Lecture_Notes_10_24

Lecture_Notes_10_24 - Pipe flow” analysis ' Energy...

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Unformatted text preview: Pipe flow” analysis ' Energy equation: 2 ' 2 Zi+ fl+£+gzl — 31+V—2+gz2 2h] m ,0 2 7 ,0 2‘ Friction‘loss' h -f£V—2 f—f(Re 3) ' l D 2 V ’1) V2 Minor losses: h] = K 7 valves, fittings, bends, etc. .‘ggsfi amasse- I, 2, -E§§§a E --fifi$§ a Enos .3 ==mes ‘ — \ ==gsg§ -» v . g a“ ‘ 7 - U 1 ii-I § ‘ . 0.02 4 . t _ i g a 232?; —II J .j :" 00004 _‘—-"'§ ' ‘ " _ ‘~.-...h -‘ oloooz 1.2: _-; -, F t ; Jfinn II “all” f f 29 I v A [om : . It Awe % II F MW 0'011-12345;9u‘v2‘314579 2345‘79 2345797 10‘ 10‘ 10‘ 10‘ 10 . Reynolds number Problems: leen: q',D F 1nd: D1v1ng force (p1, h, WS) . Given: p1 0r h or W, D Find: (iterative solution) Given: p1 or k or WS, 4'; Find: D (iterative solution) Example: Water (V = 1.14x10*6m2 /S) flows from the tank and through the pipe shown. The pipe is made of new commercial steel whose roughness is e = 0.046mm. Solve the following problems: a) If the pipe diameter is 10cm, what is the pressure [91 to cause aflow rate of q' = 0.051713 / 5 through the pipe. Modified‘Bernoulli equation between the surface 1 and the pipe exit 2. ' . p 2 2 {fl+£+g21]—{—&+Kz—+g22]= h] p 2‘ V12 << V22 and p2 = pa" V2 191 *pa = p[7+g(22 _Z1)+h1] V A = 7r(0.1)2 /4 = 7.85 3610—37112 V = q'/A —_— 0.05/7.85x10‘3 = 6.37m/S Re = VD/V = 6.37(0.1)/1.i4x10“.6 : 5.59x105 e/D = 0.0046/10 = 0.00046 From diagram f = 0.0175 L/D = (40+20+30)/0.1,= 900 h, = f(L/D)V2 /2 = 0.0175(900)6.372 /2 : 320m2 /s2 p1 — pa 4—— 9%[6372 /2 + 9.8(5) + 320] = 389,000N/m2 b) What is the flow rate q in a 106m diameter pipe if the . pressure [21 — pa = 200kPa? 2 ' 2 El+-V—1—+gzl —— i’2‘2“+—I{2—+gz2 2}]! ,0 2 p 2 V12 << V22. and p2 2 pa ' '2 2 2 a+ Z—z =—+ —#—= 1+ ———~— p g(1 2) 2 fl) 2 ( fD) 2 L/D=(40+20+30)/0.1=900 V2 200000' ‘ 1+900 ———= . +9.8 —5 2151.4 ( f) 2. 998 ( ) I V: 4—3—9‘3'8— v Re=KQ f=f(Re,e/D) 1+900f v . I Iterate: . v Re 2 100,000 —> f = 0.0203 ——> V = 3.96 —> Re 2 348,000 Re 2 348,000 —> f = 0.0182 9V 2 4.17 ——‘> Re 2 366,000 Re 2 366,000 —> f = 0.0181 ——>'V 2 4.1.8 —> Re 2 367,000 q' = VA 2 4.18%(0.1)2 = 0.0328m3 /s c) If the pressure p1 — pa = ZOOkPa, What is the diameter of the pipe to carry a flow rate of q‘ = 0.05m3s. 191-]? V2 LV L V a+ 2" =—+ —_——=1+ —_— p_ 8(122) 2. fDZ. ( fD)2 V2 200000 1+900 —= +9.8 —5 2151.4 ( f) 2 998 ( ) V: 302.8 RCZQ fzflRe’e/D) 1+_9_Of V ' D 421/351) - D: E 4 - 77V Iterate: Assume V=4.0 D: /4_q: 4(0-05):0.126 . 7W 7&4) Re : Q _: = ' V 1.14x10 f=f(Re,e/D)=0.0179 1+~ 1+ 0/0179 ' Df. 0.126( _) V 24.69 —>D = 0.117 —->Re= 481,000 —> f: 0.0177 —> V = 4.55 V: 4.55 —>D = 0.118 ’—>Re = 471,000» f: 0.0178 —> V = 4.56 D=0.118m MINOR LOSSE S separated flow _- FIGURE 9.11 Flow through round and square bends showing regions of separa— tion. From Visualized Flow, Japan Soc. of Mech. Engineers, Pergamon Press, 1988 x72» (mjnor losses) . = K 2_ TABLE 9.2 ‘ Typical Loss Coefficients for Pipe Entrances and Exits (D is the diameter (of the pipe, and r is the radius of the rounded entry) Re—entra nt 0.78 Square-edged 0.5 Entrance type Rounded (r/D = 0.02) 0.28 Rounded (r/D = 0.06) 0.15 Rounded (r/D 2 0.15) 4 0.04- Exit type. Abrupt 1.0 TABLE 9.3 Typical Loss Coefficients for Pipe Fittings Valve or fitting Gate valve (open) Globe valve (open) Angle valve (open) Ball valve (open) Standard 45° elbow Standard 90" elbow Long—radius 90° elbow Standard tee (flow through run) Standard tee (branch flow) 0.20 6.4 0.35 0.75 0.45 A 0.4 1.5 Losses xiv VA! 1/55J F1 rr/NGS) gE/vDS) expfiNSIoA/S'J (a NVRHCT/‘(VSJ .576. 77,555 DEVICES Caz/:5 D/SroRBfiA/(Es ,qm/p ,rpRB'azE/vCE WHMH REIULTS IN Losses 0F MECIfflN/CAA ENE'RG'IES «DP 7745 MEAN FLOW. _ z _ _ é=k% MM“ _ ., . COEF/C/ENT' - FREE J'ET" . v2 ' " \%\+é+}z./ . _ k . 1:0 flwfik —\§ +£fw+£4 z «321-: Viz/2. =§ K111 SUDDEN Ex FA‘NS'M M d) If the pipe diameter is 100m, what is the pressure p1 to cause a flow rate of q 2 0.05m3 / 5 through the pipe. Include minor losses: sharp edged entrance K = 0.5, 90 deg elbow K = 0.75. Modified Bernoulli equation between the surface 1 and the pipe exit 2. 2 2 [BL+“VL+3Z1]‘(£2—+K2‘+gz2j 2 hi ,0 2 V12 << V22 and p2 : pa V2 ‘ P1 ‘Pa : P[—2—+ g(Zz "21) + hi] V 4—- q/A -_—§ 6.37m/S Re 2 VD/v = 559,000 e/D = 0.00046 From diagramf = 0.0175 L/D=900 2 2 r i i 4 2' h, = fgg+ZK£é~ = 320+ (0.5 +0.75 + 0.75) 6:7 = 36lm2/52 p1 _ pa : 998F327 + 9.8(5) + 361] = 430,000N/m2 compared to [91 ~ pa = 389,000N/m2 Without minor loSses ...
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Lecture_Notes_10_24 - Pipe flow” analysis ' Energy...

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