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# Conservation Prin in BME BMED 2210

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This 0 page Class Notes was uploaded by Juliet Ryan on Monday November 2, 2015. The Class Notes belongs to BMED 2210 at Georgia Institute of Technology - Main Campus taught by Robert Lee in Fall. Since its upload, it has received 99 views. For similar materials see /class/233986/bmed-2210-georgia-institute-of-technology-main-campus in Biomedical Engineering at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

PART 1 INTRODUCTION 1 Chapter 1 Dimensions Units and Their Conversion Problem 1 1 Convert the following quantities to the ones designated a 42 ftZhr to cm2s b 25 psig to psia c 100 Btu to hphr Solution 420 ft2 10 2104 2 1hr a cn 108 cm2s hr 32808ft 10m 3600s 100 Bt 393 10394 h h b u 393 X 102 hp hr l Btu 8001bfl321741bmft 1kg 1m IN c 2 2 356 N 1bfs l2201bm32808ft1ngme Problem 12 cal Btu Convert the 1dea1 gas constant R 1987 gm01K to 1b m01oR Solution 1987cal lBtu 454gm01 1K 7 1 98 Btu gmolK252cal llbmol 18 R 39 1b m 1 R Problem 13 Mass ow through a sonic nozzle is a function of gas pressure and temperature For a given pressure p and temperature T mass ow rate through the nozzle is given by m 00549 p T05 where m is in lbmin p is in psia and T is in OR a Detennine what the units for the constant 00549 are b What will be the new value of the constant now given as 00549 if the variables in the equation are to be substituted with SI units and m is calculated in SI units 2 Chapter 1 Dimensions Units and Their Conversion NOZZLES Fluid Feed I I Air Click on image for animation Fig 1a Ultrasonic nozzle Fig 1bA conventional nozzle spraying a fluid courtesy of Misonix lnc Farmingdale of suspended particles in a flash dryer NJ Spray nozzles are used for dust control water aeration dispersing a particular pattern of drops coating paintings cleaning surfaces of tanks and vats and numerous other applications They develop a large interface between a gas and liquid and can provide uniform round drops of liquid Atomization occurs by a combination of gas and liquid pressure differences The Figure below courtesy of Misonix Inc compares the particle sizes from the ultrasonic nozzle with those from the conventional nozzle PARTICLE SIZE COMPARISON SONIMIST Spra Nozzles I I Conventional Spray Nozzles Tobacco Smokey IBacteria I Fumes I I Clouds FIP 1c IFog IMist IDrizzleIRain Dust 01 10 10 100 1000 10000 PARTICLE SIZE MICRONS F ig l c Chapter 1 Dimensions Units and Their Conversion Solution a Calculation of the constant The rst step is to substitute known units into the equation 1b 1b quot 00549 mm m R We want to nd a set of units that convert units on the right hand side of the above expression to units on the left hand side of the expression Such a set can be set up directly by multiplication 1b gt lbw in2 R0 5 minX lbf quot min 2 o 05 Units for the constant 00549 are lbm R m1nlbf b To determine the new value of the constant we need to change the units of the constant to appropriate SI units using conversion factors 00549 lbmin2 R 50454kf 147lbfin2 1min 1K p lbemin 1 1bquot 1013 gtlt 103Nm260s18 R05T 5 p m 449 x 108 m s K05 a Substituting pressure and temperature in SI units m 449 10 msK mom 1Nm2 m YET 449 X 10398 wherepisinNmzandTisin K JILL T05 4 Chapter 1 Dimensions Units and Their Conversion Problem 1 4 An empirical equation for calculating the inside heat transfer coef cient hi for the turbulent ow of liquids in a pipe is given by 0023 G08 K067 013033 hi D02 H047 where hi heat transfer coefficient Btuhr 2 F G mass velocity of the liquid lbmhrft2 K thermal conductivity of the liquid Btuhr F Cp heat capacity of the liquid Btulbm OF u Viscosity of the liquid lbmfthr D inside diameter of the pipe ft a Verify if the equation is dimensionally consistent b What will be the value of the constant given as 0023 if all the variables in the equation are inserted in SI units and hi is in SI un1ts Solution a First we introduce American engineering units into the equation h 00231bmft2hr0 8 BtuthftX F067Btulbm F033 1 GO 2 lbm fthr0 7 Next we consolidate like units I 70023Bm067 lbm08 047 1 110047 hl 7 lbm033lbm047 l6 067 02 OF0670F033 hr08hr067 I 7 Btu h1 7 0023 hr 2 OF The equation is dimensionally consistent b The constant 0023 is dimensionless a change in units of the equation parameters will not have any effect on the value of this constant Chapter 2 Conventions in Measurement 5 Chapter 2 Conventions in Methods of Analysis And Measurement Problem 21 Calcium carbonate is a naturally occuIing white solid used in the manufacture of lime and cement Calculate the number of 1b mols of calcium carbonate in a 50 g mol 0fCaCO3 b 150 kg of CaCO3 c 1001b 0fCaCO3 Solution 50 g mol Caco3i 100 g CaCO3 i 1 1b CaCO3 i11b mol CaCO3 1gm01CaCO3 1454 g CaCO3 1001b CaCO3 011 1b mol 150 kg CaCO3 12205 1b CaCO3 i 1 1b mol CaCO3 1 kg CaCO3 1001b CaCO3 330 1b 101 100 1b CaCO3 1 lb mol CaCO3 100 CaCO3 100 lb mol CaCO3 Problem 22 Silver nitrate lunar caustic is a white crystalline salt used in marking inks medicine and chemical analysis How many kilograms of silver nitrate AgNO3 are there in a 1301b mol AgNO3 b 550 g mol AgNO3 S olution 130 lb mol AgNO3 1701b AgNO3 1 kg 1002 kg or 1000 kg 1 1 1b mol AgNO3 12205 1b 550gm01AgNOq 1 170gAgNO3 1 1kg 1 gmol AgNO3 1000 g 935 kg 6 Chapter 2 Conventions in Measurement Problem 23 Phosphoric acid is a colorless deliquescent acid used in the manufacture of fertilizers and as a avoring agent in drinks For a given 10 wt phosphoric acid solution of speci c gravity 110 determine a the mol fraction composition of this mixture b the volume in gallons of this solution which would contain 1 g mol H3PO4 Solution a Basis 100 g of 10 wt solution g MW gmol mol fr H3PO410 9797 0102 0020 H20 90 1801 500 0980 psoln b Specl c graVIty The ref liquid is water pref 3 3 The density 0fthe solution is 13910 g solncm 50m 13900 g Hzocm 1 10 i111 100 g HZOcm3 1 39 cm3 1 cm3 soln i 1 g soln i 9797 g H3P04 i 2642 gal 110 g soln 01 g H3PO4 1 g mol H3PO4 106 cm3 024 galg 101 Problem 24 The density ofa liquid is 1500 kgm3 at 20 0C a What is the speci c gravity 20 C4OC of this material b What volume ft3 does 140 lbm of this material occupy at 20 C Solution Assume the reference substance is water which has a density of 1000 kgm3 at 4 C p soln kgm3soln M pref 7 kg ber 1000kgm3 a Specific gravity 150 1 3139 39d 1k 3531ft3 1401b b 1120011321 220g1b 1m3 m 150ft3 Chapter 2 Conventions in Measurement 7 Problem 25 The 1993 Environmental Protection Agency EPA regulation contains standards for 84 chemicals and minerals in drinking water According to the EPA one of the most prevalent of the listed contaminants is naturally occuring antimony The maximum contaminant level for antimony and nickel has been set at 0006 mgL and 01 mgL respectively A laboratory analysis of your household drinking water shows the antimony concentration to be 4 ppb parts per billion and that of nickel to be 60 ppb Determine if the drinking water is safe with respect to the antimony and nickel levels Assume density of water to be 100 gcm3 Solution The problem may be solved by either converting the EPA standards to ppb or vice versa We will convert the EPA standards to ppb ppb is a ratio and therefore it is necessary for the numerator and denominator to have same mass or mole units The mass and volume that the Sb contributes to the water solution is negligible Antimony 0006 mg Sb i 1L soln i 1 cm3 solni lg 6 g Sb 1L soln 1000 cm3 soln 100 g H20 1000 mg 109 g soln 6 PPh Nickel 01mgNii lLsoln i1cm3solni 1g 9gNi 1L soln 1000 cm3 SO111110 g H20 1000 mg 109 g soln 100 ppb House hold drinking water contains less than the EPA mandated tolerance levels of antimony and nickel Drinking water is therefore safe Problem 26 ine making involves a series of very complex reactions most of which are pe1fonned by microorganisms The starting concentration of sugars determines the nal alcohol content and sweetness of the wine The speci c gravity of the starting stock is therefore adjusted to achieve desired quality of wine A sta1ting stock solution has a specific gravity of 1075 and contains 127 wt sugar If all the sugar is assumed to be C12H22011 determine a kg sugarkg H20 b lb solutionft3 solution c g sugarL solution Solution Basis 100 kg starting stock solution 8 Chapter 3 Choosing a Basis 127 kg sugar 100 kg solution kg sugar 3 100 kg soln 873 kg H20 145 kg H20 1075 g solncm3i100 gH20cm3 i 1 1b i2832x104cm3 7 67 1 lb soln 10 g H20cm3 i i454gi ft3 39 ft3 soln 1075 g solncm3 i 10 g H20CII 13 i 127 g sugar i 1000 cm3 g sugar 10 g HZOcm3 100 g soln 1 L 136 L soln Chapter 3 Choosing a Basis Problem 31 A liqui ed mixture of nbutane npentane and nhexane has the following composition in percent n C4H10 50 nC5H12 30 C6H14 20 Calculate the weight fraction mol fraction and mol percent of each component and also the average molecular weight of the mixture Solution Note that the hydrocarbon mixture is liquid so that the composition is therefore in weight percent It is convenient to use a weight basis and set up atable to make the calculations Basis 100 kg kgwt fr MW kg mol mol fr n C4H10 50 050 58 086 057 n C5H12 30 030 72 042 028 nC6H14 20 020 86 023 015 100 100 151 100 Average molecular weight total mass 100 kg 66 total mol 151 kg mol Chapter 4 Temperature 9 Chapter 4 Temperature Problem 4 1 Complete the table below with the proper equivalent temperatures 0C 0F K 0R Solution The conversion relations to use are 0 F 18 O C 32 K 0C 273 OR 0F 460 OR 18 K 0C 0F K 0R 400 400 233 420 250 770 298 437 425 797 698 1257 235 390 384 698 Problem 42 The speci c heat capacity of toluene is given by following equation Cp 20869 5293 X 10392 T whereCp is in BtuLB mol 0 F and T is in O F Express the equation in calg mol K with T in K Solution First conversion of the units for the overall equation is required 20869 5293 gtlt10392T F Btu 252 cal 11b mol 18 0F 1 1b mol OF 1 Btu 454 g mol 1 K cal 20869 5293 X 10392 ToF g mol K Note that the coef cients of the equation remain unchanged in the new units for this particular conversion The T of the equation is still in 0F and must be converted to kelvin 1 0 Chapter 5 Pressure MEASURING TANK PRESSURE RS485R8232 CONVERTE Flt l I Click on image for animation SIGNAL Fig 2 The measurement of pressure at the bottom at P1 of a large tank of uid enables you to determine the level of uid in the tank A sensor at P2 measures the density of the uid and the sensor at P3 measures the pressure of the gas above the uid A digital signal is sent to the remote control room where the sensor readings and calculations for volume can be displayed on a PC The volume of uid in the tank is determined by multiplying the known area by the height of uid calculated from the pressure and density measurements Level can be determined to an accuracy of i 18 inch a value that leads to an accuracy of about 02 in the volume In a tank containing 300000 bbl of crude oil the error roughly corresponds to about 12000 in value ChapterS Pressure 11 ToF TK 27318 32 Cp 2069 5293 gtlt10392TK 273l8 32 Simplifying Cp 3447 9527 X 102 TK Chapter 5 Pressure Problem 51 A solvent storage tank 150 m high contains liquid styrene sp gr 0909 A pressure gauge is xed at the base of the tank to be used to determine the level of styrene a Determine the gage pressure when the tank is full of styrene b If the tank is to be used for storage of liquid hexane sp gr 0659 will the same pressure gage calibration be adequate What is the risk in using the same calibration to determine the level of hexane in the tank What will be the new pressure with hexane to indicate that the tank is lll 0 Solution a The liquid in full tank will exert a gage pressure at the bottom equal to 15 0 m of styrene The tank has to operate with atmospheric pressure on it and in it or it will break on expansion at high pressure or collapse at lower pressure phpg 715 0 0909 g styrenecm3i 10 g H20CII 13 i 103 kgm3i980 mszi lPa 39 m 10 g HZOcm3 l 1 gcm3 l1 kgm391s392 134 X 103Pa 134 kPa gage 5quot Hexane is a liquid of speci c gravity lower than that of styrene therefore atank full of hexane would exert a proportionally lower pressure Ifthe same calibration is used the tank may over ow while the pressure gage was indicating only a partially full tank c Newph p g 715 0 0659 ghexanecm3 i 10 g H200m3 i 103kgm3 i 98 ms2 i lPa 39 m 100 g HZOcm3 l l lkgm391s392 96900 Pa 969 kPa 1 2 Chapter 5 Pressure U TUBE MANOlVlETER gt Vacuum Vacuum Fig 3 Various forms of manometers Typical liquid manometers consist of a Ushaped tube of glass or polycarbonate plastic partially lled with what is called a manometer uid The size and height of the manometer and the manometer fluid are selected so as to measure the desired pressure over the expected pressure range Typical manometer uids are mercury water the uid in the system being measured and heavy oils with very low vapor pressure A manometer with the reference end open to the atmosphere makes gage measurements ie measurements relative to the existing barometric pressure A manometer with the reference end sealed so as to contain only the manometer uid vapor measures roughly absolute pressure but more precisely the reference pressure is the vapor pressure of the manometer uid hence the use of mercury which has an especially low vapor pressure at room temperature 2 x 10393 mm Hg 3 x 10394 kPa The sensitivity of a manometer can be increased by using special oils of speci c gravity of 08 to 10 that also have very low vapor pressures The accuracy of manometers depends on how closely you can read the meniscus in the glass tube Chapter 5 Pressure B 1 WWW xsuscdlo dcwmnnc the pressure dmp across 1 mi owmg m m We m is a sulfunc and soluuon mm a speci c d 15 mercury wnh a specfo gavxly solsari 1 55 The manomc39 Icadmgxs 5 35 mnhcs and all pans of39hc mm are at sumpcraluxc of 5m Wham the pressure dmp across me m cc mam mps anonnccmcm The 2m 517501 of 501mm Fust we mum density of and and mercury 1 250 524mm ms 7 3 PM 7 Luxxmsms 7 commm 13 56 524mm ms 7 3 PH 7 Luxxmsms 7 0490mm Clitk on gure for Animation gt at 27 and add up the mm c p a 5m mm 121 pressure ammuqu The memes othc Hg m me 1m and nghl columns below A m m tube camel each mm 50 we stop addmg at level A 277qu ngmqu Ana 9 PnhlE 92 Pnhzg PFIng m 7 P2 mm 7112 PHxhzg m p DANE PFIng m 39 P2 PILg 39 F de 31mm me dcnsmcsmthc m1 equmn 0490 oAsmmtsasm 327st2 P1 2 m2 32174mabmy521b Emma 14 Chapter 5 Pressure Problem 53 The pressure difference between two air tanks A and B is measured by a U tube manometer with mercury as the manometer liquid The barometric pressure is 700 mm Hg a What is the absolute pressure in the tank A b What is the gauge pressure in the tank A Solution Click on image for larger View Tank A is connected to tank B through a U tube and Tank B is connected to the vertical U tube The vertical tube can be used to measure the pressure in tank B and the U tube can be used to relate the pressures of tanks A and B a At Z0 pa h1 pHg g pb neglecting the effect ofair in the U tube 1 at Z1 pb h2 pHg g 2 Eliminate pb from the equations pa hipHg g h2 pHg g pa h2 hi pHgg 840 mm Hg absolute The pressure measured by this manometer system is the absolute pressure because the reference pressure above the mercury in the vertical tube is a vacuum 1 pa 840 700 140 mmHg Part 1 Additional Problems 1 5 PART 1 ADDITIONAL PROBLEMS Answers will be found in Appendix A Chapter 1 Convert the following to the desired units a 60 mihr to ms b 30 Nm2 to lbfft2 c 163 J to Btu d 421 kW to Js Change the following to the desired units a 235 g to pounds b 610 L to cubic feet c 30 gL to poundscubic feet d 147 lbin2 to kgcm2 Find the kinetic energy of aton of water moving at 60 minhr expressed as ftlbf An elevator which weights 10000 lb is pulled up 10 it between the Ist and second oors of a building 100 ft high The greatest velocity the elevator attains is 3 fts How much kinetic energy does the elevator have in ftlbf at this velocity The Colburn equation for heat transfer is iygf 0023 KCG Kk DGu0392 where C heat capacity Btulb of uid F p viscosity lbhrft k thermal conductivity Btuhr 2 F D pipe diameter ft and G mass velocity lbhrft2 of cross section What are the units of the heat transfer coefficient h Countercurrent gas centrifuges have been used to separate 235U from 238U The rate of diffusive transport is K 211Dp7 If K rate of transport of light component to the center of the centrifuge in g molscm of height D diffusion coef cient p molar density g molescm3 and 7 log mean radius r2 7 r1ln rZrl with r in cm what are the units of D The density of a certain liquid is given an equation of the following form p A BTeCp where p density in gcm3 1 temperature in OC and P pressure in atm The equation is dimensionally consistent What are the units of A B and C a b In the units aboveA 1096 B 000086 and C 0000953 Find A B and C if p is expressed in lbft3 T in OR and p in lbfin2 16 21 22 23 24 25 26 31 32 Part 1 Additional Problems Chapter 2 a How many g moles are represented by 100 g of C02 b Calculate the weight in pounds of 35 g moles of nitrogen Convert the following a 120 lb mol of NaCl to g b c 120 lb of NaCl to g mol d 120 g mol ofNaCl to lb 120 g ofNaCl to lb mol A solution of sulfuric acid at 60 F is found to have a sp gr of 122 From the tables in Perry s Chemical Engineer39s Handbook the solution is found to be 30 percent by weight H2804 What is the concentration of H2804 in the following units a lb molgal b lb 3 c gL d lb HZSO4lb H20 and e lb mol HZOlb mol total solution A mixture of liquid hydrocarbons contains 100 percent nheptane 400 percent n octane and 500 percent ipentane by weight The specific gravities of the pure components are nheptane 0685 noctane 0705 i pentane 0622 600 F a What 1s the sp gr 600E of 93 lb of th1s m1xture b How many US gallons will be occupied by 130 lb of this mixture Convert 172 ppm NH3 in water to the equivalent number of ppm of NH3 gas in water vapor Five thousand barrels of 28 API gas oil are blended with 20000 bbl of 150 API fuel oil What is the 0API API gravity of the mixture What is the density in lb gal and lbft3 Chapter 3 A mixture of gases is analyzed and found to have the following composition C02 12 0 CO 60 CH4 273 Hz 99 N2 448 How much will 3 lb moles of this gas mixture weight You have 100 lb of gas of the following composition CH4 30 H2 10 N2 60 What is the average molecular weight of this gas Part 1 Additional Problems 1 7 33 41 42 What is the composition of the gas in Problem 31 on a nitrogen free basis no N 2 in the analysis Chapter 4 Two thermometers are to be checked against a standard thermometer The standard reads 722 F What should the other two thermometers read if they are calibrated in OC and K respectively Mercury boils at 630K What is its boiling temperature expressed in 0C In 0F In 0R Chapter 5 What is the gauge pressure at a depth of 450 mi below the surface of the sea if the water temperature averages 60 F Give your answer in lb force per sq in The sp gr of sea water at 60 F60 F is 1042 and is assumed to be independent of pressure The pressure gauge on the steam condenser for a turbine indicates 262 in Hg of vacuum The barometer reading is 304 in Hg What is the pressure in the condenser in psia Examine the gure Open Ah 20 in Hey L Click on image for larger view 54 The barometer reads 740 mm Hg Calculate tank pressure in psia a An ori ce is used to measure the ow rate of a gas through a pipe as shown in Fig 54 The pressure drop across the ori ce is measured with a mercury manometer both legs of which are constructed of i in inner diameter ID glass tubing If the pressure drop across the ori ce is equivalent to 465 in Hg calculate h and h3 both in inches if h is equal to 1350 in b The right glass leg of the manometer in Fig P16D becomes corroded and is 3 replaced Wlth glass t bulg Wthh 15 8 in ID The manometer is again lled with a suf cient volume of mercury For the same pressure drop as in part a calculate h and h3 both in inches PART 3 Gases Vapors Liquids and Solids 71 Chapter 13 Ideal Gases Problem 131 A steel tank having a capacity of 25 m3 holds carbon dioxide at 30 C and 16 atm Calculate the weight in grams of the carbon dioxide Solution T 30 C 30 27315 303K p 16 atm V 25 In3 Use pV nRT The gas constant has to be chosen to be consistent with the units of pressure temperature and volume Use standard conditions to calculate R 7 100 atm 22415m3 7 athm3 R kg mol 27315K 008206 kg molk n 1608 kg mol or 1608 gmol 71608 gmol 44gC02 7 3 Welght of carbon d10X1de 7 g mol CO2 7 707 X 10 g C02 Alternate Solution Use V 22415 m3kg mol at T 273K and p 1 atm abs P1 V1 n1 T1 25 m316 atmabs273Ki 1kgmol I 44 kgCOg 1000g I latm abs I303K 22415 m31kgm01C02 1kg 707 x 103 g 002 72 Chapter 13 Ideal Gas Law Calculations Problem 132 20 ft3 of nitrogen at 300 psig and 100 F and 30 f of oxygen at 200 psig and340 F are injected into a 15 ft3 vessel The vessel is then cooled to 70 F Find the partial pressure of each component in the 15 ft3 vessel Assume that the ideal gas law applies Solution The problem can be solved by determining the number of moles of oxygen and nitrogen and solving for the total pressure at the final conditions Then the partial pressures are calculated using the mole fractions Nitrogen Oxygen V20ft3 V30ft3 p 300 psig 3147 psia p 200 psig 2147 psia T 100 F 560 R T 340 F 800 R 314720 214730 N2 RT 1073560 1051b 01 02 1073800 075 1b mol nT nN2 no2 180 lb mol Final pressure pr 2 2 682 psia moles mole fn y i pi p yi 02 075 042 286 N2 105 M 396 180 100 Ema Chapter 13 Ideal Gas Law Calculations 7 3 Problem 133 A steel container has a volume of 200 m3 It is lled with nitrogen at 22 C and atmospheric pressure If the container valve is opened and the container heated to 200 C calculate the fraction of the nitrogen which leaves the container Solution The solution involves determining the number of moles of nitrogen in the tank at the nal given temperature and pressure T2 and p2 and at the initial conditions and subtracting T1 22 C 295K T2 200 C 473K p11atm p21atm v1 200 m3 v2 200 m3 V V n1P111 1121ET2 1000 1000 n1 7 008206295 n2 7 008206473 826 kg mol 515 kg mol The fraction of N2 leaving 060 Problem 134 Chlorine gas containing 24 percent 02 is owing through an earthenware pipe The gas ow rate is measured by introducing air into it at the rate of 115 m3min Further down the line after mixing is complete the gas is found to contain 1085 percent 02 How many m3 of the initial gas were owing per minute through the pipe Solution The problem is similar to other nonreacting material balance problems It is convenient to imagine the pipeline between the point of injection of air to the point of sampling as a mixer The process may be considered to be a steady state process without reaction 74 Chapter 13 Ideal Gas Law Calculations Amo mol fr 02 021 Air Chlorine Fmo mol fr P mol Clz 0976 02 quot10 fr mol 0100 Clz Clz 1 02 01085 n2 Cick image to see enlargement x N 2 N 2 n 3 Step 5 We could use F 100 mol as the basis instead of the given owrate All ow rates can be then be converted to the basis of 1 minute at the end of the problem However it is easier to use the given ow rate in m3 assuming that the temperature and pressure are the same for all streams so that moles are proportional to m3 in each stream A 115 In3 Step 6 Let xc12 and KN2 be the respective mole fractions in P The unknowns are F P xCl2 and KNZ or n1 and n3 Step 7 The balances are C12 02 N 2 and 2x1 2 1 or Eni P so the problem has a unique solution Step 8 Total F 115 P 1 Oxygen Balance 0024F 021115 01085 P 2 Chlorine Balance 0976F xc12 P 3 Nitrogen Balance 079115 XNZP 4 2x1 1000 01085 Kg KN2 21000 5 Steps 8 and 9 Only 4 of the balances are independent Substitute the total balance for one component balance and solve 1 and 2 together to get P253m3 atTandp Vin 138 m3 at T andp Step 10 Use 2x1 as a check from 4 X1512 079115253 0359 from 3amp1 x212 0976138253 0532 from 5 0359 0532 01085 2 100 Chapter 13 Ideal Gas Law Calculations 75 Problem 135 A synthetic gas generated from coal has the following composition C02 72 CO 243 H2 141 CH4 35 N2 509 a Calculate the cubic feet of air necessary for complete combustion per cubic foot of synthetic gas at the same conditions b If 38 excess air were used for combustion what volume of ue gas at 750 F and 738 mm Hg would be produced per cubic foot of synthetic gas at standard conditions c Calculate the ue gas analysis for a and b Solution The problem is just a material balance problem with volumes of inlet and outlet gases specified instead of moles We could select ft3 or moles as a basis Steps 1 2 3 and 4 This is a steady state problem with reaction mol fr mol tr mol C02 0072 C02 Xcoz nC02 CO 0243 2 XH20 nH20 H2 0 l4l N2 KNZ nNZ CH4 0035 A 1b mol 02 on 1102 N2 0509 AF mol r 100 P 02 021 N2 079 100 Part a Calculate the required air first Step 5 Basis 100 mol F Step 4 Component mol Reaction mol reqd 02 C02 72 CO 243 C0 02 gt C02 1215 H2 141 H2 02 gtH20 715 CH4 35 CH4202 gt C022H20 70 N2 509 1000 263 Accompanying N2 263 7921 98 9 Total 125 2 76 Chapter 13 Ideal Gas Law Calculations 1252 mol air 125 mol air 125 113 air a39 F 100 mol feed 39 mol feed 100 ft3 feed both at same conditions Since the temperature and pressure are the same for both streams the mole ratio and the volume ratio are the same Next let us make a material balance for the case in which the air supplied is 38 in excess Keep the same basis A 1252 138 1727 mol Part b2 Step 6 Unknowns P and xCOZ xHZO xNZ x02 or nCO2 nHZOnN2 noz Steps 7 and 8 The element balances are Carbon balance 72 243 35 PxCO2 nCO2 1 Hydrogen balance 1412 354 2 PxHZO 2nHZO 2 Oxygen balance 722 243 021 17272 2PxCOZ PxHZO 2PxoZ 3 Nitrogen balanceN2 509 079 1727 PxNZ nNZ 4 Also xHZO xCOZ xNZ x0Z 1 5 or nHZO nCO2 nN2 no2 P Step 9 From the above equstions we get Comgonem mol C02 35 H20 211 02 101 N2 1873 Total 253 Vol ofP 750 F 738 mm Hg T 750 F 1210 R p 738 mm Hg 7 2531b moll 359 ft3 1 1210 R I 760 mm Hg 7 7 1 lb mol 492 R I738 mm Hg 7 100 lb mol 359 ft3 7 4 3 VF 11b 01 359 x 10 ft 230 X 105 ft3 VP Chapter 13 Ideal Gas Law Calculations 7 7 VP 2 230 x 105 2 m V F 7 359X104 7 641 ft3 airatsc atT andp Part c Flue gas analysis a The water and carbon dioxide are the same as in part b and there is no 02 in the ue gas mol mol fr C02 35 023 H20 211 014 N2 M m 155 100 b mol mol tr C02 35 014 H20 211 008 N2 187 074 02 m m 253 100 Problem 136 An old way of producing hydrogen gas in the laboratory was by the reaction of sulfuric acid with zinc metal H2 8041 Zns gt ZnSO4s H2g How many grams of sulfuric acid solution 98 must act on an excess of zinc to produce 120 m3hr of hydrogen at standard conditions Assume all the acid used completely reacts Solution Steps 1 2 3 and 4 You rst must determine the number of moles of hydrogen produced and then calculate the acid required Via stoichiometry This is a steady state process with reaction except for the Zn which can be assumed to exit the vessel Zn mass Zr MW 103 g mal mol Zr mol Zr Zg 1 00 H2504 098 98 100 090 l H2 39 002 120 m 3 sc H20 18 1111 0100 Gg mol 540 g mol 100 1111 100 L011 F mol 0 OO 00 P mol P g Doggogago g Zn 504 XZn 04 0 no P 8000000ng H 20 XHZO Zn xZn 78 Chapter 13 Ideal Gas Law Calculations Step 5 Basis 1hr 120 m3 at so in G 7 moles ofhydrogen 7 n 7 RT 3 vol ofhydrogen V 12390 m 120 X104L 1 atm120 x 104 2 n 008206 273 536 X 10 gmol Step 6 Unknowns are F P X21804 X zo and X21 The only unknown we want to solve for is F Steps 7 and 8 Balances We can make H S O and Zn element balances they may not all be independent plus 2X1 1 The easiest balance to make to get F is an H2804 compound balance but other reasonable balances can be used In M Generation C onsumgtz39on Accum 090F 7 0 536 7 0 0 F 596 g mol H2804 596 g mol H2804 solnl 100 g soln 11111x103 gmol98 soln 537 x 104 g 98 soln Problem 137 Polymeric membranes are proposed to be used to reduce the 02 concentration in the waste gas of a sulfuric acid plant from 15 to 001 Hollows bers made of polymeric membrane ll the separations unit The waste gas containing 15 02 is the fresh feed in the amount of 1000 m3 hr at 25 C and 1 atm 1013 kPa To meet the product gas speci cations part of the product stream has to be recycled to reduce the concentration of 02 entering the separator itselfto 110 802 in G the process feed The process is isothermal at 25 C and isobaric at 1 atm a Determine the recycle stream ow rate R in m3hr b Determine the waste stream owrate W in m3hr Solution Steps 1 2 3 and 4 This is a steady state process with reaction and recycle All the known data have been placed in the gure No reaction occurs Chapter 13 Ideal Gas Law Calculations 79 25 C w m3 at T and P 1013kPa 25 C mol fr 1013kPa so 2 0011 Others 0989 1000 G m3 at T and P Pm 3 at T and P Tm3 at T and P 25 C 0 101 3kPa 25 C molfn 1013kPa SO 2 00001 F m3 at T and P Others 09999 1000 m3 lhr 10000 39 luhuimy T mom R m3 at T and P SO 0015 250C Others 0985 M 1ooo 1013kPa so2 oooo1 Others 09999 Click on image to see animation 10000 Step 5 Basis F 1000 m3 at ZS C and 1013kPa same basis as 1 hr Step 6 Ignore the exit stream from the reactor Then G P W and R are unknowns and their compositions are known Since the temperature and pressure are constant throughout volume balances can be used mole fraction is the same as volume fraction The balances could be made in moles and then converted to the basis of 1000 m3 Steps 7 8 and 9 The system is the overall process The balances are Total 1000 P W 8021 00151000 00001 P W 1000 Other 09851000 09999 P a tie component 80 Chapter 13 Ideal Gas Law Calculations SEPARATION BY MEMBRANES Feed in Hollow thin walled plastic tube Perrneate Permeate out Retentate 1 Permeate Tubular 0 multichannel Reject Splrawound Fig 17a Tubular membrane separator Fig 17b Spiral wound membrane separator Membranes are used to separate gaseous mixtures or liquid mixtures Membrane modules can be tubular spiralwound or plate and frame con gurations Membrane materials are usually proprietary plastic lms ceramic or metal tubes or gels with hole size thickness chemical properties ion potential and so on appropriate for the separation Examples of the kinds of separation that can be accomplished are separation of one gas from a gas mixture separation of proteins from a solution dialysis of blood of patients with kidney disease and separation of electrolytes from non electrolytes Chapter 13 Ideal Gas Law Calculations 81 Two of the balances are independent Solve any two to get P 9851 m3hr at 25 C and 1 atm W 149m3hr at 25 C and 1 atm To get R we make a balance on the mixing point where F and R combine to make G Total 1000 R G SOZ 00151000 00001R 0011G Other 09851000 09999R 0989G Two of the balances are independent Solve the rst two to get R 364 m3hr at 25 C and 1 atm G 1364 m3hr at 25 C and 1 atm Step 10 Check 0151000 7 0001 9851 149 15 15 OK Problem 138 In a sulfuric acid plant sulfur is burned in the presence of excess oxygen to produce sulfur dioxide which in turn is further reacted in the next step with oxygen in a converter to produce sulfur trioxide In the plant 802 along with 10 excess air is fed into the converter which operates at 1500 C and 1 atm The per pass conversion of 02 is 75 and overall conversion is 100 If 106 m3hr of 03 at 1100 C and 1 atm is fed to the converter calculate the a ow rate of the product stream P in m3hr at 1500 C and 1 atm and its composition in mole percent b ow rate ofthe recycle stream R in m3hr at 1500 C and 1 atm Solution This is a steady state problem with reaction and recycle Steps 1 2 and3 02 702 gt 03 82 Chapter 13 Ideal Gas Law Calculations R kg mol 100 so2 P kg mol F kg mol mol fr mol 503 X503 r1303l mol in 02 X02 n 02 S 02 100 N2 XNZ N2 100 P 106 m3 Ihr 1000 C gt A kg mol mol fr 1 atm quot 25 C 021 1 atm 019 1 00 Step 5 Basis F 106 In3 at llOO C and 1013kPa equivalent to 1 hr Ste 4 We will make the balance in rnoles In3 could also be used if A and F are adjusted P to 1 atm and ISOOOC pV 1013106 1 E8314E1100EE27 3 874 kg 101502 Next calculate the value of A the entering air kg mol Required 02 8874 i 4437 Excess 02 014437 EEEE444 Total 4881 079 N 18360 2 4881 021 Step 6 The system is the overall process The unknowns are P nsog n02 and nN2 or P xsog x02 XNz Step 7 The balances are S O N and Zni P or 2X121 Steps 8 and 9 For the overall process the element balances are units are kg mol S 8874 P xsog 11503 N2 18360 PxN nN 0 8874 2 48812 88743 11022 Chapter 13 Ideal Gas Law Calculations 83 n30 8874 nN2 18360 no2 444 Thus P 8874 18360 444 27618 kg mol or 27620 kg mol a Apply pV nRT V WW 402 x 106m3hr at 150000 and 1 atm At the mixing point ofF and R we get G F R G b Make the system the reactor plus the separator to avoid having to calculate information about the converter outlet stream Use an 02 balance in kg mol Accum 0 In Out Gen Consum 1008874R 7 R100 0 7 0758874R R 2958 kg mol which corresponds 043 X 106 m3hr at 1500 C and 1 atm 84 Chapters 14 and 15 Real Gas Relationships Chapter 14 Real Gases Compressibility and Chapter 15 Real Gases Equations of State Problem 141 S ven pounds osz at 120 F are stored in a cylinder having a volume of 075 ft3 Calculate the pressure in atmospheres in the cylinder a assuming N2 to be an ideal gas b assuming the pressure of N2 can be predicted by van der Waal39s equation c using the compressibility factor method d using the Redlich Kwong equation of state Solution V 075 ft3 T 120 F 580 R Basis n 7le2 0251bmolN2 R 1073 psiaft31b mol R a Using the ideal gas relation pV nRT 7 0251073580 latm 7 p 7 0I75 m 7 141 atm 2074 pSIa b Using van der Waals equation 2 NE Vinb nRT v2 01113 gmo From the text for nitrogen a 1347 X 106 g mol 2 cm3 1 andb386 2 3 2 3776 x103 J a 2 lbmo 5086 ft J cm3 2 lb mo 1 g mol 3 2 3 2 386 cm 1 160X1021 J 3 2 b gmquot 2 06176 J 3 1b mo 1 g mol cm3 g mol a 1347 X 85 Chapters 14 and 15 Real Gas Relationships 075 7 025 06176 025 1073 580 0252 5086 0752 p 5651 0596 1556 p 2046 psi 139 atm Using the compressibility factor pV anT From Appendix D in the text the critical properties for nitorgen are pc 335 atm and Tc 1262K227 R 7 RTc 07302227 7 3 V07 pc 335 7 495 ft lb mol T 580 R T 7 a 256 139 Tc 227 R V 075 ft3025 1b mol V 061 r V 495 ft3lb mol 0 On one of the compressibility charts having the right domain determine the intersection of the V 061 and the T1 256 lines Some approximation may be required From this point you can read 2 on the vertical and p1 on the horizontal axis 2 105 p1 45 anT 1050251073580 2178 psiamg am 075 Pr PPc or use pr p PrXPc 45 335 151 atm These answers are close enough in View of the accuracy of the original data Using the Redlich Kwong Equation of State a AibRT a P T 2 V V b T35 RT where a 04278 and b 00867 6 pc pc 2 86 Chapters 14 and 15 Real Gas Relationships 2 25 a 04278 0730533 27 5286 OR 05 atm ft3lb mol2 07302 227 3 b 00867 335 0429 ft lb mol Substitute these values into the equation p i 3 7 0429 07302 580 580 2 3 3 0429 p 2096 2571 4235 p 144 atm The pressures determined by three methods are comparable although not exactly same Problem 142 Suppose 100 g mol of C02 at atemperature of 220K is placed in a Dewar vessel of unknown volume and the pressure reading made is 690 kPa What is the volume of the vessel Solution Basis 100 g mol CO2 Let us use the SoaveRedlichKwong equation which is cubic in V Calculate the coef cients the equations are listed in Chapter 15 for the equation and use Polymath or another nonlinear equation solving program to find RT a P v b vvb p 690 kPa R 8314kPam3kg molK T 220K b 297 X 10395 m3g mol a 0370 m6 Pag mol2 K 0826 A 134 Arrange the equation so that p is subtracted from the right hand side the equation then equals zero and solve for 7 238 x 10393 mSg mol 87 Chapters 14 and 15 Real Gas Relationships Problem 143 A gas analyzes 60 methane and 40 ethylene by volume It is desired to store 123 kg of this gas mixture in a cylinder having a capacity of 5 3914 X 10392 m3 at a maximum temperature of 45 C Calculate the pressure inside the cylinder by a assuming that the mixture obeys the ideal gas laws b using the compressibility factor determined by the pseudo critical point method Basis 123 kg Solution a deal Gas Law pV nRT aV mol wt EyiMWi 060 16 040 28 208 kg 5101 7 123 kg W n 4i 208 kg mixture 7 0591 kg mol R 8314 kPa m3kg molK V 7 514gtlt 1021113 T 45 C 318K 7 nRT 7 0591 8314318 7 7 V 514 X 102 30400 kPa b Pseudocritical method 1902 PciYi Tc2TciYi V02VciYi i i i Methane Ethylene Tc 1907K Tc 2831 K pc 458 atm 4640 kPa pc 505 atm 5116 kPa 7 RTc 7 m Vc 7 pc 7 0460 k g mol 7 RTc 7 m V07 pc 7 0342 k g mol Tc 060 1907 040 2831 2277K pc 060 4640 040 5116 4830 kPa L3 kgmol 7 T 7 318K 7 Tr 2277K 13940 C V 7 514gtlt 10392m30591 kg mol 70 22 0389 m3kg mol 39 Vc 7 060 0342 040 0460 7 0389 Vr A Using T and V we get p1 53 approximately so that p prp 534830 26000 kPa 88 Chapter 16 Vapor Pressure Chapter 16 Single Component Two Phase Systems Vapor pressure Problem 161 Calculate the vapor pressure of benzene at 50 C using the Antoine Equation Also estimate the normal boiling point of benzene the vapor pressure at 1 atm and compare it with the experimental value taken from a handbook Solution Ant01ne Equatlon lnp A 7 m From Appendix G in the text the coefficients are Benzene A 159008 TK B 278851 p mmHg C 75236 a Vapor Pressure ofbenzene at 50 C 278851 7 1110 159008 5236 50 273 p 270 mmHg abs b At the boiling point the vapor pressure is 1 atm 760 mmHg abs 7 278851 ln760 7 159008 7 5236 T Solving T 3533K From Appendix D in the text the normal boiling point of benzene is 35326K The two values agree well Problem 162 Prepare a Cox chart for ethyl acetate The vapor pressure of ethyl acetate is 200 mmHg abs at 42 C and 50 atm at 1260 C By using the chart estimate the a boiling point of ethyl acetate at 760 mmHg and compare with the experimental value 771 C b vapor pressure of ethyl acetate at its critical temperature of 5231K Compare with the experimental value of 378 atm Solution To construct the temperature scale the following data for the vapor pressure of water are used Chapter 16 Vapor Pressure 89 T C 300 500 1000 1500 2000 2500 patm 0042 0122 100 470 1536 3922 The procedure is as follows 1 First select a logarithmic vertical scale and place even values of the vapor pressure on the vertical scale to cover the desired pressure range 2 Then draw a straight line representing the vapor pressure of water at any suitable angle so that you cover the range of temperatures and pressures needed for the problem 3 To locate each integer value of the water temperature such as 30 50 100 etc on the horizontal scale note the corresponding vapor pressure on the vertical axis and move horizontally to the line Then drop vertically down to the temperature axis and mark the axis with the selected temperature the spacing will not be at even intervals Repeat to get a series of temperatures 4 Finally put the two points given for ethyl acetate on the chart and draw a line between them for ethyl acetate Cox Chart H20 Ethyl acetate Ixann r00 11111111111 0 10 20 30 40 50 75 100 125 150 175 200 250 TquotC The normal boiling point is estimated to be between 75 C and 80 C a bigger chart would produce a more accurate result and at the critical point 250 C the estimated vapor pressure is about 50 atm 90 Chapter 17 Saturation Condensation and Vaporization Chapter 17 Two Phase GasLiquid Systems Saturation Condensation and Vaporization Problem 171 If suf cient water is placed in a vessel containing a dry gas at 15 C and 1005 kPa to thoroughly saturate it what will be the absolute pressure in the vessel after saturation The temperature and volume of the vessel remain constant Solution H 0 Air Saturated air 62 15 C a T 15 C 754 mmHg p Once the air is saturated with water vapor the water if the water vapor is in equilibrium with liquid water exerts a pressure equal to its vapor pressure at 15 C Basis Dry gas at 15 C and 1005 kPa The vapor pressure of H20 at 15 C 17 kPa Since the temperature and volume remain constant pT pa pHZO 1005 17 1022 kPa Problem 172 A gas saturated with water vapor in a tank has a volume of 100L at l75 C and a pressure of 1062 kPa What is the volume of dry gas under standard conditions How many grams of water vapor are present in the gas Solution Sat Gas Dry Gas Water Vapor V 100L 3 T 273K T 273K T 175 C p 1013 kPa p 1013 kPa p 1062 kPa V At the initial conditions the total pressure is the sum of gas pressure and vapor pressure of the water The rst step in the solution involves determining the pressure of the dry gas assuming constant temperature and volume Look up pHZO p zo at l75 C 200 kPa pT 1062 kPa p zo pgas 200 pgas pgas 1042 kPa 100L 273K 1042 kPa 7 3 2905K 1013 kPa 0 97L at SC b One solution technique is to apply pV nRT to the water vapor 100L 200 kPal 1 atm 1 g molK 18g 17120 2905K 1013 kPa 008206 Latm 1 gmol H20 Another approach is to use pV nRT to calculate the total moles present and multiply the result 0039 g mol by the mole fraction water 2001062 0015 g H20 Chapter 17 Saturation Condensation and Vaporization 91 Problem 173 Dry air at 25 C is saturated with toluene under a total pressure of 760 mmHg abs Is there adequate air for complete combustion of all the toluene If so determine the percent excess air present for combustion Solution The solution involves rst determining the number of moles of toluene and the moles of oxygen in a speci ed volume Steps 1 2 3 C7H8 902 gt 7C02 4H20 Air toluene Air 25 C T 25 C 760 mmHg p 760 mmHg Toluene T 25 C p 760 mmHg Step 5 Basis 1 g m01C7Hg Step 4 From the chemical equation we see 9 mol of 02 are need per mol of toluene hence 9021 4286 mol of air is needed per mol of toluene The mole fraction C7Hg would be 14386 00228 Steps 6 7 8 and9 At 25 C C7Hg exerts a vapor pressure of 1111 16013715367252731 p 2822 mm Hg abs At saturation 760 7 2822 73178 mm Hg abs pail pa nauI 73178 25 93 Ptoluene ntoluene 2822 Since the molar ratio of airtoluene required for complete combustion is 429 and under the given conditions the airtoluene ratio is only 2593 the amount of air available is not adequate 92 Chapter 18 Partial Saturation and Humidity Chapter 18 Two Phase GasLiquid Systems Partial Saturation and Humidity Problem 181 A gas mixture contains 00083 g mol of water vapor per g mol of dry CH4 at a temperature of 27 C and a total pressure of 200 kPa Calculate the a percent relative saturation of the mixture b percent saturation of the mixture c temperature to which the mixture must be heated at 200 kPa in order that the relative saturation will be 020 Solution Basis 1 lb mol CH4 p zo 27 C 3536 kPa ptotal 200 kPa pHZO ptotal yHZO P a Percent relat1ve saturatlon lilo 100 PHZO 7 00083 7 YHzo 100083 0390082 pHZO 200 kPa 00082 164 kPa 164 W 100 464 PM b Percent saturatlon rel saturatlon 100 PT 1 2007354 100 460 200 7 l 64 0464 Note that the percent saturation is always less than the percent relative saturation c In the heating at constant pressure the mole fraction of the water vapor does not change so pHZO 164 kPa pliizo 020 or E64 020 pHZO pHZO The vapor pressure has to be calculated to get the temperature pHZO 82 kPa and from stream table T 315K 420C Chapter 18 Partial Saturation and Humidity 93 Problem 182 A gas at 200 F and 30 in Hg abs has a molal humidity of 010 Calculate a the percentage humidity b the relative humidity c the dew point of the gas F Solution phloem 200 F 11525 psia pT 30 in Hg 1003 atm 1474 psia molal humidity 41L or 4m or nvapor free gas Pvapor gas 134 39 pi 010 pHZO pm P gas p 134 psia PT PHZO Pgas gas pH20 pT 39 PHZO a Percentage humidity 100 279 PHZO T pH20 pH20 b Relat1ve hum1d1ty 100 116 pH20 c The dew point of the gas can be determined from the steam tables It is the temperature of which the partial pressure is equal to the vapor pressure of vapor cooling at constant total pressure pHZO 134 psia p The corresponding T is T 112 F 94 Chapter 18 Partial Saturation and Humidity Problem 183 Air saturated with water vapor at 80 F and 7450 mm Hg abs is passed through an air compressor and then stored in a tank at 250 psig and 80 F What percentage of the water originally in the air was removed during the processing Solution This problem can be treated as a steady state problem without reaction or as an unsteady state problem We will carry out the solution as a steady state problem Recall that the water vapor condenses on compression so that the compressed gas is still saturated Steps 1 2 3 and 4 19L mol F I H lb mol P lb mol air A n alr til 97 9 0 H20 P quotA 2 8OOF 745 mm Hg 80 p Liquid WIb mol 25psi9397psia H20 100 2053 mm Hg Step 5 Several basis can be considered F 100 lb mol P 100 lb mol W 100 lb mol F 745 lb mol and so on but we will take a tie component as a convenient basis Basis F 100 lb mol Step 4 Calculate the gas compositions F P pa at 80 F is 05067 psia 262 mm Hg abs p3 262 Iang pAin F 745 262 7188 mm Hg pA in P 2053 D 262 2027 mm Hg 11 pf 7188 mi pi 2027 EEW EE Steps 6 7 8 and 9 We can make two component balances and one total balance F P W of which two are independent Two unknowns exist W and P Total balance lb mol 100 P W Air balance lb mol 2 718 8 100 so that P 0977 lb mol 745 2053 W 100 D 0977 0023 lb mol 100 002300364 63 Chapter 18 Partial Saturation and Humidity 95 GAS COMPRESSORS TURBO COMPRESSORS Imoejiers DISCT BFQE voiutes Impeller inlet labyrinth seals i rr 39 T u H i v r 39 g i smtawe 75 7 met r 39Tii bearing I r Cowpresso 39 quotv a discharge g 39 nczze 7 quotT a Wetgas centrifugal compressor Fig 18a Centrifugal compressor quarter Fig 18b One half section of a wet gas centrifugal section open compressor courtesy of Sul aer Turbosystems Compressors take in gas at one pressure and release it at a higher pressure Some axial flow compressors have adjustable guide vanes such as occur in units for natural gas liquificati on A minimum flow exists below which the compressor operation becomes unstable surges and back flow can occur as well as mechanical damage A different type of compressor found in common use is the reciprocating compressor which uses pistons to compress gases to higher pressures than can be achieved with axial flow units but can only handle smaller volumes of gas Both types of compressors can be used in stages to achieve higher pressures than provided by a single unit The approximate range of functioning of compressors is in Fig 18c 1 o3 r DISCHARGE PRESSURE PISA 39o N l Tl 39 O I E O I I I 103 104 105 1 0 INLET FLOW RATE CUBIC FEET PER MINUTE 1 I Fig 18c Range of operation of compressors 1o 102 96 Chapter 18 Partial Saturation and Humidity Problem 184 Toluene is evaporated into dry air The resulting mixture at 40 C and a pressure of 1013 kPa has a percentage saturation of 50 It is desired to condense 80 of the toluene in the mixture by a process of cooling and compressing If the temperature is reduced to 50C determine the pressure to which gas must be compressed Solution Steps 1 2 and 3 This is a steady state process without reaction kg mol n F Tol n F i aquot T kg mol F Liquid P Toluene 80 of initial Step 4 Use the Antoine Equation for toluene to get pquot for toluene A 160137 B 309652 C 5367 B MP3 2 A 39 CEET atT400C p f5871mmHg abs atT5 C p f9118 mmHg abs pTotall 1013 kPa 760 mm Hg At 40 C PT01 050 p01 050 5871 2936 mm Hg Step 5 Basis F 100 kg mol Step 4me toluene condensed is T 080000 00386 2 3088 kg mol Calculate the composition of the inlet gas pgol nllbt p ot 760 Step 6 Unknown are nah ngol P Chapter 18 Partial Saturation and Humidity 97 Steps 7 8 and 9 An overall balance an air balance and a toluene balance can be written only two are independent But n n 01 P is the third equation needed Air balance kg mol 100 1 7 00386 nilI Toluene balance kg mol 100 00386 3088 n 01 Note that because of condensation the toluene in P is saturated and amounts to 20 of the original toluene or 077 kg mol The partial pressure of the saturated toluene in P is 9118 mm a s pTol 2 PM nTol 077 pTot pTot um 0779614 9691 PTot 9114 1150 mm Hg abs 153 kPa 98 Chapter 18 Partial Saturation and Humidity Problem 185 A constantvolume vessel contains dry air at 66 F and 212 psia One pound of liquid water is introduced into the vessel The vessel is then heated to a constant temperature of 180 F After equilibrium is reached the pressure in the vessel is 310 psia The vapor pressure of water at 180 F is 751 psia a Did all of the water evaporate b Compute the volume of the vessel in cubic feet c Compute the humidity of the air in the vessel at the nal conditions in pounds of water per pound of an Solution Steps 1 2 and3 Assume the liquid water occupies a negligible volume in the vessel The system is the vessel GA F lb mol Initial Final Gas Gilb mol F G H20 nHZO a a H20 nHZO Air HF T766F T7180F Air HG Sir PAir 212 psia pTot 310 psia ir Total nTot V ft3 V ft3 Total nTot Possible liquid 1 lb 1 d H O 1qu1 2 20 imi p11 66 F 03162 psia pgzo 180 F 751 psia Step 5 Basis Initial gas at conditions shown in the Figure Steps 6 7 8 and 9 First we have to determine if all the water is vaporized or not in the vessel We can calculate the partial pressure of the air in the vessel at 1800F by applying the ideal gas law to both the initial and nal conditions and using as the material balance the equality of the initial and final moles of air 111111r ngir pfmv nilIRTF piirv ng RTG G G F T l80 460 0r PAir PAirTF 212m 269 ps1a Thus the pressure of the water vapor in the air at the nal conditions is G 7 7 pHZO 7 3117 269 7 42 ps1a Because 42 lt 751 psia the air at the nal state is not saturated Chapter 18 Partial Saturation and Humidity 99 a All the water vaporizes We can calculate the volume of the vessel from the nal given conditions plus the value of the partial pressure of the water vapor in the nal state G G V nHZORT X8073640 H20 Gi 908 ft3 pHZO The volume of the water vapor at its partial pressure is the same as the volume of the vessel c The humidity is 421b mol HZOi 181b H20 l11b mol air 7 lb H20 269 lb mol air 1 1b molH20 291b air 0097 lb air 100 Chapter 18 Partial Saturation and Humidity Problem 186 A silica gel drier removes 1000 kg of water per hour Air is supplied at a temperature of 55 C and a dew point of 265 C The air leaves the drier at a temperature of 32 C and a dew point of 72 C The pressure in the system is constant at 1000 kPa Calculate the volume of the wet air at the initial conditions which is supplied per hour Solution Steps 1 2 3 and 4 Some of the data in the gure have been calculated in Step 4 below The process is a steady state one without reaction Because gas are involved the calculations will be in moles The systems is the drier P mol T 32 C Air 09898 DP kP a silica gel H20 0391 pTOt T 55 C 39 DP 265 C pTot 1000 kPa mol fr Fkg mob Air 0965 H20 0035 w 5556 kg mol 1000 kg 1000 mol fr H20 100 Step 5 Basis 1 hour Step 4 1510 265 C 350 kPa Z 72 C 102 kP sz0 a Calculate the stream compositions rst Inlet air Outlet air 7 u 7 7 a 7 sz0 7 pHZO 265 C 7 350 kPa pHZO 7 pHZO 72 C 7 102 kPa pT 1000 kPa pT 1000 kPa Step 6 Unknowns are F and P Step 7 Balances are H20 and air or total Steps 8 and9 Air F0965 P 09898 F 2214 kg mol H20 F0035 P00102 5556 V 604 X 104m3hr at 550C and 100 kPa nRT 7 22148314328 1000 Chapter 19 VaporLiquid Equilibria 101 Chapter 19 The Phase Rule and VaporLiquid Equilibria for Multicomponent Systems Problem 19 1 Assume that Raoult39s Law holds for the f lggwing mixture mo e nHexane 20 Benzene 50 Toluene 30 a What is the dew point pressure of the mixture if it is at 150 F b What is the dew point temperature of the mixture if it is at 85 psia c What is the bubble point temperature of the mixture if it is at 70 psig The barometer reads 780 mmHg d What is the bubble point pressure ifthe mixture is 150 F Solution The dew point pressure is the pressure of which the vapor rst starts to condense at 150 F The mixture is assumed to be all vapor and the condensate composition is determined by H n 7 Yi l 7 E YI 17 E 7 or i 7 pTH p PT 11 Pi Assume an ideal solution exists and use the Antonine Equation to determine pf for the pure components 7 B 7 a 1npeAicT T7150F73386K A B C Pi Yi YiPi nHexane 158366 269755 41878 68467 020 2921x 104 Benzene 159008 278851 75236 47293 050 1057x 103 Toluene 160137 309652 75367 17175 030 1747x 103 3096 X 10393 1 Yi 3096 gtlt103 PT 2 pi pT 333 mmHg b The dew point temperature is the temperature at which the vapor rst condenses when the vapor and liquid are in equilibrium For an ideal solution at 85 psia 4395 mmHg the relation to use 1s L 2i pT pi B A77 p eXp CT 7 02 05 03 4394 6 T 2697 55 2788 51 3096 52 39 e 158366739 e 159008739 e mom 739 4878T 5236T 5367T 102 Chapter 19 VaporLiquid Equilibria Solve for T by trial and error or on a computer to get T 500K pT 13105 mmHg T 425K pT 3842 mmHg T 435K T 435K pT 4321 mmHg c The bubble point temperature at a given pressure is the temperature at which the liquid mixture vapor pressure equals the total pressure The mixture is assumed to be all liquid and the vapor composition is assumed to be in equilibrium with the liquid The relation used under these assumptions for an ideal liquid is pfzpfn 70 psi 7600 mmHg 7 14696 psia 3619 mmHg pT 3619 780 4399 mmHg abs 278851 158366 5236T 03 e 4399 020 e 269755 159008 4878T 05 el 160137 30963952 5367 T Solve for T by trial and error or on a computer to get T 400K p 2372 mmHg T 425K p 4070 mmHg T 429K T 430K p 4496 mmHg 1 The bubble point pressure at 150 F 3386K is the pressure at which the liquid rst starts to vaporize when the vapor and liquid are in equilibrium For an ideal liquid pT Z 131 X1 94 A B C P1 Xi PiXi nHexane 158366 269755 4878 68467 020 13693 Benzene 159008 278851 75236 47293 050 23647 Toluene 160137 309652 75367 17175 030 5153 4249 pT 425 mmHg Chapter 19 VaporLiquid Equilibria 103 Problem 192 A natural gas has the following analysis at 400 kPa Mole Propane 780 Butane 120 Pentane 100 Its temperature must be kept above what value to prevent condensation If it were cooled what would be the composition of the liquid that rst 39 out of the gas Solution Basis 100 mol gas at 400 kPa Assume ideal vapor and liquid at the dew point Xi yiKi in 1 K1 p pT 2yiKi10r pTZyi pi 1 for an ideal solution Procedure Assume atemperature and obtain values for p from the Antonine equation or a handbook Calculate Zyi pf pT to see if the sum equals 10 If not repeat t0 bracket the value 1 1st iteration let T 310K pfkPa y pf pT 2yi pf C3 1269 6147 x 104 C4 3434 3494 x 104 4001999 x103 0768 C5 1047 9551 x 104 Z1919gtlt10 31ltPa 1 2nd iteration let T 290K p Yi pi pTZYi pf C3 765 1020 X 10393 C4 1865 6434 X 10393 400 3644 X 10393 1457 C5 505 1980 X 10393 Z 3 644 gtlt10 31ltPa 1 3rd iteraction T 300K pf Y pf pT 2Yi pf C3 9935 7851 X 10394 400 2610 X 10393 104 C4 2559 4689 X 10394 close enough C5 7376 1356 X 10393 2 2610 gtlt10 31ltPa 1 T 300K 104 Chapter 19 VaporLiquid Equilibria At the dew point temperature the liquid which rst condenses has the composition Xi YiPT pi YipT Pi 078400 kPa 0 314 C3 9935 kPa 012 400 kP C X a 0188 4 2559 kPa X 010400 kPa CS 7376 kPa 039542 2X1 104 Chapter 20 FluidSolid Equilibria 105 Chapter 20 Liquids and Gases in Equilibrium with Solids Problem 201 Yaws Environmental Engineering World MayJune 1995 p1620 suggested the following empirical correlation for the equilibrium adsorption of organic compounds on activated carbon loglo y A B loglo p C logio plz where y is the g of compound adsorbed 100 g of activated carbon p is the concentration of gas at 25 C and 1 atm in parts per million by volume ppmV ABC are constants in the correlation equation Yaws lists values of ABC for 283 organic compounds A pollutant in the exhaust stream from Unit A in a plant contains 100 ppm of toluene C7H8 at 25 C and 1 atm Ifthe exhaust stream is passed through a bed of activated carbon so that the stream reaches equilibrium with the carbon estimate the minimum amount of activated carbon needed to remove all of the toluene per m3 of the gas phase The coef cients for toluene are A 111466 B020795 C002016 aolution Step 2 The process is shown in Figure A201 106 Chapter 20 FluidSolid Equilibria A 9 carbon yo 0 g toluene 100 9 carbon V G X0 100 ppmv X1 O ppmv G 3 gt gt 3 1 m toluene toluene 1 m y A 9 carbon y1 g toluene 100 9 carbon Figure A20l Because the concentrations of toluene in the gas phase are so small G can be considered to be the solute free volume as well as the total volume of gas Step 5 Basis l m3 of gas at 25 C and 1 atm containing 100 ppmv oftoluene Steps 3 and 4 Calculate the amount of toluene in G using the ideal gas law n 1013k13a100x10 6m3 T I 83141ltPam3298K kg molK 4088 X 10392 g mol or 00376 g Calculate the amount of toluene absorbed per 100 g of activated carbon using Yaw s equilibrium relation 16g10 y 111466 020795 16g10 100 7 002016 16g101002 Chapter 20 FluidSolid Equilibria 107 At equilibrium for 100 ppm the amount of toluene removed per mass of carbon is y 2818 gtoluene 100 g carbon Steps 6 7 8 and 9 The material balance is GX0 X1 AY0 Y1 and with y1 0 00376g tolune X0 X1 1m3 gas 0 133 g carbon G yo 2818g toluene 39 m3 gas 100g carbon 108 131 133 Fresh Feed Gas at 1000C and 150 kPa 260 Lmin ofC6H6 950 Lmin of H2 Part 3 Additional Problems PART 3 ADDITIONAL PROBLEMS Answers will be found in Appendix A Chapter 13 In a test on an oilfired boiler it is not possible to measure the amount of oil burned but the air used is determined by inserting a venturi meter in the air line It is found that 5000 ft3min of air at 800F and 10 psig is used The dry gas analyzes C02 107 percent CO 055 percent 02 475 percent and N2 840 percent Ifthe oil is assumed to be all hydrocarbon calculate the gallons per hour of oil burned The sp gr of the oil is 094 In the manufacture of dry ice a fuel is burned to a ue gas which contains 162 percent CO 2 48 percent 02 and the remainder N2 This ue gas passes through a heat exchanger and then goes to an absorber The data show that the analysis of the ue gas entering the absorber is 131 percent C02 with the remainder 02 and N2 Apparently something has happened To check your initial assumption that an air leak has developed in the heat exchanger you collect the following data with a wettest meter on the heat exchanger Entering ue gas in a 2min period 47800 ft3 at 6000F and 740 mm of Hg Exit ue gas in a 2min period 30000 ft3 at 600F and 720 mm of Hg Was your assumption about an air leak a good one or was perhaps the analysis of the gas in error Or both Benzene C 6H6 is converted to cyclohexane C6H12 by direct reaction with Hg The fresh feed to the process is 260 Lmin of C6H6 plus 950 Lmin of H2 at 1000C and 150 kPa The single pass conversion of H2 in the reactor is 48 while the overall conversion of H2 in the process is 75 The recycle stream contains 80 H2 and the remainder benzene no cyclohexane a Determine the molar ow rates of H2 C6H 6 and C6H12 in the exiting product b Determine the volumetric ow rates of the product stream if it exits at 100 kPa and 200C c Determine the molar ow rate of the recycle stream and the volumetric ow rate if the recycle stream is at 1000C and 100 kPa Recycle Gas Stream R 80 H2 20 C6H6 P gas C6H6 H2 C6H12 Reactor Separator 134 A natural gas at 300C has the composition of C H4 80 C2H4 10 N2 10 Part 3 Additional Problems 141 142 143 162 171 109 If the pressure in the line is 100 kNmz what are the partial pressures of the three components If the temperature is raised to 400C will the partial pressures change If so what will they be Chapter 14 and 15 Find the molar volume in cm3 g mol of propane at 375 K and 21 atm Use the Redlich Kwong and PengRobinson equations and solve for the molar volume using 1 a nonlinear equation solver and 2 the compressibility factor method The acentric factor for propane to use in the PengRobinson equation is 01487 Also check your results with the value found in a data base or a handbook What weight of ethane is contained in a gas cylinder of 10 ft3 in volume if the gas is at 1000F and 2000 psig Use 1 an equation of state and 2 the compressibility factor method Calculate the volume occupied by 20 lb air at 735 psia and 392OF Chapter 16 Estimate the vapor pressure of aniline at 3500C from the following vapor pressure data the experimental vapor pressure is 40 atm tquotC 1844 2128 2548 2927 patm 100 200 500 1000 a Prepare a Cox chart to solve this problem b Fit the coefficients in Antoine equation using a nonlinear optimization code or a nonlinear regression code and predict the value from the Antoine equatlon Estimate the vapor pressure of benzene at 1250C from the vapor pressure data TquotF 1026 212 p psia 336 255 by preparing a Cox chart Chapter 17 Carbon disulf1de C82 at 200C has a vapor pressure of 352 mm Hg Dry air is bubbled through the C82 at 200C until 445 kg of C82 are evaporated What was the volume of the dry air required to evaporate this C82 assuming that the air is saturated if the air was initially at 200C and 10 atm and the final pressure on the airCS vapor mixture is 750 mm Hg 110 182 Part 3 Additional Problems A mixture of acetylene C2H2 with an excess of oxygen measured 350 ft3 at 250C and 745 mm pressure After explosion the volume of the dry gaseous product was 300 ft3 at 600C and its partial pressure of 745 mm Calculate the volume of acetylene and of oxygen in the original mixture Assume that the nal gas was saturated and that only enough water was formed to saturate the gas An 8007liter cylinder contains a gas saturated with water vapor at 2500C and a pressure of 1026 kPa What is the volume of the gas when dry at standard conditions Oxalic acid H 2C204 is burned with 248 percent excess air 65 percent of the carbon burning to CO Calculate the dew point of the ue gas Chapter 18 Toluene is mixed with air at 210C in such proportions that the pa1tial pressure of the vapor is 133 kPa The total pressure is 993 kPa Calculate the following The relative saturation b The moles of toluene per mole of vaporfree gas c The weight of toluene per unit weight of vaporfree gas d The percent saturation e The percentage of toluene by volume A rigid vessel which is 1 ft3in volume contains 1 lb ofNZ and 1 lb of H20 at 100F a What is the pressure psia in the vessel b What is the molal humidity in the vapor phase c What mass fraction of the water is liquid Around airports jet aircrafts can become major contributors to pollution and as aircraft activity increases and public concern brings other sources under control the relative contribution of aircraft to pollution could go up Recently federal state and local govemment pressure has speeded the introduction of new combustors in aircraft In a test for an aircraft fuel with the average composition C120H440 the fuel is completely burned with the exact stochiometric amount of air required The air is supplied at 240C and 100 kPa with a humidity of 80 percent For the entering air compute the following 1 The dew oint 2 The molal humidity 3 The relative humidity Leather containing 100 percent of its own weight of water ie if the dry leather is 1 lb the water is 1 lb is dried by means of air The dew point of the entering air is 400F and in the exit air it is 55oF If 2000 lb of leather are forced through the dryer per hour how many pounds of water are removed per hour The barometer reads 750 mm Hg The following is the solution to the problem Explain whether the solution is correct or not If not what is the solution Basis 750 lb mol wet air in pHZO at 400F 629 mm Hg p1le at 55 F 71105 mm Hg lb mol dry air in 750 7 629 7 7437 lb mol lb mol dry air out 750 71105 7 739 lb mol Part 3 Additional Problems 111 629 lb 1 m amp 114 1b H20 1 lb mol Dry air balance 1105 lb mol 7437 18 1b 7 200 1b H 0 H20 739 1 1b mol 2 H20 absorbed 200 7 114 86 1b H20 Coma mol mol tract mol tract M W l Dry Air 7437 0991 0991 29 287 H20 629 0009 0009 18 02 Total 750 10 10 289 Basis 2000 lb leather 1 hr 2000 lb 1eather1050 1b H20186 1b H20 removed 754 libhr 1 hr 11 1b leatherl 114 1b H20 in 185 Air at 300K and 100 kPa has a dew point of 289K a b c d e What is the percent relative humidity To what pressure must this air be compressed to cause condensation to start the temperature remains at 300K To what temperature must this air be cooled to remove 25 percent of the moisture the pressure stays constant at 100 kPa What would be the percent relative humidity of this air after heating to 340K at constant pressure Did the molal saturation change during the heating indicated in part d 186 Air saturated with water vapor is at 1400F and a pressure of 2968 in Hg a b C To what temperature must the air be cooled to separate 68 percent of the water in it as liquid water pressure is constant To what pressure must the air be compressed to separate 68 percent of the water in it as liquid water temperature is constant If the temperature is reduced to 1000F and the gas is compressed to 25 psia what percentage of the water separates out as liquid Soybean akes from an extraction process are reduced from 096 lb of C2HCl3 per pound of dry akes to 005 lb of C2HCl3 per pound of dry akes in a desolventizer by a stream of N 2 which vaporizes the C2HC13 The entering N2 contains C2HCl3 such that its dew point is 300C The N2 leaves at 900C with a relative saturation of 60 The pressure in the des a b olventizer is 760 mmHg and 1000 lbhr of dry akes pass through the drier Compute the volume of N2 plus C2HCl3 leaving the desolventizer at 900C and 760 mm Hg in cubic feet per minute The N2 leaving the desolventizer is compressed and cooled to 400C thus condensing out the C2HCl3 picked up in the desolventizer What must the pressure in the condenser be if the gas is to have a dew point of 300C at the pressure of the desolventizer 112 191 195 196 Part 3 Additional Problems Chapter 19 Equilibrium ash vaporization of a liquid mixture of ethane 15 mol propane 15 mol nbutane 15 mol and pentane 15 mol at 1400F and 200 psia takes place in a vessel What ratio of liquid to vapor exists when vaporization is complete Kvalues are respectively 370 138 057 and 021 A vapor composed of 30 percent nhexane 30 nheptane and 40 noctane is partially condensed at 100 kPa Assume that the vapor and liquid are ideal solutions Calculate the mole fractions of the three components in the vapor at equilibrium if 80 percent of the original vapor condenses A solution containing 12 wt percent of dissolved nonvolatile solid is fed to a ash distillation unit The molecular weight of the solid is 1230 The effective vapor pressure of the solution is equal to P P x where p effective vapor pressure of the solution x mole fraction of water p vapor pressure of pure water The pressure in the ash distillation unit is 1121 psia and the temperature is 1000F Calculate the pounds of pure water obtained in the vapor stream per 100 lb of feed solution and the weight percent of the dissolved nonvolatile solid leaving in the liquid stream Consider atank of water at 00C under a pressure of 1013 kPa at the surface of the water Would it be possible to maintain ice at the bottom of the water tank at 00C in equilibrium with the water Determine the equilibrium vapor pressure in kPa of the following liquid at 21 C Component Mole percent C2 H6 30 C3 H8 680 C4 H10 240 C5 H12 50 Assume Raoult s law applies If all of the C2 H6 were removed from the liquid cited in Problem 191 what would the equilibrium vapor pressure be What would the vapor composition be for Problem 191 Part 3 Additional Problems 198 113 Calculate the total of 1b of O2 in a gas phase of 3 ft3 in volume that exists in equilibrium with 180 lb of water at 30 C Assume Henry s law applies Calculate a the boiling point at atmospheric pressure and b the composition of the vapor in equilibrium with the liquid of the following mixture in mole n butane 50 npentane 170 nhexane 650 nhaptene 100 noctane 30 Chapter 20 Water contains a toxic solute that is to be extracted with polymer beads Five ppm parts of beads by weight in the water will reduce the toxic solute to 25 of its original value and 10 ppm will reduce it to 35 of the original value Estimate how much polymer is required to reduce the concentration to 05 of the original value Fit the following adsorption data for CO2 on silica gel at 298K using the Freundlich isotherm p mm Hg x g adsorbed 1000g gel p mm Hg x g adsorbed 1000g gel 195 1112 3879 9152 430 1637 485 1289 862 257 588 1729 137 3599 708 1808 2476 6116 763 257

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