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# Assign4_solns - 4 AE 312 Assignment 4 J C Dutton In the...

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Unformatted text preview: 4. AE 312 Assignment 4 J. C. Dutton In the analysis of generalized one-dimensional ﬂow, verify the expressions for 8P0 E dPo/Po and as E ds/cp given in the last two lines of the Table of Inﬂuence Coefﬁcients. Air is ﬂowing at a Mach number of 0.5 in a channel at a location where the area is 1 ftz. At this location the static pressure is 10 psia and the static temperature is 500°R. (a) Calculate the mass ﬂow rate through the channel. (b) What percentage reduction is area would be necessary to increase the ﬂow Mach number to 0.75? (c) What percentage reduction in area would be necessary to reach a Mach number of 1.0? (d) What would happen if the area were reduced to a value smaller than that in part (c)? Air ﬂows at 500 m/s, a static pressure of 50 kPa, and a stagnation temperature of 320 K in a 10 cm diameter duct. The duct converges to a diameter of 9 cm, where the ﬂow exhausts to an ambient pressure of 100 kPa. Find: (a) the Mach number, velocity, static pressure, and stagnation pressure at the exit; (b) the mass ﬂow rate through the system. 1 2 D1=10 cm D2=9 cm V1=500 m/s _ P1=50 kPa Pam—100 kPa T01=320 K Air is supplied from a 7 cm diameter pipe to a converging section whose exit diameter is 5 cm. Measurements at station 1 in the supply pipe show that the mass ﬂow rate is 5.6 kg/s, the stagnation temperature is 300 K, and the static pressure is 80 kPa. The back pressure is 100 kPa. Determine the Mach number, static pressure, and velocity at the exit. PB=100 kPa De=5 cm Made In U.S.A. IiL'DataCom/ WWJ #1: In 'H'xe, arm/33:3 of QWWI’RP one; dimension? I 3007”, Ver. ' 44W, DLPF‘Csﬁlth 5" 2P = dP an. ES: ii 3N n In 'f [615+ he [has of ﬂu e ”RAE! PInFIuencLGo ‘oen Fro-1N1 “Hm”, ISCmLkoptc, m‘vac-nn Ecmn P at P' “6—! _ P = P ( 1+ L4 M > O ”W”; M” GMVWHE’”) D; germ-H 0.11933 “ cm - a2 + Q m 06) 7’; - 7° (+- war/11 or‘ = KM /’2-. E 1/ 2P0 4: + ”W H Su Edinluﬁng OED-(£5: and g”) #pm MW— of «VF/Menu, 003%.» 2P0: [%]§Z:Lf1:u'_ﬁb” fgc [YWCHE'M)]% [/%:(€A +[miﬁc +042 (732)]2‘5 -74: W P ' 2 or EPD _—_ [015A "L [' 7%“? {c + [’jj X27; DMTTOAI ‘i P1208,1 (amt) For ‘Hﬁﬁ @n"TDP\j elm/rage. s+ar+ wi-Hx 'H‘ne. Tel; eqnf T43 = dk— 4P Z)? d] AP Ol\$' CPd‘T—l— .. FT. = C? T " IQ TS .. 05: LT_R> D- 0(7- .; p‘ _ 4 g“ — P ‘,F’(§)% j 51' (EM subshwna 14va 0F rag FM grands). o'-£\$= 7H 5 , [19%)léI-P +£CIWW+§M§J£g [EL'DataCom/ l-Mz ‘7‘ _ ,'.____4 ' ‘— 7 7.v 'I’,,#, ,, ”k, .__,:* '1 burn/4, , l P1208 ﬁg; A‘ is I in + Hack Bar of 0.5 ‘ ﬂap-01" msl'zma] c'rannefih’: aalgcofh'on wk??? Jrhe area is "i ii”? MT'AIS logm‘w‘on Hue, 9(ch Pressure, is 10 Psm amok ‘H’re dud-9c, +empem+ure r; €009R . - , / / / M.='b-‘5 Wan/‘4 ¥ W 5:: I SOLUTION? (D E; L“) Cairn/Lida ‘H'Ae, mass How m}: H’VDual“ 'H’W Charme" . J _ sn= w; = cagxmwmﬂ‘; « 2444922 live; r.“ 2 ‘2qu I? é—s K 1% “156) b WIN/1+ pawn"? rcAuULI'm {n mm W ulal 10¢ newsew % [norm L> 'Hﬁe. How Mad-w numbcrh 0.7§ 20 - L3 0 * (ﬁr/A") _ . — “/49an x 100 070 Area poolud'l‘on 3 _ @/A*jn=05 Lt) 'Wka-l- reducHon in area would be, nucssarqj {'0 mack a Mack Rambo“ 0? L02, D70 Arum RCAUCHO'N 3- (KL/419M205 ’QA/A")M=L (A/A*5m =05 A Pcducho’n .1 1.33% ‘1 _ wig-“Maw 1.337% X "90 " 3&4?" \ \ r = ‘ ID, IUD-Fae, chvl’ 'H'zure is Hng rcaluchon .ﬁlﬁmsganj +0 go Parka-23% ﬁzz-:1). , 2c [00 m) (at) NM Wm hap Smaller +1" an m a SyLr c/H rm lug}- su 6mm, Hue, maximum Mack number +50} ca DumN j ngﬁ; (WM n41; Hwe. area were, rtJuceo‘ 1‘1? 0. valuc. ha'l’ In Paf‘l’ LC.) 2 n be ache/col Cover M 1. Ha flo would m-ﬂwa-I- inﬂieﬁo‘pkeisnew mummﬁn aw 10ch ton The, Mack num Bars 011’ all ups+r aam 10\$)? would decrease. Fr m ku‘r rew‘ous values amt ﬁve 5!“ to}: Pressures +LmPCm ares an Aansnﬂ‘cs would all :ncrease. 50 SHE Z'I'S 22-142 100 SHE: 5 22-141 “mm” 22-144 200 SHEITS ’0). N: ‘77” if 7 DUTT‘ON P205,7#3 1 . 1 , Air ﬂows at 500 m/s, a static pressure of 50 kPa, and a stagnation temperature of ‘ 320 K m a 10 cm diameter duct. The duct converges to a diameter of 9 cm at its exit, where the ﬂow exhausts to an ambient pressure of 100 kPa. Find: (a) the Mach number, velocity, static pressure, and stagnation pressure at the exit; (b) the mass ﬂow rate through the system. z ﬁe ' 0H — (004,557- _ 6P5- 3 _ 4-K" [01A mmProL'm “ no |n+UPDIQ m 77 ' 399‘Q—Zﬁ‘737 F = HSLK‘ _' i In {EL 5"“? 9 o M1=L = 500 W; 13'752’4‘ (Tm—17' P [ll-”)9“; V émm“, a,[email protected]‘- no‘i @- CmV. t/m55-755 W5 ~ Now moVC‘i’o ‘H'tc vii: ‘ ‘ A1: A94 Iggy). Hm)=l.l485 2‘; P) _ at» M e-%%S?%= 775 5 Ta = @973 =é7”“0)(320l4 = .224 LIL V MOB—E3“!- 40314)M(29970]‘m (which IS less 'H'ia ~- “cram-deli 3 Pram abVEJP B9113, = 973.4 ”a. (9 ‘ [Q There. are, maﬁa. chow +0 awalmiem. (14' m: P,A M, {—— “5019913)“ EWQ WWW P120811} - 3 ﬂ ‘ Air is supplied from a 7 cm diameter pipe to a converging section whose exit diameter is i 5 cm. Measurementsiatstationlinthe supply pipe show that ﬂcﬂgrssﬂowrateis 5.6 kg/s, ‘ the stagnation temperature is 300 K, and the static’p’ré’s’sﬁre is 80 kPa. The back pressure is ‘g' 100 kPa. Determine the Mach number, static pressure, and velocity at the exit. PB: [00 LPG! TS 5 SQUARE T5 5 SQUARE TS 5 SQUARE unqu \$Mu1’nod: 7 We. have, enbughiniurmai'l‘m a+® +0 calmlwie. M, from Hie. mass How M: on? .‘G A NATIONAL ' J Mtéwlgmf) ___a Bio RA: X équaring bails sides: 2 ' V (gong) Fatwa)". a " 1:5} 20.35 = M,2+ azMﬂ Quadraiﬁc, in Mf‘ : 0.2%" 4. M}- 90,35 :0 .L M.“= —lt PMCDIZMDD'ZSUL 27.9% (ml “mi 0.4‘ M ”W” '. M,;2;907 ﬂ 94;»:me ‘— .'. This is noi‘ a evnverging nozﬂes Lu‘l’ m‘i’hu— 0L superwm‘c J-‘FFu j . .- 4+ M: = 2-504 5 fig. 0.038% 1 ”0,5g7436 f._. 3.530% R, , To: EL}- [Wm ' [Pé08'**‘°’"’° Max/[r13 +0 Hue, 00+ s+o+ion ugfnﬁ area m‘Ho inﬁrma‘k‘m: —- = 4e I? = wk (gramme) = [)9an, +H: swat-sonic branch 0Li,he,,,iscn+r°rl‘c— 7 Fblaﬁoﬁ? Maggy—I7 ' f5. 9: 0.1830 T3 —: (mags— Po T; E : P¢/% : D: '53 : = P P‘ w)“ (ﬁﬁﬂmm 39‘" 4 ”*4 C Tr, = [.2 T — @53¢7§)(3oon) = m D I4 =7}; 33% l, V .1 '2. ...
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Assign4_solns - 4 AE 312 Assignment 4 J C Dutton In the...

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