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327exam01-fall06

# 327exam01-fall06 - 530.327 Intro to Fluid Mechanics Su Exam...

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Unformatted text preview: 530.327 - Intro. to Fluid Mechanics - Su Exam 1 ~ 11 October 2006 Name gogdmmgs 50 points Problem 1 — Short answers. (Total 15 pts) (a) (10 pts) A steady jet of water, with cross-sectional area A1, strikes two objects: a wedge, and a. half-cylinder, as shown“ After hitting the objects, the ﬂow splits into two streams, which have cross-sectional areas A2 = A3 = 111/2. Ignore viscosity» You can quantify the drag on each object as the force, F, required to hold it in place against the water jet Which object has less drag? Explain your answer using as little math as possible“ _ A5 A1 :31 «l» é .3 \l .3 \I \f w a H 1:4,, 3 "“ﬁéf-LF %F [Lam W .3, ~H>U Al Wk'rﬂl- N? 4; 3 361’ A7, 7‘ A3 #5 a ' l““"""?‘2- ‘—..._..-~ ° CDMIDEF (“0.5: VH1") ’1; 4 V) __1___> ‘T‘D‘Js (.6th 361,003 ) I“ _, “hula “9 ‘_ ... 34273 ”05106) a \lUuiMﬁ W ”.3 = \IJA! 3 Odffwm ‘3 Zquz =2‘J3A3 .3 c) \jl-\f3=\lJ O .- memo»: hi . New wow. F -~ (mmté )N/ANMOMEN‘WM). F0?- Tmr: Vzvcitmuee, @ Theta 16 £9 ulmue m XeMoP-WNTUM } Fog. Tee wanes, “me X--MbI-LENTUM ls whom) 5mm Tee STMMS 10?.» we Fem 1H9 xuams (:9 mime Li'uneea Ms Less bye. ([3) (5 pts) Imagine you have a pool ﬁlled with water at constant density“ You need a. downward force F to hold a. volleyball under water so that the top of the volleyball is 1 foot deep. Do you need to exert more, less, or the same force to hold the same volleyball (pumped up to a preesure sufﬁcient to consider it fully rigid) under water so its top is 10 feet deep? Explain briefly. mice 2 me wﬁltm’ o? 01:3?me wrap-mg MW newsn‘f ls \neeﬁpmf /" lee Foﬁtt 3? we. To Coastal Tee '30on “Fiﬁcé. Tue 4%th or perm, to we saw Woe “fez—met w tam cases. @ 1 IDrOblem 2“ (Total 15 pts) When engineers want to do construction on the seaﬂoor (say, to build bridge supports) they often use caissons, which are chambers from which water can be pumped out, and which can be ﬁlled with pressurized air. Suppose there is a square door on the side of the caisson with side length d = 1.2112, hinged at its bottom edge“ The door is free to swing outward, but not inward. The hinge sits at a. depth h = 18m below the surface“ Assume that gravity is g m 9.81:!1/52 and that atmospheric pressure is pg = 101325 - 105 N/In2.‘ a) (10 pts) What's the minimum (absolute) pressure, 13, that you would need inside the caisson in order to force the door open? 1)) (5 pts) Suppose the absolute pressure inside the caisson is p m 1.5 139.. What’s the minimum force, F, you would need to apply to the door to force it open? “3 Kim g1: . ”mi map col—mm l6 \“Ll’rws “sew Tue were? serrate. 1.2M 7.1:) Tug Mbpasrknc, Vegas as ”ﬁle 0002. is (D Fm PM = (augme- W i = (tome-ww- + es%-a.ri§,.\1.qm)\wﬁ IM A ALT-gin} --* ’bﬁlowoSN © 4 3 WWW ..._. A1. 2; A A ‘53 2. 0‘11. ' L‘” ”F Mm“ MD is =i~v Ma «ix «as = a is? i, x T" A - 2 «v, z 2- “ K M 1t W FL bin-109p 12, :7 113011. N inﬂow "me supreme @ WM (Extra workspace for problem 2) C7) ' PM THE (4%“ WWW T’V—BESUFE P; Far—w " VA: P. LL’rLtwt?" = F? E; Mfg Tame“ canyon)“ ° \o 136quth ‘9 sewn“ To {fugue 000?. om , MANLE Togauas A’FOUND Hmoa C) mm) mm To FL : TL: FL“ (mun Hiram]: 1,330J05Nm (momma) ””3 006 To ﬁg 2 Tg: P'}»L*L*W1-O.QW\ :: ? Ouﬂol'f mg (Couﬂ- (33 mm T '1’ 5 \$30? ﬁgﬂp ; IF: 1330'“ 34"" a 1,931. I053)" Passage Haven ‘HW ”43 W?" To rage hoof. afflu- GD ’9] €owose 'P=\~5?o== L.5\%<s-\05% 4 P W 13m 0:] 1.. =3 N669 A»! mm (homagmcmxsg 1‘0ng é“ C- 1% T FOY' A GWEN TOILQUEJ ”(H-p, M‘MM‘JM W53 NOGUD gg @ M Ammo AT THE “WW” ”WWW (Wm) WM "We HKNGE, New TLETF.*FT~1.ILM ® 1.330.I05\\!.m = l,51%3-I0517_.oxebtm°’ APE-4.2% PM 690 VT 3 ﬁﬂ’llblouf N MWWVM. \$09415, To OWN 000?. W1 (’29 “7 P=\.S?D . Problem 3. (Total 20 pts) A crude ramjet engine consists of a constant—diameter (d = 70 cm) duct that draws in low—pressure, low»density air on the left side, adds fuel, burns the mixture, then purges the exhaust to right. The remjet sits on a test stand“ The inlet velocity is Vin m 200m/s, the outlet velocity is Vout m 500 m/s, and the fuel mass ﬂow rate, mi, is 1 percent of the mass ﬂow rate of the air at the inlet. The fuel ﬂows into the engine vertically The ramjet operates in steadyetete The atmospheric pressure is patm M 1001325 » 105 N /m2, and you can ignore gravity in the problem. Where appropriate, draw your control volume carehﬂlyh a) (4 pts) What is the density of the exhaust gases? b) (10 pts) What is the magnitude and direction of the force, F3, that the support needs to exert on the ramjet to hold it stationary? c) (6 pts) Suppose you hold a messless ﬂat plate perpendicﬂar to the ramjet exhaust” What is the magnitude and direction of the force, Fp, that you need to exert on the plate to hold it in place? 61; (D \r .- " " " ' .. a] CV“ "Lie JFWM (5’va now he CAI.) / ifs No.55 commence: O= AXE-(lg) L9 27':P;.4\rwA + (”fidel‘g'o'o‘ + POWVWTA l‘ A" w W»! we? 1» mm exam-3f 2'0 FUN”: (5.00 : bl Fae XvMOMWWM games: we W. elm Awe. (Extra workspace for problem ZF= ﬁjeVd’d” Jrjexgv (\1 oh) i =0 swkm©e 3) Inﬂow G) L—~> N0 mu communeso mom wemcamﬁ OWN :3) =A [VMM Pm 10m 1»: *‘QMV VGWZJ (D -11“ '1» 5N ‘ 5p! kg m 7- w-w of! mm. 40 w— 0. A! a .. 0%,»- w- ”J m) D 5 m?- a! O z I w (1.006) = 2.617» 10%) W y I 1 a] CM. my {e— ‘wa ILL, X—mmemw EQMIM: a»? = J F “X swam» =3 "F? ziﬂom’ ow 3A warm 0 o o F AN "“3 1,. =' LM‘HO N WW @ G) e—FF ( DEVIN‘ED ks 1795mm To LE3?) M k M 7' +03% £3600 "4'3 1 13on m +34 DLFEmmJCD (puma vamp—T5 m: gamer. No CnNTF-iwmﬂ 1 2 : maﬁa. . (560%) -§ (071M 2. CD Wm (no v-X mnemon) To new new wmw, WM TW/ Barron 9? CU (VI-g :0) Q) ...
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327exam01-fall06 - 530.327 Intro to Fluid Mechanics Su Exam...

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