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Rocket TA 1 Solution

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Rocket TA 1 Solution

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Of both the stages to be same. M, = aoe Fe Fabri pn é Hor M,= [Soo Sj s- M,= Mut Me t MP gg = 00 Agree Bele te) GoM Find 8€ Mus buck “a FOIE PG — MLE e lee Fen Seo ke us ead = ML = on son (ie ‘ von 7 7 po) py = 4108. Se6 A, = lose fon Undo Km Me = Ieee (BG. WA HAdD Ton - stapes d= Meas caters ne fs aA aagte aa may —+ Me ea fr Ast ste a lS 7 Moy = TSS Gp M Ue B4E~S ww,” Me oe d= Peg loo Serra, Baclese ‘I > Mo, ~lco-¥%, en) = |ssarhy 27 fiGor ater Ms) = Ms ict Wm €D ” pe Bet Mae Me Ms = Dow ky € nN DN 2 Uele (=) | a Mey Me pv cttela ( sey [Gow —8FRAT “BEMAE [or hat ety) fus Laka > Mg = ow — MD é-_"s “ oe 3 dur My = ire ey Toon ERA AG BEIAAE - [0 Arg = Alo -ws kg] fen) My = (684. stek | 1.5 va Me 2 le ase Ulo 42 T gare — Jor lé = 014r§ Ce = Mu ma -™L ce 248 AWL4Z —lo Dw Be _ 5 b(t) AV = AX sete (ee ay = CIS t WT [arent 7 ashy | Hicat release from combustion of cryogenic propellants: Detomie the bent peed fm he combsion of hg gid bogs (8) a + bk Ff temperature of 20K and liquid oxygen (LOX) st 80K at 9 mixture ratio of 6. The standard ‘heat of formation of H20 is ~ 286 kJ/mole. The following are the thenno-physical properties Of ead BOX WX = Moct= ao (Fiat bun Properties of H (in liquid and gas phases) egret ama at cee mek Sa so bnbma ret tem utd, he ries fn id og ps Za oper fs cn an bay alk 3 £Ak1CY mM = Met 2 < Wet 2X] ‘Specific heat of LOX: $4. ki-kaol K ‘Specific heat at coustaux presse of gaseous onygem: 29 kVkmol K, Latent heat of LOX: 6800 ki auol MA=6 =” L x Meet ex aC Y MA fy Pee) —a oo 8th | Gos Zo A possOy ek He (dH. (Pp Sowa bur f Lekewhe ck - Gert OBR +S, AT hey 77 Ulaacbe) 40-8 + Se(2te -24) (wor ey xb . D> 42% Bey . + 0 812 F G2 LD : =p ° os wo (aw A ke Si Ler a gh otek bush A Fae hs (ood? Fa i bw siv(4s-8) 4 68 t 24(e8 Ip v [soo Le = 1g. 3A x0-bFS « ") ? F.61eg #S ove 4 Ce F . ALEK ES OME = (U-21ES ET ek Tet 8 | Ko tw can) = Lore el] Lm = Que tes | wel = Que = (Yale = Qwo- | : . KF] wide 7 Ugg, 2 ! o ge. d on (Ke ql, haz lo aS, a> (op tt KF 3) Kerosene (Dodecane Ci2H2s) undergoes stoichiometric combustion with nitric acid (HNOs) at 25 C, The standard heat of formations of: 4c cuwrarruine Cigtye OHNE, PLEO, ELE cote at HO} cl Gelomersy | = ab <) CO is -112 ki'mole (required for problem 4) He acs me 24 +b @ Corresponding molar specific heats Cy of products (assuming H:0 to be in gaseous phase); O% 3 F120 (eas) is $8 Timole K “ = 2c & COnis 63 Jule K Nois 37 Jianole K COis 37m ese fx tio fon a@ « o 2 ec wk Determine: lo i) Heat generated ii) Stoichiometric Misture satio or mass ratio = a is) Final temperature of gases co aR b= af x iv) Molecular mass of combustion products ©) characteristic velocity C* if universal ans constants §.314 kFkmol K 2 fe N Siac 1 Bo any 7 GG Re A asl 5 soe Gy, +t (2 Ho # Ht n, 7 qu Unto S Cec tT Yt Ci, F (ws) Lnto, “7C2) lo, + Jou) te 4EYN, ? oe Ralows yet a2 Lenk Che = List + U4) (-(3he) = Feb SS 70 e (3% t (204) (16) t + ate = Ho EE Ma= > Aly ~ Me) XE), Pere) Hyl0s a q Mee Y | woe Ww) Mk wv raantahy ~ ur} L+ Ut weary (x Fax yp act) ew) aT = grasa xle — ag <0" 1axce) + (geass) + 24 GT UC Te = 200% k) z (iv) weston wen ax PCF el) dort ate) gerne F- 4 (ax “ L (2 tdout EY “ES ey! = or "> et net g/t ye () a = Re es Re SUNY =30r0 ay Tteq Wer wen = tf IY ~n) HED Sp Ej _ twas FD (ax 63) A Cows 8h |p eye oe = Yul lts¢ epee tty 4) Ifthe combustion is fuel-rich and a mixture ratio of Nitric acid to Kerosene is decreased to 4.5, determine: ‘ds i) Molecular mass of the combustion products Hrd 3 0% ii) Energy released from the combustion of | kg of kerosene = : iii) Final temperamre of gases (Adiabatic flame temperanire) Cure ix) Chatactenistic velocity C* ¥) Searlevel Specific Impulse (CC */ge) assuming chamber pressure to be 7 MPa, and the gases are expanded in the nozzle to the ambient pressure of 0.1 MPa LP, J Verify that the MW of the products and the final \Gpeste ‘of gases have decreased. The decrease in temperature is higher than the decrease in MW and hence the C* and Ix of this fuel-ich mixture are lower than that investigated in the previous problem Mee UE GD Stes Mood = VX MEL UTES 63K sf Met Tanir £20%) (a okt CAS cage a K= Ae <—fz Cu Mee F (LQ l18) Hog — acy fF HO tho FNL CMe F MIM > Co.) Oy Figo F HO KleQothod He e: (L=% 4b OD ° ) t+ pe xiang = 2a + I78 = (qt &) alan 7 aati raking wee eee aa ypa tees - 73) 1) molendeg - Sb t Met (Ast(Q tO) Cotto “poeees + Cet F Cur (ax'y) qa et 4ORF OF > as tee pele. . —— CEU (-UL UD up = 5234s) t rH yo ? C8 ES > EO FEEUEKS Mg = ISL + igattey ( -1t1¢) Me = 7 (Ary - 2.) a 26% nLeks — pruw- 1C% = koe Ca) ay = MMe [ (5:1 68) 4 Clasryee) alecayaa yt corr 33) = 24t-% pe Se iy cre TRE ge TY = Be 2 al AF (HG oer lees Fen ype a ae 7 men clay ) Gee ss | Ye (Fa ve bs AF sae () Te - CFX CH 2 CFM [BES po yet “Lay UR 29 s[fe he = ol e fe eo tula moa - ous ae CK ISNES LL 21 RS Lone 4 VIS Seras

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