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# 55 Note for MATH 411 at PSU

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Date Created: 02/06/15

Fluid Mechanics for Chemical Engineers Third Edition Noel de Nevers Solutions Manual This manual contains solutions to all the problems in the text Many of those are discussion problems I have tried to present enough guidance so that the instructor can lead a useful discussion of those problems In addition I have added discussion material to many of the computation problems I regularly assign these as computation and then after we have agreed that the computation is correct asked the students what this computation tells them That leads to discussion Wherever I can I begin a discussion of some topic with a computation problem which introduces the students to the magnitudes of various quantities and thus requires them to read the part of the text covering that topic Once we all know the magnitudes and have all read that section of the text we can have an interesting discussion of their meaning In this additional discussion I have presented reference when I could Often I relied on industry quotcommon knowledgequot folklore and gossip Ihope I got it all right If not I apologize for leading you astray Where I am not sure about the folklore Ihave tried to make that clear in the discussion Those problems whose numbers are followed by an asterisk are ones whose answer is presented in the Answers to Selected Problems in Appendix D of the book Many of these problems have been class tested Some alas have not This manual as well as the book is certain to contain errors I will be grateful to those who point these errors out to me so that they can be corrected I keep a running correction sheet and send copies to anyone who asks for it If you find such an error please notify me at Noel de Nevers Department of Chemical and Fuels Engineering 50 South Central Campus Drive University of Utah Salt Lake City Utah 84112 NoeldeNeversmccutahedu 8015816024 Solutions Fluid Mechanics for Chemical Engineers Third Edition Solutions Manual Page 1 FAX 8015859291 Many of the problems go beyond what is in the text or show derivations which I have left out of the text to make it read easier I suggest that the instructor tell the students to at least read all the problems so that they will know what is contained there Some of the problems use spreadsheets In the individual chapters I have copied the spreadsheet solutions into the text in table format That is easy to see but does not let the reader modify the spreadsheets In the Folder labeled quotSpreadsheetsquot I have included copies of all the spreadsheets shown in the individual chapters and also the high velocity gas tables from the appendices These are in Excel 40 which is compatible with all later versions Noel de Nevers Salt Lake City Utah 2003 Solutions Fluid Mechanics for Chemical Engineers Third Edition Solutions Manual Page 2 Fluid Mechanics For Chemical Engineers Third Edition Noel de Nevers Solutions Manual Chapter 1 An on a problem number means that the answer is given in Appendix D of the book 11 Laws Used Newton39s laws of motion conservation of mass rst and second laws of thermodynamics Laws Not Used third law of thermodynamics all electrostatic and magnetic laws all laws discussing the behavior of matter at the atomic or subatomic level all relativistic laws 12 By ideal gas law for uranium hexa uoride g 1 atm 352 J PM L 1b p 101 3 00130i3081 I1 RT Latm ft 39 6 0082 10 cm cm J562 27315 K mol K Here the high density results from the high molecular weight At its normal boiling point 4 K by ideal gas law helium has PM 14 1b 0012 g3076 in RT 00824 cm ft p Here the high density results from the very low absolute temperature The densities of other liquids with low values are liquid methane at its nbp 042 gmcm3 acetylene at its nbp 062 ethylene at its nbp 057 Discussion the point of this problem is for the students to recognize that one of the principal differences between liquids and gases is the large difference in density As a rule of thumb the density of liquids is 1000 times that of gases 13 p Z massZ volume For 1001bm pleOlbm 501bm 3 501bm szlozlbrsn 4496231bmft 6231bmft ft Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 1 page 1 Discussion this assumes no volume change on mixing That is a good assumption here and in many other cases In a few like ethanol and water have changes of up to a few 14 The maximum density of water occurs at 4 C not at zero The relation between the meter and the kg was de ned to have the density of water at 4 C be 100 gmcm3 However for various historical reasons it has ended up that the density of water at 4 C is about 099995 gmcm3 m m 45 1724 15 p gross tare g 3 gzlillo g3 V 25cm cm m m m 45 1724003 p ms m 2 g 3 g1109 g3 V 25cm cm Omitting the weight of the air makes a difference of 0001 01 This is normally ignored but in the most careful work it must be considered often be quoted as A B39deg API bbl where B Z 001deg API 16 This scale hasthe advantage that it 200 I I I I I places a higher number on lower density oils That matches the price structure for 150 39 39 oil where lower density crude oils have 3 a higher selling price because they are 100 39 more easily converted to highpriced 3 products e g gasoline Oil prices will 00 5 VI 0 llllllllllllllllll Theplotcoversthewholerangeof 5 HII Illmlmllmllmlg petroleum liquids from propane sg Z 39 05 06 07 08 09 1 U 05 to asphalts sg E 11 Water sg 1has 10 API specific gravity specific M as 17 For 1deal gases 5 For methane and propane the values are graVIty ideal gas air 16 29 055 and 44 29 151 Propane is by far the most dangerous fuel in common use If we have a methane leak buoyancy will take it up and disperse it If we have a leak of any liquid fuel it will ow downhill on the ground and be stopped by any ditch or other low spot Propane as a gas heavier than air ows downhill over small obstructions and depressions It often finds an ignition source which methane or gasoline would not find in the same situation Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 1 page 2 1 8 dV has dimensionJ S 1 39 ft E dy of has dimenSionJ lbf lbf ft lbm ft sec momentum 1 2 2 2 momentum ux of ft lbmsec ft sec area t1me has dimensionJ 1 lbf ft2 lbm of dVdy ls fts 19 Paints not settle in the can Inks spread when you are writing don t leak when you are not Lipstick spread when applied then not move Crayons same as lipstick Blood corpuscles don t settle Viscous resistance is small in large arteries and veins Chocolate easy to spray on wa1m or apply by dipping then does not run off until it cools Oil well drilling uids low Viscosity when pumped up and down the well high Viscosity when leaking out of the well into porous formations Radiator sealants low Viscosity in pumped coolant high Viscosity in potential leaks Aircraft deicing uids stick to the plane at low speeds ow off at high speeds Plaster spread easily not run once applied while it is setting Margarine spread easily but remain solid when not being spread k2 R2 110 a Forr R V0 J 0 For 7 V inner cylinder CR 2 2 2 2 V9w k ZJi kRJw k ZJRl k a2kR l k kR l k k d V k2 d R2 2R2 m221m 2 3 and dr r l k dr r l k r W k2 Jl ml o39r a 2 2 dr r l k r Here we can ignore the minus sign because as discussed in the text its presence or absence is arbitrary Then at the surface of the inner cylinder r rimer cylinder kR and a 4 inner cylinder 1 k2 kR2 1 k2 b D 2515 mm c Here k 5m m 09106 so that R Domquot 2762 mm 1027 min 2 J 1226 Oinner linder 39 2 Cy m1n 60s 1 09106 s Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 1 page 3 S 7102 6 S 6E2 6 111 Sphere I7 D Cube with edge E E F E Right cylinder D3 6 2 2 D2 7239D V st D 4 This appears strange but is correct If we ask what is the surface area for each of these gures for a volume of 1 cm3 we nd that for the sphere D 6 cm37I13 124 cm for the cube E 1 cm and for the right cylinderD 4 cm37t13 108 cm Thus for equal volumes the sphere has the least surface the right cylinder the next least and the cube the most 112 Crystallization of supersaturated solutions boiling in the absence of a boiling chip a pencil balanced on its eraser all explosives a mixture of hydrogen and oxygen in the absence of a spark a balloon full of air in the absence of a pin a charged capacitor 5280 m1 3 113 VD3 8000mi 3951023ft3 1121021m3 1b m pV55 624 395 1022ft3135 10251bm 61410 kg There is no meaning to the term quotthe weight of the earthquot because weight only has meaning in terms of a welldefined acceleration of gravity For the earth there is no such welldefined acceleration of gravity 2 114 a 623 lbmft3 b 623 lbmlgj k ll6lbf c same as a 13 s 393221bmft 3 393 115 mileszsg j g illl1012gal 30109galL12L14mi3 m1 231 ft ll10 gal The value in cubic miles is surprisingly small We use about 5 E 9 bblyr of which we import about half This suggests that at the current usage rate we have 30055 Z 12 years of oil reserves For the past half century this number has remained about constant we have found it at about the rate we used it In the past few years this has mostly been by improved recovery methods for existing fields rather than finding new fields In the 1950 s the US could produce about twice as much oil as it consumed now we can produce about half This is mostly not due to a decline in production but an increase in consumption Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 1 page 4 23 b W W 62 108electrons 116 coulom 96 500 gm equ1valent 778ft1bf 4184 4184kJ Btu cal kcal a joule See any thermodynamics text written before about 1960 for the use of this J to remind us to convert from ftlbf to Btu 117 J 1 This J is easily confused with the J for 3221bmft lbmtoof 118g 1 2 l 2 toof 322ft or lbfs lbf s 3221bmft 1 lbmft dnoces s g 1be2 lbfdnocesz J322 The toof and the dnoces are the foot and the second spelled backwards No one has seriously proposed this but it makes as much sense as the slug and the poundal 1b 52802 2 119 mmt pV 623 11ft J 271106 lbm ft acre k pV 9982 g3jm100m2 998106 kg m mhectare meter The acreft is the preferred volume measure for irrigation because one knows roughly how many ftyr of water one must put on a field in a given climate to produce a given crop Multiplying that by the acres to be irrigated gives the water demand The Colorado River over which arid Southwestern states have been fighting for a century ows about 20 E6 acre ftyr The Columbia ows about 200 E 6 acreftyr The south western states look to the Columbia with envy the Northwestern states are in no hurry to give it up 39 2 ft 2 lbf 2 BTU BTU 120 c2186000 5280 S 38510 s m1 3221bmft 778ftlbf lbm 2 2 c2 2998108 NS L8991016i s kgm Nm kg 121 300 seconds is not the same as 300 lbfslbm So this terminology is wrong But it is in very common usage Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 1 page 5 lbf 3221b ft ft S 9660 lbm lbfs s 15p 300 As discussed in Ch 7 this is the exhaust velocity if the exhaust pressure matches the atmospheric pressure If they are different the common case then this value must be modi ed However European rocket engineers use the quoteffective exhaust velocityquot which takes the pressure into account the same way US engineers use the speci c impulse 2 122 i lcal Btu 305ch 3600213340 Btu2 A cms 252cal ft hr hrft q 11 Btu m 2 3600s Btu Z 0317 2 A ms 1055J 3281ft hr hrft ft lbm 05 iio JL623 J 2 2 DV 3 ft lbf 123R p 5 CP 75 S 46105 u 1002cp 20910 lbfs 3221bmft 052 Egg2 m 33928 S Pam2 Nm 5 or R 3 2 4610 100210 Pas N kgs cm 1J00p 001gm atmcms2 09910 8cm2 10610 11ft2 124Darcy 1 s mm cmscp101106 gm N 125 F 2039 2 727410 3 01 m 00145 N 00033 lbf 148 gforce m Btu C 18 F lbm 2520al 10 1bm F gmca1 C 454g Btu 126 X The calorie and the Btu were de ned in ways that make this 10 Some are confused because F l8 C32 but if we differentiate 51 F 18 51 C which is the relation used above Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 1 page 6 127 One standard atmosphere 1013 bar Clearly the bar is a convenient approximate atmosphere It is very commonly used in high pressure work as an approximate atmosphere Meteorologists express all pressures in millibar Currently thermodynamic tables like steam tables and chemical thermodynamic tables show pressure in bars 128 The US air pollution regulations are almost all written for concentrations in gm3 and the models expect emission rates in gs The available auto usage data is almost all in vehicle miles per hour or day or second so to get emission rates in gs one multiplies the vehicle miless times the emission factor in gmi The USEPA would be happy to do it all in metric but the mile doesn39t seem to be going away very fast in the US lkgf 981N 104cm2 2 2 981kPa 0968atm cm kgf m 129 This is very close to one atmosphere The most common type of pressure gage testers the deadweight tester balances a weight against the pressure in the gages The direct observation is the values of the weights on the tester and the crosssectional area of the piston on which they rest which has the dimensions of kgfcmZ Ithink this is slowly losing out to the kPa but it is not gone yet 2 a 159kgf 3221bmft kg 221bf60 s 05597 I 130 454 kgm lbf s2 221bm kgf n 2 mi min This leads to the conclusion that kgkgf lbmlbf Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 1 page 7 Solutions Chapter 2 21 As the sketch at the right shows depth r l cos 0 and W 50ft gt 4000 5280ft arcsin 23674 10 23674 10 t9 arcsin05 w r arcsin r l cos0 depth 40005280 1 cos 23674 10 9 59210 5 0018 mm V rcos 0 Discussion The point of this problem is that for engineering purposes for modest sized equipment the world is at Older computers and some hand calculators do not carry enough digits to solve this problem as shown above If we use the approximations 92 9 96 9 m sint9 mtan t9 05w r and cos 9 l 2 4 6 we can drop the higher terms and write 100 2 8 4000 5280 ft On my spreadsheet this approximation agrees with the above solution to 1 part in 50000 depth wz 8r 59210395ft lbm ft lbfs3 lbf 22 623 322 623 a 7 pg 03 s2 3221bmft 03 k kN 7 pg 9982 g3 981 9792 m S m b 6236L116E 9982 21996k N 7N39 322N 39ft3 39 39 m3 k kN k 23 y pg 9982 39813 9792 9982f m S m m I don39t know if civil engineers in metric countries use the speci c weight at all and if they do whether they use the proper SI value 9792 kNm3 or the intuitive 9982 kgfm3 US civil engineering uids textbooks show the SI value 9792 kNm3 24 lb ft lbf 2 ftz I pg623 I1322 2 S 2 0423E ft s 3221bmft 144m 0 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 1 This is close to 05 psift so most ofus remember that dPdh E 05 psift 25 lb ft 6233217 2 z PP h147 sia S 0 pg p 3217 lbmft 144i 2 l47 433 1803 psia 433psig Discussion39 0433 psift E 05 psift in water is a useful number to remember One can also discuss why it hurts The pressure inside our eardrums is Z the same as that in our lungs which is practically atmospheric So a pressure difference of about 5 psi across the eardrums is painful Scuba divers have the same air pressure inside their lungs as that of the outside water The air valve that does that automatically invented by Jaques Costeau made scuba diving possible People who dive deeply wearing a face mask must equalize the pressure in their lungs with that in the face mask or the blood vessels in their eyes Will burst This is a problem for deep free divers lbf 1000 2 26 hA P lbninz 3217lbmft1442m 2244ft684m Pg 103623 3J3217 2J 1be S 27 lbm ft 623 3 3217 21450ft Pl Rpgh15psig 2 15 627 642psig443MPa 3217lbmft l441n lbtsz ft2 This is a higher pressure than eXists in any municipal water system so in any such tall building the water taken from the municipal water system must be pumped to the top Furthermore one cannot tolerate such a high pressure in the drinking fountains on the ground oor they would put out people s eyes Tall buildings have several zones in their internal water system with suitable pressures in each and storage tanks at the top of each zone Ns2 Pam2 kgm N 28 P PM pg 1013kPa 1039982k j98122j 11000 m In S Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 2 Fab 1013105 1109108Pa 11105105kPa 1096 atm16115psia P 1109105kPa 1095 atm 16100 psig gauge lbf 10000 2 29 p mz 1441n 32l7lbmft96Olbm1538g ft ftz lbf 2 ft3 3 g 3217 2 15000ft S m S Discussion When I assign this problem I sketch the ow in an oildrilling rig for the students Highpressure drilling uid is pumped down the drill pipe and ows back up the annulus between the drill pipe and the wellbore This drilling uid cools the drill bit and carries the rock chips up out of the well Ask the students how one would get such dense drilling uids The answer is to use slurries of barite barium sulfate sg 4499 In extreme cases powdered lead sg 1134 has been used The great hazard is that high pressure gas will enter the drilling uid expand lower its density and cause it all to be blown out Most deep drilling rigs have mechanical quotblowout presentersquot which can clamp down on the drill stem and stop the ow in such an emergency Even so one of the most common and deadly drilling accidents is the blowout caused by drilling into an unexpected zone of highpressure gas If the gas is rich in HZS the rig workers are often unable to escape the toxic cloud 210 The pressure at the gasolinewater interface is 2 2 Ppgh072623m3217 20ft lbfs 3 sz 32171bm ft 1441n2 2093 psia 435 kPa gauge 1449 kPa abs 623 psig The increase in pressure from that interface to the bottom of the tank is 2 2 APpgh623lb I3n32l7 220 2 ft s 3217lbm ft l441n 865 psig So the pressure at the bottom is the sum P bottom 1488 psig 2958 psia 10265 kPa gauge 20395 kPa abs Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 3 The gure is sketch roughly to scale at the right 20 Depth ft 40 623 1488 Pressure psig 211 a See example 27 The result is exactly onehalf of that result or 235105 lbf for internal pressure and exactly onefourth of that result of 1175105 lbf for internal vacuum b Almost all pressure vessels can withstand much higher internal pressures than vacuums An internal pressure blows the vessel up like a balloon straightening out any nonuniformities An internal vacuum collapses the tank starting at a nonuniformity and magnifying it Most students have see the demonstration in which a rectangular 5 gallon can is lled with steam and then collapsed by cooling to condense the steam The maximum vacuum there is 147 psig An internal pressure of the same amount will cause the sides and top of the can to bulge out somewhat but the result is far less damaging than the vacuum collapse You might suggest the analogy of pulling and pushing a rope If you pull a rope it straightens out If you push one the kinks are increased and it folds up 212 Here let the depth to the bottom ofthe can be M the height ofthe can be M and the height to which water rises in the can be h3 Then after the water has risen at the interface the pressure of the water and the pressure of the air are equal PW Pa quot J 13sz Pmrzpazm39 2 w a pghl ha hrhj The easiest way to solve the problem is to guess h3 and solve for both pressures on a spreadsheet Then one constructs the ratio of the two pressures and uses the spreadsheet39s root finding engine to find the value of h3 which makes that ratio 100 The result of that procedure using Goal seek on an excel spreadsheet is h3 02236 ft Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 4 and Pinterface 1893 psia Before we had spreadsheets we could have done the same by manual trial and error or done it analytically If we equate the two pressures Pm 413 1 ngi 3 Pm 413 PgUa h Patmhj hl hjh2 h3h25 hlhj hlh3 hj which can be factoredto pg P h ks hz hl j hzhl 0 which is a simple quadratic equation To simplify Pg let Pl Ai A2 4hlh2 h2 hl M A so that h3 2 Now we can insert values pg lbf 147 2 P 144 3221b ft hZ a mA 1ft 10ft 1b 12 1 m 3398ft Pg 623 m322 ft lb ft3 s2 3398ft J 3398ft 2 410ft1ft and h 2 0223ft 268in The liquid pressure is P 147 39 10 0223ft 623lbm 322ft z lbf39sz 1894 39 Sla 513 W p if s2 144ml 3221bmft p while the gas pressure is 1ft 1894ps1a P 147psia l 0223f gas This has an elegance which the spreadsheet does not 213 a See the solution to problem 28 P 16099 psig b dP pgdh p0g1 MIDmph P dP h 1 P PU 1 P PO peg0 dh pgh Elnlp73 Pojpn Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 5 eXp p gh l 1n P11 pogh P P l l From App A9 03 10 5 psi so that 75 eXp p0gh jbm 62332395127 11000 3281 ps1 ft2 lbfs2 2 10494 l441n 3217lbmft and 10494 l P P0 f 16477psig 03 10 ps1 The ratio of the answer taking compressibility into account to that which does not take it 16 477 into account is Ratio 1023 099 Discussion Here we see that even at the deepest point in the oceans taking the compressibility of the water into account changes the pressure by only 23 One may do a simple plausibility check on this mathematics by computing the ratio of the density at half of the depth to the surface density Using the data here the density at a pressure of 8000 psi is about 1024 times the surface density For the depths of ordinary industrial equipment this ratio is almost exactly 100 M 214 PZPIeXpL g Z RT 2 Z 42173229 lbm 10000 2 s lbmol 32171bmft l441n latmeXp lbf 1073iDZ 3 400 R bmol latm eXp 04692 0625 atm dP dz 215 PP o pgz dt png dP 1b ft ft lbf 2 ft2 39 kP 0075 I13217 22000 S 2 104 pSI 718 a dt ft s m1n 3217lbm l441n mm mm Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 6 E P P k T P k 216 kconst k const P1k Z 2 p E T1 Pl RT dP M M P dPpgdzPMgd g 1 k dz RT P RT RT P 571 gM Ll P L k gM P k dP Plkdz 2 Az RT1 Pl k RTI k k lgAAz H P P1 Eq 217 2 1 El k IAM Wig 1 k RTI 59 697 F 000356 R 217 a T519 R z519 R zabz 36150ft ft dP gM gM dz P gM 2 g T d 1 1 P RT Z R abz nP0 bR 1naszo bR nT0 P T 57 Here Po To gM 321791b11n S 11106 M 10 73M 5272 0 2 0000356 mZ 32 lbm 144m lbmol R lbfsz ftz b At the interface 3900R 0572 PP0 m 130752 0222P0 c In the stratosphere T constant so we use the isothermal formula M 2 2 RT stratosphere k 218 Using Eq 218 we have T2 0 when k 1 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 7 lbf 1073 2ft3 Lu 0 AZRT1 k lbmoloR 519R 144m2 32171bmft 96783 2 2 gM k l 32172229 lbm ft 1be s lbmol 04 This is the height which corresponds to turning all internal energy and injection work Pv into gravitational potential energy adiabatically The fact that the atmosphere extends above this height shows that there is heat transfer in the atmosphere which contradicts the adiabatic assumption At this elevation the predicted pressure from equation 217 is P 0 219 Taking values directly from the table in example 24 gMAz 4300ft a P Pex 1atm 0036l6 4300 1 J 1 J 0856 atm b is given in Table 21 0697 atm 29 028 ft 0350 atm 1000 ft MAZ c Pzg 02m P1 PL gij 1atm 003616 1 These are based on To 519 R the standard atmosphere If one uses 528 R the answers are 0858 0701 and 0356 atm Discussion you might ask the students about the barometric pressures which are regularly reported on the radio for places not at sea level These are normally quotcorrected to sea levelquot so that for example in Salt Lake City the reported barometric pressures are normally about 299 inches of mercury which is about 5 inches higher than the actual barometric pressure ever gets Those values are used to set altimeters in airplanes which have a dialin for the sea level barometric pressure which they compare to the observed pressure to compute the altitude 220 a dT E 000356F dz AZ 36150 ft ft Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 8 ft lbm dT k l gM 04 32172291me 1be2 ft2 R lbf Z 000536 dz R 14 3 3217lbmft 144m 2 k 1073413 lb mol R Discussion These are called lapse rates and the minus sign is normally dropped so that the standard lapse rate is 356OF 1000 ft and the adiabatic lapse rate is 536 0F 1000 ft These terms are widely used in meteorology and in air pollution modeling 221 a From Table 21 0697 atm b T2Tli EEampJJT1 0361651908967 k RT 14 4654 R57 R k T 71 a P2 E 147psia 089640 4 1004 psia 0683 atm 1 c See the solution to Problem 217 R T 59 000356J104ft 234 F 4831 R T 757 4831 R 72 PPO 1 atm 390 J latm 0685 1007 psia T0 519 R 222 Here the pressure at sea level is due to the weight of the atmosphere above it from which we can easily conclude that there are 147 lbm of atmosphere above each square inch of the earth39s surface Then AE47rDz 1471bi147r4000mi 118610191bm 1n m 528011 12in2 m A m1 llbf 14439 2 2 23 FPAi4r120 2 2 ll 1631061bf724MN 1n This is normally enough to crush such atank which explains why such vents are of crucial importance and why prudent oil companies pay high prices to get one that always works Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 9 2 2 2 224 P I8OOkg 9821ms Ns Pam A 02m kgm 8892 kPa 1281 psig Discussion This small value shows how all hydraulic systems work A small pressure acting over a modest area can produce an impressive force The small air compressor in the service station easily li s the heavy car fairly rapidly You can ask the students whether this is psig or psia You can also ask them about buoyancy Is the piston a oating body Answer not in the common sense of that term If you look into Archimedes principle for oating bodies there is an unstated assumption that the top of the oating body is exposed to the same atmospheric pressure as the uid is Here for a confined uid that is clearly not the case If an immersed body is in a con ned uid that assumption plays no role and the buoyant force computed by Archimedes39 principle is independent of the external pressure applied to the uid k N 2 P 225 a P pgh 9982 g3981m2230 m S a 2 225 MPa 326 ps1 m s kgm Nm b See Ex 28 kg m 2 2 9982 981 76m230 m 2 F ngh 13 52 NS 197101 N4421091bf 2 2 kgm 2 26 a The width W is linearly related to the depth 20 m at the top zero at h 10 In so that W 20 le iJ 10 m b Here as in Ex 28 the atmosphericpressure terms cancel so we may save effort by working the problem in gauge pressure Direct substitution of this value for Wleads to 2 3 10m Fpg20mjhl jdhpg20mh h 0 m 2 310m 0m k 10 2 10 3 N2 cF 9982 9813220 m m m S326 MN 221061bf m s 2 310m kgm 0 This is exactly 13 ofthe answer to Ex 28 d We begin by writing Eq 2M for the gauge pressure of a constantdensity uid hdA F ng hdA Multiplying by AA and rearranging we find F pgAT Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 10 But lh dAA is the de nition of he the centroid of the depth measured from the free surface so this equation may be simpli ed to F pgAh Wh h e F pgAh A little algebra shows that this is identical to the formula in part b 7 3 227 aF 2ng deptw r2 depth2ddepth 2Jrz depch 0 29982 m 9815 50m N S Pa m2 816 108N 1831081bf 3 kgm N 17239 2 2D U 2 3 F 1 D b 3814 c P 2 4 3 pg 12 3pgr the same as in part a If this doesn39t convince the students of the convenience of the centroid method one wonders what would 228 See Example 29 There at 60 ft the required thickness was 00825 in The required thickness for the assumptions in the problem is proportional to the gauge pressure at the bottom of each band of plates which is proportional to the depth a The values are Depth ft Pressure psig Design t in 60 229 0825 50 190833333 06875 40 152666667 055 30 1145 04125 20 763333333 0275 10 381666667 01375 b The calculated thickness for the top band is less than 025 inches so a plate with 025 inches thickness would be used Some designs call for 516 inch It is not asked for in the problem but the same industry source who gave me these values also told me that the maximum used is 175 inch because if you make it thicker than that you must heat treat the welds which is prohibitively expensive for a large tank c The rst two columns of the following table repeat the rst and third columns of the preceding table However in the middle column the lowest value 01375 has been replaced by 025 because the top plate has a minimum thickness of 025 inches The rightmost column shows the average thickness of the tapered plate except for the lowest entry which is not tapered So for example the lowest plate is 0825 inches at the bottom and 06875 inches at the top for an average thickness of 075625 inches Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 11 Depth ft Wall thickness inches Average thickness inches uniform thickness plates tapered plates 60 0825 075625 50 06875 061875 40 05 5 048 125 30 04125 034375 20 0275 02625 10 025 0 25 We can then compute the average thickness of the walls of the whole tank by summing the six values in the second and third columns and dividing by 6 nding 0500 inch and 0452 inches Thus the tapered plates reduce the weight of the plates for equal safety factor to 0904 times the weight of the constantthickness plates The same industry source who gave me the other values in the problem also told me that the tapered plate solution is quite rare It is apparently only done when there is an order for a substantial number of very large tanks all of one size 229 As a general proposition the larger the tank the smaller the cost per unit volume Surprisingly the volume of metal in the wall of the tank is Z proportional to the cube of the tank diameter for all three types of tank because the surface area is proportional to the diameter squared and the required thickness Z proportional to the diameter So it is not simply the cost of the metal but the cost per unit stored of erection foundations piping etc that decrease as the size increases I thank Mr Terry Gallagher of CBI who spent a few minutes on the phone and shared his insights on this problem He said that the biggest at bottomed tank he knew of was in the Persian Gulf 412 ft Z 125 m in diameter and 68 ft Z 20 m high Compare this to Example 29 Here the depth increases by 6860 and the diameter by 412 120 so the calculated wall thickness would be I 0825in 321 in 815 cm 60 120 which is thick enough to be a problem to handle and would require heat treating of all the welds He said that they stressrelieved this tank in the field after it was complete by building an insulating tent around it and heating to a stress relieving temperature That costs but the economics of having this big a tank apparently made it worthwhile For spherical tanks he and Ithink that the difficulties of erection become too much as the size goes over about 80 ft I have seen an advertisement for Hyundai showing one of their liquid natural gas tankers which had some huge spherical tanks in it LNG is transported at a low enough temperature that one must use expensive alloy steels so the cost of those spherical tanks must have been justified Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 12 For sausageshaped tanks the limits are presumably the difficulty of shipping where roads and railroads have maximum height width and length speci cations The diameter may be set by the largest diameter hemispherical heads which steel mills will regularly produce The heads are onepiece hot pressed Ithink Mr Gallagher who refers to this tank shape as quotblimpsquot indicates that there are some fielderected tanks of this type which are larger than the values I show Most of the tanks of this type are factory assembled with a significant cost saving over field erection and then shipped to the user Two interesting web sites to visit on this topic are wwwchicagobridgecom and wwwtrinitylpgcom 230 See Example 29 1000331 f t M 005ft 060 in 152 cm 210000 2 in From the larger table in Perry39s equivalent to Appendix A3 we see that a 12 inch diameter schedule 80 pipe has a wall thickness of 0687 inches It would probably be selected 231 The plot below shows all the values from App A2 Clearly one should not fit a single equation to that set of data If one considers only the values for 25 inch nominal diameter and larger then 1 01366 in 00240 Dinside R2 0994 Thus for these pipes A Z 01366 inch andB 0240 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 13 C Ox 0 0 u 0 4 lll O 56 o L O O O 0 N Pipe wall thickness inches 0 O 0 quot0 N O 5 10 15 Pipe inside diameter inches However if one ts the sizes from 18 to 2 inches nominal one nds t 00719 in 00454Dinside R2 0897 One can get R2 0976 by tting a quadratic to this data set showing that the values for the smaller pipes do not correspond very well to this equation If as shown above A Z 01366 is the corrosion allowance then the Eq 225 can be rewritten t 01336 in 0024D 123 1 from which P 039 ll owable 0048 Comparing this Oallowable will the de nition of schedule number in App A2 we see that this corresponds to Sch 48 not Sch 40 The difference is conservatism on the part of those de ning the schedules 232 Hooke s law says that the stress is proportional to the strain which is the change in length divided by the original length If as stated the increase in diameter is the same for all diameters while the original diameter is greater the further one moves from the center then increase inJ Young 5 diamter AD stress k or1g1nal D diameter modulus Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 14 and the stress decreases as lD from the inner to the outer diameter For D D 15 the stress at the outer wall would be ll5 23 ofthe stress at the outer inner inner wall 233 a Example 210 shows a wall thickness of 07500 inch The thinwalled formula shows x I P39r C 259PSI395 006297ft07556in SEJ 06P 20000 ps1l 06 250 ps1 b The thickwalled shows 12 12 SE P 20 000l250 15 rlCC5ftm 5ft0006290ft07547in J a c The plot is shown below We see that below about 2000 psi the three equations give practically the same result Above that Eq 225 gives values which increase linearly with pressure while the thinwalled and thickwalled equations give practically the same value as each other but larger values than Eq 225 50 39 i 40 Gm A g 393 30 3 3 2 20 39 5 7g 5 I I B 10 0 0 2000 4000 6000 8000 l 104 Internal pressure P psig P 50000 390ll39 234 a I r C p51 1 011m SEJ 06P 80 000 ps1 06 50 000 ps1 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 15 12 12 t SE 13 r C 011 inw 011in 01189in SE P I 80 000 50 000 b I 4 50 000 pSl0llln 220 in 3920 000 psi 06 50 000 psi 12 s rtofa 20000150000 0I11inq I J 20 000 50 000 negatlve Clearly this does not work For Eq 227 safety PD 4 80 000 psi022 in 0 88 in 403mg 4 20 000 psi 39 t40llin 2 factor Which is much larger than the values shown in part c safety I 05 074 022 39 c thickendofbarrel 236 and faCtor lfthin walled formula 1 1 m safetyj tmick end ofbarrel 05 074 022 in Ithick walled formula 01189 in The simple formulae used for pressure vessels give some indication about the design of guns but this example shows it is only and indication factor 235 The pressure at the bottom ofthe tank is 277 psig From Eq 225 tw 00922 ft 11076 in 2 30 000 psi This differs by 14 from the value calculated by the API method 236 a For the cylindrical container remembering that the length of the cylindrical section is 6 times its diameter V D3LEJD26Dn 6 4 6 13 6 3 13 D V 20000 gal 799ft 107239 107239 748 gal PD 250 psi 799 ft 2 20000 psi 00499 ft 0599 in tcylindrical 2 tensile The hemispherical ends have exactly onehalf this thickness Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 16 volume of shell shell 2 m S 39 tspherical 71D tcylindxical i 39 6D metal shell thicknes surface 2 EDZ 39tsphencal 6tcylindn39cal 65771 tcylindn39cal 657r799 ft2 00499 ft 6513 ft3 lb mmll pV 79623 r 6513ft3 320501bm b For a spherical container 3 6 13 6 V D3 D Vj 20000gal 6 7239 7239 13 J 1722 ft 748 gal PD 250 psi 1722 ft 00538 ft 06456 in 4039 420000 ps1 t tensile volume of shell shell DZ metal shell N thickness surface t 000538ft 7239 1722ft2 501 ft3 lb mmll pV 19623 501 ft3 246501bm The spherical container requires only 77 as much metal exclusive of corrosion allowances foundations etc However the sausageshaped container is small enough to be shop assembled and shipped by rail or truck while the spherical container could be shipped by barge but at 17 ft diameter it could probably not be shipped by truck or rail 237 For this size pipe the wall thickness is 0258 in From Eq 225 2t039 20258in 10 000 psi P 1022ps1 D 50471n But as shown in Prob 231 the sizes of Sch 40 pipe seem to include a corrosion allowance of 01366 inches Reworking the problem taking that into account we find 2m 2025801366in10 000 psi P 481ps1 D 50471n This suggests that a new 5 inch Sch 40 pipe would have a safe working pressure of 1022 psi but that at the end of its corrosion life it would have one of 481 psi The rule of thumb cited in App A2 suggests a safe working pressure of 400 psi Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 17 238 Comparing Eq 233 and 235 we see that the axial stress in a cylindrical pressure vessel is exactly half the hoop stress for the simple assumptions in those equations Groovedjoint connectors Perry s 7e Fig 10139 use this fact They cut a circumferential groove halfway through the pipe to be joined The external clamp which ts into the groove supports the pipe against external expansion so the doubling of the hoop stress is not a hazard The clamp only operates on half the thickness of the wall but that is enough to resist the axial stress which is only half the hoop stress 239 The buoyant force equals the weight of the water displaced BF 50 4725 0275N F 0275 N k Thus the mass ofwater displaced was m 28032 10 2 kg 0 98132 N S So the density of the crown was F m 5N981 2 k m k i S g 18149 g 18149i3 v v 28083210 5m3 W F cm and its gold content was 10 Id 18149105 V0 ogo pallo psllver 0869 869 100 pgold psilver 193 3910395 The corresponding wt mass gold is 924 If one computes the buoyant force of the air on the crown in the weighting in air one nds 000033 N If one then substitutes 500033 N for the 5 N in the problem one nds 867 vol gold Discussion This assumes no volume change on mixing of gold and silver That is a very good but not perfect assumption 240 If you assign this problem make clear that one should not assume that air is a constantdensity uid but must insert the perfect gas expression for its density Then Payload BF weight Vgp2m pg Vg Mm Mgas Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 18 nRT But V so this becomes Payload rzgMajr Mgas Thus the answers to parts a b and c are the same Viz ft 322 10 lb 29 lb 41b Payload 1b m lbsz fl m m 6251bf 278 N I 2 lbmole lbfs 241 Starting with the solution to EX 211 29 Z j Payload 34 lbf mole mole 367lbf 29 4E mole mole This is an increase of 76 which is signi cant but probably not enough to justify the increased hazard of hydrogen combustion in most cases As far as I know no one uses hydrogen in balloons today Payload Vg 242 Payload Vgpmr pgas pgas pm Here the students are confronted with the fact that the weight in kg is not an SI unit One must make the distinction between kgf and kgm to solve here nding p 120 kgm ZOOkgf 981ka 120 004771152g gas 3 3 m k f 2 3 m 20m 981 2 g S m 6 s 120 Tgas TnL J 29315K 3053K 3214 C 899 F pg 1152 243 The volumes of the lead and the brass weight are 2501b 25 V1d nllb 3551210 3ft3 mess 4721 10 3ft3 623113 m 62385 3 and the difference in their volumes is AV 472110 3 355110 3 117010 3ft3 a The differential buoyant force is Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 19 ft lbm 322 2 3 3 72 ABFpAV 623 3 lbm 117010 ft 72910 lbf ft 322 2 lbf s so the indicated weight is Indicated weight 2500 729 102 2573lbf 32 b here the differential buoyant force is ABF 0075 51170103 877103951bf so the indicated weight is Indicated weight 2500 87710395 249991 lbf Buoyant PB 2 PT ghgasoline pgas hther pw pang1m Agh p hamtad hm h pwood hwood hwood hwater gasoline hwater pwater hwatei pwood pgasoline SGwoool SGgasoljne h l SG 244 F T FEW T FT0p l APB PT pwoodgA hm wood pwater pgasoline gasoline Discussion this shows that Archimedes principle does indeed give the right answer for two uid problems and why the density of the gasoline does appear in the answer 245 Gravity Force i Buoyant Force T Treading Force 085 GraV1ty Force mmm g Buoyant Force Vii laced pg mmm pwhisk 5quot 099pm y FT F Fb 2mmquot g mm 085 pwhiskez g aquot 099 pm ft 092 325 2 1501bm 1 085 3151bf 1402 N 099 32ft lbm s7 lbf 246 Payload pw p10gs V payload m 500kgf k 981kg2m 250451113 gPw Hoes 98171 089982 g3 kng S m k Massoflogs Vp 25045 In3 08 9982 g3 2000 kg m Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 20 247 a Buoyant Force weight g mGs Vs mship g m i 1 1 Kanks Sh Khip mship water pwater psteel 8 1071b 1 1 T ln 1084106ft3 6233 103 079 ft b The crosssectional area of the cable would be M 4000 in2 and its diameter stress 20000 lbfin2 A 4 4 2 D 4000 m 7141n m 6 ft wh1ch 1s not a quotcablequot in the ord1nary 7239 7239 sense Whatever was pulling on it would have to have a buoyant force large enough to support its own weight and also the weight of the battleship which it was raising c For a steel cable 1000 ft long hung by its end in water the stress at the top of the cable due to the cable39s weight is F N SG 1Pwae1LAg 039 Z A Z 79 1 623113n 1000 ft mgk 4310 3000 psi ft s 32 lbmft ft This is 15 of the allowable stress For deeper sunken vessels the fraction is higher To the best of my knowledge the fist time this was ever tried was when the CIA attempted to raise a sunken Russian submarine for intelligence purposes during the cold war They lowered a clamping device on multiple cables and tried to pull it up It broke up while they were doing it so they only got part of it but that was said to have revealed lots of useful technical information In 2001 the Russians successfully raised the submarine Kursk this way with multiple cables 3 lbm 322 ft s2 248 FB Vg 5 20 30ft 623 3 322 lbm lbfsz 1871051bf 831105 N This is a serious problem in areas with high water tables empty swimming pools are regularly ruined by being popped out of the ground by buoyant forces 249 BF weight 200 lbf Vgpzm pheuum Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 21 BF 200 lbf 32 lbfft V 3093103 3 876 m3 gpair phelium 32 00751bm1ij 1bez s2 ft3 29 V 3093 ft3 mm 1 7365 ft3 D3 42 42 6 13 D 7365ft3 52ft158m 7239 250 In the boat the part of the boat39s displacement due to the block is 1001b Vl m3 1605 ft3 pwam 623 lbm ft In the water the displacement due to the block is V m 1001bm 3 0203 ft3 psteel 79623 lbm ft So the volume ofthe pond decreases by AV 0203 1605 1402 ft3 and its and elevation falls by 3 dz Ade 140ft AP p 4 p10 f This is a counterintuitive result which has made this type of problem a favorite puzzle for many years 00179ft 0214 in 251 a P 13 pgh1 h 193 6231b m i ZL 44 839 ft z 362 I 2496kP 3 321bm 1n12in144in2 ps1g7 agauge lbfs2 b P2 361psig147psia183psia 1263 kPa abs 252 P2 P1 pwgz1 22 P3 P2 pHggZZ 23 P4 P3 pwgz3 Z4 Pz P4 adding these equations canceling and grouping produces 0 pwgZl 22 Z3 24 pHggZZ Z3 but Zl Z4 A2 and z2 23 Ah so that 0 pngz pHg pw gAh and AhAZL pHg Pw Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 22 Adding these and canceling like terms Pb Pa ngAh p1ghl hi Pb Pa gAhp1 p2 For maximum sensitivity one chooses p1 p2 as small as possible However as this quantity becomes small the two uids tend to mix and may emulsify So there is a practical limit on how small it can be made 254 Do part b rst AP gAZ AP orifmlbmf 144 2 Az m1leffS t0231 277in70mm Pg 623 332 2 ft s Thendoparta A A2 27739 Zsint9AL Ho11072in272mm AL s1nt9 s1n15 Discussion this shows the magni cation of the reading obtained with a quotdraft tubequot and hence why these are widely used for low pressure differences 255 AIALI AZALZ 1 Z 2 in A D AL2 1AL1 1 L1 8 10in0039in A2 D2 21n Which is small enough that it is regularly ignored 256 lbf 14699 2 39 2 1b ft 144 ah lbrlrfll ft32 1112 11 2493 2929in760mm Pg 136624322 1be t s This is the set of values for the standard atmosphere which is de ned exactly as 101325 kPa That corresponds to 1469595 psia To get values in terms of heights of mercury one must use the density of mercury at some temperature The values above which lead to the values shown inside the front cover of the book are for mercury at Z 0 C At 20 C the density of mercury is 1352 gcm3 compared to 136 at 0 OC Laboratory barometers Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 23 are supplied with conversion tables allowing one to correct for the thermal expansion of the mercury P 39 2 1b ft 14439 2 1 i23391 407in1034m Pg 624 32 2 1be ft ft s which is impractically tall c The vapor pressure of water at 20 C 68 F is 03391 psia This is 0023 atm 23 of an atmosphere Thus the above calculation would replace the atmospheric pressure with 14696 03391 reducing all the computed values by 23 A water barometer would need a temperature correction not only for the thermal expansion of the liquid but also for the vapor pressure of the liquid which is about 0006 atm at 0 C 257 The pressure at ground level for a nonmoving atmosphere must be 2 2 PM PM 320 2 120 pgdz 20 Egdz avggZtop Here the average is over the elevation from the ground to ztop which is the elevation below which all the mass of the atmosphere is We now differentiate dP PM 1J AP AP 47 g2 2 dT R PT gzw P p RT W AP Ang avg 1320 The troposphere contains Z 77 of the mass of the whole atmosphere and has an average temperature See Fig 24 of535 F Z 455 R Z 20 C P P 1025 0995 t Here M m 0003 so that 20 latm AT T ii 455 R0003 l4 R avg avg Z 0 An average decrease in average temperature through the troposphere of l4 F would account for the difference in pressure between a high and a low Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 24 Changing moisture content changes the average molecular weight Following exactly the same logic we could say that if the temperatures were the same between a high and a low the difference in molecular weight would have to be AMv Mav E 29i0003 0087i g g PZ0 mol mol At the average temperature of 4550 R the vapor pressure of water is slightly less than 1 torr Z 00013 atm If the atmosphere were saturated with water at 20 C its change in average molecular weight would be i i AM1V 0001329 18 002 g mol mol which is about 14 of the required amount If one runs this calculation at 20 C where the vapor pressure of water is 0023 atm one nds a change in M avg of 026 but most of the atmosphere is much colder than it is at the surface Thus it appears that the principal cause of atmospheric highs and lows is the difference in average temperature between air masses 258 The pressure at the bottom of the dip tube is ft 32 2 1b 2 ft pgh 2psig00813nj S 6rt 2 20033psig PM Pmquot 321bm 39 144m lbfsz lbmf t P 2 0033psig 321be 2 144 inz a h T Z 4808ft 1465 m Pg 60 3 32 2 s That is the depth at the end of the dip tube The total depth is 4808 05 5308 ft b The difference is 0008 ft or 015 of the total If we use 2000 psig the calculated depth is 4800 ft Discussion One can discuss here the other ways to continually measure the level in a tank One can use oat gages pressure difference gages etc If the gas ow can be tolerated this is simpler 259 If we ignore the gas density then AP gm uid 0ghmammeter h 15 k k 1b W m 9982 14973 g 9345 31 h id 10 m m m ft Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 25 If we account for the gas density then AP gh uidp uid pgas ghmanometerpman0meter pgas kg hm hm p id Rik J h 1 pg 14973 15 1120 14967 f f which is a difference of 004 Discussion one can determine densities with hydrometers or with some sonic gages This system is simple and continuous P M M 250 See EX 216 PA PB gZI Kj j J uegas R T a T ft lb lb 32 29 28 s2 100 X147PS31a lbmol lbmol z 39 0 0 2 32lbmft 1073 ps1aft 530 R 760 R l441n lbfs lbmol R 00170 psi 047 in HZO 0ll7 kPa Discussion This introduces the whole idea of the use of stacks for draft Currently we use stacks to disperse pollutants more than we use them for draft but historically the height of the stack was determined by the required draft to overcome the frictional resistance to gas ow in the furnace 23961 Ptop Pbot poilgh Pbot pwatergh ft 32 2 Sz 4 lbm ft Pmp ghpwm poil 321bm 3910 ft103 623 55 3 144in2 lbf s2 637 psig 439 MPa gMAz 262 P2 P1 eXpLVJ 32 216i0 104ft gMAZ s lbmol R 01954 RT lbf 3 lbm ft 2 32 1073 13 lbf s2 lbmol R P2 10147psianp0l954 1234 psia 1219 psig 8400 kPa gauge Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 26 b For constant density MA P 131 gRT Z10147ps1al19541213ps1a1198ps1g 8257 kPa gauge Error w 17 1234 263 This is part of oil industry folklore the pipeline was from the Central Valley of California to the Coast and crossed the Coast Range of mountains When the oil was slowly introduced it owed over the water and bypassed water leaving some in place This led to situations like that sketched in which the uphill AP leg of a rise was lled with water while the downhill leg on the other side was lled with oil oil in AZ water in pipe pipe For any one such leg the pressure difference is AP hAp 32m 2 200 ft 1 0 8962 3 lbm 17 3 39 51 g 3221bmft1bfs2 ft3 p g and for 10 such legs it is 173 psig 264 PA P8 gAz ft lbm 32 2 2 147ps1a 623FW40ft144 Z 147 173 26 ps1a 32 in 1be2 This is an impossible pressure the water would boil and the pressure would be the vapor pressure of water which at 70 F is 037 psia This low a pressure will cause a normal pressure vessel to collapse In oil industry folklore it is reported that this has happened several times Normally a newly installed pressure vessel is hydrostatically tested and then the workmen told to quotdrain the tankquot They open valve B and soon the tank crumples See Noel de Nevers quotVacuum Collapse of Vented Tanksquot Process Safety Progress 15 No 2 74 79 1996 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 27 The point of this problem is to show the students how one calibrates springtype pressure gauges Common cheap ones are not locally calibrated but simply used with the factory design values In the factory they are not individually calibrated because each one coming down the production line is practically the same at the next For precise work pressure gauges are calibrated this way 266 No This is a manometer and if bubbles are present they make the average density on one side of the manometer different from that on the other leading to false readings If all the bubbles are excluded then this is an excellent way to get two elevations equal to each other d2 ft ft 267 PM mg d fJ h322 2 322 0 Z S S Z 268P phg d fj dz lbf 5 d2 P 1b ft 144 2 ft ft f g322i2 322 2465 3221437436 dz ph 623 118 lbfs ft p s s ft The elevator is moving upwards The horizontal part of the device plays no role 269 At spill the elevation difference h from one end of the tank to the other is twice the original freeboard or 2 ft dzx h 1 Pg P dtz 2 d hg2 322g332 t dt 1 20ft s s Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 28 270 As shown in the sketch at the right the interface is a plane passing through B and D We compute the pressure at C two ways A 2 dzx dzx PC ng1 12 p2 F PA PA 12p1 W ngl1 The pressure is the same e1ther 2x 2x way we calculate it so ng1 12 12 ngZ1 or dzx dzx 2 gllp1 p2 ZZFQI p2 and A dj tang The d1v1s1on by pl pl 2 is only safe if this term is not zero Physically if the two uids have exactly the same density then they will not form a clean and simple interface Most likely they will emulsify The numerical answer is 12 Z 2 t9 arctand t arctan S arctan 003106 rad 003105 rad 1779 g 322 2 S 15m m2 271 Page pmzjlm rdr p K151n2 l41n2 g 2 lbm 623 2 2 2 2 3 1000 lbf ft 27r 225 196in2 S 2 2 min 60s 3217lbmft l441n 1485psig 1024 kPa gage 272 A rigorous solution would involve an integral of P dA over the whole surface Here P is not linearly proportional to radius so this is complicated If we assume that over the size of this particle we can take an average radius and calculate the quotequivalent gravityquot for that location then we can use Archimedes39 principle as follows Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 29 lbm 3 0153lbf 0683 N 27r1000 min 60s 1be2 1n 32l7lbmft Z 144 in2 a BF przr 623 min 001in3 b This points toward the aXis of rotation Ins l 273 See the sketch at the right Work in gage pressure P10 Pz Plpghh 2 PPp 5 4 B Pspgh4 hj Adding these and canceling like terms B pigoh hm r32 ml 2 10 10472l 60s s lbm ft 1 10477 2 ZJ 1be2 ft2 P 623 322 4ft 22ft 24ft 4 f i s2 2 s 3221bmft 144m2 623 1288 5044 1053 si 726kPa au e 32 2 144 p g g g 274 The gasolineair interface is a parabola as shown in example 219 Using the sketch at the right we compute the pressure at C two ways nding Gasoline By path DEC 2 a PC PD pgasolinegAZ pwater Ar 2 By path DAC 2 a PC PD pgasoline pwatergAZ Subtracting the second from the rst 0 pwater pgasoline2mz pwater pgasoline kAZ Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 30 I 2 2g Which is the same as Eq 238 for air and water Two parabolas are the same or A2 2 75 Yes it will work If one solves it as a manometer from the top of the funnel to the nozzle of the jet one will see that there is a pressure difference equal to the difference in height of the liquid in the two bottles times g times density of water density of air This is clearly not a steadystate deVice as it ows the levels become equal and then the ow stops Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 2 page 31 Solutions Chapter 3 31 Migrant agricultural workers airline pilots who have homes or apartments at both ends of their regular route military personnel on station truck drivers passing through the state rock artists in the state for a concert traveling salespersons construction workers visiting drug dealers 32 Boundaries USA including money in bank vaults in people s mattresses etc Accumulation number printed number destroyed either intentionally by treasury department which destroys old bills or unintentionally for example in a re in a supermarket ow in across borders mostly with tourists some by interbank shipments ow out across borders Currently not too many 1 bills circulate outside the USA but the US 100 bill is a common medium of exchange outside the USA including both real and counterfeit bills 33 Accumulation amount refined in sugar plants amount consumed both by people who put it in the coffee or cereal and by food manufacturers who add it to processed foods ow in across state borders normally in trucks or railroad cars ow out across state borders either in bulk carriers or packaged for sale in stores 34 Boundaries the skin of the balloon including the plane of the opening Accumulation ow out The boundaries are not xed in space nor are they fixed in size but they are readily identifiable 35 Boundaries The auto including all parts with boundaries crossing the openings in the grille and the exhaust pipe Accumulation 0 0 ow in carbon dioxide and other carbonbearing gases in in owing air carbon in bugs smashing into the windshield or in road oil thrown up on body bird droppings falling on car ow out carbon dioxide and carbon monoxide in exhaust gas fuel leakage oil leakage antifreeze leakage tire wear paint aking off carbon dioxide and other carbon compounds in driver and passenger s breaths leaking out through windows carbon in paper and trash thrown out the windows by litterers The point is that there is one main ow out the carbon dioxide in the exhaust which equals the negative accumulation of carbon in the fuel and then a lot of other minor terms Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 1 36 Here de ning the boundaries is the big problem If we choose as boundaries those atoms originally in the recracker then accumulation 0 If we choose the topologically smallest envelope that includes all the solid parts then accumulation ow out of gaseous products which may move faster than solid products ow in of air engulfed in the expanding envelope 60 1 1 1 W 3 7 QVA1 10ft50ft S fa 2244105amp 500 14162 s m1n ft mm s s A ft 500ft2 ft V2 Vl 11 20476 0145m A2 s 7150ft s s 3 W 38 Q107 435104 43510 yr acreft yr 3 43510 V2 2gtl 217108 69 47 A 2000ft yr s r 3 45000 39 a V2 SZ1119 A 87r48ft s 3 45 000 b V2 S 225 1534E 2102 A 200 ft8ft s hr s The purpose of the arti cial ood was to mimic the spring oods which regularly passed down the canyon before the construction of the Glen Canyon Dam 1963 That was supposed to reshape the beaches and in other ways make the canyon more like its pristine state Those who organized the ood pronounced it a great success QLO V27rrdr 310 Vavme 2 A 711 a Substituting Eq 321 and simplifying Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 2 2 2 77wa11 I r w Q Ito XL r2 szdr 2V 22 4 w V V W m w r max 2 4 avenge A 727mll rmll 2 4 0 2 Vm 2 b Substituting Eq 322 and simplifying 17 Q 1 mp 4 ZW r 17 Vavm 25 ml lLJ W S A Wall rwall W0 rw 1 1 w 2 1 2V r r 2 49 2m will vi IV 1 Vmax Vmax 08166Vm rwall 2 1 2 l 0 7 7 r0 KW e lr39nax 2 21L 1224Vwe 6 08166 mg r F me Q I 0 max wr J 2727dr 110 7 w max rrmu r c 2 2 Lo l W A Wwan rwall VW 21 ll raw r 10 r 10 2V 1 1 2 200 1m yin rwl rvlv 1 1 Km Knax Knax wall 2 l 2 1 231 10 10 10 10 r0 V Vmax uswm 6 08568 mg 311 a Substituting Eq 321 in Eq 323 and simplifying mm r2 r2 3 average kinetic J 0 Km W 2 rdr r V3 wa11 W 6 4 2 2 4 6 energyaper 2 quotw J 0 rW 3rwr 3rwr r rdr rmnl vemge IQngme 7 unit mass V3 8 l 3 3 l W l 8 39rwa11 39 V r 2 4 6 8 V 8 avg wall avg Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 3 But for laminar ow Km 21Vg so that average kinetic g 2 av 1 Kw energy per g V2 2 E unit mass laminar ow Substituting Eq 322 and simplifying 17 3 rrvall r average kinetic J 7 Km 1 rdr 37 r70 rw V3 J wan r J max 1 rdr ener er gy p rvtall Vaverage Kvg rvjall KO rw un1t mass 2 1 raw r r V3 1 7 1 7 1 V3 49 V3 V3 i 39Vian g 3 3 m quotquot X02882 m Vavgrwan 2 1 2 1 Vavg 170 Vavg Vavg 7 7 7 r0 But from the preceding problem Km 1224Vavmge So that average kinetic 2 V3 122 V energy per 02882amp X 02882g g 05285ijg 1057 25 unit mass jgbf gvggf39 avg avg approximation b The total momentum ow in a pipe is given by total momentum rmquot r ow 2 lwhole ow area Vdm 2 Lo VPV ZWdr 3quot Substituting Eq 321 we nd 2 total momentum raw 2 2 V2 2 Wm JI Vmaxrw r p2nrdrJO r2zrdr OW 70 rw rw anaxfnp rvirz 2r rquot 16rrw S z rvjpl r 2 4 6 be 6 w Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 4 But for laminar ow Km 2V so that avg 9 avg total momentumJ 2 2 1 2 1 2 2 ow sznrwp 21452me 1333 717pr laminar ow Substituting Eq 322 we nd r 271 r 1 2 1 1 total momentum rawquot r m 2 2 Vw 12 j Vm 1 pzmdr sznprmn ow r0 l 49 anax Z prvtall 2 2 Vnzmx2 prw2anm 0393403Vrzax2 prwzall 1 7 7 But from the preceding problem Vm 1224Vavemge 2 avg Z prvtall 10137pmvla11laiemge total momentum Jturbulent ow 034036224V ow 17 power approximatin These are the values in Tab 31 7239 2 PM MT 1 312 rXQp D V V2 4 RT 7r4D PM The velocity increases as the pressure falls This has the paradoxical consequence that friction which causes the pressure to fall causes the velocity to increase This type of ow is discussed in Ch 8 313 Assuming that the soldiers do not change their column spacing 4mi 12 mi Qp1V1A1P2V2A2 V2 Vlf39 48 d r 314 770Zrgtfng Q3Q1 Qz 7239 2 ft 721 2 ft 7r 7 ft3 1ft 5 7 5 255 Q3 4 s 4 2 s 4 4 s Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 5 1b ft3 1b 1 pQ3 623255 159 In S S 3 255i V3 3 S 13 A3 E l s 4 2 315 Pz z M u M dt um dr MV 3 073 pm 530 R101b m dP lbmol K hr p51 M196h 6 10ft29 r lbmole Students regularly try to nd a way to include the current value of the pressure 100 psig in their answer It plays no role The answer would be the same if the initial pressure were any value as long as the pressure was low enough for the ideal gas law to be appropriate 0001lb I3n 316 From Eq 3AC Psteadysme 1 atm lbm 00133 atm 0075 317 a Start with Eq 317 take the exponential ofboth sides rX psys nal J Q0 4 exp V At a where a is used to save writing in Psys inital QM sys Then psys nal EL 0sz5 inital ampJ Q a 1 apsys initiall psys nal out out out M a Q M Here p5ys in 00751bm ft3 and out a y psys m 0001 0075 lbm ft3 3 1 a eXp 151132 72 min 000074659 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 6 W P M Q P aP0 P RT dt RT dP RT RT P RT QPa PO PPl 2 a JJa aP dt V W V W W P dP 1 J ajdt t 0 ln a PE a a 1ft3 a g aRT min 5 103941bm 07302 atmft3 530 R V W 10ft3 minatm lbmol R 10 329 lbm lbmol 0100667 t a 010 6672 10 666721042 mm mm B 6662810 3 atm 0 t7 1 n001atm 0006628atm 56 5mm 010067 latm 0006628atm 39 min At steady state P 3 a 0006628 atm 0006628 atm If one assumes that the leak rate is 00005 lbmmin independent of pressure then repeating EX 38 we nd a nal pressure of 000666 atm instead of 000663 calculated here and a time to 001 atm of 5697 min instead of the 5650 min calculated here All real vacuum systems have leaks whose ow rate is more or less proportional to the pressure difference from outside to in assuming laminar ow in small ow passages but this comparison shows that replacing Eq 3AN which is a more reasonable representation of the leak rate with a constant leak rate makes very little difference in the Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 7 computed nal pressure and time to reach any given pressure The reason is that the pressure in the system falls rapidly to a small fraction of an atmosphere and thus the leakage rate quickly becomes Z independent of pressure 3 319 l d V l Mm 7XM10000 8000 20003 p dt dt p s m3 dh ldV 2000 2105m 72mm 28in r1s1n d A dt 1001061112 s hr h g dV 320 All the ows are incompressible so 0 and out and sys ft ft3 ft QM Qm Q0 05 212 J 03 216 J 12 S S S The minus sign indicates a ow out from the tank through the vent and r ft3 1b 1b k M Qp 12 JK0075 T 0090 m 00408 g s ft 5 S 321 d chomt01p l dt m1n dt V dt 3 B a r l 1 39OImin 1758 10011101 dt quot RT dt 1073psi ft3 5300R 39 min lbmol R r r 1b k a M 1758 10quot 29 00051 In 00023 g mm min d Vc 322 i cQin 0Qout where c is the concentration Here dc Ky 1s constant and cout csys and cin 0 so that V Qc 2dr 1n Qt 0 and c V 00 V k 10 3 39 c c0 exp 2At 10 g3exp wtj V m 1000 m Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 8 which is the same as Eq 316 after taking the exponential of both sides with concentration replacing density and the Q V one tenth as large as in EX 37 One can look up the answers on Fig 35 if one takes the vertical scale as c c0 and multiplies the time values on the horizontal scale by 10 d 323 VE Qc K K dissolution rate ibng ml jbgw c KQ V co KQ V c c0 K Qexp QtKK Q Comparing this solution linebyline with EX 38 we see that it is the same with the variables renamed Its analogs also appear in heat and mass transfer 3 324 Vsysa bt b1 HIJ m1n ch d bt 7357 5 462 go Qc but go 0 so that dc dc dc dc dt b bt bt b 1 ad c d Qc a dt Q mc Q am 1 bt 1 ll iml c a c lnc O Qout bzln m a c0 3 ft 325 Vsys a br 61 10ft3 b 01 mm dpV S dpa bts S W L m Qp 611 611 Comparing this to the solution to the preceding problem we see that they are the same with the variables renamed so that we can simply copy 12 17 1bat p0 lbat ppOW a l 10 3 l t 7 1 f 1 1 P Po Qb 01min 1074i11 3min01 3minj Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 9 6406min 100minL lJ 278256 326 Assume that the earth is a sphere with radius 4000 miles that it is 34 covered by oceans and that the average depth is 1 mile That makes the volume of the oceans be 3 3 10001 Z47r52804000ft2 5280ft m J 631020 liters maquot 3281ft m molecules 1000 gm 23 molecules 25 602310 33410 liter lggm mole mol This ignores the differences between pure water and ocean water which is negligible compared to the other assumptions Then a randomlyselected liter contains 334 1025 W 53 104 molecules which were in the liter that Moses examined The assumption ofperfect mixing ofthe oceans since Moses39 time is not very good The deep oceans mix very slowly The Red Sea does not mix much with the rest of the world ocean except by evaporation and rainfall 327 The mass of the atmosphere is lbm 2 47r4000528012 in211861019 lbm 1n 454gm mol lbm 29 gm mass 147 mols mass 1857 1010mols Caesar lived 56 years so he breathed in his lifetime 365 day 24hr 60min 10breath lL mol 131107 mols yr day hr min breath 224L 56 years Thus the fraction of the atmosphere that he breathed was 131107 Fraction he breathed 20 707 1014 185 10 At 1 atm and 20 C one liter contains 250E22 molecules so that the number of molecules in one liter of the atmosphere that he breathed must be molecules breathed by Caesar 250 1022 molecules 70710 177109 breath liter Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 10 I have carried more signi cant gures than the assumptions justify because it is so easy to do on a spreadsheet The logical answer is Z 2 billion The perfect mixing assumption is much better for the atmosphere than for the ocean in the preceding problem Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 3 page 11 Solutions Chapter 4 In working these problems I have used the and Keyes 1969 Steam Table for English Unit problems and the NBSN RC Steam Tables for SI problems For Freon 12 I have used DuPont publication A973 41 The plots will be straight lines on loglog plots The gz term will have slope l and the V22 curve will have slope 2 Forz 100 ft we have ft Btu lbf 2 Btu kJ PEgz322 2100ft 0129 0299 s 778 lbf 322lbmft lbm kg and for V 100 fts we have 2 2 2 KEL100 s Btu 1be 20200Btu20464g 2 2 778ftlbf 322lbmft lbm kg 100 a 3 a m E5 10 lt12 E E g 8 o 8 1 5 Potential E D S M I Illll I IIIIIIIl I Illllll 10 100 1000 104 Elevation ft or velocity fts 42 ft 2 a KE 002 1bm 1242103ft1bf16841d 2 2 3221bmft ft V 2000 b VV0 gt V0whent fi6211s g 322 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 4 page I 1 ft 1 ft 2 det Vet ng 2000 621s E3227621s2 62110quot S S 2 c PEmgz0021bm322g621104 amp s 3221bmft The fact that the initial kinetic energy and the potential energy at the top of the trajectory are the same is not an accident Air resistance complicates all of this see Prob 694 621103 lbf1684 kJ 2 43 W Fdx mgAz 201bm6 210ftamp 373 ftlbf505 J s 3221bmft V2 44 Our system is the ball dimEu gz 0 0 0 def sys Here the left hand term is zero so there is no work done on the ball This appears paradoxical but the ball is simply converting one kind of energy into another Work was done on the ball as the airplane lifted it from the ground If one applied the above equation with the same system from takeoff until the ball hit the ground then we would have 2 V2 Alm ugzV7M 0 0 Aan sys with the final KE being equal to the work done by the airplane in lifting the ball V2 45 a System the water dim u gz 0 0 0 dVKf sys but for this system the changes in internal potential and kinetic energies are all Z 0 so that def m 0 This seems odd but the water is merely transferring work from the pump to the piston rack and car Here the def is the algebraic sum of the work done on the water by the pump and the work done by the water on the piston rack and car V2 b System water piston rack and car 6434 gz 0 0 0 dVK f sys with this choice of system the change in potential energy is significant def mgAz 4000ftlbf 5423 N m 5423 k c System the volume of the hydraulic cylinder downstream of the pump 2 dim ugzV7JJ hindmin 00dWnf sys Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 4 page 2 def almusys hm aimin if we assume no change in temperature a good assumption then usys uin constant and def Pvin aimin here def is the work done by this system on the car rack and piston 4000 ft lbf Solving for Vin dmin dVK f 4000ft lbf ft2 v dm m m 2 200282ft3793E4m3 P 1000 1471bf1n 144m Here we work the problem in gauge pressure Ifwe had worked it in absolute then we would have needed to include a term for driving back the atmosphere as the car rack and piston were driven up 46 System the water path through the dam from inlet to outlet steady ow 062W h 000062 kWh 2810 4 kWh kg kg lbm 0622W h 000062m 2810 4w kg kg lbm This is negative because it is work owing out of the system 47 System the pump steady ow hm hem E 39799 80794 2 412E dm dm kg The work is positive because it is work done on the system The enthalpy values are from the NBSN RC steam tables If one does not have access to those tables one can use Ahz CPATLW120kJ4is4 o z20401z403 p 962g3 m bar kg C kg m which is only approximate because the equation used is approximate and the values of the density and heat capacity are only approximate But it shows how one would proceed if one did not have access to a suitable steam table 48 System the power plant steady ow 12 dW V2 ft 4 gAz 4 322 2 80 ft 2 S dm Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 4 page 3 235m 0302 0702 bm lbm kg The minus sign goes with the new sign convention work ows out of the power plant r WV 49 System power plant steady ow Po rX quotf dm m 2 m 2 15 dW AVZ j 9 j 2 LgAz 7 98140m 3924720 320 2 S S dm k 2 N 2 J w PoSOOOiJL320J J 1602MW s kgm Nm J s This is power leaving the system hence negative according to the sign convention 410 System nozzle steady ow 0 h J Lh J tweetwen AhCPAT 2 out 2 ft2 ftlbf Bt 195510 7 607E4 780J S lbm lbm l 1955106 2 AT B Sz 1be Btu 2600F 03 3221bmft 778ftlbf lbm F T 600 260 340 F 800 R 444 K out 411 System the tank up to the valve unsteady state Here the students must use the fact stated at the beginning of this set of problems that for a perfect gas u and h are functions of T alone and not on the pressure Idmu 0 howlaim t IdQ 0 AQ Amu lthmout sys but Amsys Am0 t so that AQ Mullah u AmoutCP CVV AmoutRT Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 4 page 4 V Anom P1 Pl here we sw1tch from mass to moles and use molar heat RT capacities V 3 lbf 144 A P P RTVP P lft 80 l48Bt 1562kJ Q RT1 2 1 2 inz 778 u 412 System contents of the container adiabatic claimys hindm m9 and for ideal gases C P kTin This is the classic solution which appears in almost all V thermodynamics books and in the first two editions of this book see page 118 of the second edition Unfortunately the adiabatic assumption cannot be realized even approximately if one tries this in the laboratory Even in the few seconds it takes to ll such a container the amount of heat transferred from the gas makes the observed temperature much less than one calculates this way The amount of heat transferred is small but the mass of gas from which it is transferred is also small so their ratio is substantial This is explored in Noel de Nevers quotNonadiabatic Container Filling and Emptyingquot CEE 331 2631 1999 413 See the discussion in the solution to the preceding problem In spite of that this is a classic textbook exercise which is repeated here System contents of the container adiabatic J Wu 11 dm muf mu hmf m CVmef 7117 CPTinmf 7 C m m E PT Tf 7quot P P 1 L Substitute and rearrange to CV mf VL Pfji RT CV f Tf If P 0 this simplifies to the solution to the 1 f 1 I 9 Pf T CV preceding problem 414 See the discussion with problem 412 Using the solution to problem 413 we have ET T 2 CV 14029315K f P 1 C 05 1 1x TPT 1 29315 1429315 P T 1 CV m 10 29315 f Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 4 page 5 29315K 342K 689 C 156 F 1 05 04 415 czdm dQ d 141012 1 Btu dm Q u 14410 31bm 065 g C 3851013 252031 lbm If this won t convince the students that 02 is a large number I don t know what will Z 416 41 1800 2 lbm V 21800B mj 778 lbf 323921b12n ft 9497it 18 2899 lbm Btu lbf s s s s Such velocities are observed in meteorites earth satellites ICBMs and in the kinetic energy weapons proposed in the Star Wars defensive systems To stimulate class discussion you can tell the students that hydrocarbon fuels like gasoline have heating values of Z 18000 Btulbm and this problem shows that high explosives only release about one tenth of that Ask them why Some will gure out that the heating value is for the fuel plus the oxygen needed to burn it taken from the air and not included in the weight of the fuel High explosives include their oxidizer in their weight They do not release large amounts of energy per pound compared to hydrocarbons What they do is release it very quickly much faster than ordinary combustion reactions The propagation velocities of high explosives are of the order of 10000 fts compared to 1 to 10 fts for hydrocarbon ames That is so fast that the reaction is complete before there is significant expansion and the solid or liquid explosive turns into a high temperature gas with Z the same density as the initial liquid or solid You can estimate the pressure for this from the ideal gas law finding amazing values 417 This is a discussion problem Fats are roughly CH2n Carbohydrates are roughly CH20n For one C atom the ratio ofweights is 1430 047 So iffats have 9 kcal gm we would expect carbohydrates to have 9047 42 E 4 kcal gm This simplifies the chemistry a little but not much You might ask your students what parts of plants have fats Some will know that the seeds have fats the leaves and stems practically zero Then some will figure out why The assignment of a seed is to find a suitable place put down roots and put up leaves before it can begin to make its own food It is easier to store the energy for that as a fat than as a carbohydrate Our bodies make fats out of carbohydrates so that we can store them for future use in case of famine As fats an equal amount of food takes about 47 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 4 page 6 as much space and weight as the same food as carbohydrates Thermodynamics explains a lot of basic biology 418 System contents of calorimeter unsteady state Note that the metal walls of the calorimeter are outside the system dmu 0 0dQ0 012 l 10 1 02a 5000g5 C fa gC 500g5 C 300 25 000 25 300 cal AQ 25 300 cal cal Btu Au 2 6325 11395 m 4g g lbm Sources of error heating of gases in calorimeter condensation of water vapor produced on combustion accurate measurement of small temperature difference The value here is between that for fats and for carbohydrates It corresponds to a good grade of coal 419 System earth unsteady state The heat ow is 2 0 Q kA j 103m 7r 8000mi 528 03902 F 11210 dt dx hr F ft m1 ft hr The amount of mass converted to energy is d E 1121014 m dt hr 1b 7 2 Btu 29h t C 3851013 r 420 System the boundaries of the sun assuming negligible out ow of mass This latter is not quite correct because particles are blown off the sun by solar storms but their contribution to the energy balance of the sun is small compared to the outward energy ux due to radiation 2 c almsys dQ 421 System the power plant from inlet to outlet steady ow 2 der AhgAzALJ dm 2 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 4 page 7 ft 2 ft 2 2 50 400 13Z75 Btu 1be s S 1 1 m s 778 ft lbf 39 322 lbm ft 2 3977839 322 Btu 1d 1 0096 3144 224 l07 lbm kg 422 a System valve and a small section of adjacent piping steady ow hf h b System one kg of material owing down the line Au0AW uf uc Bvl vaf hfh The results are the same as they must be What nature does is independent of how we think about it 423 None of these violate the rst law all violate the second law We may show that they do not violate the rst law by making an energy balance for each Bt ft Bt lbf 1 a Au gAz 0 001284 11 322 210 1b 3 778 lbf 322ft lbm 001284001284 0 Bt Bt Bt b ume1 01051bm l4334 uJ 08951bm12322 uj 1088 u which lbm lbm lbm is equal to the initial internal energy c The upstream and downstream enthalpies are the same as they must be for an adiabatic throttle These all violate the second law They all violate common sense If you doubt that put a baseball on a table and watch it waiting to see it spontaneously jump to a higher elevation and cool Be patient Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 4 page 8 Solutions Chapter 5 51 du gz0 2 Au gAZ 42232 1000 wk i 12 E s 3221bm ft 778ft lbf lbm 1 29 a AT CA lem 11 F Vstee1 012 m lbm 0F A 029 b AT C 113m 13 F Vwater 10 tu lbm F Discussion we do not observe friction heating unless it is concentrated e g smoking brakes spinning tires on drag racers hot drill bits or saw blades boy scouts making fire by friction AP AVZ dW F has dimension 52Theheadformis Az zL th 2g gdm g of lbf has dimension E 3221bm pg of lbm 1be2 rt2 ft3 s2 V 2 V2 has dimension 2g 0f 2 def hasdimenSiOH ftlbf 3221bmft of WH gdm F hasdimension 2 g of kg m 2 9982 8 V2 3X j N 1 P 53a APp m S S a m3194kPa464psi 2 2 kgm Au0 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 1 b AP 09 answer from part a 287 kPa 428 psi 01AP 01 3194 kP N J J Au revrslble k a Z 320 p 9982 83 Pam Nm kg m J c AP 0 Au320 kg l J 100 V2V2 2 2 2 54 APp 1 2 I10 J p S 2 2 a For water 1b 4 ft2 lbf 2 ft2 AP 62331100 J S 060 psi 41 kPa s2 3221bmft 144in2 b For air AP above answer 075 3 7210 4psi 00050 kPa P 55 The answers are the answers in Example 52 multiplied by 1 with all v Pl 130quot 2 pressures absolute For 1 psig this factor is P 15 7 psia 1 P1 13 J 157 psia 147psia J 1016 and the calculated velocity is 2 2 V 34021016 34552 s s Below is a table comparing the three solutions carrying more significant gures than are shown in Table 51 delta P psi V simple BE V Ch 8 V this fts fts Problem fts 001 35094 35090 35100 01 110640 110747 110827 03 190352 190992 191311 06 266546 268384 269198 1 339697 343602 345239 2 465799 476269 480405 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 2 3 554136 572306 579222 5 678100 713154 725709 From the table we see that while the simple approach in Example 52 underestimates the velocity the approach in this problem overestimates it but by a smaller percentage than the simple approach in Example 52 If one is not going to use the approach in Ch 8 then this approach is more accurate than the simpler one in Example 52 56 The plot will be a straight line with slope 12 on loglog paper For a height of 100 ft ft ft V 2gh 2 322 2 100 ft 802 s s The gure is shown below 1000 10 I 10 100 1000 hft ft ft m 57 V2 2322 12ft278 847 S S S ft ff 3 QVA278 2ft2 556 157 S S S 58 The density of the uid does not enter Torricelli s equation so VJ2g 2322 230 ft 440 ft 134E s s s Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 3 Discussion ask the students if you dropped a can of water from 30 ft with zero air resistance what the velocity at h 0 would be The ask about a can of gasoline 1 ft A2 2 2 answer to Prob 57 27398 ft m 59 V Zgh l 303 924 1 A1 Z zjz 5 1 016 S S 139 2 5ft ft 3 3 Q 303 2 2 606 4713 S S S 510 Vngh 4232232 726 ft 216 6592 47 S S S r Discussion the dam is fairly thick at its base so the pipe would be long and friction would be signi cant Torricelli s Eq is not reliable here See Chapter 6 ft 511 EJ293132 10m 1401m 46 S S S 3 3 Q VA 14012 51112 7003 2474 s s s I m sure all of you liked the movie quotTitanicquot particularly the scene in which the boat39s designer resigning himself to death says that the design allowed for the boat to survive ooding of two compartments but not four or five Apparently the safety doors divide the boat39s hull into some number 7 on the Titanic of compartments all of which are open at the top So if the hull is punctured as in this problem and the doors close properly one compartment will fill with water up to the level of the water outside but the other compartments will remain dry Apparently the impact with the iceberg opened the hull to more compartments than the ship could survive even with the safety doors closed Presumably if the compartments were sealed at the top then even that accident would have been survivable but it is much easier to provide closed doors on passageways parallel to the aXis of the boat in which there is little traffic than on vertical passageways stairs elevators etc on which there is considerable trafflc 512 From Eq 512 V 42g 4392 which is absurd s From Eq 522 V IZgh l 0 which is quite plausible pair Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 4 AP AV2 AP 513 gAz 0 V2 2 gAz p 2 p APP2 Pl ngAZ V2 lm A2 1 He The densities are proportional to molecular weights so V2 quot2 32222 40ft 2749 1 127 39E S S S This looks strange because the velocity is higher than the velocity from Torricelli39s equation However it is correct The reason is the very low density of the helium which makes the difference in atmospheric pressure seem like a large driving force dh 5 14 V V Zgh Q VAoutlet Kank Atank Amnk 6h surface tank dh A ft Zgh001quot2981 10m014 046 dt twk Atka s s s Zgh Some students will say you should use V That multiplies the above 2 1 A2 A1 answers by 100005 which is clearly negligible compared to the uncertainties introduced by the standard Torricelli39s assumptions AP 2 7gAZ but AP PgasounegAZ so that ft ft V 2g Az 1 413 2 3222 30 1072 2233 7092 PW s S S This is the same as EX 55 with different uids AP P P 516 V2 f 1 2 pw pw 30ft glamquot P2 10 ft 8PM 20ft gpou Pz ZO 39gpwam p011 V 220ft 322 2 1 09113 346 S S S 515 V2 0 So Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 5 AP AV2 517 F gAz TJ Assuming the ow is from left to right p 8lbf 2 F W144in 322 220ft0 623 11 S ft S ft ft ftz ft2 2 595 2 644 2J 486 2 451m 2 S S S S The negative value of the friction heating makes clear that the assumed direction of the ow is incorrect and that the ow is from right to left 518 V2 Z E gAz 201M 3221b ft 14439 2 ft ft V2 2 11 tl Izn in 5rt322 2 574 175E 623 m lbfs ft s s s 3 AP AV O P VZ O 519 gAZ 2J0 L 1gh 2 J20 p 2 p 2 Z 100 V2 lbm ft 2 1be2 ft2 P h 4 623 322 5ft S 1 pg 2 if s2 2 3221bmft 144 in2 2 16 067 l49 psig 103 kPa gage This indicates that at that velocity there is a vacuum in the top of the vessel lbf 50 2 39 2 520 V 2 gAz 2 W144n 322 230 p 623 3 ft S ft 7425 2262 s s Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 6 521 Let the waterair interface be 1 the mercurywater interface be 2 and the outlet PH m l m m ft 28m981 2 981 21m 558 183 s 136 s s s 522 Let the waterair interface be 1 and the outlet be 2 Then V gzlz 10M 39 2 3221bmft l44in2 lbm 39 lbf 2 ftz 623 S ft ft m 2 322 210ft 462 l4l s S S 523 122 f 2ip APjpm2dr V p2r22r12 mzr22r12 2 39 2 2000 21in2 20in2 13412 112E 342 S S S W2 2 22 r2 r1 min 60 s 524 In Bernoulli39s equation devices orifice meters venturi meters pitot tubes the ow rate is proportional to the square of the pressure difference These deVices most often send a signal Which is proportional to the pressure difference If this is shown directly on an indicator or on a chart the desired information the ow rate is proportional to the square of the signal One solution to this problem is to have chart paper with the markings corresponding to the square of the signal For example if one inch of chart covers the range from zero to 20 of full scale then four inches of chart will correspond to the range from zero to 40 of full scale The chart manufacturers refer to these as quotsquare rootquot charts meaning that in reading them one is automatically extracting the square root of the signal Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 7 With recent advances in microelectronics many new ow recorders extract the square root of the signal electronically and use a chart with a linear scale 525 Vm J2g quot2322 10 254 17377E S S r S These have actually been sold boat fans will buy anything For speedboats the pitot tube is normally connected directly to a bourdontube pressure gage which is marked to read the speed directly in mihr or equivalent 2gh 526 V p uid pair pair ft 05ft 623 0075 ft 39 Vm 2 322 2 472 322 1442 s 12 0075 s hr s Lowspeed ows of gases are hard to measure with Bernoulli39s equation devices because the pressure differences are so small Moving blade or moving cup anemometers are most often used for lowspeed gas ows e g meteorological measurements The students have certainly seen these in weather stations 527 see preceding problem V Zgh pmanometer uid 1 pgasoline ft m l 046 014 s s 39 322 lbmft 144 in2 ft mi m a V 2 2 1926 131 587 0 0751bm lbf s ft s hr s 3 b V Above answer 039075 2209it 151 6732 57 s hr s This is one of my favorite discussion problems On the dashboard of an airplane private or commercial is an quotindicated air speedquot dial The indicated airspeed is Vppsea1 level12 which is found by putting this pressure difference across a diaphragm and using a linkage to drive the pointer around a dial This indicated air speed is the Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 8 velocity of interest for pilots The lift is directly proportional to the indicated air speed so this is also effectively a lift indicator see Secs 614 and 76 You can ask the students why commercial airliners y as high as they can For a given weight there is only one indicated air speed at which they can y steadily in level ight As the air density goes down the corresponding absolute velocity goes up So by going high they go faster This means fewer hours between takeoff and landing That means less fuel used fewer hours to pay the pilots and crews for and happier customers who do not like to sit too long in an aircraft 529 hr 3600 s km N s2 Pa m2 AP 1002 3 m 2 kg m 2 60km hr 1000111 kg 1392 kPa 202 psig 530 Equation 5BN predicts that for a VP which we would call AP of 1 inch of water 003615 psig the velocity is 4005 ftmin Using the methods in this book we find for that pressure difference lbf 003615 2 39 2 144 322lb ft ft ft V 2 1bm 2m Izn6685 4011 0075 11 ft lbfs s min This is 10016 times the 4005 in Eq 5BN The form ofthe equations is the same V proportional to the square root of AP so for any value of AP the prediction of this quotpracticalquot equation is within 016 of the value from Eq 516 531 Example 58 shows that for a pressure difference of 1 psig the volumetric ow rate is 249 ft3s We can get other values by ratio eg for 1 fth 1 3 2 S J 0161psig Q APAP 1 p 51g 249ft3s 0 Q0 and similarly for other values Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 9 The plot is shown at the right in 1 0 two forms The upper one on loglog coordinates is probably the more useful because the fractional uncertainty in the reading is practically the same over the whole range Q cfs The lower one on arithmetic coordinates is probably easier for nontechnical people to use and would probably be selected if technicians were to use it Here I have chosen the pressure drop as the independent variable because that is the observational instrument reading Delta P psi lbm lbm 0075 ft 2 322 2l ft 623 s 2 h 532 V2 2 Q g pmwnm I p m if re A 1 D2 D1 1 05 2392 7282 S S One may compute that the Reynolds number at l is E 37 E 5 From Fig 511 one would estimate Cv Z 0984 so that if we take this correction into account we would report V 235 7162 In most common work we would probably ignore this difference s s 533 We use Eq 519 ft lb 2322 2 1ft 623 13n 1 072 S ft ft m 6230721 05 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 10 Q V2242 517 05ft2j 101 00293 S S S 2 P P 1 2 2gzl 22 534 V2 2 i 1 A1A2 lbf 275 2 gm 3221brznftl4421n 2322 2 2 072623 3 1be 5 ft m 4 240 733 l 05 s s ft ft3 Q VA 240 05ft2 472 0132 S S S 535 Let 1 be in the pipe opposite the top leg of the manometer 2 be in the vertical section opposite the lower leg of the manometer See Eq 519 Here DZDl 05205 ft 2 1b 2 322 ftj 136 1623 I3n 12 ft ft 2 5 lbm 4 120 366E 6231 05 S S Q 14A 120 05rt2 60 0173 S S S From the gure there is no way to tell which way the water if owing 536 V CV 2 P1 Pz 2 pwepmmmy 132 pm g2in pl A2 A1 mm mercuryiair PI Pmercurywter pHgg 39 01 in interface interface Pl P2 2in pwg0lianggpwg2l36in ft V 10 lbm 123 3752 S S 0075 1 012 Q VA 123 1 2 123 3483 S S S Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 11 2 rX AV V 537 a AF A A A A A PA amp MF AFVFpF VF 2 pp pp We cancel the pressure difference because both the incoming air and the liquid gasoline 2 in the reservoir are at the same pressure Then A A and F F 1 1 l 2 DF pi pF DF pF b As the above equation shows the airfuel ratio should be constant independent of the air ow rate This explains why this type of carburetor was the practicallyexclusive choice of automobile engine designers for about 80 years With a very simple device one gets a practically constant airfuel ratio independent of the throttle setting The rest of the carburetor was devoted to those situations in which one wanted some other airfuel ratio mostly cold starting and acceleration for which one wants a lower air fuel ratio quot rich ratioquot lbm 2 2 0075 D D D 1 c 15 im 41915 L0052 DF 072 623 DF D2 1915 d At 5280 ft the atmospheric pressure is about 083 atm The air density falls while the fuel density does not so the air fuel ratio would become AF15 083atm 21371bm J latm lbmfuel and the engine would run quotrichquot High altitude conversion kits smaller diameter jets are available to deal with this problem Rich combustion leads to increased emissions of CO and hydrocarbons there are special air pollution rules for autos at high elevations Demands for higher fuel economy and lower emissions are causing the carburetor to be replaced by the fuel injector In a way it is sad the basic carburetor is a really clever simple selfregulating device 538 By Bernoulli39s equation the velocity of a jet is proportional to the square root of the pressure dropdensity The density is proportional to the molecular weight For the values shown the predicted jet velocity in the burners will be the same for propane and natural gas Because of the higher density of propane the diameter of the jets will normally be reduced to maintain a constant heat input But the velocities of the individual gas jets are held the same to get comparable burner aerodynamics Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 12 539 V2C 2AP ApPVZ124 V D 4 D0 ft 7239 2 1 21 in ft V2 4 4 25 S 02in 4 lbm ft 2 4 55 325 1 02 2 2 AP S 2 1be 1031if709kPa 206 322 lbm ft 144m2 in2 A22 4 540 Here for D2 D1 08 1 1 08 0768 A 4 and for DzD102 1 14 12 1 02 09992 So the curves for C would simply be the corresponding curves for Cv divided by these values or multiplied by their reciprocals For 02 the multiplier is 10008 or practically 1 while for 08 it is 130 This is a simple scale change for each curve on Fig 514 541 This is simplest by trial and error First we guess that D2 D1 05 Then 2 2 V 4E and D2 s 05 s 4 2 AP kg 21 D4 2 C lbm ft 2 1 05 2 3577 2 062 322lbmft 144m in This is done on a spreadsheet We then ask the spreadsheet s search engine to nd the value oszDl which makes the press drop 3 psi Doing this by quotgoal seekquot on excel one nds D2D1 05205 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 13 542 a 2 2 1b m 322lbmftl44n 630 t192m 623 m ft S S V from b V2 2 JJM630quot 411 125E parta 20 20 s s 543 gAz AVZ 0 Az p 2 pg 2g lbf ft 2 AZ 41 147 322 lbm ft 144ml 10 39 Sz l ft 623 3 322 2 239 322 2 ft S S 3l67 l55 30ll ft 918 m 544 The velocities will be the same as in EX 512 Solving Eq 5AX for 22 21 we nd P P V2 22 21 2 2 pg 2g lbf ftjz 147 034 lbm 144M 253S 1b ft 3932392 2 2 ft 623 m322 1be 2322 3 Si Si 3319 994 2325ft 709m 545 This follows EX 512 2 ft 12 ft m V2gh1 h3 232210ft 253 771 s s s One may make the next step applying BE from 1 to 2 or from 2 to 3 Either gives the same result Working from 2 to 3 we have 2 2 z zmn 0 p 2 A2 Pz P3p23 22gp V51 51 J 2 A2 2 2 6231b115 322fzt 2 ft s 3221bmft 144m Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 14 623 3 2 ft 10ft 3221bmft 1441n ft 2 lbm 253 5 115 22 1be2 ft2 2 lbf 216 538 7547 754 psig 716 psia 494 kPa abs 1n 546 Take the ocean surface as 1 the tip of the propeller at its highest point as 2 21131 132 me p 2gZI Zz lbf 2147 026 2 2 3221b ft 144 ft ft lbmm I in 2322 4 491 150E 623 3 lbfs ft p s s ft ft 491 V 104 a Lax S 62RPM 547 See solution to Prob 546 Clearly the greater the depth the higher the speed at which the propeller can turn without cavitation so this is not as severe a problem for submarines submerged as it is for surface ships However the noise from propeller cavitation is a serious problem for submarines because it reveals the position of the submarine to acoustic detectors Submarines which do not wish to be detected operate with their propellers turning slower than their minimum cavitation speed 548 See EX 514 As in that example take 1 at the upper uid gasoline surface and 2 at the outlet jet Take 2a at the interface between the gasolinewater interface AV2 gAz T 0 Here the veloc1t1es Then applying BE from 1 to 2a we have pgasoline are both negligible so that PM pgasoline gz1 2 Then applying BE from 2a to 3 we find Z Z Z paasolmeg 1 2 gzm ZS 2 Ti 2 Water At this point we can observe that as long as the gasolinewater interface is above the nozzle the pressure at 2a will always be that due to 20 m of gasoline which is the same as that of 144 m of water If we replaced the 20 m of gasoline with 144 m of water then the ow at any instant would be unchanged So this is really the same as EX 514 with the initial height being 244 m and the final height being 144 1 154 m Thus this is a plugin Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 15 2 154111 2 244111 2 At 2 n 41m 11 410 m2 2 458s 076 min An alternative which may please some students better is to write V2 asoine Z Z a i w gzm z3 a gzm 23 where a 1s a constant as water long as the interface is above the outlet Then one repeats the whole derivation in EX 514 nding the same result One can also compute the time from when the interface has passed the eXit to when the surface is 1 m above the exit nding the same time as for water because in this period the density of the owing uid is the same as that of gasoline The period when the interface either gasolinewater or gasolineair is close to the nozzle is not easy to predict by simple BE because the AZ becomes comparable to the diameter of the opening and the nonuniform ow phenomena discussed in Sec 511 come into play See Prob 553 I have the device described in that problem I regularly assign the problem then run the demonstration One can estimate well down to an interface one or two diameters above the nozzle but not lower 549 See the solution to the preceding problem The easiest way to work the problem is to conceptually convert the 10 ft of gasoline to 72 ft of water Then this is the same as EX 513 asking the time for the level to fall from 172 ftto 72 ft At A 2 4172 J72 365s 24E 2 39 100 s2 550 a See the preceding two problems 20 psig is the equivalent of 4619 ft of water so that 12 2 14619 54619 At f 7lls0l2min 41 2 2 322 2 7r 410 ft2 s b In this case we have P P0 E 100 147psiaamp 1147 psia 1 V A6 hft 6 hft where h is the height of the interface above the nozzle initially 5 ft finally 1 ft The final pressure is 2294 psia 824 psig This gives the answer to part c the 5fold expansion of the gas lowers its absolute pressure by a factor of 5 which would produce a vacuum and stop the ow if the initial pressure were 20 psig Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 16 AP At any instant the velocity is given by VIM 2 gh where AP the p above absolute pressure minus 147 psia If we substitute that equation in the velocity expression we get an equation of the form VIM b dh constants and an integral of the form a This might be reducible to one of ch ch where a b and c are b h the forms in my integral table but not easily Instead we proceed by a spreadsheet numerical integration For time zero we have lbf 100 2 39 2 1b ft ft ft Von t 2 322144 322 2 5 ft 12266 m lbfs ft s s 623 3 ft dh A ft ft and JquotVu et 001 12266 123 The time for the surface to fall dt Atmk s s Ah 025ft from 5 to 475 ft above the outlet would be At W 0203 s ifthe velocity 123 dt s remained constant over this height But as the following table shows the velocity declines so we cover this height step using the average of the velocity above and the velocity at h 475 ft which is 107 fts Then we make up the following spreadsheet h ft P psia P psig Vinst fts det fts delta t Cumulativ e t 5 114700 100000 122664 1227 0000 475 91760 77060 107813 1078 0217 0217 45 76467 61767 96639 0966 0245 0461 425 65543 50843 87778 0878 0271 0733 4 57350 42650 80482 0805 0297 1030 375 50978 36278 74302 0743 0323 1353 35 45880 31180 68949 0689 0349 1702 325 41709 27009 64227 0642 0375 2077 3 38233 23533 59997 0600 0403 2480 275 35292 20592 56159 0562 0430 2910 25 32771 18071 52636 0526 0460 3370 225 30587 15887 49368 0494 0490 3860 2 28675 13975 46310 0463 0523 4383 175 26988 12288 43422 0434 0557 4940 15 25489 10789 40673 0407 0595 5534 125 24147 9447 38033 0380 0635 6170 1 22940 8240 35479 0355 0680 6850 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 17 Finding a time of 685 s The original spreadsheet carries more digits than will t on this table One may test the stability of this solution by rerunning it with smaller height increments For increments of 025 ft shown in the table the time is 68498 s For increments of 01 ft it is 68563 s Both round to 685 s c See the discussion at the top of part b 53951 Km d2gl l Qout AexitIbut AtankE h Ak 50 250 2mj 50 ft225fthabh where a and b are constants z A 2gb A0mJ2gJ0thLZ bJj aZdh dt abh wimp 1 a b 3 AF Jz h2 A 2g 1 3 2 2 h0 1 50 ft2 25ft 3 2 1 Jzo 3 20ft 743s 1ft J2322S 2 2 2 552 The instantaneous depth in the tank is h Let h39 h1 h Then this becomes the same as EX 513 2JZ Jh 1 iJz g A1 At 2 NZ 997i JZ A 05 I ft 4 4J2g 2322 14 20 s2 553 This is the same as EX 514 with much smaller dimensions which match a Plexiglas demonstrator which I have used regularly in class Following EX 514 2 1m 11in At 4 03in2 ft 239 12in 278521301111 S 6in55in At 1220 h1h1 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 18 When I have run this in a classroom with a sink the times are typically 84 to 88 s Much of the time is spent in the last inch where the assumption that the diameter of the nozzle is negligible compared to the height ofthe uid above it becomes poor see Sec 511 When I do this I assign the problem and ask each student to write herhis name and predicted time on the blackboard before Irun it Some students like that others don39t 554 a See Ex 55 I ft 75ft 29 ft VC Zgh pm l 06 2322 2 1 343 pgas s 12 16 s Here we introduce the ori ce coefficient because the hole drilled in the lid of the can is much more like a flatplate ori ce than like the rounded nozzles in the Torricelli examples b For totally unmixed flow the densities above and below the interface between gas and the air which has owed in from below are constant so that V f h V 2 E We then square both sides and differentiate wrt finding 2 initial 2V dV 1 dh 22 2 Then using the same ideas as in Ex 514 we write that initial dt h dt dh Aoutlet E 4 We subst1tute this in the preced1ng equation and s1mp11fy tank cross section 2 1Viuitial I Amulet d1 2 Ainkcnsssectnn which we can then separate and integrate from start to any time finding to V2 1 lionE39na1Aout1ett V miginal 2hoAtnnk Cross section V2 vs I Its value is zero when This is a straight line on a plot of 2quot Jl 6i ll tank Cross section 2 migiualeutlet 3 43 For the totally mixed model we have Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 19 pair pgas mixture in can 1 instantaneous pair 1 pgas mixture in can initi a1 2 initial By material balance taking the can as our system In this problem two Vs appear the volume of the can and the instantaneous velocity Following the nomenclature both are italic When a Vis the volume of the can it has a quotcanquot subscript dp Kan dt ruin Mam T 4iaiiilet pair pmixmie Which we rearrange to dpmixture KAoutlet dt pair pmixture Kan and integrate to 1 lpaii pmixture 2 J Aoutlet th npair pmixture nitja J 2 is the number of can volumes which have owed into and out of the tank can can where can since time zero We take the exp of both sides and rearrange nding Q pmixmie pair pair pmixturemiqja1 exp V We then substitute this into the velocity ratio equation and simplify to V2 pair V V27 initial 1 1 pgasinitialJ expQJ pair Vcan In principle one should be able to eliminate Q from this relationship by A 2 AM ngt to get Vcan 0 Vcan initial been able to do so nor has anyone shown me how Try it you ll be impressed However it is easy to solve this on a spreadsheet The solution is shown below The rst line is obvious at time zero there has been no ow in and V V0 3431 s Most ofthe spreadsheet carries more signi cant figures than are shown here as an explicit function of I but I haven t For QV 01 we can easily compute that eXpQV 0905 Then L 16 290905 Vmim 1 1 16290905 V 0916 3341ft s 3145 ft s We get deltat by differentiating the material balance above finding 0916 and Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 20 f 75ft 65 jz V A V 01 0 Aamp Q can4 12 122 1285s Acme V2mge 0533413145 4 12 s Each number in the rightmost column is the sum of the one above it and the one to its immediate left Q V eXpQ V VVo V fts delta 1 s tcum 0 1 1 3431 0 01 0905 0916 3145 12850 12850 02 0819 0845 2899 13983 26833 03 0741 0782 2684 15136 41969 04 0670 0727 2495 16316 58285 05 0607 0678 2326 17527 75812 06 0549 0634 2174 18776 94588 07 0497 0594 2037 20065 114653 08 0449 0557 1912 21399 136052 09 0407 0524 1797 22784 158836 1 0368 0493 1692 24221 183057 11 0333 0465 1594 25717 208774 12 0301 0438 1504 27275 236049 13 0273 0414 1420 28899 264948 14 0247 0391 1342 30593 295541 15 0223 0370 1269 32363 327904 16 0202 0350 1201 34213 362117 17 0183 0331 1137 36147 398264 18 0165 0314 1077 38171 436435 19 0150 0297 1020 40289 476725 Itested the stability of this solution by reducing the increment size from 01 to 005 and found that the time to QV 19 changed from 476725 s to 476994 s which is certainly a minimal change c From the above table we may interpolate that 11 fts corresponds to Z 420 s which is a tolerable match with the observed Z 325 s For the plug ow model the agreement is 1 1 not good as good We can see that the required value of 39 0321 We solve 0 for it by V2 2 0321 1 Vzmginaleutlett t 1 0321 ZhOAtankcrosssection 2 miginal 2 hDAtank cross section V originaleutlet 75 ft 1 03212 6 Sin 2 t 12 167 S 343 025 in S Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 21 We always tell the students that the totally mixed and plug ow models will normally bracket the observed behavior of nature That happens here In making up this solution I found two errors in 7 I hope the version here is error free As of spring 2003 you could watch a film clip of a version of this demonstration at httpchemmoviesunleduChemistryDoChemDoChem049html The combustion specialists in Chemical Engineering at the U of Utah like this demonstration very much For me it is a ne demonstration of unsteadystate ow They see all sorts of interesting combustionrelated issues in it 555 Replacing the owing liquid with a gas heavier than air requires us to combine Ex 514 with Ex 55 We see that the instantaneous velocity is given by 2 29 I2gh1 il nghquot1 0584 Zgh pm 44 This is true at all values of h so we can simply substitute this value and nd that the required time is the answer to that example divided by 0584 577 minutes I give this problem as an introduction to a safety lecture Propane is by far the most dangerous of the commonlyused fuels If there is a large leak of natural gas buoyancy will take it up away from people and away from ignition sources If there is leak of any liquid fuel gasoline diesel heating oil it will fall on the ground and ow downhill But the ground will absorb some and ditches dikes or low spots will trap some or all of it But released propane forms a gas heavier than air which ows downhill and ows over ditches dikes and low spots looking for an ignition source and then burning at the elevation where people are See quotChemical Engineers as Expert Witnesses in Accident Casesquot hem Eng Prog 8416 2227 1988 and quotPropane Overfilling Firesquot Fi Journal 81 No 5 8082 and 124126 1987 556 At the periphery where the air ows out its pressure must be the same as that of the atmosphere subsonic jet By material balance assuming that the width of the ow channel between the cardboard and the spool is constant the velocity is proportional to lradius so the velocity decreases with radial distance Thus by BE the pressure must be rising steadily in the radial direction In the center hole the pressure must be higher than atmospheric in order to give the gas its initial velocity and to overcome the frictional effect of the entrance into the channel between the cardboard and the spool Thus the figure is as sketched below This is a very simple portable cheap dramatic demonstration to use in class Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 22 Cen rline radius of disc radius of hole Pressure above atmospheric Atmospheric pressure Pressure below atmospheric 1 L L 557 Q 2 05 The area ofthe periphery is s s A n39Dh n39 35 mm 02 mm 220 mmz so the velocity at the perimeter is 0 5 L 39 106 3 VQ 2 22736 227 A 22 mm L s s The velocity at the edge of the center hole is D 35 Vl39nole erimeter enmeter mm 1 Dhole s 71 mm s Then we apply BE from the edge of the hole to the perimeter nding p 13920 g3 m 2 m 2 st Pa m2 AP V2 V2 m 227 112 2 periphery hole 2 S S kg m N 72 kPa 105 psi At the periphery the pressure is 000 psig Since the pressure increases by 105 psi from center to periphery the pressure at the edge of the center hole must be minus 105 psig ie a vacuum of 105 psi Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 23 558 The numerical values are shown below P2P1 incompressible isothermal adiabatic 1 0 0000 0000 105 005 0049 0049 11 01 0095 0097 115 015 0140 0143 12 02 0182 0187 125 025 0223 0230 13 03 0262 0272 These values are plotted below with smooth curves drawn through them It must be clear that at low values of the pressure ratio the three curves are practically identical 035 IIll 03 025 02 MR T dWdm 015 01 005 0 IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 095 1 105 11 115 12 125 13 135 P2P1 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 24 W 559 The pressure difference needed to support the load is AP Z ED2 4 2AP The velocity under the skirt is V p thus 2W 001 I 8 5000 lbf 3221b ft ft3 QVAV7rDh7zh ftn39 mm 11 61 vp l2 00753 lbf s s 4 ft rdW RT P P Po rX quotf p 11n 2Plan 2 dm M P1 P1 W 5000 lbf ft2 pl ZT144 2 044psi305kN 10ml 1 4 3 39 Z P0l47g61 Jln147044144m hps in 147 ftz 550ft1bf As far as I know these devices were quite popular for a while but have recently declined in popularity As far as I know there are still ferry boats that use this system to stay above the water regularly crossing English Channel 069 hp 052 kW s 560 a The gauge pressure inside the structure must equal the weight of the roof per unit area P 101bfin2 0007 psig 0036 in H20 00479 kPa lbf ZAP 2 3221bmft ft m b V lb 2 293 893 p 0075 I311 S S ft 3 3 Q VA 293 5ft2 1465 8760 cfm 4153 S S S ft3 lbf 1462 10 AP 2 HP c Po Q S S 0355 HP 0264 kW 77 075 550ft1bf These structures are widely used as winter covers for outdoor tennis courts and some other applications There are two put up each winter within ve miles of my house 561 a using Eq 522 with the coef cient described below it 067WJ2g 3 Q h2 32 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 25 2 2 3 Q 3 311001113 3 063W 2 2 S m 1007m33ft 39 g 06750mJ2981 2 S 10033 Q s m b V 0 1 A 550m2 c Here looking at the equation 5BI we see that the average value of Zgh is m 0 1007 Zghavg 2 98ls 2 2 m 98832 S m2 008 2 s Ratio while the value of V122 is constant at 008 mZsz so the ratio of the neglected term to the one retained is 2 0008 08 988m 2 S 562 Vtop quot2 32230 025 4377ft S ft Konm 2 32230025 4414 S L 09917 bottom Clearly we make a negligible error by ignoring this in the standard Torricelli39s problems 564 See the preceding problem I have the solution on a spreadsheet so I can simply duplicate that and ask the spreadsheet s numerical engine to nd the value of A2 for which Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 26 DZD1 01 025 04 nding A2 0594 ft By hand we rearrange the above equations to 2 V D 4 1 025 in AZ2 Og 1 2 322ft 01in 2 059m SZ At rst I was puzzled that E 06 ft produces D2 D1 04 while 1 ft produces D2 D1 035 However we see that D2 D1 2 AZ14 so that the diameter decrease is rapid at the nozzle and becomes much slower as the stream accelerates due to gravity 565 I got this quote out of a book somewhere I wish I remembered where As the following calculation shows by simple BE one would estimate a much higher velocity However choosing 1013 bar as atmospheric pressure may be too high for a major hurricane the pressure at which the velocity is negligible may be substantially below that In any event V ZAP 2l013 0850105Pa Nm2 kgm kg P N 2 p 120 3 a S m 1642 5382 367g s s hr Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 5 page 27 Solutions Chapter 6 For all problems in this chapter unless it is stated to the contrary friction factors are computed from Eq 621 61 For nonhorizontal ow one simply replaces the P1 P2 term with P1 Pz gpZl 22 which makes the equivalent of Eq 64 become rP1 PZgpZl Zz 2Ax and the equivalent of Eq 69 becomes Q D3 PlPz pgAZJ f u 128 Ax For vertical ow AzAx is plus or minus 1 for some other angle it is cos 0 so that Eq 69 D4 becomes Q E 0 cosg 128 ix 6 2 Assuming a transition Reynolds number of 2000 74 VR y 2000 0018235111672 10 lbm387E1182 Dp J 00753 ftscP s s 12 ft AP 4f V2 16 0008 Ax Dp 2 f R lbm ft 2 E 400080075387 lbf 82 Ax ij a 322lbm ft 144m2 12 39 kP 47105E00011 a m These values are low enough that air is very rarely in laminar ow in industrial pipes 63 Again assuming a transition Reynolds number of 2000 2000 1002 cP 672 103941bm ft V 0259 00792 A 6231bm ft s cP s s 12 1 13 ggig 2091e 51bfsft2 Ax 7239 D 7239 ft12 00250 if 00001735E 00039 ft ft m or taking f 16 R 0008 and using the friction factor formulation Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 1 2 g 4 0008 623 0259 1 i imwn igz ps1 Ax 2112 322 144 ft mile These values show that water is occasionally but not often in laminar ow in industrial sized equipment It is often in laminar ow in laboratory or analytical sized equipment AP 7239 i AP 1 2 64 Vavg 2 W 5 From Eq 68 A 7270 Ax u 8 AP 1 2 V Km i so that 3 2 which is Eq 610 Ax 4 Vmax 3 U 2 2 kg Vrdr Ax 4 0 ro rrdr 5 2 Vrdr 2 on 02 V2 mos 3r4r2 3r2r4 r ydr 68 1 2l 4 7 0 0 0 0 2 4 6 8 i 7 etc n etc 1 1 etc 162 04 24 2 2 4 Vm2 Vavgj which is Eq 611 A2 D4 65 u Take In of both sides and differentiate nding d d d Az 461D d d Ax dlnpoQQ p A2 Do Q Ax d d d a and b iOI Q or p so that a 10 error in Q or p causes a 10 error p in u 61 dB c 01430 so that a 10 error in D0 causes a 40 error in u 1 0 This marked sensitiVity to errors in diameter shows why this type of Viscometer is almost always treated as a calibrated deVice 2 9398 012m J N S 177i s Nm kgm kg 66 F gAzAu Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 2 J 177 4k 7 AT CV 214kJ705510 0709910 F kg C For much higher viscosities this temperature rise can be signi cant The corresponding viscometer for very high viscosity uids replaces gravity with a pump and attempts with cooling to hold the whole apparatus and uid isothermal All serious viscometry is done inside constant temperature baths see Fig 15 4 2 4 67 Q AP n3 1bf1n 7239 210121b 3221bmft ft AX J ft 12816567264 m 144111 fts 3 3 000133fi 3751053 s S 3 V2M 006083 00185m A 7r42ft12 s s One may check nding R 001 so this is clearly a laminar ow 3 1041 68 war 55001273 3 8 410 m m kg 10 3miO0127 J1050 J DV 3 P 2 N2 R p S 1 am S044 u 00303 Pa s N kgm If the uid density does not change then the lowest viscosity is one would correspond to R 2000 and R 4 303c Mmquot 2000 p 2000 00067cp App A1 shows that this low a viscosity is rarely encountered in liquids so that this kind of viscometer can be applied to most liquids Cryogenic liquids however have very low viscosities so there might be a laminarturbulent transition problem using this particular viscometer on them 69 The volume of uid between the two marks is V Az D2 1 cm l cm2 cm3 4 4 4 The volumetric ow rate in the example is for the level at the top The volumetric ow rate is proportional to the elevation so at mid elevation of the test Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 3 1150m 8m3ll5 8m3 0 095810 Q QExsz 120m s 12 s and cm3 3 AtK4 L582 Q 0958 40433 106 cm s Many common industrial viscometers are of this quotread the time between the marksquot type and the viscosities are reported in quotsecondsquot eg Saybolt Seconds Universal SSU which is the standard unit of viscosity for high boiling petroleum fractions dV AV I 610 F ma m m here Av 40 40 80 so that 611 At hr 5 mi ft 161bmj80 hr 52803 MS 10 S 36001 322 ft hr F 0ll41bf 611 If the balls are thrown in the I direction then the I component is independent of the y component The x component force depends on the speed of the train and how often the balls are thrown back and forth but not on their velocity 612 Start with Eq 64 which is a general force balance applicable to any kind of ow and equate the shear stress to the value shown in the problem A1 2f122 2m p 2 4 Ax Then for horizontal ow use Eq 6B with both sides divided by Ax F 1 AP Eliminate between these two equations nding Ax p Ax Ax V2 D F F f p j which is rearranged to f 2 2 4 Ax 4AxDV 2 Which is Eq 618 This shows how the 4f appears naturally in the Fanning friction factor D2 AP 613 f L In Poisueille39s equation Vavg Ax p AP Eliminating A between these two we find x Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 4 f32uVMJ D 16y 5 2sz DpV R avg avg 614 a The ow is laminar so doubling the ow rate doubles the pressure drop to 20 psi 1000 ft b At this high a Reynolds number we are on the at part of the lines of constant relative roughness for all but the largest pipe sizes so that the pressure drop is proportional to the velocity squared Doubling the velocity quadruples the pressure drop to 40 psi1000 ft One would think students would all see this They don39t Assign this problem after you have discussed laminar and turbulent ow and you will be appalled at how few of the students can solve it After they have struggled with it they will be embarrassed when you show them how trivial it is Maybe that way they will learn about the different relation between pressure drop and velocity in laminar and turbulent ow 615 fDmyWeismch 4mmn g See Eq 619 Poisueille39s equation can be written as fDarcyWeisbach 4fFanning 4 39 If you look at the equivalent of Fig 610 in any civil or mechanical engineering uids book you will see that the laminar ow line is labeled f 64 R Jl ll zs t TJ 616 R lbm 327105 1002 cp67210 4 ft s cp 0001839 5 m 00003 f 00042 D 60651n lbm ft 2 E 4000426237 lbfsz z Ax 6065 ja 322 lbm ft 144 in2 12 P 11210 2E 112amp 0253 a ft 1000ft m AP psi by linear interpolation in Table A3 m 111 Ax 1000ft Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 5 617 a The easiest way to work this problem is to use Fig 612 As sketched at the right one enters at the bottom at 30 psi 1000 ft reads diagonally up to the 08 sg line then vertically to the 5 cs line and then reads horizontally nding about 175 gpm 5cs 175 gpm 08 30 psi1000 ft b To do it using Fig 610 one sees that it is the same type ofproblem as EX 65 the Type 2 problem in Table 63 which leads to a trialand error solution This is very similar to that shown in Table 65 and is shown below Variable First guess Solution D ft 0256 0256 L ft 1000000 1000000 delta P psig 30000 30000 6 inches 0002 0002 r lbn ft3 49840 49840 n cs 5000 5000 m cp 4000 4000 f guessed 0005 0006 V fts 8447 7685 R 40044302 36428733 eD 0001 0001 f computed 0006 0006 f computedfguessed 1189 1000 Q galmin 176746 The answer 1767 gpm is slightly more than the 175 we got in part a but no one should believe any calculation this type to better than i 10 as discussed in the text 618 This is easily solved on Fig 612 One enters at 200 gpm reads horizontally to the zero viscosity boundary which corresponds to the at part of the curves on Fig 610 then down to sg 075 and diagonally to Z 23 psi1000 ft This is less than the available 28 so a 3 inch pipe would be satisfactory Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 6 To solve using Fig 610 or Eq 621 we compute 200 ft V 870 we calculate that R 15106 and eD Z0006 so that 230 S fts f000456 and lbm ft 2 AP 4000456075 623 1870 lbf 2 z Ax 3068J 2 3221bmft 144in2 00271E 271amp ft 1000ft This is adequate agreement with Fig 613 and indicates that a 3 inch pipe is big enough 2 0256ft 54211 13n 6523 150 ft 619 a hit 652 R lbm S 89650 3i 5 15672e 4 fts ft s f 053 so lbm ft 2 AIL 4f sz 4X000531000ft087623652E lbfsz z D 2 3068 j 2 3221bm ft 144m2 12 205 psi b by chart look up on Fig 613 the pressure drop is 19 psi Fig 613 generally predicts pressure drop values for turbulent ow somewhat less than does Fig 610 most likely because it uses a lower value for the absolute roughness The most likely reason for the lower roughness in Fig 613 is that most petroleum products are slightly acid and will smooth out a steel pipe in use while most waters will rust or leave deposits on steel pipes so that they become more rough over time The value for the roughness of steel pipe Table 62 is presumably more in accord with the experience with water in steel pipes than the experience with various petroleum products in steel pipes 2 620 F Au gAz 322 2j 20fti k s 778 ft lbf 3221bm 0026 0060E lbm kg Au 0026 M A 0043 F 0024 C CV 06 Btu lbm 0F Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 7 Here again we see that the temperature rise is negligible 621 The table is shown below The first three columns are the same as those in Table 65 except for the number of digits shown The bottom 4 rows are in addition to what is shown in Table 65 The right column is made by copying the column to its left inserting 30 in place of 10 for A2 and then using the spreadsheet s numerical solution engine quotGoal Seekquot on Excel spreadsheets to make thefcomputedguessed l00 by manipulating the value of f guessed Variable First guess Solution Prob 621 D m 01 01 01 L m 100 100 100 A z m 10 10 30 8 inches 00018 00018 00018 p kgm3 720 720 720 I cp 06 06 06 f guessed 00050 00044 00044 V n s 44294 47102 82232 R 53153363 565222127 986784836 eD 00005 00005 00005 f computed 00044 00044 00044 f computedf 08870 10007 10001 guessed Q n s 00348 00370 00646 ft3s 12277 13055 22792 gals 91838 97659 170497 galmin 5510298 5859539 10229792 With this number of significant gures chosen to fit the page it would appear that the value off did not change from Ex 65 to Prob 621 Looking at the original spreadsheet we see it went from 0004425 to 0004352 which is practically a negligible change indicating that we are on an almost at part of Fig 610 If there had been no change in f then the volumetric ow rate in this example would be 3 times the value in Ex 65 or 1015 gpm vs the 1025 gpm computed here Again remember the i 10 uncertainty in all such calculations 622 The table is shown below The first three columns are the same as those in Table 66 except for the number of digits shown The bottom 2 rows are in addition to what is shown in Table 66 The right column is made by copying the column to its left inserting 2000 in place of 500 for Q and then using the spreadsheet s numerical solution engine quotGoal Seekquot on Excel spreadsheets to make the AP computedguessed l00 by manipulating the value of D guessed Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 8 Variable First guess Solution Prob 622 Q cfm 500 500 2000 D ft 1 066700631 111911808 Lft 800 800 800 5 inches 000006 000006 000006 p lbmft3 008 008 008 1 cp 0017 0017 0017 V fts 106103 238489 338872 R 743018538 111396028 265572884 eD 0000005 74962E06 44678E06 f computed 000465346 00042 00035 AP calculated 001446184 01000 01000 Allowed AP 01 01000 01000 AP calcAP allow 014461836 10000 10002 D calc m 02033 03411 inches 80041 134294 Here f declines substantially mostly because eD declines due to the larger diameter If there were no change inf then quadrupling the volumetric flow rate would cause the required diameter to double As shown here it increases by a factor of 168 623 This is the same as EX 64 with different numerical values By trial and error one nds V 105 fts R 646 E5 f 00039 Q 1639 gpm ft ft To use Table A3 we compute F gAz 322 2J 200 ft 6440 Z s s Then for horizontal ow 62 3lbm AP 39 3 ft2 lbf 2 ft2 39 39 L z 644 T S200173E173 Ax Ax 5000ft s 322 lbmft 1441n ft 100ft Then entering the table for 8 inch pipe we find for Q 1600 gpm dP dx 165 psi100 ft and for Q 1800 gpm dP dx 208 psi100 ft By linear interpolation 173 psi100 ft corresponds to Q E 1630 gpm 624 See the solution to Prob 623 With the same pressure drop 173 psi1000 ft we enter Table A4 and find that to get 10000 gpm we need an 18 inch pipe 625 Assuming the ow is from the first to the second tanks AP gAz F p Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 9 10E F ml M 144ml 08562 3lb m 11 st ftz 2 ft ft 322 2 20ft 876 644 232 2 S S This negative value of the friction heating indicates the above assumption is incorrect the ow must be from the second tank to the first Then we compute the equivalent pressure gradient for a horizontal pipe taking the ow from right to left 0 85 62 3lb mj AP p 39 39 3 ft2 lbf s2 ft2 psi psi F 23 2 000529 529 Ax Ax 500 ft s 322 lbm ft 1441n ft 1000 ft 100 and read from Fig 613 at V E 1176cs Q E 18 gal min This is only an approximate read because to the need to interpolate the 1176 cs line The result is in the laminar ow region so we could solve directly from Pouisueille39s equation 5 29kf in2 7239 3068ft124 ftz cP 144 in2 03 gal Q 2 0038 172 1000ft 128 100 cp 209e 51bfs ft s m1n 0 68 JL4o Jw 0 39 3 626 R DV 12 5 95105 lbm ft s cp From Fig 610 or by trial and error in Eq 621 for this Reynolds number and friction factor I 1002 cP 67210394 0000065 8 00000653068in 000020 in See Table 62 This new pipe is not as smooth as drawn tubing 8 000006 in but is smoother than any of the other types of pipe shown in that table 627 There are easier and most satisfactory ways for finding them The density is easily measured in simple pyncnometers to much greater accuracy than it could possibly be measured by any kind of ow experiment The viscosity is measured in laminar ow experiments as shown in Ex 62 Furthermore in a ow experiment we would presumably choose as independent variables the choice of uid the pipe diameter and roughness and the velocity We would measure the pressure drop and compute the friction factor If we were at a high Reynolds number ie at the right side of Fig 610 we would see that the friction factor was independent of the Reynolds number so the measurements would be independent of Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 10 the viscosity The friction factor would also be independent of the density which is part of the Reynolds number although the calculation of the friction factor from the observed pressure drop would require us to know the density in advance We may restate this by saying that for really high Reynolds numbers Figure 610 really only relates three variables f D and 8 628 For EX 65 the equations to be combined are Eq 6Q and 621 Eliminatingf between these we have 12 V 2g A2 51b 6 13 40001375 120000E 10 J D DVp The only unknown in this equation is thich appears on both sides to the 16 power on the right I doubt that this has an analytical solution it would be tolerably easy numerically but the procedure shown in EX 65 is certainly easier For EX 66 we combine Eq 6X and 621 g 106 1 Ax V2 APpm 40001375 1 20000 D DVp D 2 and then replace both V s with Q 7239 4 D2 The resulting equation gives AP as a function of D and variables specified in the problem Again I doubt that this has an analytical solution it would be tolerably easy numerically but the procedure shown in EX 66 is certainly easier 629 If you assign this problem you may want to give some hints otherwise the students ounder with it First you should point out that Fig 612 is a loglog plot but that it is not plotted on normal loglog paper It has a scale ratio of about 15 vertically to 1 horizontally If one replotted it on normal loglog paper one would see that it has three families of parallel straight lines the laminar ow lines with slope 1 the turbulent ow lines with slope about 055 and the transition and zero viscosity lines with slope 05 a for the zero viscosity boundary we estimate dD 000182 00009 The from Fig 610 we readf 00048 Then for a velocity of 1 fts and sg 100 we compute lbm ftz 4000486231 lbfsz z 39E23p72 2806 3221bmft144in2 12 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 11 7psi 077 psi ft 1000 ft 00007 From Table A2 we see that 1 fts in a 2 in pipe corresponds to 1045 gpm so we plot this pressure gradient at that volumetric ow rate We draw a line of proper slope through it completes the zero viscosity boundary b We can get both the laminar ow region and the transition region from one calculation by choosing our point as a Reynolds number of 2000 and a viscosity of 40 cs We compute 2 V w 200mm cs10810 5 5022 Then from Table A2 D 2067ft scs S 12 Q 1045 502 524 gpm Then for sg 1 u 40 cP and ft AP 8 850240cP lbm lbfsz z 2 2 67210 2 Ax r0 2067ft ftscP 322lbmft 144m 24 E3L3 p51 forsgl00 ft lOOOft 00313 This point lies on both the 40 c8 laminar curve and on the transition curve We draw a line with slope 1 through it for the laminar region and a line with slope 05 through it for the laminarturbulent transition Then we complete the laminar region by drawing lines parallel to the first line for various viscosities In the laminar region at constant ow rate the pressure drop is proportional to the viscosity so for example the 20 cs line is parallel to the 40 cs line but shifted to the left by a factor of two and the 80 cs line is parallel to the 40 cs line but shifted to the right by a factor of two c For the turbulent region we again need to calculate one point We must guess a value of the kinematic viscosity for which the line will fall between the transition and zero viscosity lines From Fig 612 it seems clear that the 1 cS line is likely to meet that requirement For 1 fts Q 1045 gpm Then for sg 100 2067ft 1ft R 12L2159104 f00073 1cS10810395 s cS 4 00073 132 62 3lb m quot 39squot 3 lbfsz ftz Ax 22067ft 3221bmft 144m2 39 12114 39 000114p Sl39 pS1 ft 1000ft Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 12 We then draw a line with slope 055 through that point We could compute similar lines for other kinematic viscosities if we wished d is taken care of in b above The problem does not ask the students to compute the lower section which corrects for differences in density You might bring that up in class discussion The lines there would have slope 1 on an ordinary piece of loglog paper The values calculated above are all for sg 100 which corresponds to the bottom ofthe gure sg 100 If for example the true s g is 05 then that bottom section shows that the pressure gradient will be 05 times the value for sg 100 This is obvious for turbulent ow where we set the density to 623 lbmft3 For laminar ow it is a bit more subtle If we rewrote Poisueille39s equation replacing u with pv we would see that the pressure gradient is proportional to the kinematic viscosity times the density In making up the laminar part of the plot we used the kinematic viscosity with an assumed s g 100 Thus in this formulation the laminar pressure gradient is also proportional to the speci c gravity Here I have used 2000 as the transition Reynolds number The authors of Fig 162 and the others in that series used 1600 see Prob 631 630 Here lbm 001801 cSt623 v lbm 150cSt P 0075 3 CF ft Entering the chart at 100 gpm and reading left to this kinematic viscosity interpolating between 10 and 20 in the turbulent region and then reading upward to the head loss scale on the top of the chart we find AP head loss 39 J Ax 40 m 0040 ft 1000 f t 1000ft Then the pressure drop is 2 2 amp00400075H I1322 1t 2 Ax ft s 3221bmft1441n 5psi 00208 psi ft 1000ft It is most unlikely that anyone would use Fig 612 to solve this type of problem But this problem shows that one could If one solves this by standard friction factor methods one finds R 6872f 00088 and dPdx 0021 psi1000 ft 208107 6 31 The 20 cSt line enters the transition region at 31 gpm and exits at 87 gpm The corresponding Reynolds numbers are Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 13 3068ft 3lgpm 12 230 ft s ansmm 2 1595 m 1600 20 cSt 108105 s cSt and 87 Rm above gpm 4477 m 4500 31gpm 632 The lowest velocity value for 12 inch pipe 03 gpm has a calculated Reynolds number of 1346 taking into account that the viscosity at 60 F is 113 ofthat at 68 F For this value we can compute the pressure gradients for laminar and turbulent equations nding 0062 and 0073 psi100 ft The value in the table is 0061 showing that this is a laminar value One can also see from a plot of APAx vs Q that there is break in the curve for 12 inch pipe between this value and all the others indicating that it is laminar A few of the other smallest velocity values are also laminar All the other values are for turbulent ow 633 One may compute that the friction factor falls from 00036 to 00031 over the range of values which corresponds to Reynolds numbers from 056 million to 278 million These match Fig 610 reasonably well for an gD of 000007 Comparing them to the vales from Eq 621 we see that for the lowest ow the ratio ofthe value n Tab A3 is 104 of that from Eq 621 For increasing ow rates this ratio diminished For velocities greater than 958 fts it is 100 i 1 As a further comparison I ran the friction factors from Eq 621 using gD 0 finding that the values in table A3 are 117 to 131 of the smooth tube values Clearly even at this size pipe we do not have smooth tube behavior 634 Solving the problem either by Fig 610 or by App A3 we have Ax V2 2 F 4f g Az 981210m 981m 2 D 2 s s Either way the value of the friction heating per pound and of MD are the same so if we assume an equal friction factor we would conclude that the velocities were equal Then for water m2 kg 981 9982 2 N P kP p 52 m3 5 a20979 a432 p51 Ax Ax 100 m kg m N m m 100 ft In App A3 we must interpolate between 425 and 450 gpm finding about 436 gpm compared to the 386 gpm in EX 65 Here the ratio of the velocities is proportional to the Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 14 square root of the ratio of the friction factors This is tolerable agreement The main reason for this is that in turbulent ow the volumetric ow rate is proportional to D5 D 403 I Thus for the two different size pipes M 3 94 in 10228 and 1n EX 65 5 D 11196 Thus our best estimate of Q from Table A3 is Ex 6 5 436 gpm Q 11196 389 gpm which agrees well with the 386 gpm in EX 65 This problem shows that the friction factor is practically the same for gasoline and water in this geometry so that we can solve the problem using data for water if we take the difference in diameters into account 635 Entering Fig 614 at the left at 1000 cfm reading horizontally to the 12 inch line and then vertically to the pressure drop scale we find 02 inches of water per 100 ft so for 1000 ft of pipe the pressure drop is 2 inches of water 636 Drawing a horizontal line on Fig 614 at 100 cubic meters per hour and a vertical line at 1 Pa per meter we see they intersect between the 01 and the 0125 meter diameter lines By visual interpolation the required pipe diameter is about 0115 m 6 37 Drawing a vertical line on Fig 614 from 5 Pam to the 0125 m diameter line and then a horizontal line from there to the right hand of the figure we find a ow rate of 300 cubic meters per hour 3 1000 m1n ft 638aV mm 2 831 From App A1 u 0009cP 7239 6065ft 60 s s E 12 A1 is hard to read but my HPCP 71e shows a value of 0009 at 300 K so this reading is close to right lbm 2 lbm p 0075 00052 ft 29 ft 639065 83130005211311 R12 51m1 36104 80003 f00058 0009 cP 672 10 4 fts cP lbm ft 2 E 400065 00052831j M S z Ax 6065112 3221bm ft 144 in2 12 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 15 39 39 HO 39 HO P 0000177E00049m 2 0491n 2 401 a ft 100ft m b For air at the same pressure gradient and pipe size we draw a horizontal line from 1000 cfm to the 6 inch line and then down vertically to read about 65 inches of water per 100 ft If the friction factors are the same we would expect the pressure gradients to be proportional to the densities Taking the ratio of the densities as the same as the ratio of the molecular weights we estimate 65inHzO 3 045inHzO Ax 100 ft 29 100ft The answers to parts a and b differ by 10 which is within the range of uncertainty of any friction factor calculation The principal reason for the difference is that the kinematic viscosity of hydrogen is Z 7 times that of air so the Reynolds number is 1 7 that of air Ifwe repeat part a for air we find R 26105 andf 00043 Thus this is only an approximate way of estimating the behavior of hydrogen For gases with properties more like those of air it works better 639 a From Fig 61 before Itried Eq 621 f00048 b By Colebrook the solution is a trial and error shown in the following spreadsheet First Guess Solved values fguessed 05 000482514 left side 141421356 143961041 right side 150676463 143909702 ratio 009385763 100035674 I intentionally made a bad first guess but the quotGoal Seekquot routine on excel when asked to make the ratio ofthe left to right sides 100 had no trouble findinng 00048 c By straightforward plug into Eq 660 we find f E 00049 d By straightforward plug into Eq 621 we findf 00048 e From Eq 661 which I found in the source listed and for which that book gives no reference by straightforward plug in one finds f E 00045 All of these give practically the same answer Observe that in the original publications give the equations for the darcyweisbach friction factor 4 fanning friction factor Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 16 AP V2 640 a E 4 p If everything on the right except Vis constant then this should plot as a straight line with slope 2 on that plot b For 5 inches and less the curves are parallel straight lines as best the eye can see To find the slopes I read the 4 inch line as beginning at 15 and ending at 430 Then AP Q l 2 l 2 10 1000 lo 4301 z205 oglAPJ nongJ n g g 5 1 1 This result is sensitive to chart reading but the exponent is very close to 200 c For all the pipe sizes greater than 5 inches the curves are concave downward The curvature seems to increase with increasing pipe size d Those making up the charts concluded that for small diameter pipes the normal ows were on the at part of the constant eD curves on Figure 610 However as D increases with constant 6 one goes to lower and lower values of eD and hence lower curves on Fig 610 These do not atten out until higher values ofthe velocity so the curves correspond to a value of f which decreases slowly with increasing Q giving the curves shown ft3 ft lb 1000 min 1ft2122 0075 11 e V mm 2122 R l32los 10rt2 605 S 0018cP67210quot 4 ftscP 2 Jl 23221ft144003615 f 4 A 1b S 2 m 2 000496 m 40075 2122 Dp 2 ft2 s Substituting this value in Eq 621 and solving we nd eD Z 000050 6 00005 ft 1 From Table 62 we see that this corresponds to the value for quot galvanized ironquot which is Z 3 times the value for commercial steel From the caption for Fig 614 we see that it is for quotgalvanized metal ductsquot etc So this matches Table 62 fairly well 641 From EX 612 we know that for the quotKfactor method the pressure drop due to the ttings is 31 psi independent of pipe diameter For the equivalent length method we must calculate the Reynolds number eD f and the equivalent length for each pipe size For all sizes LD is constant at 1089 The computations are easy on a spreadsheet Considering the 12 inch pipe we nd that 724100 eD 000015f00062 Lequiv 1089 ft and dP Z 31 psi The plot covering the range from 3 to 12 ft is shown below Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 17 50 40 Equivalent length d 30 g Kfactor 2 5 a 20 g m 10 39 0 39 I I I I I 39 2 4 6 8 10 12 14 Pipe diameter inches We see that the equivalent length estimate is higher than the quotKfactorquot estimate for small pipes but that the two values cross at about 1 ft I ran the value for 24 inches on the same spreadsheet program finding dP E 65 psi I don t know which is best other than to follow Lapple39s suggestion in the text Another peculiarity of this comparison is that the estimate by the quotKfactorquot method is unin uenced by a change in uid viscosity as long as the ow is turbulent But the equivalent length method is sensitive to changes in viscosity I reran the spreadsheet that generated the values for the above plot taking the uid viscosity as 1 cP instead of the 50 cP in those examples That raised the Reynolds numbers by a factor of 50 reducing the computed fs by almost a factor of 2 That reduced the estimated deltaP by roughly a factor of 2 going from 237 psi for 3 inch pipe to 174 psi for 12 inch pipe In this case the curve on the above plot ends up nearly horizontal the computed value for 24 inches is 152 psi AP 0466 39 642 From Table A4 we read that V 610 fts and pSI Ax 100ft For the equivalent length method 1002 Axeqmlem 50ft ft 2 30 135 212 ft 12 AP 0466 39 AP J Axe pSl212 ft 0992 psi Ax 1 100 ft For the Kmethod using the values from Table 67 2K 2 074 2 348 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 18 AP 0466 psi Alimemly Ax mm m 50 ft 0233 ps1 1b ft 2 2 623 12n 61 2 AP ZKpV 348 S 1 0870E ttings 2 2 ft in2 1be2 and APtotal 0233 0870 1103 psi 761 kPa We nd very good but not exact agreement between the predictions of the two methods You might ask your students what a check valve is Normally there will be one visible in the vicinity of your building as part of the lawn sprinklers which you can point out 643 Let 1 be the upstream end 2 be the 2 inch pipe as it enters the sudden expansion 3 be the 3 inch pipe as it leaves that expansion and 4 be the far end of the pipe Writing BE from 1 to 4 and multiplying through by p we have V V3 2 The p T term has three parts the rst pipe the second pipe and the expansion We can simply look up the two pipe terms in Table A3 for 100 gpm in each ofthe two pipe sizes multiplied by the lengths nding 759 and 0525 psi For the expansion term PAPFpF D 2067 39 we rst observe that 1 0674 Then we read Fig 616 or use the D 3068 1n 2 equation printed on it to nd that K 0298 E 03 From Table A3 we nd that the velocities before and after the expansion are 956 and 434 fts Then for the expansion 9 562 1b 39 lbf 1 ftz pF AP03623 S S 2018psi ft 2 322lbmft 1441n Finally jz jz 434 956 V2 V2 1b lbf 2 ftz p 4 1 623 S S S 2 0487psi 2 ft 2 3221bmft 1441n So that P4 P1 756 0525 0180487 781psi 2AP 2 644APF L4f 1ltexp Kmmj V L P 2 D p 4fBKeKC Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 19 The expansion and contraction coefficients are 10 and 05 The solution is a trial and error in which one guesses f and then does a suitable trial and error This goes well on a spreadsheet Here 5 0006 and our rst guess off 0004 Then 39 Z 30 144n 322 t in ft lbfs ft g 1b 10 458 623 111 40004 1 05 S ft3 3068 12 This corresponds to R109e6 andf000458 Subsequent trial and error easy on a spreadsheet leads to V449 fts R107e6 andf000458 and Q 230 449 1032 gpm 645 Writing BE from surface to outlet we have Ax V2 K V22 V 2 g AZ Ax 14 K fD Here g 0006 and Ris large sofz 00045 and for a large diameter ratio Koontraction m ft 2 323 20 ft ft m V S loft 242 736 140004 05 S S 3086ft 12 3 Q mo j 2422 556 gpm 02623 ft s s s To be safe we check the Reynolds number nding R 574105 andf 000467 which is close enough to the assumed 00045 to make another iteration unnecessary The three terms in the denominator under the radical have values 1 070 and 05 Each contributes to the answer 646 From table A4 for 500 gpm in a 6 in pipe Ax 100ft AP AP Ax F p Ax p g i 0720 psi ft3 s2 321bm ft 144 in2 Ax Ax pg 100ft 623lbm 32 ft lbf s2 ft2 ft 166 ft 0 0166 00166 ft 100 ft Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 20 647 This is a demonstration which our shop made for me with a clear plastic window to let the students watch the falling level In a lecture room with a sink one can ll the tank with a piece of tygon tubing and let the students watch the results I assign this as a homework problem and ask each student to write herhis answer to part a on the board with herhis name Then we do the calculations then run the test a See the solution to Prob 643 V exit let 21 22 h 20 013 Then see EX 514 from which we have directly At T2 Ax 1K 4 e fD I 2 A 65 4 1n 0016 fz0013 2508 K 05 1 i A 0104in2 D Students argue with me about relative roughness The above value is for galvanized pipe as shown in the problem They use the value for steel pipe getting a much lower value 2min fi j 250 l2 12 ft 2 322 S2 At 24 in l05 4 0013 03641n When I rst did this in class it took about 75 sec for that change in elevation As it was used more and more and the pipe rusted the time climbed slowly to 85 sec Cleaning it with a wire brush got it back to its original value b For a Reynolds number of 2000 we nd Z 2000 l077 10 5 ft m V S 071 022 0364ft S S 12 Ax 2 V21K 4f 3jj 0713 A2 S 491 0039 ft 046 in 0012 m 2g 2322 2 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 21 This would suggest that the transition would not be seen But the transition region is entered at a Reynolds number of about 4000 for which one would calculate Sometimes when I have used the demonstration this has worked well and been easily observed others not The fundamental cussedness of inanimate objects To show this clearly I replaced the pipe in the problem with a threefoot length of 18 inch sch 40 steel pipe for which the calculated elevation for R 2000 is 12 inches This works well one can see several oscillations in the ow rate One can sketch Fig 62 on the board and show that this oscillation represents a horizontal oscillation between the two curves in the transition region 648 By steadystate force balance dP dV PIPZ2ly 21Axry 1 Jy y x dy 2 IdV d Pjijydy V 1y C dx u dx u 2 dP 1 h 2 dP 1 h 2 yJ V0 C l J 2 dx2u 2 dx2u 2 649 This is derived in detail on pages 5154 of Bird et al rst edition and page 5356 ofthe second edition 650 The ow rate is proportional to the third power of the thickness of the leakage 1 h Q 393 1 path so that 4 L 353 1508 hi Q h 00001 in 1508 000015 in True leakage paths are certainly more complex than the uniform annulus assumed here but this description of the leakage path is generally correct Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 22 In that example Itreated the annular space between the stem and the packing as if it were a rectangular slit You might ask bright students how much error is introduced that way I checked by comparing the slit solution to that for an annulus with inside diameter 025 inches and outside diameter 02502 inches using Eq 629 The computed ow for the slit is 10004 that for the annulus 651 Here DO 02502 in The other values are taken from EX 613 D2DZiDZDZ J Ax u 128 0 l 0 39 lnD0Di lbf 100 2 2 02502 02500 mzi025022 025002n2 025022025002 in 11n06cP 128 1n02502 02500 Pft2 14439 2 39 3 c 5 2m 75210 5 20910 lbfs ft s To see the difference between the two solutions one must copy more signi cant gures 7515 8 QEquot613 l0004 Q1111 problem 75188 This problem shows that this is one of a large class of problems which can be greatly simpli ed and thus made much safer from error by replacing some other geometry with a planar geometry off the spreadsheets nding 652 a Here the l 721 and h D0 Di 2 Making those substitutions in Eq 628 produces Eq 630 b Dividing and canceling like terms produces D 2 2 sto 128 2 Q6213 12 D2 D2 D2 D 2 D2 12 0 l 0 lnD0 Di The D in the numerator can be taken as D without much problem c The values ofthis ratio for DoDi11101100110001 are 0954 09951 09995 100004 For smaller values of DODi the spreadsheet produces values with oscillate between 132 and 159109 almost certainly because it does not carry enough signi cant digits to handle the 00 limit of the last term in the denominator I tried to reduce this ratio algebraically with no success and then tried by introducing D0 Q A and simplifying with and without dropping higherorder terms in A again Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 23 without much success If one of you will work that out satisfactorily and send me a copy Iwill be grateful 653 This is ow in a slit for which Q J1m3 dx l2 lbf 001 1 1 cPfts 3221bmft 24in10 3jn3 74 2 2m 12 0018 cP 67210 lbm lbfs 12m 222 10 6E 628 4043 V g 001332 000402 avg s s s s this is a very lowleakage window A gap of 000 inches makes a very good seal The ow is laminar 654 This is really the same as the last problem after one sees that the ratio of the vessel radius to the length of the ow path is large enough that one may treat it as a linear problem rather than a cylindrical one lbf APJI 1M3 min 2 1 1 10n 105in3 Q Ax 12 lin 1002cP 12 2091075 lbf ftscP 3 77 3 74 isga1 79E l2510 4510 09310 3510 s hr mm s This low ow rate shows why we can use mating surfaces which are practically smooth as seals All real gaskets are the equivalent of this the ow rate through them is not zero it is simply too small to detect 655 Ar ea Wetted per1meter l 7239 D2 D l a HR L n39D D 4 2 7 ii 2 2 4 D b HR 1 wh1ch 1s the same as for a circle 4 72D 2 D2 D c HR 4 Z wh1ch 1s also the same as for a c1rcle Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 24 E 2 d HR4D2 DIZZMZDZDI 7rD2Dl 4D2D1 4 6 56 As in EX 65 this is a trial and error We first assume that a square duct with 8 inch sides will be used 2 HR 8 2in 4HR8in 481n ff 500 39 ft ft 000006 V2A11251875 e A 8ft122 mm s D 8m 8 ft lbm 1875 J0080 3J 12 s ft 0017cP 672 10 lbm ft s cP 0000075 0 R 87510 fz000447 ft 2 1b 40004471875 008 I1800 2 2 AP S 1be ft Zt j 3221bmft 144m2 12 0065 psi Then we trialanderror n D to nd the value which makes AP 01 psi nding D 731 in V 224 fts R 957104 andf 000438 31416 Dsnare 31416 For equal crosssectional area A Dsquwe 4 Dim 3L 4 0886 circle Perimeter 4D 4 4 I sgaxeswJJ l128 Perzmeter 7TB 7239 4 circle circle This is also the ratio of weights for equal wall thickness In this problem the computed 731 square diameter is T 0914 of the corresponding c1rcular duct somewhat more than the equal area value of 0886 This shows that the friction effect of a square duct is more than that of a circular one and we need a little more than equal area 657 The velocity is proportional to 1square root of the friction factor so fuse in ac deglg Ml Kim 428fts 1049 festimatedinEx Iactual design 389 S 615 and Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 25 fmdm m 00024 1049 000252 design From Eq 621 we nd that this corresponds to a relative roughness of 207e5 which corresponds to an absolute roughness of 00014 ft compared to the 0001 estimated in the example Az 2 HR Az 658 a Combining the equations we have V C HR g Ax f Ax These are the same if C E f 2 HR n 2 0012 2 322 sz ft b C ga a 16 g 1 122 f n HR f 1709 ft 00024 s 659 Here the velocity will be much too high for us to use any of the convenient methods so As a rst trial for 8 D 00018in 3068in 00006and a large Reynolds number try f 00040 Then 39 Z 2 1000lbi 1441n 323921bm 322 10 ft 1n ft2 lbf s2 s ft V 303 rst guess S 10 1 40 004 3068 12 The corresponding R 7198e6 and from Eq 62lf 000453 Then using the search engine on my spreadsheet we find the velocity at which the guessed and calculated values off are equal nding V 2957 fts R 7102e6 and from Eq 62lf 000453 AP AVZ dW AV2 660 a gAz F quotf Here is zero so p 2 dm dVV APum AP d mf 2LEF 7gAZ39 AppumppFAPlZgAZ172 From Appendix A 4 for 150 gpm in a 3 inch pipe Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 26 224 psi 100 ft 2300 ft 515 psi pF 1b ft lbf 2 2 APum 515psi20psi6233227230ft 2 p 1 ft 2 32lbmft 1441n 51520 995 171 psi 118MPa AP b Po dWquotfn gtI u 39quot Qp4P 9 dm p p p lbf 1 ft3 14439 2 h 39 1717150 g2 1 pmm 150 hp111kW 1n m1n 39 748 gal ftz 39 33 000 ft lbf According to the sign convention in this book this is power entering the system in this case by driving the pump 661 V LgAZC As a rst trial for D 00018in 5047 in 000036and a large Reynolds number try f 00040 Then 23223210 ft S Kim guess 1 40004 L S 5047 12 The corresponding R 189e6 and from Eq 621f 0004094 Then using the search engine on my spreadsheet we find the velocity at which the guessed and calculated values off are equal nding V 4814 fts R 188e6 and from Eq 621f 0004094 662 Let 1 be the air inlet and 2 the eXit from the stack Then applying BE from 1 to 2 assuming that 1 is far enough from the furnace that the velocity is negligible we nd P P V V L L J F K 4 p gZz Zl 2 2 fumace f Ck P P The rst term is J L p Lgz2 21 so that p pstack pair Viz L J Z Z 1 1K 4 g 2 1 E p J 2 fumace D nk stack Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 27 12 p 2gzz 21 m J psmk Here L J 810R 2 L 1 0589 1Kfumace 4f pstack 28 528 R stack L 100 ft Knme 30 and 20 D stack 5ft To solve the problem we need an estimate off We assume that the stack is made of concrete a common choice so from table 62 we estimate that 8 002 in and 8 D m 000033 This leads to a first guess Off rst guess 0005 Then 12 ft 2 322 2 100 ft 0589 ngms S 294 1340005L0ft S 5 ft stack This corresponds to R 13 107 andf 00040 We estimated the viscosity ofthe stack gas as 0025 cP Z the value for air at 350 F from Fig A l and the gas density from the ideal gas law Then with our spreadsheet s numerical engine we seek the guess off which makes the guessed and calculated values equal nding V 297 fts and R 13107 andf 00040 Students are generally uncomfortable with the idea that the pressure drop though the furnace is estimated by a constant times the kinetic energy based on the ue gas temperature For the highestquality estimate we would follow the ow through the furnace taking into account the changes in density viscosity and velocity from point to point in the furnace That is now possible with CFD programs But the simple estimating method here is widely used and reasonably reliable AP 663 Assume the ow is from left to right gAz F p lbf 30 2 2 3221b ft 144 ft ft F in 322 2 30ft 2318 966 1352 2 60 m lbfs ft s s 3 The minus sign indicates that the assumed flow direction is wrong ow is from right to left The equivalent dPdx is 2 lbm 1352 60 2 2 Cl P i Si g lbf S 00175E dx equivalent p Ax 1000ft 39322 lbm ft 39 144ml equival ent If we enter Figure 6 13 at 175 psi1000 ft read diagonally up to SG 096 then vertically to 100 cSt and horizontally we find Q E 57 gpm We also see that this is in the Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 28 laminar ow region which we could solve by Poisueille39s equation first computing a viscosity of963 cP nding lbf 175 in2 7239 3068ft 12 ft2 cp 144 in2 ft3 gal Q 2 0l26 567 1000ft 128 963cp 209e 5 lbf s ft s m1n 664 a We must specify the maximum ow rate in this case 200 gpm b For all conditions AP 61 AP F ampAZ pg pump gdm pg g The highest required head will correspond to the highest elevation and highest pressure in vessel 2 and the lowest elevation and pressure in vessel 1 The required head will be highest for the lowest specific gravity 080 and for the highest kinematic viscosity 5 cSt The quotequivalent lengthsquot are 6 30 4 l3 l 340 572 so that 3086 Leffective 672 572LTJ 774 ft On Fig 612 we enter at the right at 200 m read horizontally to the 5 cSt line and then loss in headgj vert1cally to the top findlng ft per 1000 ft m 05 Z from wh1ch F 105 ft J 773ft J81ft g max 1000 ft 81 8lbf 2 2 AP 11 s 322lbr1ftl44n 12721 81 Pg pump dams t 32 1be ft 2ll10681398 ft l2lm The entrance and exit losses which we ignored are E 065 psi Z 0 So we would specify a pump with a design point of 200 gpm and 398 ft ofhead This is a high head for a centrifugal pump at this volumetric ow rate we would probably have to specify some other type of pump 665 The maximum corresponds to the highest pressure and elevation in vessel 2 and the lowest in vessel 1 and to the lowest values of the specific gravity and viscosity The minimum corresponds to the opposite of those conditions Here the open globe valve adds 340 x 3068 l2 87 ft so the effective length ofthe pipe is 860 ft In both parts Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 29 of the problem we have a trialanderror for the velocity The values shown are those at the end of that trial and error For both parts 2 V F 4f 1 D 2 For the maximum Z 2 81 8ps1 322lbmftl441n 322ft106 s 0 8 62 3 lbm lbf 52 z 2 Vmax 860ft 173952 1 4000488 S 3068 ft 12 230 ft QM 175 402gpm ft s s For the minimum 47 20 ZLQ 322 l44 322100 43 Km 2 085623 86 2103 1400057 S 3068 ft 12 230 ft QM 103 238gpm ft s s 666 We begin by assuming that the pressure atA will not be low enough to cause boiling Then we solve for the velocity applying BE with friction from water level to outlet Here the adjusted length of the pipe is 10020 ft Ladjusted 60 ft 2 20 933 ft By BE with no pressure difference 12 2 2 wehave gAzV 2 F Ke4f jV 2 orV LAQL 2 D 2 1 1 K 4 f5 2 g 000018 and Ris large sof 00034 and Ke l2 so ft 2 32573910 ft ft 246 gpm 933 146 QT 1 05 400034 39 S S 1012 At this point we check finding R ll106 and by numerical solution find that f 00036 and V 143 fts 444 ms and Q 3526 gpm Then using those values ft 146 3592 gpm s Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 30 2 ug2a Zl V F VKe 4f J p 2 2 D V Lu P Pl plge zl71Ke 3 jz 146 147 62391 322E 20ft S S 4166 1 05 4000364 1012 lbf s2 ft2 3221bm ft144in2 This siphon would probably work satisfactorily although the absolute pressure is low enough that dissolved air might come out of solution and cause problems 147 117 30 psia 667 This was actually done in the 1950s At that time high quality plastic pipe had not developed to the extent it has now so the pipe was made of aluminum One can show that in terms of strengthweight high quality aluminum alloys are still the best which is why we make airplanes of them If you walk down the trail from the North Rim to the river you can see the occasional valve boxes which allow the pipe which is buried by the trail to be shut down and drained for maintenance and repair when needed Whig arm Ax p menm Ax Ax friction P AP AZ lbm ft 3000 ft lbf s2 ft2 pg 623 2 322 2 2 Ax Mon Ax ft s 14 5280 ft 322 lbm ft 144 in 39 176 39 00176p S1 A From Table A3 we see that this requires ft 100 ft pipe larger than 6 inches and smaller than 8 inches I don t know if they had a special pipe made or used an 8 inch pipe lbm Pm pgAz 623 322S f7000 2500ft 1be2 ftz 3221bmft 144m2 1948 psi 134 MPa PD thickness 2 for an 8 inch diameter pipe and a fairly conservative value of 039 1947 psi 8 in 2 10 000 psi 039 10 000 psi we compute thickness 078 in 20 cm 668 This is EX 615 of the rst edition pages 193195 where a trial and error solution based on Table A3 is show through four trials The nal ow rates are A 285 gpm B 60 gpm C 225 gpm Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 31 The following spreadsheet solution is longer because it requires nested numerical solutions but is probably more illustrative The actual spreadsheet carries and displays more signi cant gures than are shown in this abbreviated table P P We begin by de ning a 1 2 z1 22 J pg Then we write BE for each of the three pipe segments and solve for V V V A 105 4fLD B 105 4fLD and V C 1 054fLD individual pipe sections Then we guess a value of a solve for the three ows and compute the algebraic sum of the ows into point 2 The correct choice of amakes that algebraic sum zero One need not know the individual values of the two components of a If the pipes were exible and we raised or lowered point 2 the ows would be unchanged because the two parts of awould change but their sum would not 0 5 where the values off L andD are those for the The second column of the following table is the solution for ac 10 ft For each of the three pipe sections we have a trial and error for the velocity done using the spreadsheet s numerical engine In this case the guessed value is Q and the check value is the ratio of the Vbased on that guess to the one computed from BE as shown above We see that for a 115 ft the algebraic sum ofthe ows into point 2 1 16 gpm Z 0 and the three ows are 304 gpm 64 gpm and 241 gpm These are somewhat higher than those shown in the example in the rst edition but within the accuracy of interpolations in Table A3 This is a somewhat clumsy way to solve this class of problems Real solutions would use nested DO loops in FORTRAN or the equivalent P2rho gz2 10 20 12 115 guessed ft Section A L ft 2000 2000 2000 2000 D ft 05054 05054 05054 05054 eD 00003 00003 00003 00003 Q guessed galmin 2811204 4134696 3112119 3039138 V guessed fts 31236 45941 34579 33768 Re 146583 215593 162273 158468 f equation 621 00040 00037 00039 00039 V BE fts 31235 45941 34579 33768 V BEVguessed 10000 10000 10000 10000 Section B L ft 2000 2000 2000 2000 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 32 D ft 02557 02557 02557 02557 eD 00006 00006 00006 00006 Q guessed galmin 665264 449489 626858 636623 V guessed fts 28925 19543 27255 27679 Re 68663 46392 64699 65707 f equation 621 00047 00052 00048 00048 V BE fts 28924 19543 27255 27679 V BEVguessed 10000 10000 10000 10000 Section C L ft 1000 1000 1000 1000 Dft 03355 03355 03355 03355 eD 00004 00004 00004 00004 Q guessed galmin 2483668 1984037 2390666 2414192 V guessed fts 62719 50102 60370 60964 Re 195377 156074 188061 189912 f equation 621 00037 00039 00038 00038 V BE fts 62719 50102 60370 60964 V BEVguessed 10000 10000 10000 10000 Sum of ows to point 337729 1701170 94594 1 1677 2 gpm 669 See the solution to Prob 668 Here the ow in section B is from the reservoir 0 5 toward point 2 In the above spreadsheet we must write VB so 1054fL D that the term in brackets will be positive Then in the sum of ows into point 2 we must take ow B as positive rather than negative With these changes the trialanderror on a is the same type as in Prob 668 with solution a 75 ft The three Q39s are 2394 204 and 2596 gpm The algebraic sum ofthe ows into point 2 017 gpm Z zero 6 70 Writing BE from 1 to 2 we see that there is no change in elevation so The velocity out the individual tubes is proportional to the square root of the pressure in the manifold below them Thus if AP is positive the far tube with squirt the highest while if it is negative the near tube will squirt higher For multiple tubes one rewrites these equations for successive pairs oftubes In all cases the velocity is declining in the ow direction because each tube bleeds off some ofthe ow So for all cases AV2 is negative Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 33 a If F 0 then AP is positive and the water will squirt highest out the farthest tube b If the absolute value of F is greater than that of AV2 2 then AP is negative and the water will squirt highest out the nearest upstream tube One can build the device to the dimensions show in the article and see that these predictions are observed c First maintain your humility Second if the experienced workers in some facility you are new at are goading you into betting on something don t bet very much 671 We guessedf 00042 For 200 gpm in a 324 in diameter pipe R 96 500 eD 00005 and from Eq 62lf 00051 If we take the ratio ofthis value to 00042 to the 16 power we nd that the economic diameter would be 1035 times the value in that example For an assumedf 001 it would be l 155 times the value in that example The point of this problem is that because of the 16 power the computed diameter is quite insensitive to modest changes inf Annual C12 Ax C12 Ax 672 L JAmB r ABAx D2p r cost D2 4 EU 4 4 dannual cost ZABAerpD ZCIZrAx 0 dB 4 EU 4 D C5 7r AB 4 p This is for low voltage transmission where the resistive losses are the only serious factor In modern high voltage transmission lines corona losses are apparently more signi cant so this formula no longer applies It is however an interesting historical example of the apparently first use of simple economics to find an optimum Inside our houses and other buildings we set the wire diameter for safety reasons We want the maximum AT due to resistive heating to be low enough to pose no fire hazard total annualJ C 2 fo4 722 16 4711 cost 673 a P 2 D5CCPPDAxf p Substituting this value for f in the first term and simplifying we have total annual r 8 Ax 4 2 Powccpp DAx cost p Taking the derivative with respect to D and setting it equal to zero we find Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 34 39 2 FOWCOPPN p PC 791732 4 2 7 Dim 2 WK 72 but 2 Q2 Substituting this and taking the fth p CC PP p root produces Eq 663 b For 200 gpm 2000 cP sg l and the economic values on Figure 623 004 200 gal 2 5 3272392000 cP D kWh mm mquot 004 2 yr in ft 15 8760 hr ft3 2 kWs 2091051be ft yr 748 gal 7376 ft lbf cP ftz 12 in 0582ft 698in From Fig 623 for 200 gpm and 2000 cS we interpolate between the 6 in and 8 in lines nding E 65 inches which is Z the same as calculated here 2 15 04 c For constant v1scos1ty Eq 663 becomes Decon 0c Q Q and Km oc QAWZ Q1 which is the slope on Fig 623 D Q d The density does not appear in Eq 663 674 a Based on Table 68 we would conclude that the velocity should be about 40 fts This is not quite right because it is for pipes with LD much larger than 100 Here we will see that LD is of the order of 25 so that this is a short pipe for which the entrance loss is signi cantly greater than the 4fLD term However 40 fts is still a plausible estimate 4083mi22000ft 5280ft 2 hr 11 ft3 b Q 15810 24hr mi 60 min m1n 3 ft l5810 D i2 i g 9158ft V 7 4060 m1n AP V2 L V2 L c 2 F 4f KejV22 APp 1Ke 4f j p 2 D 2 D Here R 2109 and gDm 161078 for steel pipe so that fz00015 and 4f 400015WJ 017 D 9158ft Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 35 ft 2 lbm 40 lbf s2 ftz AP0075 3 S 105017 2 00022psi ft 2 322 lbmft l441n r ft3 1b 1b d rgt15810 0075 I11185101 l1 m1n ft min For an isothermal compressor See Prob 558 dW RT P 9871b Bu110R 5280R l4 7 0 022 Bt nf1n1 m0 1 39 39 0053 u dm M P1 29 lbm 147 lbm lbmole Po 1185 1010 1003 Bull hp hr K60 m1 dm m1n lbm 2545 Btu hr l49107hp 11 100 MW e These values are all wildly impractical How would you build a pipe with that diameter The power requirement is roughly that needed for a population of 5 million people including residential commercial and industrial uses The only practical solution is to prevent emissions at the source 675 Based on economic velocity calculations See Fig 623 or Table 68 a velocity of about 7 fts would be used This may be a bit low for this large a pipe because the friction factor will be lower than that in typical industrial pipes but as shown by the solution to Prob 671 this makes little difference Then the pipe diameter would be ft3 4107435010 4Q yr D V 711 365243600s S 50l ft For this diameter and velocity R 329108 and eD Z 3106 From Eq 621 we compute that f 0002 although for this large a pipe that equation may not be very accurate The pressure drop would be L V2 Ap 4 pr 2 Z lbm 10005280 7 lbfsz 144ml 400020 623 3 2 278ps1 ft 50 2 3221bmft ft and the pumping power ft3 lbf hp min 144 in2 yr PoQAPQO74356104 278 2 1 in 33 OOOft ft 365 24 60min 4 100 106hp m 746 MW Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 36 As an engineering problem this is fairly easy Los Angeles draws about half this amount from the Colorado River and pumps it about a tenth as far in several parallel pipes none this large in diameter If we did this project it would probably involve several parallel pipes and several pumping stations The politics of such large water movements is much more difficult and much more interesting At one point Senator Warren Magnuson of Washington managed to pass a law forbidding the federal government from even studying such a transfer 676 Here we can use the result from EX 618 and the ratio of PP values 1 1 Dem 324m j6 221 in 10 Probably we would choose at 25 inch pipe From App A3 we can compute M5000 584 psi 0 ft And the pump horsepower is 3 39 39 2 Po QAP200 gall 584 f ft hp mm 144 681 hp mm in2 748 gali33000 ftlbf ftz These latter two values are not asked for in the problem statement but are included here to show the full example 677 The product of the velocity and the cube root of the density ftslbmft313 goes from 234 to 169 as we go from top to bottom of the table The largest value is 140 times the smallest This is not quite constant but certainly close The reason for the decline seems to be that as the density falls the product of the density and the economic velocity falls faster than the viscosity falls so the economic Reynolds number falls which causes the corresponding friction factor to increase which raises the economic diameter and lowers the economic velocity The effect is small but real 678 The 324 inches calculated in EX 618 corresponds to avelocity of 778 fts for 200 gpm From Table 68 one would interpolate a velocity ofperhaps 65 fts From Fig 623 one reads a value of 7 fts These are close to the same The disagreement between Fig 6213 and Table 68 results from two factors Fig 621 is for a uid of specific gravity 08 almost exactly 50 lbmft3 and Fig 628 uses a lower value of the pipe roughness than does Table 68 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 37 679 We wish to minimize the sum of the mass of the fuel line the mass of the pump and the mass of the fuel which must be expended to pump that fuel through the line The mass of the pump is probably practically independent of the pipe diameter although its mass depends weakly on the pressure it must develop which is a function of the pipe diameter Leaving it out of consideration we can say that mpipe ppipeDL pipe wall thickness mfuelto pump Volume of fuel to pump APpipe Heating value of pump fuel 77pump 1 For a given allowable pressure the pipe wall thickness is proportional to D and Ma C C so that mpipe CIDZ and mpump fuel D Zs mtotal CiD2 D d SC SC 7 m 2C1D 52 Dminimum combined mass 2 dB D 2C1 61V 18 V VdV 680 a Follow1ng the instructions in the problem u D2 p dx 0 IL 1Csmkessmppmg V dV D2 IO dx carrying out the integration and inserting the limits P leads to Eq 666 The stokes stopping distance appears often in the fine paiticle literature m kg Isz 10 10 m2 2000 3 b xgmkessmppmg J 0 001 k 0061 mm 0002 m 18 180018cp 39 g m s cp This startlingly small value shows that for paiticles this small about the size of air pollution interest the air is very stiff and particles come to rest quickly This is example 85 page 225 of Noel de Nevers quotAir Pollution Control Engineering 2equot McGrawHill 2000 There it is shown that for a particle ofthis size C E 112 so taking the Cunningham correction factor into account raises the computed value by 12 The behavior of paiticles small enough to be of air pollution interest is discussed in more detail in that book c If we return to Eq 663 and separate variables and integrate we find VdV 18 t V 18 Z jdt ln ut Va V Dp 0 V Dip from this we see that V 0 corresponds to infinite time For 1 of V0 kg 5 Dip V 10 my 2 00 3 5 I ln4 ln100 mln1002810 s 18 V 1800180pM 0 mscp While the paiticle theoretically moves forever it loses 99 of its initial velocity in 28 microseconds Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 38 d From example 69 we know that the gravitational settling velocity is 605105 ms so AzV Ar60510 5 2810 5 sl710399m terminal gravity S This is 310395 times the stokes stopping distance for this example and clearly negligible ft lb 00001 in2322 2J100 623 I3n ft 681 V S lbm 2 69610 6 06 181002cP 672 104 144 In S day ftscP One may check that RP Z 5 e6 so the stokes law assumption is safe On the time scale of geology this particles settles rapidly but on a human time scale it practically doesn39t settle 682 As a rst trial we assume Cd 04 Then copying from that example 785 k 4002 m98132j 785 j9982 g3j s 62 3 m V 39 m 184 3L9982k JLEJ04 S m 623 002m 9982E 184E R 2 623 s 2581 F 80010 3Pas 39 From Fig 624 se see that for this particle Reynolds number Cd E 15 Using the approximate equation in the caption of that gure we nd C 24 581 1 01458107 1406 which is within our ability to read Fig 624 ofthe 15 above Then we use our spreadsheet s numerical engine to nd the value of C d in the above equation which makes the assumed and calculated values equal We nd V 0776 ms 255 fts RP 244 and Cd 227 Be cautioned that the approximation formula is only applicable to spheres although the discs curve is very close to the spheres curve and only in the RP range shown 2F 7239 l2 4 pCd 7239 2 V2 633 F D V 4 pCd 2 4 As a rst trial we assume C d 04 Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 39 V m 2011erIn 3221brznft 216 ml Z10 20075 J04 1be 5 ft3 ft 10 165 J s 2 l6l310 4 s R 102 105 From Fig 624 we see that this corresponds to P ft 04 V2 ft 148 Cd 4 05 so that V ns E 148 and R 102105J 092105 second guess S S For this particle Reynolds number Cd m 05 so we accept the second guess g lwo 52803 12 3600 s 2 1613 1041 S 684 a RF 22 105 From Fig 624 we read thathm05 7239 2 V2 F D 4 pC 2 7239 29 2 lbm 1005280ft 2 0075 05 j 4 12 ff 3600 s lbf s2 057 lbf255 N 2 39322 lbm ft 7239 D2pC 61V dV 0 b m mV F 4 V2 dt dx 2 EDZpC dV 1 4 dx V 2 m 5122 C Ax 20 0751b m0 5x60 ft 4 p D 4 12 39 1 13 39 1nK 0161 V0 2 m 2032 lbm mi ft m V V0 eXp 0161 85 125 38 hr s s Stitching on the ball spin gravity and wind all in uence the speed and curvature of the ball39s ight path Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 40 362 0223 m 685 As the ball began its ight Rp 5 2 54 105 From Fig 624 we see 148810 5 m S that this is very close to the particle Reynolds number at which the Cd curve has a sudden change At higher particle Reynolds numbers Cd 01 at lower particle Reynolds numbers Cd 05 If we assume that this transition occurred halfway from the kicker to the goal we can then compute that at that halfway point see preceding problem solution that 7239 2 7T 2 kg D pCDAx 0223 m 12OEJ01135 m V ln 4 0074 V0 2 m 20425 kh and V 362 eXp 0074 3342 S S For the second half of the ball s ight we substitute Cd 05 nding V m m In 0372 and V 334 eXp 0372 230 s s 0 In the first half of its ight the ball lost about 7 of its initial velocity In the second half it lost about 31 of its remaining velocity The result was apparently very dramatic 686 The world record speed for the 100 m dash is Z 10 ms while that for the 100 m freestyle swim is Z 2 ms so the ratio is Z 5 If we assume that the characteristic dimension for both running and swimming is Z l m then we can compute the particle Reynolds numbers for both finding 06106 and 2106 This suggests that for both running and swimming we are to the right of the sudden transition on Fig 624 but see the preceding problem so that Cd 0 l for both This says that we are in the ow regime in which the viscosity plays no role but the drag force is Z proportional to the uid density The density of water is Z 833 times that of air so the difference is mostly due to the difference in densities The projected areas are quite different because the swimmer lies at in the water while the runner stands mostly upright If we assume the sprinter has four times the projected area of the swimmer and assume that the muscular power output is the same for the sprinter and the swimmer and all goes to overcome uid resistance then we would expect Po AAPV cszpV3 to be the same for both Surprisingly the values are quite similar kg In 3 P0 12 3 10 sEinter 4 m s 0 60 Po kg m 3 I swimmer 100 3 2 m s This may be a coincidence I think that the competitive swimmer is mostly expending muscular energy to overcome water s uid mechanic resistance while the sprinter while Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 41 sensitive to aerodynamic resistance is mostly expending energy to overcome the internal resistance of herhis own muscles and joints If readers have comments on this I would be interested to hear them This was actually tested in Aug 2003 quotSwimmers take slimy dive for an experiment at Uquot Minneapolis Star Tribune Aug 19 2003 Professor Ed Cussler ofthe U ofM had swimmers timed for 25 yards in water and in a swimming pools whose viscosity had been doubled by dissolving guar gum in it The swimmers had to modify their position holding their heads out of the water because of the gum The swimming speeds were the same within experimental error 637 Here we cannot use Eq 658 because it is speci c for a sphere We derive its dV V2 2 equivalentforageneral falling body m 0mg CdApm4 V amp dt 2 CdApajr dV CA 1 VV t V V V g4pV2 zizdt 1n 2g2tLlnLJ dt 2m K V 2V V V V 2g V V dV dV VdV VdV l VZ V2 V dx J ml Lj dt dx CdApV Vt V V V 2 0 V 2m V2 W x4ln 2 2 g 2 V 2020 lbf 322 lbm ft ft VP 2 mm 071ft 0075 ft 384 V V V 199V t ln J S ln 3l6s 2g KV 2322 2 0011 S ft 2 384 V V j 1 4ln 2 2 Sln 8954 2g K V 2322 1 001 39 2 s b Repeating the calculation withA 6 ft2 and Cd 15 leads to 107 fts 88 s and 696 ft 688 Venom ngx 23223 10ft 254it 773E S S S Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 42 2mg CdApair ft 21501bm322 2J S 2 1b 133 ftz 124 m2 m 1 10075 See the preceding problem K 2mg 2 Cth par 13254 S D iA 311133 ft2 13 ft 40 m 7239 7239 A 689 a lb 705280 ft 2 2 65ft20075 n 03 2 ft 3600 s lbf s F Amed 1101bf 2 2 3221bmft 705280 2 hp s b PoFV1101bf 3600 s 550ft1bf 206 hp 154 kW This is not asked for in the problem but can be used for discussion If this auto gets 30 mi gal at 70 mph then the fuel consumption is fuel 39 L Jziyw mgzmkm consumption 30 m1 hr gal hr fuel ener 1b Bt Bt The energy ow in the fuel is gyJ m 14 m 19 000 11 266 000 11 ow hr lbm hr If we assume a 30 thermal efficiency for the engine then Btu hp hr Poen me u 03 266 000 314 hp 234 kW 3 hr 2545 Btu and roughly 2 3 of the power output of the engine is going to overcome air resistance 690 If we guessed Stokes law and did the calculation we would find a velocity much higher than the correct value Then on evaluating the Reynolds number we would have a higher value than the real value so we would be certain to conclude we were outside the Stokes law range Thus the procedure shown in those examples is conservative it can never lead to an incorrect answer if used as shown 691 In those examples the calculated terminal velocities were 210394 and 621 fts For Ex 619 for a 1 micron particle with sg 2 setting in air we read 1 z 24 10 4 ft s The difference between the two is partly due to the Cunningham correction factor see p222 ofNoel de Nevers quotAir Pollution Control Engineering 2equot McGrawHill 2000 If we applied it to the result of Example 619 we would have found 1 z 22 10 4 ft s The rest of the difference is inaccuracy in making up the chart For Ex 620 we must interpolate between the sg 5 and 10 lines and must extrapolate offthe right side of Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 43 the gure to 20000 microns Doing so we nd V m 6 ft s which matches the result in EX 620 Fig 626 is really useful students should become familiar with it 692 0001 in 254 microns From Fig 626 we read directly a velocity of 006 fts By direct substitution in Stokes law one finds a velocity of 0064 fts and a Reynolds number of 003 Droplets this small occur in fog and clouds which settle very slowly Most drops which we call rain have diameters at least ten times as large Droplets this size are spherical while larger ones are deformed from the spherical shape 2 V CdApV d 2m 693 x dt dt 2m CDAp V 2 V 20027 1b 14439 2 100 Ax m 1n 1 m 1 ln 145ft442m Car1P V1 01 J05 in26231b mj 1000 4 ft3 My firearms consultant Ray Cayias says that pistols and submachine gun bullets are 800 to 1000 fts so this velocity is plausible for the submachine guns normally involved in James Bond movies Ri es and hunting guns fire 25003000 fts He also tells me that if you can see a fish in the water and you have a ri e you can kill it with a bullet You must aim below where you see the fish to correct for the refraction of the image at the airwater interface ft V V 2700 0 694 a For zero air resistance V V gt t 0 S 845 a 322 s ft f 322 t Si 2 zzoj thzoj 2700 322tdt02700 t t 0 0 S 2 3222 ft 2 Z 2700 84s 2 5 84s2 113 000ft S With no air resistance it would take just as long to return to earth and it would return with its initial velocity b With air resistance and an assumed constant drag coefficient 7239 sz V2 2 dV D dt abV t g 4 mp2 7 where a and b are constants introduced to reduce our bookkeeping Then Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 44 dV dV 2 dt39 bdt this is a standard form I can look up in my a bV g V2 V V dV b b integral tables nding J a tan 1VJ bt to or V V2 a a VB b l b b tan 1 tan 1 VFJ for V 0 this becomes ab a a tfortop of 39tanTILlOJZJ trajectory ab a The unknown Cd is buried in b I doubt that there is an analytical solution of the above equation for a known I But numerically it is quite easy First we guess that Cd 06 Then a 322 fts2 and 7239 2C 7239 03ft 2 06 lbm 000051 b D p 0075 3 4 m 4 12 002141bm ft ft 1 b and tfmopof tan lV0JJ 776stan 11080ll5s trajectory ab a We then use the spreadsheet s numerical engine to nd the value of Cd which makes this value 18 s nding Cd 0225 We may check this by calculating the height using this constant value of Cd We must use t Z 20 0mm 0 ZVangt and Euler39s rst method on a spreadsheet Thus for the rst 01 second we compute that ft V315 V0 a bV2At2700 14420125557 S and Az05 V0 V015At 262 Continuing this on a spreadsheet switching to l s intervals after 2 s we nd that V becomes zero between 16 and 17 s at an elevation of 8880 ft which is an excellent check on the reported values The initial particle Reynolds number is 42105 which is close to the transition on Fig 624 so without any data we would assume the drag coef cient would start at E 01 and change to E 05 during the ight Thus the 0225 value is in reasonable accord with what we would estimate from Fig 624 For going down we cannot use this high a value of the drag coef cient because there is no supersonic ow As a rst approximation we assume that the velocity at ground level represents a terminal velocity and solve for 2 7T 2 Vtexminal m DC g 4 up 2 ft 00211bm322 28 Ca 8 mg 2 0408 S Z Z 2 quotD p mm 4 0 0075 1b13n300 12 ft s Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 45 We can test this value by repeating the above numerical integration changing the sign on the drag term Using 1 s intervals we nd that after 31 s the velocity is 302 fts and the bullet has fallen 7500 ft This is a reasonable but not perfect check on the observed 300 fts and 9000 ft in 31s For the downward trip the particle Reynolds number begins at zero and ends at 46500 for which we would read a drag coefficient of 045 from Fig 624 Thus we could have calculated this part fairly well from Fig 624 alone 695 a The sketch is shown at the right In the left figure the ow is around a single particle and extends to in nity in all directions In the right figure there is a regular array of particles so that the local 0 0 ow between them must go faster than that around a single particle and the ow 0 O O O leaving one particle must turn to miss the particle behind it These two effects cause the net velocity of uid relative to particles to be smaller b For the particle by itself we look up the result on Fig 625 finding 00015 fts The term on the right in Eq 665 is 1 cquot 1 044 65 0093 so that the expected 0001520093 0000142 s s setting velocity is V hindered settling Solutions Fluid Mechanics for Chemical Engineers Third Edition Chapter 6 page 46

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