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Unformatted text preview: 52. 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! = MmeJ. =%_°324(zoo>=0011
JP
5 =(a.596) [20.900 ﬁm 5‘.___°°°’f _ 18 am fn‘) “and H1 Mfuel =2 ELI my": 753 (20°)E‘ [7.7
’5’!» 5 = {o.536)zo,_9oo In $5? 1. 5 mi. 56. 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'toaswlﬂoa .. Malina! Flea. " l:Ioa. Warehm} _= 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
Ram3g: 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—Pléa __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‘ZﬁoooleRﬁELGfmfﬂ = I112. m/S
4': ngTwEs) = 14:35:04.72 .___ 0.0176
"' «9g , 0.0:”); (1000) __ 0.02.52 Ks/MS
TSFC z r.o:76(nn7+3aS == 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: Servantwire 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 {liceJ, 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.3tﬁ
ﬁc— l'l‘f'f B and = JG+F+BXI+F+JetI>is n ® ’l‘lrirus'l: ratHo (Frcim Eqs.® earning»
:5: =__ s)([+F+B)Cl+F+BE) U; I+i' + B‘l’l'ﬁ
a tugHm .p $0.025 B __= 13.6: 3.5'0K Tm=TooK @202 5+ 8 )C t. 025+ 9(3563/5'00 ._..._—= I: 1.008
O
:3; L015 4 8 17:5 518. The performance of a series of ramjet engines, each operating under design
conditions, is to be calculated as a function of flight Mach number. The en
gines are to fly at an altitude of 50.000 ft. where T. = 205 K. p, = 11.6 kPa.
The fueiheating value is 45000 kakg, and the peak temperature is limited to
3000 K. According to the Aircraft Industries Association. a reasonable esti
mate of ramjet diffuser losses is given by PM P0: in which M is the flight Mach number.
The stagnation pressure ratio across the flameholders. o to is 0.97, and the stagnation pressure loss in the combustor, Q) to , may be esti
mated to be 5%. The nozzle and combustion efficiencies are 0.95 and 0.98. respectively, and the combustion chamber exit Mach number is 0.5. For the
nozzle A... E A.. Assume the propellant is a per (D © @ G) ©©
@ 55'..<
Q \hu._m feet gas having R = 0.287 kJ/(kg‘ K) throughout, but having 7 = 1.4 from Q to © and y = 1.3 from © to Check to see that the fuel—aiwailg does
not exceed stoichiometric (about 0.0667}. Calculate TSFC and specific thrust
for flight Mach numbers of 1.0. 2.0, 3.0. and 4.0. =!  0.1(M  I)”. M >1. AAA M 1 2 ‘5 '+ T03 = “Eff 1* 1? M”) K 246 .363 574 as:
3'55": (‘+%N2)‘£‘ Lass 7.324— 36.75 151.9
P01 " ' “1L5  o. altBo m — 1 0 1C1“! ) 1 o 5 m
{at = g: %(o.37)o.35 L7“. (.483 zit.27 6le}
— . 0.0667 0.066 7 0.06 (8
.F = :5; “ET; #0 our 0.066?
a H 55 3000
Toni. 1» 350:. K 22I7 256.3 27 i842 2098 aﬂm 734.5 “+57 ‘37th =(+Oueu Ns/ig 560.5 3342 iioa.'+ 10731
T5FC= motif/(37$) Ry“; 94:50 09630 0.060'f 0.0572, No allowance Made ‘For Variable Cp ardisﬁoct'a‘f'l’on a g. M Egg13 19 zBZaM ...
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This note was uploaded on 08/02/2010 for the course MAE 155a taught by Professor Staff during the Fall '08 term at UCSD.
 Fall '08
 staff

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