HW3Sol - 5-2. Given that a commerical airliner has a range...

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Unformatted text preview: 5-2. Given that a commerical airliner has a range of 6000 miles using 200 tonnes of JP4 fuel. estimate the range of the aircraft burning either gaseous or liq- uid hydrogen and a. The same mass of hydrogen as “’4, b. The same volume of hydrogen as 1P4. The dry mass of the aircraft is 800 tonnes. The fuel properties are as follows: Density Heating value Fuel (kg/m!) (kJng) JP4 800.0 451110 H2 (gaSeOus. SIP.) 00824 [20900 H; (liquid, l atm} 70.8 120900 5' 70 D 3.: WE 7_1_p,L_—_ .. ‘5 —. m -._-o.533 My 3 o 62'; 3.2% mamhggg £175 1: (b Some: Fuel Uot’ume a s J'PLI‘ (f) SQSEOus H1 (STP‘) Hague! = Mme-J. =%_-°324(zoo>=0-011 JP 5 =(a.596) [20.900 fim 5‘.___°°-°’-f _ 1-8 am fn‘) “and H1 Mfuel =2 ELI- my": 75-3 (20°)E‘ [7.7 ’5’!» 5 = {o.536)|zo,_9oo In $5? 1. 5 mi. 5-6. For a given work input per unit mass of fluid and given adiabatic efficiency 7", how does the actual pressure ratio for a compressor differ from the ideal (iSCfllI'OpiC) pressure ratio? Derive an expression that relates the actual pres- sure ratio to the ideal one and 11¢. As mentioned in Chapter 5, the adiabatic efficiency m is usually.r defined by = ha}; _ ho: fan ' hi . Show the dependence for m = 0.8, 0.9, and ideal pressure ratio 10 < pmfpoz C 30 and 7 = 1.4. For a diffuser. the adiabatic efficiency cannot be written as a work ratio since the adiabatic work is zero. Using the appropriate definition of adiabatic diffuser efficiency Tl: _ huh _ he ho: fl he ’ show how the actual ratio of exit stagnation pressure to inlet static pressure depends on the diffuser efficiency and the ideal ratio. Show the dependence for 11.; = 0.9, 0.95, and flight Mach number 0.5 < M < 0.9. Adiabq‘tic. ComPY‘QSSIOH: flfi= l'toaswlflo-a .. Malina! Flea. -" l:Ioa. Ware-hm} _-= M 733—7132. Tia But , respectively) The turboger pressure ratio 13 12 and the maximum allowab ture IS 1400 K For the ramp: the mammu 3’ li i 7: 6° 5: n K °€ an 5* =1.5;IL‘+C‘ZB7)2 as = +30. SM/s .f: T5FC = (L+_F)uc ._._U. 51 Ram-3g: T01: T30+r€im9=zosft+1?] 2237:? K T64: 2500 K mmfiz .P C To -- o _. 1(2500—29252 3.0.0469 = K — qs’ooo 2 CPToa[1~(§%)a ]= = 4/ zaooo)zsoo [1—(1.+ 553)“) = 12 4—6 m/S o.o'i~8‘3 000°? _= 0.0559 K%N.s Tsr-‘c = W Tirbqjgi; T”:- 297.3 as bear & 'Léoth7 K = ?- £2 — Tog= 132(§§§)r 2: if) Assummgr :C—Pl-éa __-1'3 2') -_-, [goo—(60%”? 29 ) T35 = 0+ = |O$2.G K. x -t 5;; =Cr_r_fi)¥—t= (35266)”: 0.4499 91+ 3'5 I . 99:18.95 :0 = Poi. Po: %5: if}; =[5+!-Si7 [2()D‘H BLT: = 6:5[rf—fi3f-‘iv‘1 z ‘I‘ZfioooleRfiEL-Gfmffl = I112. m/S 4': ngTw-Es) = 14:35:04.72 .___ 0.0176 "' «9g , 0.0:”); (1000) __ 0.02.52 Ks/MS TSFC z r.o:76(nn7-+3a-S == 5‘ [3. Preliminary design calculations are being made for a turbofan engine with a bypass ratio of 8:1. Concept A has core engine exhaust and bypass streams separately ex- panding to different final velocities. The bypass stream (after the fan) has stagnation pressure p0 = 0.16 MPa and stagnation temperature To = 350 K. The core engine exhaust has stagnation temperature To; = 700 K and the same stagnation pressure as the bypass stream and f = 0.025. Concept 13 would have the two streams mixing at constant pressure be- fore expansion through the same pressure ratio pry/'p. as in concept A. Determine the ratio of takeoff thrusts (flight speed E 0] of concepts B and A. Assume ambient pressure of 0.! Ml’u in the nozzle exhaust plane. The ratio of specific heats may be taken to be {.4 for both streams, as a first approximation. A: Servant-wire 71022 has : 3:: “ac-(HF) UC- + We“; using C ~For core. engine and «F {or Q=G+¥)Uc+ E304. “Fan; nit 1’5 cur Flow rate m . of c =Q+n lamp—(g3. )'—j + a/zcnaip-(_g5>e] eLm’WN ' “(f-“)1? +5 Tail—($3)? <0 6: Mixed {lice-J, one nozzle: :75:- @_+-F +B)n:1¢ fzchm D or S- ... .M. -- P '4 airfi—G*‘*B?J 12. [' r3“ny bui'l’h adiaba‘l‘a'c, mixing 6++)13c + 8T“. =(f+-F+B )nm 5 ° Tom_ i+-F+B'l.3-tfi fic— l'l‘f'f B and = JG+F+BXI+F+JetI>is n ® ’l‘lrirus'l: rat-Ho (-Frcim Eqs.® earning» :5: =__ s)([+-F+B)Cl+-F+BE) U; I+-i'- + B‘l’l'fi a tug-Hm .p $0.025 B __= 13.6: 3.5'0K Tm=TooK @202 5+ 8 )C t. 025+ 9(3563/5'00 ._...-_—= I: 1.008 O :3; L015 4- 8 17:5 5-18. The performance of a series of ramjet engines, each operating under design conditions, is to be calculated as a function of flight Mach number. The en- gines are to fly at an altitude of 50.000 ft. where T. = 205 K. p, = 11.6 kPa. The fuei-heating value is 45000 kakg, and the peak temperature is limited to 3000 K. According to the Aircraft Industries Association. a reasonable esti- mate of ramjet diffuser losses is given by PM P0: in which M is the flight Mach number. The stagnation pressure ratio across the flameholders. o to is 0.97, and the stagnation pressure loss in the combustor, Q) to , may be esti- mated to be 5%. The nozzle and combustion efficiencies are 0.95 and 0.98. respectively, and the combustion chamber exit Mach number is 0.5. For the nozzle A... E A.. Assume the propellant is a per (D |© @ G) ©© @ 55'..< Q \h-u._m feet gas having R -= 0.287 kJ/(kg‘ K) throughout, but having 7 = 1.4 from Q to © and y = 1.3 from © to Check to see that the fuel—aiwailg does not exceed stoichiometric (about 0.0667}. Calculate TSFC and specific thrust for flight Mach numbers of 1.0. 2.0, 3.0. and 4.0. =! - 0.1(M - I)”. M >1. AAA M 1 2 ‘5 '+ T03 = “Eff 1* 1? M”) K 246 .363 574- as: 3'55": (‘+%N2)‘£‘ Lass 7.324— 36.75 151.9 P01 " ' “1L5 - o. alt-Bo m — 1 0 1C1“! ) 1 o 5 m {at = g: %(o.37)o.35 L7“. (.483 zit-.27 6-le}- — . 0.0667 0.066 7 0.06 (8 .F = :5; “ET; #0 our 0.066? a H 55 3000 Toni. 1» 350:. K- 22I7 256.3 27 i842 2098 aflm 734.5 “+57 ‘37th =(+Oue-u N-s/ig 560.5 3342 iioa.'-+ 1073-1 T5FC= motif/(37$) Ry“; 94:50 09630 0.060'f- 0.0572, No allowance Made ‘For Variable Cp ardisfioct'a‘f'l’on a g. M Egg-13 1-9 zBZaM ...
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HW3Sol - 5-2. Given that a commerical airliner has a range...

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