ࡱ> #` Hybjbj5G5G W-W-(Q     $B jjjPvlB 8D2!L~!~!~!H%%4&7777777$:h<8 ($^H%((8 ~!~!l8s,s,s,( "~! ~!7s,(7s,s,1 r 1~! yOj)1~7t80818=)P=(11\= J24*&Lv&6s,&,&)*&*&*&88,^*&*&*&8((((B B B B B B B B B   INCLUDEPICTURE "http://www.csun.edu/~pubrels/NewLogoSm.JPG" \* MERGEFORMATINET  College of Engineering and Computer Science Mechanical Engineering Department Mechanical Engineering 390 Fluid Mechanics INCLUDETEXT include.doc firstPage \* MERGEFORMAT Spring 2008 Number: 11971 Instructor: Larry Caretto April 22 Homework Solutions 8.37 The following equation is sometimes used in place of the Colebrook equation.  EMBED Equation.3  for 10-6 < e/D < 102 and 5000 < Re < 108. An advantage of this equation is that given Re and e/D, it does not require an iteration procedure to obtain f. Plot a graph of the percent difference in f as given by this equation and the original Colebrook equation for Reynolds numbers in the range of validity of the above equation with e/D = 10-4.- The calculations were done on a spreadsheet that has been placed on the course web site. You can download it at  HYPERLINK "http://www.csun.edu/~lcaretto/me390/Colebrook.xls" http://www.csun.edu/~lcaretto/me390/Colebrook.xls. The results are shown in the figure below.  For the relative roughness of 10-4, it appears that the formula would be accurate beyond the upper limit of Re = 108, but is rapidly losing accuracy at the lower limit of Re = 5000. 8.39 Two equal length, horizontal pipes, one with a diameter of 1 in., the other with a diameter of 2 in., are made of the same material and carry the same fluid at the same flow rate. Which pipe produces the larger head loss? Justify your answer. We know that for fully turbulent flow the head loss varies as 1/D5; and for laming flow the head loss varies as 1/D4; thus we expect the smaller pipe to have the larger head loss in these two regions. In intermediate regions, the head loss is given by the equation  EMBED Equation.3  The ratio of the heat loss in the two-inch pipe to that in the one-inch pipe, when Q and L are the same, depends only on differences in diameter and friction factor.  EMBED Equation.3  On the Moody diagram, the friction factor ranges from 0.008 to 0.1; the maximum possible value of f2/f1 is 0.1/0.08 = 12.5. Even this maximum possible range does not offset the factor of 1/32. Thus, we conclude that the larger pipe will have the smaller head loss. 8.47 Water flows at a rate of 0.040 m3/s in a 0.12-m-diameter pipe that contains a sudden contraction to a 0.06-m-diameter pipe. Determine the pressure drop across the contraction section. How much of this pressure difference is due to losses and how much is due to kinetic energy changes. A sketch of the contraction (from the solution manual for the Munson text) is shown at the left. We can write the energy equation for between the flow before the contraction (1) and the flow after the contraction (2). The head loss for a contraction is written in terms of the loss coefficient times the velocity after the contraction.  EMBED Equation.3 We can find the velocities from the flow rate, which is constant, and the areas.  EMBED Equation.3  The loss coefficient is found from Figure 8.26 on page 440 with A2/A1 = (D2/D1)2 = [(0.06 m) / (0.12m)]2 = 0.25: KL = 0.4. Since the contraction is horizontal, z2 = z1 and our energy equation becomes.  EMBED Equation.3  The first term on the right represents the kinetic energy change and the second term is the loss term. Substituting in the velocity data, we can evaluate each term.  EMBED Equation.3  Multiplying the pressure head terms by the specific weight of water, 9.80 kN/m3from Table 1-6. gives the following results for pressure drop:  EMBED Equation.3  8.54 Natural gas (r = 0.0044 slugs/ft3 and n = 5.2x10-5 ft2/s) is pumped through a horizontal 6-in-diameter cast-iron pipe at a rate of 800 lbf/hr. If the pressure at section (1) is 50 psi(abs), determine the pressure at section (2) 8 mi downstream if the flow is assumed incompressible. Is the incompressible assumption reasonable? Explain. The energy equation for the pipe flow is.  EMBED Equation.3  This is a horizontal pipe so z1 = z2 and for incompressible flow with a constant diameter, V1 = V2 so the only terms left in the energy equation are the pressure drop and the head loss.  EMBED Equation.3  The pipe diameter = 6 in = 0.5 ft and we have data for a weight flow so we find the velocity and the Reynolds number as follows.  EMBED Equation.3   EMBED Equation.3  So the flow is turbulent and we find the roughness for cast iron pipe as e = 0.00085 ft from Table 8.1 so e/D = (0.00085 ft) / (0.5 ft) = 0.0017. From this relative roughness and Re = 7.587x104 we can find f = 0.0245 from the Moody diagram. Using this f on the right side of the Colebrook equation gives the following result.  EMBED Equation.3  Using this friction factor we can find the pressure drop.  EMBED Equation.3  If the initial pressure is 50 psi(abs) the downstream pressure is 50 psi (294.6 psf) / (144 psi/psf), or p2= 48.0 psi. 8.66 Water flows steadily through a 0.75-in-diameter, galvanized iron pipe system shown in the figure at the right at a rate of 0.20 cfs. Your boss suggests that friction losses in the straight pipe sections are negligible compared to losses in the threaded elbows and fittings of the system. Do you agree or disagree with your boss? Support your answer with appropriate calculations. From the diagram we see that the flow path goes through (6 in + 6 in + 4 in + 1 in) = 17 in of straight run pipe. In addition it flows through two elbows, the branch of a tee and a reducer before exiting to the atmosphere. The total head loss is the sum of the major and minor losses. If all the minor losses are based on the same velocity, V, the velocity in the straight run pipe sections, we can write the total head loss as follows.  EMBED Equation.3  We can find the water velocity from the flow rate of 0.20 ft3/s and the pipe diameter of 0.75 in = 0.0625 ft.  EMBED Equation.3  The velocity head that is used to compute the head loss is computed from this velocity.  EMBED Equation.3  Next, we find the Reynolds number, using the value of n = 1.21x10-5 ft2/s for water from Table 1.5.  EMBED Equation.3  So the flow is turbulent and we find the roughness for galvanized iron pipe as e = 0.0005 ft from Table 8.1 so e/D = (0.0005 ft) / (0.0625 ft) = 0.0017. From this relative roughness and Re = 3.367x104 we can find f = 0.038 from the Moody diagram. Using this f on the right side of the Colebrook equation gives the following result.  EMBED Equation.3  With this friction factor we can compute the major head loss due to the straight run pipe that has a total of 17 in.  EMBED Equation.3  The minor loss coefficients are 2(1.5) = 3 for the two 90o threaded elbows, and KL = 2 for the threaded tee branch flow. There are no data in the text for the reducer, so we can approximate the loss coefficient as that of a sudden contraction; this should overstate the loss that we would expect from the more gradual reducer. Since the reducer reduces the diameter from 0.75 in to 0.60 in the area ratio is [(0.60 in)/(0.75 in)]2 = 0.64 and the loss coefficient is 0.15 for this area ratio according to figure 8.26. Since the loss coefficient for the sudden contraction uses the downstream velocity to compute the head loss and we want to have all the loss coefficients based on the same (upstream) velocity we can modify this loss coefficient as follows.  EMBED Equation.3  So the sum of the loss coefficients is 3 + 2 + 0.366 = 5.366, and the head loss due to the minor losses is  EMBED Equation.3  So the total head loss is 3.544 ft + 0.556 ft = 4.100 ft, with 13.6% coming from the straight run pipe. You should probably tell your boss that she was basically correct that the minor losses are much more significant than the major losses, but the major losses do contribute 13.6% of the total loss and you think that you should include them in your calculations. 8.75 The pump shown in the figure to the right delivers a head of 250 ft to the water. Determine the power that the pump adds to the water. The difference in elevation of the two ponds is 200 ft. We know the pump head is 250 ft and the relation between the pump head and the pump power is  EMBED Equation.3  In order to apply this equation we have to know the flow rate. We can see if we have enough information to compute the flow rate from the energy equation. We begin by writing the energy equation for a control volume between the surface of the lower pond, point 1, and the surface of the upper pond, point 2. The difference in elevations, z2 z1 is given to be 200 ft, and since both ponds are open to the atmosphere at their surfaces, we have p1 = p2 = 0 and V1 = V2 = 0.  EMBED Equation.3  Setting V1 =V2 = p1 = p2 = 0 gives  EMBED Equation.3  Entering the given data for z2 z1 and hs reduces the problem to one of finding V.  EMBED Equation.3  This needs to be an iterative solution since f depends on V. Substituting known values for the sum of the loss coefficients from the diagram (0.8 + 5 + 1.5 + 1.5 + 1.5 + 1.5 +1 = 12.8) and the other known terms in the head loss gives the following computational equation.  EMBED Equation.3  Use the kinematic viscosity of water from table 1.5, QTUV     " # > ? @ A B H ̿̿󡖎{{skdZh)h)"5CJ h15CJ h1CJaJh6VOJQJh[OJQJhzdh[OJQJh}xOJQJhnih}xOJQJjhnih}xOJQJUh156CJOJQJaJh8 56CJOJQJh156CJOJQJh15CJ aJ h15OJQJh1OJQJh1jh1U UV? 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Guess a value for Vpipe 2. Compute Re = 6.198x104Vpipe 3. Find the friction factor from the Moody diagram or equations 4. Use the value just found for f in the equation for V2pipe to find Vpipe 5. Continue the process in steps 1 to 4 until two values of Vpipe are sufficiently close. If we start with a guess that Vpipe = 5 ft/s, we find Re = 3.099x105 and f = 0.0144. Solving the V2pipe equation gives Vpipe = 11.99 ft/s. For the next iteration we can use the value just found, Vpipe = 11.99 ft/s, as our guess. That gives Re = 7.431x105 and f = 0.0123 and Vpipe = 12.39 ft/s. For the third iteration we again use the value just found, Vpipe = 12.39 ft/s, as our guess. That gives Re = 7.677x105 and f = 0.0122 and Vpipe = 12.40 ft/s. This is very close to the value we used at the start of the calculations and we will use this as our final value. We are finally able to find the power input to the pump.  EMBED Equation.3      Jacaranda (Engineering) 3333 Mail Code Phone: 818.677.6448 E-mail:  HYPERLINK mailto:lcaretto@csun.edu lcaretto@csun.edu 8348 Fax: 818.677.7062  REF date \h  \* MERGEFORMAT April 22 homework solutions  INCLUDETEXT "include.doc" headerCenter \* MERGEFORMAT ME 390, L. S. 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?3"`?2"O\h9od]`!U"O\h9od(8X #xڥ=hQ;}wzǝ #.b!&H.E"l0)8Q`؊;KDW"ͼ}įFo?3o[!w$lz[(}bP[  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~78Root Entry FOiData LWordDocumentObjectPool'gwOO_1238601342FgwOgwOOle CompObjfObjInfo  #$%&'(+./01234569<=>ADEFGHIJKLMNORUVWXYZ[\]^_`abehijklmpstuvwz}~ FMicrosoft Equation 3.0 DS Equation Equation.39qQ8 f=1.325ln)D3.7+5.74Re 0.9 ()[] 2Equation Native _1238601891 FgwOgwOOle CompObj f FMicrosoft Equation 3.0 DS Equation Equation.39qQ{Я\ h L =f!DV 2 2g=f!D12gQA() 2 =f!D12gObjInfo Equation Native  _1238602532FgwOgwOOle Q)D 2 4() 2 =2f!gQ 2 D 5 FMicrosoft Equation 3.0 DS Equation Equation.39qQÛؿ\ h L 2CompObjfObjInfoEquation Native _1238603176 FgwOgwO  h L 1  =f 2 f 1 D 15 D 25 =f 2 f 1 1in() 5 2in() 5 =132f 2 f 1 FMicrosoft Equation 3.0 DS EqOle CompObjfObjInfo!Equation Native "uation Equation.39qQÈ\ z 2 +p 2 +V 22 2g=z 1 +p 1 +V 12 2g"h L =z 1 +p 1 +V 12 2g"K L V 22 2g FMicrosoft Equation 3.0 DS Equation Equation.39qQ8lu V 1 =QA 1 =Q)D 12 4=4)0.0_1238604216"FgwOgwOOle )CompObj*fObjInfo,Equation Native -T_1238604187FgwOgwOOle 7CompObj 8f4m 3 s0.12m() 2 =2.182msV 2 =4QD 22 =4)0.04m 3 s0.06m() 2 =8.727ms FMicrosoft Equation 3.0 DS Equation Equation.39qQtL! p 1 "p 2 =V 22 "V 12 2g"K L V 22 2gObjInfo!:Equation Native ;_1238604497$FgwOgwOOle ?CompObj#%@fObjInfo&BEquation Native C_1238610194P)FgwOgwO FMicrosoft Equation 3.0 DS Equation Equation.39qQز\ p 1 "p 2 =V 22 "V 12 2g"K L V 22 2g=8.727ms() 2 "2.182ms() 2 29.80665ms 2 "0.48.727ms() 2 29.80665ms 2 p 1 "p 2 =3.640m+1.553m=5.193mOle PCompObj(*QfObjInfo+SEquation Native T FMicrosoft Equation 3.0 DS Equation Equation.39qQf8&\ p 1 "p 2 () KE =9.80kNm 3 3.640m()1kPa"m 2 kN=35.6kPap 1 "p 2 () loss =9.80kNm 3 1.553m()1kPa"m 2 kN=15.2kPap 1 "p 2 () total =9.80kNm 3 5.193m()1kPa"m 2 kN=50.9kPa FMicrosoft Equation 3.0 DS Equation Equation.39qQËl3 z 2 +_123861274316.FgwOgwOOle cCompObj-/dfObjInfo0fEquation Native g_12386126103FgwOgwOOle nCompObj24ofp 2 +V 22 2g=z 1 +p 1 +V 12 2g"h L =z 1 +p 1 +V 12 2g"f!DV 22 2g FMicrosoft Equation 3.0 DS Equation Equation.39qQ) p 1 "p 2 =p 1 "p 2 g=f!DV 2 2g!p 1 "p 2 =f!DV 2 2ObjInfo5qEquation Native rE_12386133748FgwOgwOOle x FMicrosoft Equation 3.0 DS Equation Equation.39qQE}l^ V=2WAg=2W)D 12 4()g=4ft 3 0.0044slugCompObj79yfObjInfo:{Equation Native |a_1238613514,@=FgwOgwO)800lb f h0.5ft() 2 s 2 34.174ft1slug"ftlb f "s 2 h3600s=7.995fts FMicrosoft Equation 3.0 DS EqOle CompObj<>fObjInfo?Equation Native Fuation Equation.39qQ*$\ Re=VD=VD=7.995fts0.5ft()5.2x10 "5 ft 2 s=7.587x10 4_1238617481FjBFgwOgwOOle CompObjADfObjInfo FMicrosoft Equation 3.0 DS Equation Equation.39qCp 8 .  @=&@= & MathType -$h -OlePres000CEEquation Native _1238614427ZHFgwOgwOOle -FFz@-- i -  $+-+r-z ,A ((-(u)-u )h)h)-!-Times New Roman w)w 0w f- 2 N:02482 9.22 .902 2 %135082 0.52 065 2 _*02452 _:*.22 _z)022 _ &10 2 _2#58752 _".82 _"782 '512 e'.12 &212 +712 +2.12 +r31 2 u00172 .02 U00 2 Jlog72 f0o2 .o2 F2o2 MRe2 512 .12 212 +l 712 + .12 +L 31 2 log72  0o2 .o2 2o2  1oTimes New Roman w)w 0w f-2 '4o2 `1102 `10Symbol 7 w)w 0w f-2 8=02 302 .=02 -02 -02 -02 -02 {02 {02 {02 {02  +02 u-02 H=02 \02 \02 \02 \02 5 02 5 02 5 02 5 02  +02  e02 /-02 =0Times New Roman w)w 0w f-2 6f02 _%x02 M8f02 U D02 Mf0 & "Systemܨ f !-NANIQ! 1 f  ="2.0log 10 )D3.7+2.51Re f  ()="2.0log 10 0.00173.7+2.517.587x10 4  0.0245  ()=6.350!f=0.0248CompObjGIfObjInfoJEquation Native O_1238619063MFgwOgwO FMicrosoft Equation 3.0 DS Equation Equation.39qQ3hd( p 1 "p 2 =f2!DV 2 =0.024828mi0.5ft5280ftmi0.0044slugft 3 7.995fts() 2 1lb f "s 2 slug"ft=294.6lb f ft 2 FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjLNfObjInfoOEquation Native QÊ095 h L =f!D+K L " ()V 2 2g FMicrosoft Equation 3.0 DS Equation Equation.39q_1238619251;yRFgwOgwOOle CompObjQSfObjInfoTEquation Native B_1238619260WFgwOgwOOle CompObjVXfQ&Pr V=QA=Q)D 2 4=4)0.020ft 3 s0.0625ft() 2 =6.519fts FMicrosoft Equation 3.0 DS Equation Equation.39qQ h L =V 2 2g=6.519fts() 2 232.174fts 2 =0.6604ftObjInfoYEquation Native  _1238617229\FgwOgwOOle CompObj[]fObjInfo^Equation Native V_1238617618aFgwOgwO FMicrosoft Equation 3.0 DS Equation Equation.39qQ:wD! Re=VD=VD=6.519fts0.0625ft()1.21x10 "5 ft 2 s=3.367x10 4 FMicrosoft Equation 3.0 DS Equation Equation.39qBp 8 .  @;&; & MathType Ole CompObj`cfObjInfoOlePres000bd     "%&'(),/01234569<=>?BEFILMNOPQRSVYZ[\]^adefghknopqrstuvwxyz}-$h --FFz@-- i -  $+-+r-z ,A''-'u0(-u8(((,2!0,Times New Roman w)w 0wfr- 2 803712 V8.32 703 2 /18912 -/.82 m.58 2 _)03812 _h).32 _(032 _N%10 2 _`"36712 _".62 _@!362 &512 6&.12 v%212 +,712 +.12 + 31 2 u00812 .02 U00 2 Jlog12 f0o2 .o2 F2o2 MRe2 512 .12 212 +l 712 + .12 +L 31 2 log12  0o2 .o2 2o2  1oTimes New Roman w)w 0wfr-2 &4o2 `1102 `10Symbol5 ش w)w 0wfr-2 o6=02 202 L-=02 `,02 `,02 `,02 `,02 {02 {02 {02 {02  +02 u-02 H=02 \02 \02 \02 \02 5 02 5 02 5 02 5 02  +02  e02 /-02 =0Times New Roman w)w 0wfr-2 S5f02 _$x02 M8f02 U D02 Mf0 & "Systemrfr !-NANIEquation Native _1238619871gFgwOgwOOle  CompObjfh!fQ! 1 f  ="2.0log 10 )D3.7+2.51Re f  ()="2.0log 10 0.0083.7+2.513.367x10 4  0.038  ()=5.189!f=0.0371 FMicrosoft Equation 3.0 DS Equation Equation.39qQ00 h L Major  =f!DV 2 2g=0.0371()17in0.75in0.ObjInfoi#Equation Native $L_1238618617_KlFgwOgwOOle *6604ft()=0.556ft FMicrosoft Equation 3.0 DS Equation Equation.39qQ K L V 22 2g=K L 2CompObjkm+fObjInfon-Equation Native .._1238619844UeqFgwOgwOgV 1 A 1 A 2 () 2 =K L A 1 A 2 () 2 V 12 2g=K L* V 12 2g!K L* =0.151.64() 2 =.366Ole 7CompObjpr8fObjInfos:Equation Native ; FMicrosoft Equation 3.0 DS Equation Equation.39qQ  h L Minor  =K L " ()V 2 2g=5.3660.6604ft()=3.544ft_1238624256vFgwOgwOOle @CompObjuwAfObjInfoxC FMicrosoft Equation 3.0 DS Equation Equation.39qQj 2W shaftnetin =Qh sEquation Native D_1238620750o{FgwOgwOOle GCompObjz|Hf FMicrosoft Equation 3.0 DS Equation Equation.39qQ z 2 +p 2 +V 22 2g=z 1 +p 1 +V 12 2g+h s "h L =z 1 +p ObjInfo}JEquation Native K_1238620867FgwOgwOOle T1 +V 12 2g+h s "f!D+K L " ()V pipe2 2g FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjUfObjInfoWEquation Native X_1238621291~FgwOgwOQÐ4 z 2 =z 1 +h s "f!D+K L " ()V pipe2 2g!f!D+K L " ()V pipe2 2g=z 1 +h s "z 2Ole _CompObj`fObjInfobEquation Native cB FMicrosoft Equation 3.0 DS Equation Equation.39qQ&x f!D+K L " ()V pipe2 2g=h s "z 2 "z 1 ()=250ft"200ft=50ft FMicrosoft Equation 3.0 DS Equation Equation.39qQOR f!D+K L " ()V pipe2 2g=f5_1238621532FgwOgwOOle iCompObjjfObjInfolEquation Native mk_1238622103tFgwOgwOOle {CompObj|f00ft0.75ft+12.8()V pipe2 232.174fts 2 =50ft666.7f+12.8()V pipe2 =50ft()232.174fts 2 =3217.4ft 2 s 2 !V pipe2 =3217.4ft 2 s 2 666.7f+12.8 FMicrosoft Equation 3.0 DS Equation Equation.39qQH0P Re=VDObjInfo~Equation Native d_1238624455FgwOgwOOle =V pipe 0.75ft()1.21x10 "5 ft 2 s=6.198x10 4 sft()V pipe FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjfObjInfoEquation Native 1Table:=Qp 2W shaftnetin =Qh s =V pipe Ah s =V pipe )D 2 4()h s =62.4lb f ft 3 12.40fts40.75ft() 2 250ft()hp"s550ft"lb f =155hpOh+'0  0 < H T`hpx Larry Caretto      !"#$%&'()*+,-./01234569;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXjk* *hkQ21VSɹs$ڻk[jؑb߂oA ] s߷RSNas>GM;+o#TƧ[i>vc,CzՇ?v/ N-qN}Q͛lB(>xhpG}pVic),Ҩ~DpN,'􉺮67|r_Lِ-:gljA[,>s=4 ȱ_O;l]&bֹ-$8ׯNϼo,Qg{$'( rzNϼ'yioOtjyЍ<۹b W #{_#a׮@f֙K[dꀎ (<DyK yK |http://www.csun.edu/~lcaretto/me390/Colebrook.xlsyX;H,]ą'c Dd _(CC0  # A"5 $SMNu Ә@= $SMNu6$dmI x pU^xyD1OC̘hMQZt A)g@AJtu,SAQis=rsG 39{ݽ{@-Ir;  ,KBGC_Eos%Rxv0A[+m=V2 xtw3^4?L1Pe>>kzڍdoI&|CzkH[u}eL7˺w8VÖ{>'6{G IP076aS>OъVIIߧ}W|_c? 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