{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture_Notes_10_24

# Lecture_Notes_10_24 - Pipe ﬂow” analysis Energy...

This preview shows pages 1–7. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

Lecture_Notes_10_24 - Pipe ﬂow” analysis Energy...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online