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Lecture_20_notes

# Lecture_20_notes - ChE 374eLecture 20*Pipe Networks 0 As we...

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Unformatted text preview: ChE 374eLecture 20*Pipe Networks 0 As we often consider ﬂow in single pipes, pipe networks are the rule not the exception. 0 Series and Parallel Pipes. 0 Series Flow — Total ﬂow rate is constant by mass conservation. — Total pressure drop is that through each individual section. * Includes minor losses. ~ For expansions and contractions, losses “belong” to the smaller pipe. — Type I problem: Know D, L, Flow, —> get APL. >I< Calculate Re, f, APL, in each section and add upp the pressure losses in each. — Type II problem: Know APL, D, L, find * Produce a system demand curve for each pipe: APL vs. l2. l/ is constant and APL : 2, AP“. >1: * Add the curves vertically to get Vtot curve and read off PM for given pressure loss. * Or, write equation AP], = Zi APM explicitly and solve nonlinear equations. — Type III or IV problems: Find D or L. This doesn’t count: not unique. * Mathcad example for Type II problem. 0 Parallel Pipe Networks. — Total ﬂow rate is the sum of the ﬂow rates in individual pipes. * Pressure drop is the same through each pipe connected to the same two junction points. 7 Type I problem: Flow rate known in a given pipe, APL unknown: * Compute AP], in known pipe as usual, then with APL known in the other pipes, compute ﬂow rate as a usual Type II problem. — Type II problem: Pressure drop known, compute ﬂow rates. >I< Compute a type II problem in each pipe as usual. — Type I propblem: Total ﬂow rate known, pressure drop unknown (hard). * Draw a system demand curve for each pipe. Add horizontally to get the total ﬂow rate versus pressure drop curve. Then for known total ﬂow, pressure drop is known and read off the ﬂow rates through each pipe at that pressure drop. Or, once pressure drop and total ﬂow are known, solve type II problem in each pipe. * Can also solve system of nonlinear equations, equating pressure drops through each pipe and the expression for pressure drop in terms of ﬂow rate and total ﬂow rate is the sum. 0 Can devise complex networks, applying the above principles — Like Kirchoff’s laws for electrical networks, but harder since we have nonlinear systems of equations. * Use a multi—dimensional version of Newton’s method (ChE 541). Law“ up ‘?"Pr “MARC, ‘ Flu/Lt) 'Déhf‘at—L‘FW \$ys¥M -/'\v-( i ’ “346/3 r (Nita » m‘l , /7%nml.m (30);”) / PMaHtl_ o 445/75? {0 eltt4HI-(d' Chad’s 2 Priﬁcwt/o: . {23 CQMJ‘ W55 dowevomcvax ﬂ] : 731 I r o a? t: ‘7 1% E A; 4' EﬁPﬁ’G/DM / C94+facglvlaw A {>051 J {1" 3W Pbﬁ Tyre I pm; L, I), ’5 ~—--» It.» A? 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Normal HviArial 23:10 B I Q ‘3 E E 55 3:- :EWIESSJIHIIﬁvS 33 aw in mite ._ -_ ~— . - 3 Pt .— 320000 p .- 998 ll ‘- 1-002 10 ﬂ ;= 1101 el 1: 0.00024 L1 := 100 D1 := 0.05 Guesses a := 0.01 e2 := 0.00012 1.2 := 150 D2 := 0.045 ‘33 1‘ “3002 L3 := 30 D3 := 0.04 a := 0'01 Q := 0.1 Given 2 2 2 L1 4 4 :rADl :r-D2 Tr-D3 v M) L + 2 10 el + 2.51 = U V p 2 7:01 L + 24° e2 + 2.51 = U [I 2 :r-D2 ﬂ . . p. i .Q 1‘ 1-032 W Re := Find(Q,ﬂ ,fZ,f3) + — 3 2.64074 x 10 = 0.03141 0.02716 0.03148 KCR-dy) ’34 W/FK YT ‘ f. ‘ 0f“ 3 A}; r A? _ J 2"“ PA} is 9/“ ['7 F’F‘" I 7.. 3e 7 [B I: 9”“ ﬂ” P'po I’ 7' 3 5/17..) 22% Iowa A? k 9!“ n0h~§¢~ (I Fir-93 compwt A? H mm ff,» M “NJ '14". / w/ &P' (c‘*-‘PL«-7’( ﬁﬁﬁw 3.3254 In 5%“ ' Frog/w 3D,?“ A)? Krowvl/ 5/03 .944) MRLMVJ, [an-779.1 77;;(‘17 53fth J—w rack Phw A Q at '59... 1 3W - ‘ii/‘J. C M, W Z a) A? t 4"}:qu M \I: Q4 7, D1 1: ‘ :nf’i 3 A V1 7,0, s/ E 1 {BE—1L3 W“ P :Krzz .2} fm an L fl ‘ 11M!) 0/ ﬂat 7S?- (é‘> Q'LIG‘ +& 4Q m é €7N/mwf AWN)” N! 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'4; a" 0 "I ‘3 “Mathew - [ParalléLpipesxmcd] ' 7 ﬂ File Edit mew Insert Format Tools Symbolics Window Help -.b~g gay 3 n ma: 15.95110030ng132) Normal ' Ariel 10 B I g a; g 5— g— 3] MYSite (>60 ._ -_ ~_ , -3 Qt.— 0.01333 p.— 993 p.— 1.002 10 ﬂ := 0m Q1 i: 0.004 81 ;= 0.00024 L1 2: 100 1312:0135 Guesses 3:: 0.01 Q22= 0.004 e2 := 0.00012 L2 := 150 D2 := 0.045 e3 := 0.0002 L3: 80 D3: 034 f3:= 0.01 Q3_:= Qt—Ql —Q2 Given L1 401 2 L2 4122 2 ﬂ.D_.£. =9._.£. 1 2 7“312 D2 2 “022 L1 1 4-Ql 2 L3 4-Q3 2 ﬂ'ﬁé' 2 “35% 2 3-D! 7r-D3 i— + 2-10 81 + = U \[H 013.? p.01 4-Q1 _\(ﬁ 11 2 7r-Dl Lmo *2 +___2-51_ =0 \{ﬁ 132-3.? p-D2' 4'Q2 VIE p 2 Tr-DZ L +240 83 + =0 \[E D337 p-D3. 4113 .45 p 2 :r-D3 Qt. = Q1 + Q2 + Q3 0.030663 0.026613 0.03113 ;= Fmd(ﬂ,f2,f3,Ql,Q2,Q3) *3 _ -3 Q1 ‘ 5.775203x 10 + Q2 3.88944? x 10' 3 Q3 -3 U ml 2 3.66535x 10 .. .= .__.£. ' 3“" ﬂD12[ 2] AP=2.647x105 :r-Dl ...
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