT The plot on the preceding page was used to analytically size the wagon wheel configuration for a 4g lift-off thrusto It should be noted that the configuration with W = 5395" and z = 1" is a design point and this was the

.9 ^^/ go | ~-0- B.P.i vs. ei for 4g take-off /^^ -^-~ | B.P.j vs. Ej for W = 4".~~/^ 1~~~ --- | 3.P.P vs. e. for W = 3.9511 -- - | B.P.i vs. Ei for W = 3.9".6 12 13 14 15 16 17 BURNING PERIMETER (FT.) Figure 4.

-71 - design point chosen for the configuration as it was felt that z = 1" was a reasonable value to keep the internal velocity at a reasonable magnitude and still maintain a fairly high loading factor. This plot could be extended to cover various g take-off conditions and more values of W so as to make a complete design chart for a given size propellant chamber and the five spoke wagon wheel configuration. It should also be noted from the plot that the value of W = 3.95" is a very good choice for a 4-g liftoff as it comes very close to giving this for a range of values of e and z~ DETERMINATION OF BURNING PERIMETER AS FUNCTION OF TIME Using the five spoke wagon wheel with a web of 3.95" and z = 1" the burning perimeter as a function of x is: B.P. = 10 {(14.05 + x)(.3435) + x( -.3435 +.628) + (6.36 - x[.3249]) + (6.71) - 1.7x} = 10 {4.825 +.3435x + 1.8553x + 6.36 -.3249x + 6.71 - 1.7x} 12 = 10 {17.9 +.1739x} ft. Now, for C =.757 the total propellant weight is 20680 lb. and the mass flow through the nozzle is given as: 2,/7 2 7+1/2(y-1) 7~tAtPt - 7APt - 7(t3) AtPt - 7(77l) __ tPt at at 2 1/2 ac a 7~1 a the rate of generation of mass is: Acr pp = B.P. (40) aPc PP and these must be equal so (B.P.)(4o)(.00oo629)P4 (96.8)(1/12) =.7408 At Pc (144).(l.26)(52.2)( )(53160)

-72 - at t = 0, B.P. = 14.91 fto, Pc = 1000 PSI, so, (14.91) (40)(.00629)(96.8)(1/12) = (37408) (1440)At (lo00) 6(322) 35000 At =.4195 ft2 and then from the expressions on Page ~ F = (1532)(.4195)(144) = 93,200 lb. so the motor weight will be 1332 lb and the total weight will be 25,295 lb. so the lift-off will be at 4g's as expected. Meanwhile, back at the burning perimeter and hence chamber pressure as a function of time -- the governing equation is: (B.P.)(40)aPk p = Pt ac or, rearranging: l-n ac ~ 40 - a pp (B.Po) P =...... c r At Now, as the geometry is fixed and the combustion is assumed to take place at constant temperature the only variables in the above equation are PC and (B.P.), Taking the logarithms of both sides: 1-n Log Pc = Log {ac~ ao Pp} + Log (B.Po) F At where the first term on the right hand side is a constant. This constant may be evaluated by substituting in the values for (BoPo) and Pc at time t = 0. Hence (1-n) Log Pc = C + Log (B.P.).6 Log (1000) = C + Log (179)

-73 - Therefore C = 1.8 - 2.25285 + 10 - 10 = -.45285 so,.6 Log P = -.45285 + Log (B.P.) Log Pc = -.75475 + 1.66667 Log (B.P.) Now the procedure for determining the chamber pressure as a function of time will be as follows. At time t = 0 P is known and a burning rate c can be determined from this pressure. This burning rate will be assumed constant for a five second time interval -- thereby giving the distance, x, that the propellant has burned over the interval. Then a burning perimeter corresponding to that x can be determined and from the above equation a new value of pressure. t = 0 -> 5 seconds: x = 0 ->.5" B.P. = 1017.9 +.0869] = 179.87 in. and Log P -.75475 + (1.66667) Log (179.87) C = -.75475 + (1o66667)(2.25496) = -.75475 + 3.75826 = 3.00351 Therefore: P - 1008 PSI c5 and small pressure change will be assumed to have a negligible effect on r t = 5 - 10 seconds: x =.5" -> 1" B.P. = 10117.9 +.1739] = 180.739 in.

-74 - and Log Pc = -.75475 + (1.66667) Log (180.739) = -.75475 + (1.66667)(2.25706) = -.75475 + 3.76176 = 3.00701 Therefore: Pc = 1016.3 PSI and this pressure will again ca-use negligible change in burning rate. t = 10 -> 15 seconds: x = 1" -> 1.5" B.P. = 10[17.9 +.2608] = 181.608 in. and Log Pc = -. 75475 + (1.66667) Log (181.608) = -.75475 + (1.66667)(2.25914) = -.75475 + 3.76523 = 3.01048 Therefore: PC = 1024.4 PSI and this pressure corresponds to an r of.1007 in., in./sec. will be used. t = 15 -> 20 seconds: x = 1.5" -* 2.005" B.P. = 10[17.9 +.3487] = 182. 487 in. /sec. so r =.101 and Log Pc =-.75475 + (1.66667) Log (182.487) = -.75475 + (1.66667)(2.26123) = -.75475 + 3.76872 = 3.01397

-75 - Therefore: P = 1032.7 PSI + r =.101 in./sec. C20 t = 20 -~ 25 seconds: x = 2.005" -4 2.510" B.P. = 10117.9 +.4365] = 183.365 in. and Log P = -.75475 + (1.66667) Log (183.365) = -. 75475 + (1.66667)(2.26332) = -.75475 + 3.77220 = 3.01745 Therefore: P = 1041.0 PSI 4 r =.1013 in./sec. =.101 in./sec. 25 t = 25 -> 30 seconds: x = 2.510" - 3.015" B.P. = 10117.9 +.5243] = 184.243 ino and Log Pc = -.75475 + (1.66667) Log (184.243) = -.75475 + (1.66667)(2.26539) = -.75475 + 3.77565 = 3.02090 Therefore: P30 = 1049.3 PSI ~ r =.102 in/sec. t = 30 - 35 seconds: x = 3.0151" -> 3.525i B.P. = 10[17.9 + -6130] = 185.866 in. and Log P = -.75475 + (1.66667) Log (185.130) = -.75475 + (1l66667)(2.26748) = -.75475 + 3.77913 = 3.02438

-76 - Therefore: P = 1057.8 PSI r =.1017 t = 35 -4 39.150 seconds: x = 3.525" -> 3.948" B.P. = 10[17.9 +.6866] = 185.866 in. 4r =.102 ino/sec. and Log Pc = -.75475 + (1.66667) Log (185.866) = -.75475 + (1.66667)(2.26921) = -.75475 + 3.78202 = 3.02727 Therefore: P = 064.8 PSI r =.102 in./sec. c39.15 t = 39-150 - 39.167: x = 3o948" -- 3.95" B.Po = 1011.8553x]= (18.553)(3.95) = 73.284 in. and Log P = -.75475 + (1.66667) Log (73.284) C = -.75475 + (1.66667)(1o86501) = -.75475 + 3.10835 = 2.35360 Therefore: PC = 225.74 PSI 39o167 and the r may be computed as: Log (Pc)' 4.94144 4 (Pc)'4 = 8.7386 r = (.00629)(8.7386) =.054 in./sec. t = 39. 167 - 42.5: x = 3-95" - 4l13" and from the plot on the following page: B.P. = 10[(4.13)(1.25)] = 51.625"

-77 - and Log P = -.75475 + (1.66667) Log (51.625) C = -.75475 + (1.66667)(171286) = -.75475 + 2.85477 = 2.10002 Therefore P42.5 = 125.9 PSI and the r -may be computed as: Log Pc- =.84001 Pc4 6.9185 r = (.00629)(6.9185) =.044 in. /sec. t = 42.5 -> 44.5: x = 4.13" -* 4.22" B.P. = 10 [(4.22)(1.098)] = 46.3" and Log P = -.75475 + (1.66667) Log (46.3) C = -.75475 + (1.66667)(1o 66558) = -.75475 + 2.77596 = 2.02121 Therefore: Pc44 5 = 105.1 PSI and. the r may be computed as: Log P'4 =.80848 P4 = 6.434 c C r = (.oo00629)(6.434) = o040 t = 44.5 -> 45: x = 4.22" ~> 4.24" B.P. = 10 [(4.24)(1.053)] = 44.65 and Log P =-.75475 + (1.66667) Log (4465) -= -.75475 + (1.66667)(1.64982) = 1.99495 Therefore: PC = 98.85 PSI

-78 - I I Figure 5. Burning Grain Geometry.

10 8 co *0 x (/ 6 a: 4 2 0 I__ _ _ _ _ I_ _ I I__I__I__I_ __-__ _ _I I \0 I - -___~___,,!zz_.1 E 0 5 10 15 20 25 30 TIME (SECONDS) Figure 6. Chamber Pressure with Time. 35 40 45

-80 - DBTERINATION OF PROPEIIANT. CROSS-SECTIONAL AREA AND WEIGET AS A FUNCTION OF TIME The cross-sectional area is given by the expression on Page as: A = 5 {( W + x)2 ( - W)} X +}-10 A =A ~2 2... w 2.. {( w cos a) Z -x t Cos} x tanp cos (p-e) s-in tand 2 and for the particular configuration chosen this reduces to: A = 770.63 - 5 {(14.05 + x)2-(14.05)2}(.3435)-5{l.8553}x2 - 10(6.7l)x x - 10(.6882)x2 - o10{8.06 - 1.7 - 1.3764x -.3249x}x = 770.63 - 1.7175 {28.1x + x2} - 9.2765x2 - 67.1x -6.882x2 - (80.6 - 17)x + 17.013x2 = 770.63 - 178.962x -.862x2 and evaluating at the following times at t = 0 A = 770-63 in2 =.757 at t = 5 Ax = 770.63 - 89.481 -.216 = 680.93 in2 e =.669 at t = 10 Ax = 770.63 - 178.962 -.862 = 590.81 in2 E =.580 at t = 15 Ax = 770.63 - 268.443 - 1.94 = 500.25 in2 e =.492 at t = 20 Ax = 770.63 - 358.7 - 3.46 = 408.47 in2 e =.401 at t = 25 Ax = 770.63 - 449.5 - 5.43 = 315.70 in,2 =.310 at t = 30 Ax = 770.63 - 539.5 - 7.83 = 223.3 in2 e =.219 at t = 35 Ax = 770.63 - 630.5 - 10.70 = 129.43 in2 E =.127

at t = 39.15 A = 770.63 - 707.0 - 13.43 = 50.20 in2 e =.0493 at t = 39.167 A = 770.63- 707.0 - 13.43 = 50.20 in2 =.0493 at t = 44.8 Ax = 5020 {(232 } 423 - 395}{.8 10 where g = 60.4" = 1.053 radians Ax = 50.20 - [(.5265)(2.29) + *224] 10 = 35.91 in2;. =.0353 and similarly the weight as a function of time is given as: W = 23.295 + (-.757)(27350) so, at t = 0 seconds W = 23,295 lb. t = 5 seconds W = 20,888 lb. t = 10 seconds W = 18,455 lb. t = 15 seconds W = 16,045 lb. t = 20 seconds W = 13,555 lb. t = 25 seconds W = 11,075 lb. t = 30 seconds W = 8,585 lb. t = 35 seconds W = 6,055 lb. t = 39.15 seconds W = 3,935 lb. t.o. = 44.8 seconds W 3,545 lb. THRUST FOR VARYING Ps AND PA We know from Equation (12) on Page 5 8 that: F = r PAx (X )1/2 - EX At A

2.5 to - w X: I 0O Ro I 0 5 10 15 20 25 30 35 40 TIME (SECONDS) Figure 7. Weight Versus Time. 45

-83 - now, multiply by At and divide by P0 F Pt 1/2 PEX PA P =7 P MEX (TEX/T) At + ( ) C "PC PCAtEX and now the first two texrs on the right hand side are known constants from the propellant specifications and nozzle geometry and may be evaluated: F = (l.26)(.5532)(3.678)(.64)(60o.4) + (.00735)(13.05)(6o.4) P C - A (13.05)(60.4) PO so, F = (99.0 + 5.8) PC - 788.5 PA = 104.8 Pc - 788.5 PA for a check, consider lift-off: F= (lo4.8)(l000) - (788.5)( (14.7) = 104,8oo00 - 11,600 = 93,200 lb. and this checks vwith the previous alculation of this value found in the section dealing with the variation of burning perimeter as a function of time. PERFORMANCE EQUATIONS Consider a rocket vehicle in a vertical L 8 trajectory. Assuming drag negligible as specified for performance calculations, and computing the velocity and altitude relations in time increments, it was shown in the previous design LU L ~J

-84 - study that: gTAt Wo f = vo Wo'Wf H t and Wf At - = o + o Wo-W Wf 2W Where: f represents values at the end of the interval o represents values at the beginning of the interval At represents the time interval It should be noted that the average mass flow over the interval is being used. Although this is more accurate than taking the mass flow at the beginning of the interval the change is very slight due to the almost constant P and the average is being ued primarily for ease of computation. The thrust is evaluated at the beginning of the interval. Because of the nearly constant Pc 5 second intervals will be used as it is doubtful that any significant amount of accuracy would be added by considering smaller increments: From t = 0 -* 5 seconds Vf Vo + W {ln Wo Wf - gt V5 = 0 + 6230 {in 23,295 - in 20,888} - 161 = 6230 {10.05599 - 9.94693} - 161 = 6230 {.10906} - 161 = 680 - 161 = 519 ft./sec.

-85 - and hf = ho+ VgAt + f WfAt ln(W ln(Wo) At} - g 2At ~ ~ Wof { Wfl ) - (Wo)} + 2 t hg = o + o + 6230 {43.3 {-.g0906} + 5} - 402.5 = 6230 {-4.73 + 5} - 402.5 1682 - 402 = 1280 ft. 1280 ft. PA = 14.03 PSI, Pc = 1008 PSI F = 104.8 (1008) - 788.5 (14.03) = 105,700 - 11070 = 94, 630 lb. and at From t = 5 -~ 10 seconds VO = 519 + 161 94,630 {in 20,888 - In 18,455} - 161 lo 9 (20,888 - 18,455) = 519 + 6260 {9.94693 - 9.82309} - 161 = 519 + 775 - 161 = 1133 ft./sec. hio = 1280 + (519)(5) + 6260 {37.95 {-.12384} + 5} -402.5 = 1280 + 2595 + 6260 {-4.695 + 5} -402.5 = 1280 + 2595 + 1910 - = 5383 ft. and at 5383 ft. PA = 11.99 PSI, Pc = 1016.3 F = 104.8 (1016.3) - 788.5 (11.99) = 106,300 - 94.45 = 96,855 lb. From t = 10 - -15 seconds V15 = 1133 + 161 96,855. {n 18,455 - n 16,045} - 161 (18,455 - 16,045) = 1133 + 6470 {9.82309 - 9.68315} - 161 = 1133 + 905 - 16l = 1877 ft./sec.

h?5 = 5383 + 1133(5) + 6470 {33.35 {-.13994} + 5} -402.5 = 5383 + 5665 + 6470 {-4.67+5}-402.5 = 5383 + 5665 + 2135 - 402 = 12781 ft. and at 12,781 ft. PA = 9.060 PSI, P. = 1024.4 PSI F = 104.8 (1024.4) - 788.5 (9.o6o0) = 107350 - 7145 = 100,205 lb. From t = 15 -4 20 seconds V20 = 1877 + 161 ' 100,205 {in 16045 - in 13555} - 161;- (16,045 - 13,555) = 1877 + 648o {9.68315 - 9.51451} - 161 = 1877 + 1093 - 161 = 2809 ft./seec h20 = 12781 + 1877(5) + 6480 {27.2 {-.16864} + 5} - 402.5 = 12781 + 9385 + {-4.58 + 5} 6480 - 402.0 F + 2781 + 9385 + 2720 - 402 = 24,484 ft. and at 24,484 ft. PA = 5.574 PSI, Pc = 1032.7 PSI. F = 104.8 (1032.7) - 788.5 (5.574) = 108,250 - 4395 = 103,855 lb. From t = 20 -* 25 seconds Va:= 2809 + r6t 103,855 {ln 13555 - ln 11075} - 161.... (13555 - 11075) = 2809 + 6740 {9.51451 - 9.31245} - 161 = 2809 + 1362 - 161 = 4010 ft./sec.

-87 - and h2 = 24,484 + (2809)5 + 674o {22.34 {-.20206} + 5} - 402 = 24,484 + 14045 + 6740 { —4.515 + 5} - 402 = 24,484 + 14045 + 3267 - 402 = 41,394 ft. i at 41,394 ft. PA = 2.545 PSI. PC = 1041 PSI F = 104.8 (1041) - 788.5 (2.545) = 109,050 - 2006 = 107,044 lb. From t = 25 - 30 seconds V3 = 4010 + 161 1 07,01 1 {ln 11075 in 8585} - 161 30 (11075 - 8585) = 4010 + 6950 {9.31245 - 9.05777} - 161 = 4010 + 1770 - 161 = 5619 ft./sec. h30 = 41,394 + (4010)5 + 6950 {17.22 {-.25468} + 5} - 402 = 41,394 + 20050 + {-4.390 + 5}(6950) - 402 = 41,394 + 20,050 + 4240 - 402 = 65,282 ft. and at 65,282 ft. P% =.808 PSI, Pc = 1049.3 PSI. F = 104.8 (1049.3) - 788.5 (.808) = 109,900 - 637 = 109,263 lb. From t = 30 - 35 seconds V35 = 5619 + 161 109263 { in 8585 _ in 655} - 161.... (8585 - 6055) = 5619 + 6960 {9.05777 - 8.70864} - 161 = 5619 + 6960 {.34913} - 161 = 7888 ft./sec.

h35 65,282 + (5619)5 + 6960 {11.97 (-.34913) + 5} - 402 = 65,282 + 28095 + 6960 (.825) - 402 = 98,715 ft. and at 98,715 ft. PA =.1641 PSI Pc = 1057.8 PSI. F = 104.8 (1057.8) - 788.5 (.1641) = 110,900 - 129 = 110,771 lb. From t = 35 -> 39.167 seconds V39 167 = 7888 + (4.167)(32.2)(UO,771) {ln 6055 - ln 3905} - 135 39.167 (6055 - 3905) = 7888 + 6910 {8.70864 - 8.2700} - 135 = 7888 + 3030 - 135 = 10783,ft./sec. h39.167 = 98,715 + 7888 (4.167) + 6910 {7.57 (-.43863) + 5} - 280 = 98,715 + 32820 + 11620 - 280 = 142,875 ft. and at 142,875 ft. PA = *026 PSI, P = 225.74 PSI F = 104.8 (225.74) - 788.5 (.026) 23,650 - 20 = 23,630 lb. From t = 39,167 seconds -> 42.5 seconds v42 5 =10783 + (3']333)(32.2)(23,630) {ln 3905 - In 3620} 107 V' (3905 - 3620) = 10783 + 8900 {8.27001 - 8.19423} - 107 = 10783 + 674 - 107 = 11,350 ft./sec. 11425 = 142,875 + (10783)(3.333) + 8900 {42.3 (-.07578) + 5} - 111 = 142,875 + 35900 + 15930 - 111 = 194,594 ft.

12 10 8 0 J I 6 4 2 0 WEB BURNOUT I________I_________________X___ k_.. I 10:: I 0 5 10 15 20 25 30 35 40 45 TIME (SECONDS) Figure 8. Thrust Versus Time.

12 10 _ 8 o) x n 0 4 C BURN - OUT/~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~,a C~~~~~~~~~.j,, I...,... I,,,i i 0 1 ) 5 10 15 20 25 30 35 TIME (SECONDS) Figure 9. Rocket Velocity as a Function of Time. 40 45

25 20 1'0 x U. w I_J 15 10 NOTE: FROM VARYING GRAVITATION CHART / THE SUMMIT ELEVATION WOULD BE 500 MILES..I AX.. I I 5 C) 0 5 10 15 20 25 30 35 40 45 TIME (SECONDS) Figure 10. Rocket Altitude Versus Time During Powered Phase.

-92 - and at 194,594 PA is negligible and Pc = 125.9 F = 104.8 (125.9) = 13,200 lb. From t - 42.5 -> 44.8 seconds (Burn-out) V44 8 = 11,350 + (2'3)(32.2)(13,200) f{n 3620 - in 3545} - 74,440 ~ (3620 - 3545) = 11,350 + 9870 {8.19423 - 8.17329} -74 = 11,350 + 198 - 74 = 11,474 ft./sec. h44.8 = 194,594 + (11,350)(2.3) + 9870 {108.8 {-.02094} + 5} - 85 = 194,594 + 26100 + 26800 - 85 = 247,409 ft. UTATION OF MACH 1MBER AND DYNAMIC HEAD At t= 0 seconds M=0 q= 0 At t = 5 seconds c = 1112 ft. /see. M = = 466 1112 - p = (.9632)(.002378) =.00229 lb.sec. ft,4 q = 1/2(.00229)(519)2 = 309 lb./ft.2 At t = 10 seconds e = 1096 ft./sec. M = 1 - 1.035 p = (.8518)(.002378) =.002024 lb.1 ec.2 ft.4 i = 1/2 (.002024)(1133)2 = 1300 lb./ft.2 At t = 15 seconds c = 1067 ft./sec. M = 1877 1.76 1067 - p = (.6761)(.oo002378) =.oo001608 lb se =q-~~~~ 1/.068 83l ft.* q. = 1/2 (.ool6o8)(1877)2 = 2833 lb./ft.2

-93 - At t = 20 seconds c = 1018 ft./sec. M = 28 276 -- 1018 --- p = (.456)(.002378) =.001085 lbse2 ft.4 q= 1/2 (.o001085)(2809)2 = 4277 Ib./ft.2 At t = 25 seconds 4010 c = 968.5 ft./sec. M= 9685 = 4.15 p = (.2302)(.002378) =.000547 lb see 2 ft,4 q= 1/2 (.000547)(4010)2 - 4395 Ib,/ft.2 At t = 30 seconds c = 968.5 ft./sec. M = 5619 5.8 --- 968.5 -- p = (.07305) (.002378) =.0001738 lb.se.2 ft,4 q= 1/2 (.0001738)(5619)2 =2745 lb./ft.2 At t= 35 seconds O = 1002 ft./sec. M 8 = 7.87 - 1002 lb. sec.2 p = (.01407)(.002378) = (.00003343) fts - q = 1/2 (.00003343)(7888)2 = 1040 lb./ft.2 At t = 39.167 seconds c = 1086 ft./see. M= 10783 = 9.93 lb. sec. 2 p = (.001967)(.002378) =.000004675 l q = 1/2 (.000004675)(10783)2 = 272 lb./ft.2 At t = 44.8 seconds,......,,.....;...,u Too high an altitude for values to be significant. NOTE: The values of density and speed of sound used above wre interpolated from the Pratt and Whitney handbook.

-94 - 10 - 8 -., 6 -m:3 2 -D 4 - 0 0 5 10 15 20 25 30 35 40 TIME (SECOND) Figure 11. Mach Number Versus Time.

4 3 o. CD - w z Q 2 I 0 0 5 10 15 20 25 30 35 40 TIME (SECONDS) Figure 12. Dynamic Head Versus Time.

-96 - DRAG CALCUIATION As was specified, drag effects were ignored in the performance evaluation. Using the results of the performance evaluation the drag will now be calculated and the amount of additional fuel necessary to counteract the negative impulse due to the drag determined. Assuming an e/D = 3 for the nose cone: L = 40 + 35(53) = 49 Hence a characteristic dimension of 60' will be assumed in computing the Reynolds Number -- this should be conservative. The drag is given by the relationship: where S will be At t = 0 D = q CD S conservatively assumed to be 7.1 square feet. seconds D = 0 seconds R _ PVL Re = At t = 5 so, and from CD a nose cone R = e plot with CD = (.00229)(519)(60) = 193.2 x 3.693 x 10-7 handed out in class it can be i/D = 35: =.175 r may be evaluated: (.175)(7.10)(309) = 35853 b. 106 seen that for and then the drag D =

-97 - At t= 10 4 —. At t=15 At t 20 seconds R = (. 002024) (l83)(6o) 381 x 3.610 x 10l7 CD = *295 D = (.295)(710o)(1300) = 2720 o1 seconds (. oo16o8)(1877)(60) Re - o3853x-1 = 525 x CD. *255 D (.255)(7.10)(2833) = 5130 lb. seconds R (.00o1085)(2809)(60) - 567 e ~' 3.218 x 10-' (. CD =.215 D = (.215)(7.10(4277) 6530 lb. seconds. *-(. ooo 7. (oo)(_ 445 Re... = ~ 445 ' e 2.961 x 10-7 CD.= 18 D = (.18)(7.10)(4395) = 5610 lb. seconds Re (.0001738)(5619)(60) 198 x 2.961x 10-7 CD = o155 D = (.155)(7.10)(2745) = 3025 lb. seconds Re = (."00003333)(7888)(60) 53 2.961 x 10-7 CD =.16 D =- (.16)(7.10)(l0o40) = 1L80 lb, 106 106 106 x 106 At t = 25 x 106 At t- 30 106 At t = 35 106

-98 - At t = 39,167 seconds R = ooo000004675)(10783)(6) 748 x 106 4.032 x 10-7 CD -.19 D - (.19)(7.10)(272) = 366 lb. The total impulse (drag) is equiivalent to 134 seconds of sea level thrust or an additional (1.34)(481.4) = 645 lb. of propellant. This impulse as obtained by integrating the area under the following curve. ACCELERATION COMPUTATION The accelerations may be computed by the second law credited to some obscure physicist, This was used in the derivation of the performance equations and is in general stated as: T D W= W a g But as we are neglecting drag: T =W= - a g a =[T-l1g and computing the accelerations: At t = 0 seconds a = [932 1] g -- 23,295 = [4.00 1] g = 3.0 g At t = 5 seconds a = [94630 1] - 20,888 = [4,53 - 1] g = 3.53g

7 6 5 4 0 -c cJ C) I \10 \10 IO 3 2 I - 0 0 20 25 TIME (SECONDS) Figure 13. Drag Versus Time.

At t = a At t = a At t = a At t = a At t = a At t = a 10 seconds [r96855 1] g ~ 18, -55 [5.25 - 11 g= 15 seconds [100,205 ]. 16,045: [6.25 - 1] g= 20 seconds [103,855 13,555 ] - [7.66 - 1] g= 25 seconds - [107i] 11,075 = [9.67- 1] g = 30 seconds. [109.263. 8585 = [12.73 1] g: 35 seconds - 110,771, 1 6055 = [18.3 - 1] g 39.15 seconds 111., _ 4.] 3935 = [28.3 - 1] g= 4.25 g 5.25 g 6.66 g 8.67 g = 11.73 g At t = a g 17.3 g g 27.3 g

30 l 25 U/) W / Q ' WEB BURNOUT - 0 5 0 5 10 15 20 25 30 35 40 45 TIME (SECOND) Figure 14. Rocket Acceleration as a Function of Time.

-102 - At t= 39.167 second a [23'630 ] 3930 = [6.02 - 1] g At t = 42.5 seconds a = [13 200 3620 [3.65 - 1] g At t = 448 seconds a = [10,40 8 1] 3545 = [2.96.- 1lg g = 5.02 g g = 2..65 g g = 1.96 g DISCUSSION OF STORAGE TERM As stated on Page the basic differential equation governing continuity in solid propellant rockets is: d ( MtAttP Acrp = (p V) tAt + andy the mass flow out the nozze can also be written as: NMtAtPt PeAt..: at ac where: F7+1 r, = r^ = 7( ) 2(-1) and introducing the equation of state: Pc = PcRTc and defining the abbreviation: Pp = ppRT

-103 - The equation becomes: A r PP d (pcvc) + rPAt C RT0. dt RTC + a assuming T. is constant: - A ~rP = d_ (PcV,) + r —.tR. dt ac = d (Pcv ) + r PAt fRT7 dt now consider the storage term: t Vc =Vo + f r ACdt dVc - r A. dt and * d (Pcy) = Pcve + Vc c dt dt dt dPt = e3r Ac + e dt so the equation can be written: Vc = Acr (PpPc) - rPcAt cadt C A c t rc Po c} = A {r(Pp-P )At- P A t}RT C Now, introduce the expression for the burning rate: r = a Pc Then, the equation becomes: Ve t - = Ac {aPc n(PpP) r At Rc} Now, P can be neglected in comparison to P in the term (Pp'Pc) as Pp is of the order of a h is of the order of a hundred times as large (i.e. for asphalt potassium

perchlorate propellants it is 125,000 PSI and for ballistite it is 2530,000 PSI) - then: At V c g AC I{aPpPn IMrp and it can be seen that the above equation is not easily solvable for P0. Consider now the case of equilibriua neutral burning so dP~/dt = 0, then: aAcPP - rPc i AC 1-Pn.. aAaPP e r at,A * CT C aAP rAt now consider the equation when the storage term is neglected: AcrPp = PPcAt ART (from Equat. * on previous page) or A aPn = rP A BRT c c C t so, 1-n aA.PP P - 0 rho A RT. so it can be seen that neglecting the storage term will have no effect on the pressure solution as long as there is neutral burning (dPc/dt = 0) -- hence it would have negligible effect on the pressure trace of this design up to the point of web burnout. At web burnout its effect would be as shown in the sketch below:

-105 - including storage term C t ~\ / That is, it would not allow an instantaneous change in pressure - consequently the rocket would maintain a higher thrust level but burn out sooner. DETERMINATION OF VELOCITY AT END OF GRAIN The maximum velocity at the back of the grain will occur at the instant of starting as the mass flow is relatively constant, actually increasing slightly for the first 39.167 seconds of burning -- but, the port area is increasing more rapidly so the critical time will be starting. The mass flow is given by: 0 m = P cApV = 481.4 lb./sec. 2 where: Ap = 1018 - 770.63 = 247.37 in. = 1.718 ft. and from the equation of state: P (1000looo)(144) lb. ~R lb. That =...=....,ot allo........... in. pesse.. PC ( 22LO315)T 1544 A2 hRT m s-f)(316o) ft. b.OR ft.2 =.649 lb./ft.3

-106 - Hence, 481.4 lb. ft.5 V 4= —. ----l ll l....... =..2ft*/sec (V 649)(1.718) see.lb.ft 2 ft./sec. and this velocity will vary linearly to its value of O at the front of grain. COMPARISON WITH LIQUID PROPELIANT ROCKET DESIGN ** CoMparison Liquid Solid Payload Weight - lbs. 125 * 125 Lift-Off Weight - lbs. 25.000 23,295 Mass Ratio 20/1 6.58/1 Lift-Off g Loading 2.95 4.0 Maximum Acceleration 38.65 g 27.3 g Burn-Out Velocity - ft./sece 16,151 11, 474 Burn-Out Altitude - ft. 326,298 247,409 Summit Altitude - ft* 5.73 x 106 2.64 x 106 * Although 125 lb. was the specified payload weight the design analysis showed that this figure could be easily doubled for the performance shown above as there was a weight margin of 404.5 lb. From the above comparison it is obvious that the liquid propellant design is the better of the two. The two things that hurt the solid design most are high motor weight and unburned propellant. The high motor weight is due to two factors: 1) the high thrust level necessary to attain a 4g lift-off, and 2) the relatively high ratio of motor weight ** The comparison presented above between the Liquid Motor System and Solid Motor System uses a liquid motor design comparable to the one presented here but different in detail. The comments are pertinent.

-107 - to thrust produced, which is necessary as there is no way to cool the motor. Although the unburned propellant only represented 35.535% of the volume of the propellant chamber it contributed 965 lb. of dead weight which severely hampers the performance. It would be an interesting study to consider other grain configurations to minimize this factor. SUMMARY The design of a solid propellant rocket to meet the specifications listed on Page was carried out, Preliminary calculations were based on a loading (or packing) fraction of.80. After consideration of several grain configurations the five spoke wagon wheel was chosen as it provided a nearly neutral burning configuration that would meet the lift-off requirements for a fairly high packing fraction. An analytic procedure was developed to determine the optimum sizing of the wagon wheel to meet the thrust requirements. An analytic procedure was also determined to compute the burning perimeter as it was felt that this would be more accurate than graphical methods. From this procedure the chamber pressure can be determined as a function of time and the performance then analyzed. The performance analysis was carried out ignoring drag effects as was specified. The total negative impulse due to drag was computed (with aid of the "rambler" CD plots) and the amount of propellant necessary to counteract this impulse at sea level thrust conditions specified.

-o108 - The results of the performance calculations were plotted and are included in the previous pages. Although this is at best a rough analysis as it does not consider such things as heat transfer or propellant stability it is hoped that some of the methods of analysis presented herein would be useful in the more thorough consideration of more sophisticated systems such as grain designs to increase propellant utilization and determination of optimum g-take-off conditions*