ࡱ> LNIJK#` @bjbjmm |4(`S J J J T $ @G@G@GPGdH #m IzKLKKK:SZS$Skkkkkkk$nh5qk) UFR:SUUk` ` KKtl]]]U(` RK Kk]Uk]]db " @cKI h@GCXbdl0#mc8r'Z.r(@c@cJr cpS0S"]T:TSSSkkU]pSSS#mUUUU $.d . ` ` ` ` ` `   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 February 5 Homework Solutions 2.5 Bourdon gages (see Video V2.2 and Fig. 2.13) are commonly used to measure pressure. When such a gage is attached to the closed water tank of Fig. P2.5 (copied from the text at the right), the gage reads 5 psi. What is the absolute pressure in the tank? Assume standard atmospheric pressure of 14.7 psia. The absolute pressure at the gage is is 5 psi + 14.7 psi = 19.7 psia. The fact that the gage is 6 in above the measuring line is irrelevant. The only difference in depth that contributes the pressure at the gage is the 12 in difference between the gage level and the air. Thus we can write that pair = 19.7 psia + gwater(12 in). Using gwater = 62.4 lbf/ft3 from Table 1-5 on the inside front cover gives the following result for the air pressure.  EMBED Equation.3  pair = 20.1 psia 2.11 In a certain liquid at rest measurements of the specific weight at various depths show the following variation: h(ft)010203040560708090100 EMBED Equation.3 7076849197102107110112114115The depth, h = 0, corresponds to a free surface at atmospheric pressure. Determine, through numerical integration of Eq. 2.4, the corresponding variation in pressure and show the results on a plot of pressure (in psf) versus depth (in feet). We have to numerically integrate dp/dz = -g or p = p0   EMBED Equation.3 , where p0 = 0 (given). We can use a simple trapezoid rule EMBED Equation.3  where N = (b  a)/Dx. In this problem the variable of integration is z and the step size, Dz =  10 ft. The values of f(z) are the specific weight taken from the table. Some example integrations give the following results:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  The complete results and graph, done in Excel, are shown below. h(ft)g (lbf/ft3)p (psf) 070010767302084153030912405409733455010243406010753857011064708011275809011487101001159855 2.26 A U-tube manometer contains oil, mercury, and water as shown in Fig. P2.26 (copied at the right). For the column heights indicated what is the pressure differential between pipes A and B? We see that there is a line of equal pressure near the bottom of the U-tube at the lower arrow for the dimension of the 12 in. Designating the two equal pressures on either side of the manometer at this point as pleft and pright allows us to write the following equations.  EMBED Equation.3  Take the data for specific weights from table 1-5 on the inside front cover (assuming the oil is SAE 30 weight oil and ignoring differences in temperature) are gwater = 62.4 lbf/ft3, gHg = 847 lbf/ft3, and goil = 57.0 lbf/ft3. Rearranging the equation to solve for pB  pA and substituting the data gives.  EMBED Equation.3  pB  pA = 5.57 psi 2.40 The differential mercury manometer of Fig. P2.40 (copied at the right) is connected to pipe A containing gasoline (SG = 0.65), and to pipe B containing water. Determine the differential reading, h, corresponding to a pressure in A of 20 kPa and a vacuum of 150 torr [use torr instead of mm Hg] in B. Call the pressure at the gasoline-mercury interface p1 and the pressure at the water-mercury interface p2. The difference between these two pressures is due to the column of mercury with height h.  EMBED Equation.3  We can relate p1 and p2 to the pressures in A and B as follows.  EMBED Equation.3  EMBED Equation.3  Combining the two equations above gives the following expression for pA  pB:  EMBED Equation.3  EMBED Equation.3  From the first equation we have p1  p2 = gHgh. Substituting this result into the equation above for pA pB gives.  EMBED Equation.3  EMBED Equation.3  Solving this equation for h gives  EMBED Equation.3  EMBED Equation.3  The data for specific weights of water and mercury are taken from table 1-6 on the inside front cover (ignoring differences in temperature): gwater = 9.80 kN/m3 and gHg =133 kN/m3. The gasoline has a specific gravity of 0.65 and, assuming that its reference specific weight is that of water just found, we find the gasoline specific weight as (SG)(gwater) = (0.65)(9.80 kN/m3) = 6.37 kN/m3. The pressure in A is a gage pressure. The pressure in B is given as a vacuum of 150 mm Hg. A vacuum is the difference between the atmospheric pressure and the absolute pressure. By definition, pvacuum = patmospheric pabsolute and pgage = pvacuum. Thus the 150 torr vacuum is a gage pressure of (150 torr)(101.325 kPa)/(760 torr) = 20.00 kPa. Substituting this pressure for pB, the given value of pA = 20 kPa, and the specific weight data from the previous paragraph into the equation for h gives  EMBED Equation.3  EMBED Equation.3  h = 0.384 m 2.42 The manometer fluid in the manometer of Fig. P2.42 (copied at the right) has a specific gravity of 3.46. Pipes A and B both contain water. If the pressure in pipe A is decreased by 1.3 psi and the pressure in pipe B increases by 0.9 psi, determine the new differential reading of the manometer. There is a common pressure on both sides of the manometer at the point of the lower water-gage-fluid interface. Writing the two equal pressures at this point in terms of the pressures in A and B gives the following equations.  EMBED Equation.3  For the configuration shown above, hwater,left = 2 ft, hgage = 2 ft, and hwater,right = 1 ft. We can solve this equation for pB pA to obtain the following result.  EMBED Equation.3  Use the specific weight of water as 62.4 lbf/ft3 from Table 1-5 on the inside front cover for both the value of gwater and the reference specific weight for the gage fluid. This gives the specific weight of the gage fluid as (3.46)(62.4 lbf/ft3) = 215.9 lbf/ft3. Substituting these values and the original gage readings into the equation for pB  pA gives  EMBED Equation.3  The new value of pB  pA is found as follows.  EMBED Equation.3  With the increased value of pB, the water-gage-fluid interface on the left will become lower by a value of Dh so that the new value of hwater,left will become 1 ft + Dh. This change will be reflected throughout the manometer system. The water-gage-fluid interface on the right will rise by the same value Dh so that hwater,righy becomes 2 ft  Dh. Both of these differences will change the height of the gage fluid by 2Dh to a new value of hgage = 2 ft + 2Dh. Substituting these new values of the gage heights and the new value of pB  pA into the original equation for pB  pA gives  EMBED Equation.3  The products of specific weights and heights on the right-side of the equation are seen to be simply the original values for pB  pA = 494.2 lbf/ft2. This gives the following steps for obtaining Dh.  EMBED Equation.3  Solving this equation gives Dh = 1.03 ft. Adding two times this value to the original value of hgage = QTUV     " # > ? @ A B I K ̿̿󡖎{{skd]V h@5CJ heA5CJ h15CJ h1CJaJh6VOJQJh"GOJQJhzdh"GOJQJh}xOJQJhnih}xOJQJjhnih}xOJQJUh156CJOJQJaJh8 56CJOJQJh156CJOJQJh15CJ aJ h15OJQJh1OJQJh1jh1U!UV? 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The mercury pressures are equal at the mercury-water interface and the point in the incline that is at the same level as this interface. These two equal pressures can be written as follows. (In this equation we use a variable height for the following distances in the diagram: hwater= 18 in, hoil = 12 in, and !Hg = 6 in = the difference between the water-mercury and the oil-mrecury interfaces.)  EMBED Equation.3  Solving for the pressure difference gives  EMBED Equation.3  When the pressure in A increases to pA,new, the water-mercury interface will drop by a value Dh and the mercury-oil interface will rise a distance, D! along the inclined length. These changes will cause hwater to increase by Dh and hoil to decrease by D! sin30o. The vertical difference in the mercury column will increase by the sum of Dh and D! sin30o. Even with all these changes we will still have the same relationship between the pressure difference and manometer measurements. This will give the equation below.  EMBED Equation.3  If we subtract the equation for the original pressure difference from this equation we obtain the following result for the increase in pressure in A.  EMBED Equation.3  We appear to have two unknowns, Dh and D!. However these two are related because the volume of the mercury is constant. When the mercury drops by a height change of Dh, a volume of mercury equal to prwater2 is displaced from the vertical tube to the inclined tube. The volume, V, of fluid in an cylinder inclined at an angle q with the horizontal plane is given by the equation V = pr2(! + r/tanq), where ! is the distance from the bottom of the cylinder to the closest location of the inclined plane. If the value of ! is changed by D!, keeping r and :xyyyyyFz\z^zbzlzzzzzL{N{P{R{x{z{|{~{{{{{{̴̴̴§lcZOjQ`I h|~UVh|~OJQJ\h%8OJQJ\$jh|~h|~EHOJQJU\jP`I h|~UVhbh|~OJQJ\ jhbh|~OJQJU\h hgq9OJQJ^Jh h H*OJQJ^Jhy3>OJQJ^Jh h OJQJ^Jh OJQJ^Jh%8OJQJ^JhSh5OJQJh%85OJQJ4xyP{{{|Jvv PĖ $a$gdN $a$gdN $a$gd5gddgd=+ $a$gd=+gd5 $a$gd|~ $a$gd gd|) & p@ P D%0`0gd%8{|||N|X||| }},}.}j}l}n}}}}}}}}}}}~~~~~~~~~~~ɾɱɧɱɝɧɝɾɱzɝmbh5H*OJQJ^Jh=+h5OJQJ^Jh=+H*OJQJ^Jh5h5OJQJ^Jh5H*OJQJ^Jh5OJQJ^Jhy3>OJQJ^Jh=+h=+OJQJ^Jh=+H*OJQJ^Jh=+OJQJ^Jh|~OJQJ\ jhbh|~OJQJU\$j h|~h|~EHOJQJU\$BDFHJvx68:DF68„Ąλ岨ٝ岀ssssh]ss]sh)_H*OJQJ^Jh)_H*OJQJ^Jh)_h)_OJQJ^Jh)_OJQJ^J$jl'h|~h5EHOJQJU\j`I h5UVh=+OJQJ^Jh=+OJQJ\$j#h|~h5EHOJQJU\j`I h5UVhbh=+OJQJ\ jhbh=+OJQJU\h|~OJQJ^J"܅ޅ^`bdfȐ`bnptvxBP巭tj`ShdhdOJQJ^Jhy3>OJQJ^Jh=+OJQJ^J%j*hdhdEHOJQJU^Jjo`I hdUVjhdOJQJU^Jh)_h=+OJQJ^JhdOJQJ^JhdOJQJ^Jh)_OJQJ^Jh)_H*OJQJ^JUh)_h)_OJQJ^Jh)_OJQJ^Jjh50J OJQJU^Jq constant, the difference in volume is simply pr2D!, just as it would be for a vertical cylinder. Equating the two displaced volumes gives the following relationship between Dh and D!.  EMBED Equation.3  Before substituting this result into our equation for the pressure change, we have to get the necessary data. Table 1-5 in the inside front cover gives the values for gwater = 62.4 lbf/ft3 and gHg = 847 lbf/ft3. The oil has a specific gravity of 0.9 so we find its specific weight using the value of gwater = 62.4 lbf/ft3 just found: goil = 0.9(62.4 lbf/ft3)r = 56.16 lbf/ft3. We can now substitute the equation Dh = D!/4 into the pressure increase equation, set sin30o to its value of 0.5, and substitute the specific weights just obtained to get the solution for Dh.  EMBED Equation.3   EMBED Equation.3  The mercury level drops 0.304 ft.  Taken from http://www.lmnoeng.com/Volume/InclinedCyl.htm web site accessed January 19, 2008.     Jacaranda (Engineering) 3333 Mail Code Phone: 818.677.6448 E-mail:  HYPERLINK mailto:lcaretto@csun.edu lcaretto@csun.edu 8348 Fax: 818.677.7062  REF date \h  \* MERGEFORMAT February 5 homework solutions  INCLUDETEXT "include.doc" headerCenter \* MERGEFORMAT ME 390, L. S. 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?3"`?2}~CviۣFY`!Q~CviۣF H(`\xcdd``.d 2 ĜL0##0KQ* WtEAJRcgbR vI*ĒʂT@D22BaJ`*` 011nX  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHMPpQRSTVUWXYZ[\]^_`badcfeghjiklmnoqrstuwvxyz{|}~Root Entry F?#hO@RData 97WordDocument|ObjectPool'01h?#h_1231064558F01h01hOle CompObjfObjInfo !$%&'(+./012369:;<=>?BEFGHILOPQRSTUVWXYZ]`chklmnqtuvwz}~ FMicrosoft Equation 3.0 DS Equation Equation.39q( p air =19.7lb f in 2 +62.4lb f ft 3 12in()Equation Native D_1231065006 F01h01hOle  CompObj fft 3 1728in 3 FMicrosoft Equation 3.0 DS Equation Equation.39qA lb f ft 3ObjInfo Equation Native ]_1231065450F01h01hOle  FMicrosoft Equation 3.0 DS Equation Equation.39q<x dz z 0 z +" FMicrosoft Equation 3.0 DS EqCompObjfObjInfoEquation Native X_1231065855 F01h01hOle CompObjfObjInfoEquation Native  uation Equation.39q f(x)dxH"x2fa()+2fa+kx() k=1N"1 " +fb()[] ab +"_1231065981F01h01hOle CompObj fObjInfo" FMicrosoft Equation 3.0 DS Equation Equation.39q-x p 10ft =0""10ft270lb f ft 3 +76lb f ft 3 []=730lb f ft Equation Native #I_1231066159"F01h01hOle )CompObj *f2 FMicrosoft Equation 3.0 DS Equation Equation.39q…Pu p 20ft =0""10ft270lb f ft 3 +2ObjInfo!,Equation Native -_1231066238$F01h01hOle 476lb f ft 3 ()+84lb f ft 3 []=1540lb f ft 2 FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj#%5fObjInfo&7Equation Native 8_1231067440c)F01h01h p 30ft =0""10ft270lb f ft 3 +276lb f ft 3 +84lb f ft 3 ()+91lb f ft 3 []=2405lb f ft 2Ole @CompObj(*AfObjInfo+CEquation Native D{ FMicrosoft Equation 3.0 DS Equation Equation.39q_ p left =p A + oil 7in()+ Hg 12in()=p right =p B + H 2 O 3in+12in() FMicrosoft Equation 3.0 DS Equation Equation.39q p B "p A = oil 7_12310679516.F01h01hOle JCompObj-/KfObjInfo0MEquation Native N#_1231068981,3F01h01hOle [CompObj24\fin()+ Hg 12in()" H 2 O 3in+12in()=57.0lb f ft 3 7in()+847lb f ft 3 12in()"62.4lb f ft 3 15in()[]ft 3 1728in 3 psia"in 2 1lb f FMicrosoft Equation 3.0 DS Equation Equation.39qOаo p 1 =pObjInfo5^Equation Native _k_12310689448F0h0hOle a 2 + Hg h FMicrosoft Equation 3.0 DS Equation Equation.39q  FMicrosoft Equation 3.0 DS EqCompObj79bfObjInfo:dEquation Native e#_12310689961@=F01h01hOle fCompObj<>gfObjInfo?iEquation Native j#uation Equation.39q p 1 =p A + gasol h+0.3m()p B =p 2 + water h+0.3m()_1231069426BF01h0hOle oCompObjACpfObjInfoDr FMicrosoft Equation 3.0 DS Equation Equation.39q p A "p B =p 1 "p 2 " water h+0.3m()" gasol h+0.3m()Equation Native s_1262234767|OGF0h0hOle xCompObjFHyf FMicrosoft Equation 3.0 DS Equation Equation.39qW< p A "p B = Hg h" water h+0.3m()" gasol h+0.ObjInfoI{Equation Native |_1231069949;YLF0h0hOle 3m()= Hg " water " gasol ()h"0.3m() water + gasol () FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjKMfObjInfoNEquation Native =_1262234875QF0h0h! h=p A "p B +0.3m() water + gasol () Hg " water " gasol FMicrosoft Equation 3.0 DS EqOle CompObjPRfObjInfoSEquation Native @uation Equation.39qW$$ h=p A "p B +0.3m() water + gasol () Hg " water " gasol =20kPa""20kPa()[]kNkPa"m 2 +0.3m()9.81kNm 3 +6.37kNm 3 ()133kNm 3 "9.81kNm 3 "6.37kNm 3_1231072268VF0h0hOle CompObjUWfObjInfoX FMicrosoft Equation 3.0 DS Equation Equation.39q…()4 p left =p A + water h water,left + gage h gage =p right =p B +Equation Native _1231072546T^[F0h0hOle CompObjZ\f water h water,rigfht FMicrosoft Equation 3.0 DS Equation Equation.39q, p B "p A = water h wObjInfo]Equation Native H_1231073242`F0h0hOle ater,left + gage h gage " water h water,right FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj_afObjInfobEquation Native _1231073250JweF0hhp' p B "p A =62.4lb f ft 3 2ft()+215.9lb f ft 3 2ft()"62.4lb f ft 3 1ft()=494.2lb f ft 2Ole CompObjdffObjInfogEquation Native T+, FMicrosoft Equation 3.0 DS Equation Equation.39q8 p B "p A () new =p B,old +p B "p A,old +p A ()=p B "p A () old +p B "p A =494.2lb f ft 2 +0.9lb f in 2 144in 2 ft 2 ""1.3lb f in 2 144in 2 ft 2 =811.0lb f ft 2 FMicrosoft Equation 3.0 DS Equation Equation.39q p B "p A () new =81X=r4 W&00q|bرsA V0Z^( b#~#}L {)<&? 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