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TAMU MEEN 344 - Exam Solutions MEEN 344-Fall 20139

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No.Me ~:__------:--:----Closed Boo!<; C los~c:J Note$ of" e. pr.vve. Ge.( p <;;~e.et ) (33 piS) 1h e. s+-rearw'\ ~1.-t(\(.h.t?Y".S ~o\ 0. 8\'·Je.A'\ + lo!.U ~ \·cJJ () re: tyl = "X'J. Y2 Lp:l ~ z a) ·lJ)ho.-r i:;:. ~e ';e.. I vc..i+·: \)ec t:::>r? / 'b) J:s ~e +low l D1 ~D1 '3D? .:t:"s it: st-eady S"h:A~? c..) ':!=~ ~ +\ow-ro+-trt-l~Y1~ I? A~ b oes . +-\..e. l}f2 \o~ it/-f"'h?_,-y\-," ( ( 0) ex.,st-? I~ ri-a!= Cu h:At-IS It ea ~0 I t-o? e) :rs ~e. ~ \ou.r <'.nmpr~s; b le. : .t:) For /~visciJ ~\our ct.-0 flo loa:!)' l\orce 1 ~-low oloe_s the p resstJ r e vo r; 'Tl spttc e 7 ··~(33 p-B) Water C,=~o..q ~~W\3} \s. 5 \).fp\,~cl To 0. Y\Dt~ \e 6'1 a. f>' pe wrt·~ o. ~,qme-rer b, : \.5'"0 r'f'IM at-a. p\essvr-e e>~ J M Pa_.()~\l.l+e ~e()OZt\e.. \s bo\teJ on ~e ~ 0~ ~e r·re· Fo \ 0-f\ o 2 2-\ e. e.~; t d '~ VY\ e. -ter- o ~ ~ JOMM ~ete.f-fV\,~e ~e. ~t;!low·,~ ·• o.) W h~t-is ~e veioc.·~ o\ J.e W<l+e;-e~hj ~ .. flo>:::eb "? b) t\ow h;~h o\oe5 ~e. wt:>t"e< shoot-o.tp 7 c ') [)) ho r '"' ft,e ~ rce reou' r<>d TO h.IJ, +he 1'1 0 Zo ~\e. 01'\ tJ.e p ·fe ? ----~--------~--~--"f>2 = 0 ~ ar1 e f>, = 1.0 fl1 j::>a A \,s =-O,\ /IA~~bs v, = v'2. A7.. ~ -k v. -z._ ..L. v -z.::. q oo,.,9 A I -/ CA a. & I ~ 'Va~ (I-[~ Y}=-900. 'i "'~" 'V:} (o .. l..\~~) ~ 9ro.,c:r M~--l ___ ____l_ ___ j_------• --~-~--~-6 P1 := 10 ·Pa 5 P2 := 10 ·Pa PI-P2 m 2 ---= 900.901 -p 2 s 01 := 150·mm TI 2 Areal := -·Dl 4 2 Areal= 176.715·cm Area2 [ 2] 0.5· 1 -(--) = 0.4999901 Areal V2 := V22 - = 91.837m 2·g PI-P2 ---= 91.835m p·g Area2 m VI:= V2·--VI = 0.189-Areal s D2 := 10·mm kg ~:= 999·-3 m TI 2 Area2 := - · D2 4 2 Area2 = 0.785·cm PI-P2 m V2 = 42.448-s 4 (PI - P2)·Areal = 1.59 x 10 N 2 p·Vl ·Areal = 0.628N 2 p·V2 ·Area2 = 141.374 N 2 2 4 -(PI-P2)·Areal-p·Vl ·Areal+ p·V2 ·Area2 = -1.576 x 10 N3(3~ f+5) A C. u \::e !D~Ji. lOc~r~ ?<. I 0 c. M W.•~ a MCA.sS ~!f.. ?. ~ \s ~\G':C.ec!_ O!'\ o. 3 o 1 '1\ch~e c...o \JereJ. w '"""~ o;) ~= '0.,.03 Nc sec..jtv? .. ), ~e. ~ '\ VV1 o ~ D·, \ bet-wee f"1 ~e.. c.. I}. be a r"id \~c l ,~eJ S \A{'-~~ ce \ S 0. O~M M ~ \ vk. A \ .,.5" rY'/.5 J e. 1-ot uJQ-\-ef' <P= 99~ ~/~?. ~t-has o. dicvYieter" (\:>) o.\ 3 .. 0Cn'\ ~~pac. ts t~e c.ube pe.<-r~d\c.v \qr -ro ~e s;de. . • a) be.r<ve O.Yl e<f>uo.+to.., ~~ ~e a.ccelera;t"l~tf1 0~ t'he c.ube us:"~ symb0YsJ ~.e. ~= t(M,h,\/<l~+yi-'1As,Je1 Dj~t,U(c.ube~\d) ~) Wh~+\s ~e acce.\er~-hoVl. o~ ~e c,)ce +or~ cube S pe eJ o ~ L) = '\ m)s do r_,J"' +tte ~ r') c. I~ ne ~& Lf,J,. L.f..,x- rrra)( DJ-f co ,) ,_ rrr .f!VJ'<~ .. )) v.~ tJ V-JA ~ 1..1 )) .r / JF" (J . ?(YJ c.s 1tJ ../ ,_v 1 7-*As,o\e +f'v\~51~ ?,oo-M~~~:: 0 -( Vj +U)( -j{VJ-tAl ~ t}-) ~ ~ = ~ [/J-1 ~ Aso4e +II'W; 7.,30'-_/) ( ~ -[)) ~ !f]M := 2·kg Aside:= lO·cm·lO·cm N·sec ~:= 0.03·--2 m h := O.OS·mm . m VJet := -1.5·-sec Djet := 3·cm m U:= 1·-s kg p := 999·-3 m m ~:= 9.81·-2 sec Fsx(~,U,h,Aside) := -~· U·Aside h Fsx(~,U,h,Aside) = -6N Fbx := M·g·sin(deg30) Fbx = 9.81 N Fjet(p, Vjet,U,Djet) := p·(Vjet-u? 7r ·Djet2 4 Fjet(p, Vjet, U,Djet) = 4.413N f h .d . . ._ Fsx(~, U,h,Aside) + Fbx-Fjet(p, Vjet, U,Djet) are ( M, ~. U, , As I e, Vjet, Djet) .--....:.!.....:...._.:.....:. _ ___;:__ __ __::._..:..:....:...._..::.._.:..._:.........::~ M aref(M, ~. 1· ~, h, Aside, Vjet, Djet) = -0.302 m KC 2 sMFEN 31.\ L-\ Pa\ l aot3yYt: 0-+-(A.+Q+O-=D t -\ .. 0 -b+o -=o L: '3-~40+C.=O 0 JN\'. 0 +q_ +D~O =-0 O.:::o c.·. -tx ~o-b+o ;..o b::--~ L : '2.. -)A ~o+c -:::o c = -;;... 0 rvt l ~~ + o+o ::a-o t -~ +o-~+o:::o L' 2-~"""o+c.=o Q=-o f:'> b-4 lr.-~3 c:: -3 I Nb M1zo ~ A/p.l f ~'lD ~})C), / sorV)EEN 3'-tli \-q\ \ 8J.J ~~ Na £'t'IC ~ . __ C fo.s•",l \:_/,'t~-, closeJ. ~ot-a~-One. \tiel p ~-~1-:.::~ :.._ ~or \qr'/l;v'"'ot:. +~ \\y ,J ~··:Jt2iof:;~)J: Cons+av-+- o\0'\G;i-y c_ons+-o/·t , • _ I, .. .1 "'\' jJ.ri--h · 1 \1\sc.os,+;· cti('<:'~'J·..:,.....--J.0·:··~=·1-,..,t· L:·r=~ ·-.!'Jl",... •• 1 ,,,,) ".,,.:1-. \J ,. '/ -..,.; I ~·-I ' . .__., ~ , ,.. .,., .,..J V(./ . ~-.._. ~ .f ' t,._\ \ • I..J • r\ :: f '"'\• (\II t --· \)Jno,-t-1..s -~-t<:::.. rq,-t-to ~1 p/H:: ;ri10:i·1e:"ruw; -~u;< o-\' TV~e. \)IScous ~ lo tv TD 0. s; loc.u {{;ctvl the-$0i'Y1e t-o+a i tloto rcrt-e but-C{ 1At':i·~orr/1 Vt:loc...i--t/ f)f'·::>'\ ie. rY\ OfY\~L{M +-\ut< = )~ u r lA. d AlV\E~N 3 L4 t.i Fo\1 ;}.013 " tl.oo I<? 4 +-CJq'l '%{a. -n. s ;. t.. 2.. 4 +-~3 q. i;] M /s z 5 --= 7~ ~ xJO Pa = 7 ~0 K ~()(D := 2.54·cm 20 liter Q:= 0·-min Q Vel:=--TI 2 -·D 4 Jw:= IO·m ~:= O.l5·mm 5 kg P2 := 2·10 ·Pa p := 999·-m Vel= 6.578-s 3 m Le _over_ D _Elbow := 30 Le over D Globe Valve:= 340 - - - -2 -7 m v := 1.9·10 ·-sec Vel·D ReD:=--v 5 Re D = 8.794 x 10 Turbulent Flow e -3 - = 5.906 X IQ D f ·~ 0.0055t +0000 ~ + ~,~j] f = 0.033 ( L Vel2 Vel2 Vel2J PI:= P2 + p· f.-.--+ f-3-Le over D Elbow·--+ f·Le over D Globe Valve·--D 2 ---2 ---- 2 5 PI = 7.799 x 10 Pa ( L Vei2J m2 f·-·--= 277.471-D 2 2 s Vel2 m2 f·3·Le over D Elbow·--= 63.43-- - - 2 2 s Vel2 m2 f·Le over D Globe Valve·--= 239.624----- 2 2 sQuiz Ch8 MEEN344 Fa112013 Name: ------------------Closed Book & Notes, One Help Sheet An air conditioning duct in a large building has a cross-sectional dimension of 1 ft x 3 ft. The air has a density of 0.076 lbrnlft3 , a temperature of 60 F, and a volumetric flow rate of 1600 ft3/min. The sheet metal duct has a surface roughness of0.00015 ft. Determine the pressure drop for a duct with a length of200 ft. Neglect all minor head losses.b := 1·ft h := 3·ft k;= 200·ft 4·Area Dh:=---Perimeter Vel:= _g_ Area kg p := 1.225·-3 m Dh = 1.5 ft ft Vel= 8.889-sec Ibm p = 0.076-ft3 Area:= b·h 2 -5 m v:= 1.3·10 ·-sec Perimeter:= 2·(b + h) ft3 Q:= 1600·-min Re_Dh := Vel·Dh v 4 Re Dh = 9.529 x 10 ~:= 0.00015·ft e -4 - = 1 X 10 Dh f := 0.0055·[1 + (20000·~ + ~J~J Dh Re_Dh f = 0.018 L Vel2 ~p := p·f·-·--Dh 2 in Hg in H20=---- 13.5 LlP = I 0.948 Pa -3 …


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TAMU MEEN 344 - Exam Solutions MEEN 344-Fall 20139

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