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# ChemBio Engr Process Analysis CHE 2011

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This 216 page Class Notes was uploaded by Kaela Funk on Wednesday October 21, 2015. The Class Notes belongs to CHE 2011 at Tennessee Tech University taught by Joseph Biernacki in Fall. Since its upload, it has received 11 views. For similar materials see /class/225688/che-2011-tennessee-tech-university in Chemical Engineering at Tennessee Tech University.

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Date Created: 10/21/15

Chapter 1 Introduction 11 Read the MSDS Material Safety Data Sheet and write a oneparagraph description of the hazards associated with dimethyl mercury AE methanol FJ ethyl ether KO benzene P T phosgene U Z Upper case letters refer to the last names of students Information is available on the Internet at your campus library and on the Web page of your local department of environmental health and safety 1 1 One convenient Web page is httpsiriuvmeduindexhtml The Safety Information Resources Inc SIRI is supported by the University of Vermont The website is linked to OSHA and EPA as well as commercial sites where MSDS of a large number of chemical components can be easily found There are five answers to this question and it would be nice if each one were answered by means of a different Web page At UC Davis the Web page for the Department of Environmental Health and Safety will lead one to MSDS information and it would be good to identify this as a source since there should be a comparable source on every campus I hope that Brian could supply the answer to this homework problem Chapter 2 Units Section 22 2 1r Convert the following quantities as indicated a 5000 cal to Btu b 5000 cal to wattsec c 5000 cal to newtonmeter 2 1 The solution requires the simple use of conversion factors that are contained in 5000 cal 5000 cal 1 a 5000 cal 3m 252 cal 198 Btu 5000 cal 5000 cal 1 b 5000 cal xm 252 cal Btu 20933 watt s 5000 cal 5000 cal C cal 20 935 N m 2 2 At a certain temperature the viscosity of a lubricating oil is 0136 10 3 lbf sftzr What is the kinematic viscosity in m2s if the density of the oil is p 0936 gcmsr 2 2 The kinematic viscosity is given by H 0136x10 31bfs 2 V 7 p 0936gcm3 Use of the appropriate conversion factors leads to V Z 0136X10 31be 232171bmft 4536g ix in X m2 0936gcm3 1 lbfs2 1bm 12m 254cm104cm2 Chapter 2 and the final result is given by 0136x10 332174536 2 m7 696x10 6m2s S 093612254104 23 The density of a gas mixture is p 13 kgmsl Calculate the density of the gas mixture in the following list of units a lbmft3 b glen c glt d lbmin3 2 3 Parts a and b can be determined directly in terms of the conversion factors listed in Table 2 4 leading to 3 a p 1 31 lbw 3 08121bmft3 m 16018kgm 3 b p 1 31 13gtlt10 3gcm3 m 10 kgm Parts c and 1 require the use of several conversion factors as indicated by 3 3 3 3 C p 13kg 10 gXIO cm m 13glt rn3 kg 1t 106cm3 3 3 d p 13kg lbm 10 g 03048m 470X1051bmm3 m3 4536g kg 121n 24 The cgs system of units is commonly used in science What is the unit of force in the cgs system Find the conversion factor from this unit and a Newtonl Find the conversion factor between this unit and a lbfl 2 4 The letters c g s represent centimeters grams and seconds and the unit of force in this system is a dyne as indicated in Table 2 4 Table 2 4 also provides the conversion factors given by 1 dyne 10 5 Newton 2248x10 6 1bf ldyne Units 3 25 In rotating systems one uses angular velocity in radians per second How do you convert revolutions per minute revmin to rads 2 5 A revolution represents 3600 or 21c radians thus the conversion factor that we need can be expressed as 1rev 3600 21Eradians Use of this result leads to 1 revmin min rev 21Eradxmin rev 6 01047rads 0s 26 The monetary unit of the American economic system is the dollar An ounce troy of platinum used as a catalyst in many chemical processes and in automobile catalytic converters costs 100 If a catalytic converter has 5 grams of platinum what is the value in dollars of the platinum in the catalytic converter 2 6 In this case the conversion is given by 1764 converter 5g 5g lbm gtlt16ouncegtlt 100 converter converter 4536 g lb ounce m 27 In the textile industry filament and yarn sizes are reported in denier which is defined as the mass in grams of a length of 9000 meters If a synthetic fiber has an average specific gravity of 132 and a filament of this material has a denier of 50 what is the mass per unit length in pounds mass per yard What is the cross sectional arw of this fiber in square inches The specific gravity is defined as specific gravity lens Hy dens1ty of water 2 7 To begin this problem we note that there is a conversion factor hidden in the problem statement ie 1 denier 1g9000 m 1 or 1denier 111gtlt101 gm 2 The answer to the first question is given by mass per 5d 555gtlt101 3 unit length emer gm 4 Chapter 2 and in order to obtain the desired units we make use of the proper conversion factors to express Eq 3 in the form massper 555x10 4gf 1bm 254X10 2m 36in unit length m 14536 g in yard 4 112gtlt10 61bmyard To determine the cross sectional area we first note that the mass is given by mass densitygtlt volume 5 2 p A L gt The density of the yarn can be expressed as p YPHZO 6 where y is the specific gravity Substitution of Eq 6 into Eq 5 and solving for the area leads to A illL 7 YPHZO This represents the solution to the problem and we can obtain the answer by making use of the values for the various parameters to obtain 41 2 l 2 A demer 111gtlt10 gX10 mx in 1321 gcm3 denier m cm 254 cm 8 0652x10 6 in2 28 Adopted from Safety Health and Loss Prevention in Chemical Processes by AlChE The level of exposure to hazardous materials for personnel of chemical plants is a very important safety concern The Occupational Safety and Health Act OSHA defines as a hazardous material any substance or mixture of substances capable of producing adverse effects on the health and safety of a human being OSHA also requires the Permissible Exposure Limit or PEL to be listed on the Material Safety Data Sheet MSDS for the particular component The PEL is defined by the OSHA authority and is usually expressed in volume parts per million or ppm Vinyl chloride is believed to be a human carcinogen that is an agent which causes or promotes the initiation of cancer The PEL for vinyl chloride in air is 1 ppm ie one liter of vinyl chloride per one million liters of mixture For a dilute mixture of a gas in air at ambient pressure and temperature one can show that that volume fractions are equivalent to molar fractions Compute the PEL of VC in the following units a moles of VCm3 b g of VCm3 0 moles VCmole of air Units 5 2 8 We begin the formulation of this problem with the definition of PEL in the form of an equation Permissible Exposure PEL 1 cm3 of VC in 1 m3 of of 1 ppm Limit of Vinyl chloride VC a mixture of air and VC and note that this can be expressed as 12 cm3 of VC VC m3 of mixture a To determine the PEL in terms of moles of vinyl chloride we make use of the standard molar volume of an ideal gas to construct the conversion factor mole mole mole volume 24141 002414 m3 and this allows us to express the Permissible Exposure Limit as PEL of vc cm3 of vc mole 106m3 in InOIGS m3 of mixture 002414m3 cm3 414gtlt10 5 moles of vc m3 b To determine the PEL in terms of grams of vinyl chloride we make use of the molecular mass of vinyl chloride mass 645 g mole mole and this allows us to express the Permissible Exposure Limit as PEL of VC 414gtlt10 5 moles of VC X 645g 267 X10 3 g of VC in mas s m3 mole m3 c In an ideal gas mixture volumes of the individual species are additive and we have 1 cm3 of vc 103 of vc vc m3 of mixture 1 10 6m3 of air 103 of vc and for practical purposes this takes the form 6 Chapter 2 121 cm3 of vc 10396m3 of vc VC m3 of mixture In3 of air Since the volume is directly proportional to the moles in an ideal gas this takes the form 12 M 6 moles of VC VC m3 of mixture moles of air 29 A liquid has a specific gravity of 0865 What is the density of the liquid at 20 C in the following units 21 kgm3 b lbmft3 c gcm3 d kglt 2 9 Specific gravity is the ratio of the density of the liquid with respect to the density of water at 20 C which is given by pH20 99821 kgm3 Given this value we can compute the answers according to a The density of the liquid is pliq yszo 0865 99821 kgm3 86345 kgm3 b In Table 2 4 we have the conversion factors that provide p1 863453 H1 3 lbm 00283 m3 x 3 5387 lbmft3 04536 kg ft c In SI units a kilo is 1000 times the quantity and a centi is 1100 times the quantity Use of this nomenclature immediately leads to 3 0112 86345k g 1000ggt m 086345 gcm3 m3 kg 100 cm d There are 1000 liters in one cubic meter thus the density of the liquid is given by 3 pliq 86345Ig m 086345 kglt 3 10001 Units Section 23 210 In order to develop a dimensionally correct form of Eq 29 the appropriate units must be included with the numerical coefficients 585 and 1683 The units associated with the first coefficient are given by Eq 210 and in this problem you are asked to find the units associated With 1683 2 10 In order for Eq 2 9 to be dimensionally homogeneous the coefficient 585 must be replaced with the quantity indicated in Eq 2 10 and the coefficient 1683 must have a certain set of units associated with it To begin with that coefficient must contain the units of length In addition it must have the units necessary to cancel the units associated 045 with 3 cpg Alittle thought will indicate that the dimensionally correct form of the drop diameter must be expressed as 05 5856 g cm2 U M s2 dyne D32 045 045 045 15 168310 6m 2 Pasfo 45 2 i 2 1000518 3 Jopg CIn cm G P2 From Table 2 4 we can extract the conversion factor given by Pa cm2 1 0 dyne and this leads to 05 585JE gcm2 J D 32 UM s2 dyne 045 P 2 045 d 0 45 L 15 168310 6m 2 Pasfo 45 acm 7g yne 2 1000718 4 Jopg 10dyne cm G pg which can be arranged in the form 05 585 gcm2 U1 pg s2 dyne D32 L35 045 15 597gtlt104cmg W J lOOO pi 52 dyne VGPZ P2 8 Chapter 2 This means that the coefficient 1683 in Eq 2 9 must be replaced with 045 1683 gt 597gtlt10 gs2c1yneT in order to create a dimensionally correct equation 211 In the literature you have found an empirical equation for the pressure drop in a column packed with a particular type of particle The pressure drop is given by the dimensionally incorrect equation 015 085 135 u H p V Ap 47 0112 J p which requires the following units Ap 2 pressure drop lbfft2 u uid viscosity lbmft s H 2 height of the column ft p 2 density lbmft3 V superficial velocity fts d p 2 effective particle diameter ft Imagine that you are given data for u H p v and dp in Sl units and you wish to use it directly to calculate the pressure drop in lbf ftz How would you change the empirical equation for Ap to obtain another empirical equation suitable for use with Sl units Note that your objective here is to replace the coefficient 47 with a new coefficient Begin by putting the equation in dimensionally correct form ie find the units associated with the coefficient 47 and then set up the empiricism so that it can be used with Sl units 2 11 In order for the equation for the pressure drop to be dimensionally homogeneous the coefficient 47 must have a certain set of units associated with it To begin with the coefficient must contain the units of lbf ft2 In addition it must have the units necessary to cancel the units associated with it A little thought will indicate that the dimensionally correct form of the equation for the pressure drop must be expressed as 7015 7085 7185 015 085 185 AP 471 1I1bm rlll i j E at XW ft2 E ft3 s df Since this result is dimensionally homogeneous we can use any set of units for u H p v and d p If we wish to use SI units the following conversion factors must be employed llb 04536 kg 1 ft 03048 m m From these two relations one can construct the following groups Units 9 015 015 0672 lbim E 1 0942 lbim E 1 fts kg fts kg 712 3281E 1 02403 m m 3 085 3 0 85 006241b7r3 jm7 1 00946 Lr3nj H 1 ft kg ft kg 185 185 3281EIij 1 9005 1 S In S In We can use these results in our equation for the pressure drop to obtain 7015 7085 7185 015 085 185 lb kg 71 kg m 12 p Hp V A 297 if 7 7 7 X p ft2 ms m m3 S J m diiz Once again we note that this equation is dimensionally homogeneous and we can use any set of units for u H p V and d p If we wish to use this result as a dimensionally incorrect equation suitable for only SI units we express it in the form 015 085 185 AP 297 HF1 2V d p Keep in mind that this is a dangerous equation that will only produce correct values of Ap in terms of lbf ft2 when the parameters are given in terms of the following units 14 uid viscosity in kgms H height in m p uid density in kgm3 V superficial velocity in ms d p effective particle diameter in In Any departure from these requirements will lead to incorrect results 212 The ideal gas heat capacity can be expressed as a power series in terms of temperature according to C17 A1 AZT A3T2 A4T3 A5T4 Chapter 2 In this dimensionally incorrect equation the units of C17 are joulemol OK and the units of temperature are degrees Kelvin For chlorine the values of the coefficients are A1 2285 A2 006543 A3 12517x10 4 A4 11484gtlt10 7 and A5 40946x10 What are the units of the coefficients Find the values of the coefficients to compute the heat capacity of chlorine in calgr C using temperature in degrees Rankine 2 12 To obtain a dimensionally correct form for the heat capacity each term in the representation must have the same units Given the molecular mass of chlorine MW 7091 gmol and the conversion factors from Table 2 4 1 cal 4186 joule and 1 K 95 R we can follow the procedure outlined in the previous problem to obtain A1 2285 101116 C alx 01 x5 007698C al molK 4186oule 7091 g C gC A2 006543 J ez cal m 1 x5 K 1225104 cal molK 4186oule 7091g 9C gCR 2 A3 12517101 101116 Cal x m01 x 5 K 130210 7 cal molK3 4186jou1e 7091g 9C gCRZ 3 A4 11484107 1011164 Cal 01 x 5 K 663410 11 Cal3 molK 4186oule 7091g 9C gCR 4 4 A5 4094610 101116 Cal gtlt 01 5 K 131410 cal molKS 4186jou1e 7091g 9C gCR Use of these coefficients in the representation for the heat capacity will provide a dimensionally correct equation in which one can use any units for the temperature If one Wishes to use that equation as a dimensionally incorrect empiricism it could be represented as Cp A1 AZT A3T2 A4T3 A5T4 Where the coefficients are given by A1 007698 A2 1225 104 A3 1302 10 7 A4 6634 10 11 and AS 1314 10 14 This will produce correct values of C p in terms of calg C only when the temperature is given in degrees Rankine If any other units are used for the temperature incorrect values for the heat capacity will result Section 24 213 A standard cubic foot scf of gas represents one cubic foot of gas at one atmosphere and 29815 K This means that a standard cubic foot is a convenience unit for moles This is easy to see in terms of an ideal gas for which the equation of state is given by Units 11 11V nRT The number of moles in one standard cubic foot of an ideal gas can be calculated as p one atmosphere n pVRT V one cubic foot T 29815 K and for a nonideal gas one must use an appropriate equation of state In this problem you are asked to determine the number of moles that is equivalent to one scf of an ideal gas 2 13 To solve this problem we need a value for the gas constant and one convenient value is given by R 8205cm3atmmolK which leads to pV 1 atm1 ft3 n E 8205cm3atmmolK29815 K From Table 2 4 we have 1 ft 3048 cm and this allows us to calculate the number of moles in one standard cubic foot sci as 1 atm1 ft3 3048 cm 8205 cm3 atmmol K29815 K1 ft 3 j 1158 moles 214 Energy is sometimes expressed as v2 2g although this term does not have the units of energy What are the units of this term and why would it be used to represent energy Think about the fact that pgh represents the gravitational potential energy per unit volume of uid and that h is often used as a convenience unit for gravitational potential energy Remember that pv2 represents the kinetic energy per unit volume of uid where v is determined by v2 v e v 2 14 We could begin this problem by noting that 1 gravitational potential the grav1tat10nal potential 7 Pg 9amp2 convenience unit for Pg energy per unit volume energy per unit volume In this case pg represents a convenient scaling factor that allows one to discuss the gravitational potential energy per unit volume simply in terms of the height h above some reference plane If we use this same scaling factor with the kinetic energy per unit volume we obtain Chapter 2 for the kinetic energy 1 per unit volume pg convenience unit i 98 2g kinetic energy pv V per unit volume The scaled kinetic energy v22g has the units of length Under certain circumstances to be explore later in a course on uid mechanics v22g becomes a convenient representation for the kinetic energy of a uid Section 25 215 Use MATLAB and the scaling vectors to find the conversion factors for the following cases a lbm ftsAZ to Pa b lbm ftA2sA2 to Joule c lbm ftAZSAB to watts 2 15 Need help from Ramon 216 Use MATLAB and the unitsivectors to find the dimensions of the following product pD v u where p is the density of a uid D is the diameter of a pipe v is the velocity of the uid inside the pipe and u is the viscosity of the uid Re 2 16 Need help from Ramon 217 A useful dimensionless number used in characterization of gasliquid ows is the Weber number defined as Db U172 P s We where p is the density of the uid Db is the diameter of a bubble U is the velocity of the bubble with respect to the surrounding liquid and o is the interfacial gasliquid tension Use MATLAB and the unitsivectors to verify that the Weber number is dimensionless 2 17 Need help from Ramon Section 26 218 Given the following arrays Units 13 3 5 4 2 1 3 A 6 1 9 B 1 2 5 4 3 2 3 5 2 Find a The sum of the arrays A B b The product of the arrays AB c The term by term product of the arrays A B d The transpose arrays AT BT e The product of the transpose arrays AT BT f The term by term product of the transpose arrays A39iB39 2 18 Need help from Ramon Miscellaneous 219i Write an expression for the volume per unit mass 17 as a function of the molar volume I7 that is the volume per mole and the molecular weight MWi Write an expression for the molar volume V as a function of the density of the component p and its molecular mass MWi 2 19 Use SI units I MW kgmole kg mole m A r7 MW 1 V7 Vi 39 MW 9 p 220 Adopted from Safety Health and Loss Prevention in Chemical Processes by AlChE Trichloroethylene TCE has a molecular weight of 1315 so the vapors are much more dense than air Density of air at 25 C and 1 atm is pajI 1178 kgm3 while the density of TCE is pro 537 kgm3i Being much denser than air one would expect TCE to sink to the floor where it could be relative harmless However gases mix thoroughly and at toxic concentrations the differences in density of a toxic mixture with respect to air are negligible Assuming that the volumes of a gas mixture are additive compute the density of a mixture of TCE and air at the following conditions a The timeweighted average of PEL for 8 hours exposure 100 ppm b The 15 minute ceiling exposure of 200 ppm c The 5 minute peak exposure of 300 ppm 2 20 Need help from Ramon Chapter 2 Chapter 3 Conservation of Mass for Single Component Systems Section 31 31 In Figure 31 we have illustrated a body in the shape of a sphere located in the center of tube The flow in the tube is laminar and the velocity profile is parabolic as indicated in the figure Indicate how the shape of the sphere will change with time as the body is transported from left to right Base your sketch on a cut through the center of the sphere that originally has the form of a circle Keep in mind that the body does not affect the velocity profile Figure 31 Body owing and deforming in a tube 31 The circle created by a cut through the center of the sphere would deform as illustrated in Figure 3 la Figure 31a Body owing and deforming in a tube 32a If the straight wire illustrated in Figure 32a has a uniform mass per unit length equal to Em the total mass of the wire is given by mass in L If the mass per unit length is given by x the total mass is determined by the following line integral mass J xdx 160 For the following conditions 2 Chapter 3 00 a0 on x Lz OO0065kgm 0c00017kgm3 Ll4m determine the total mass of the Wire Figure 32a Wire having a uniform or nonuniform mass density 32a The mass contained in the wire can be expressed as mass x L22 fix 1 x and evaluation of the integral leads to 3 mass I OL 2 j 12 Making use of the data that are given provides mass 00017 kgm314m312 3 and the mass is calculated to be mass 00095 kg 4 32b If the at plate illustrated in Figure 32b has a uniform mass per unit area equal to we the total mass of the plate is given by mass VOA we LIL2 If the mass per unit area is given by x y the total mass is determined by the area integral given by yL2 XL1 mass Illd4 I Jwxya5cdv A y0 0 For the following conditions xy we 00065 kgmz or 000017 kgm L1 14 m L2 27 m determine the total mass of the plate Single Component Systems l 141 i Figure 321 Flat plate having a uniform or nonuniform mass density 32b In this case we express the mass ofthe plate as yLz Xlq mass wo ocxy dxdy y0 x0 and evaluate the integrals to obtain y 2 2 mass dy y0L1L2 ocLilLi2 2 2 y0 Making use of the data yields mass m 000017 kgm414m227m24 and the mass is given by mass 002518kg 32 If the density is a function of position represented by n 7 1 7 1 2 P Po0x 711 By 3L2 develop a general expression for the mass contained in the region indicated by 0 x L1 0 y Ll 0 z L3 32 The mass contained any xed volume is given by W V m and for the particular volume under consideration we have 216 FL Fla p dx dy dz z0 y0 x0 When the density is given by 1 1 2 p p on x jlq B y il2 Eq 2 takes the form y 0 x 0 Integration with respect to x yields y 2 m Po L1L213 L3 L151 y and integration with respect to y provides the nal representation for the mass m pom219 Male ad9 Chapter 3 1 2 3 4 5 6 If the density were a constant p0 both a and would be zero and the mass would simply be equal to the density times the volume Section 32 33 To describe flow of natural gas in a pipeline a utility company uses mass ow rates In a 10 inch internal diameter pipeline the flow is 20000 lbm hr The average density of the gas is estimated to be 10 kgm3 What is the volumetric flow rate in ft3s 7 What is the average velocity inside the pipe in ms 33 The volumetric rate in m3s can be calculated by Single Component Systems 5 q mk gxi 1 s s p kg where Kg K8 1 E m S mhrgtlt04536 b X3600 S 2 Then the volumetric ow rate in m3s can be transformed to ft3s as asked by means of the proper conversion factor 3 m3 fi quoti q s Q s X030483 Then the volumetric ow rate in ft3s is mg LA 252 E s r 3600 s s 3 3 im 0252 g S 3 m3 kg 2252 QS SWIG k 11 md C 891 S S m S The average velocity inside the pipe v in ms can be calculated 3 m m 97 v 4 S A m2 where the cross sectional area can be calculated 7239 gtltD2 A 5 a 4 Then Ac is 1 2 and v 6 Chapter 3 3 02523 m v 32 4973 0507m s 34 For the coating operation described in Examples 31 and 32 we have produced an optical fiber having a diameter of 125 micrometers The speed of the coated fiber at the takeup wheel is 45 meters per second and the desired thickness of the polymer coating is 40 micrometers Determine the volumetric flow rate of the coating polymer that is required to achieve this thickness 34 In Example 32 the crosssectional area of the coating polymer where the coated ber leaves the system is given by A poQo polymer 1 polymer p1 polymer ltVgt1 takeup whee Since we have no information about the polymer density at the entrance and exit of the system we are forced to assume that the density is constant in order to arrive at A Q l pelymer ltVgt1takeup wheel Here D0 is the outer diameter of the coating and D is the inner diameter which is equal to 125 micrometers We are given that the thickness of the coating is D 0 D 80 micrometers and this allows us to solve for the volumetric ow rate to obtain 8 3 Q0 polymer 7 933X10 ms 35 Mono Lake is located at about 6000 ft above sea level on the eastern side of the Sierra Nevada mountains and a simple model of the lake is given in Figure 35a The environment is that of a high cold desert during the Winter a thirsty well during the Spring runoff and an cornucopia of organic and avian life during the Summer Mono Lake is an important resting place for a variety of birds traveling the flyway between Canada and Mexico and was once the nesting place of onefourth of the world39s population of the California gull Single Component Systems 40 rninitialradius r1 nnlradius h initialdepth h1 mldepth Figure 3 5a Assumed Mono Lake profile The decline of Mono Lake began in 1941 when Los Angeles diverted the water from four of the five creeks owing into the lake and sent 56000 acrefeet per year into the Owens River and on to the Los Angeles aqueduct By coincidence the surface area of the lake in 1941 was 56000 acres The fall of Mono Lake was apparentlylt secured in 1970 with the completion of a second barrel of the alreadyexisting Los Angeles aqueduct from the southern Owens Valley This allowed for a 50 increase in the flow and most of this water was supplied by increased diversions from the Mono Basin To be definitive assume that the export of water from the Mono Basin was increased to 110000 acrefeet per year in 1970 Given the conversion factor lacrefoot 43560ft3 one finds that 479gtlt109ft3 of water are being removed from the Mono Basin each year In 1970 the surface area of the lake was 185x106 rm2 and the maximum depth was measured as 50 m If the lake is assumed to be circular with the configuration illustrated in Figure 35a we can deduce that the angle 6 is given by 9 03730 In this problem you are asked to determine the final or steadystate condition of the lake taking into account the flow of water to Los Angeles The control volume to be used in this analysis is illustrated in Figure 35b evaporation 7512 gaund water rig 39 water to Los Angels ts Figure 351 Fixed control volume for the steadystate analysis of Mono Lake Details concerning the ght to save Mono Lake are available at httpmonolakeorg 8 Chapter 3 35 In this problem we wish to determine the final or steadystate condition of the lake taking into account the ow of water to Los Angeles The control volume to be used in this analysis is illustrated in Figure 35b and in terms of this control volume the principle of conservation of mass is expressed as d pdV pvndA 0 1 dt V A Since we are concerned only with the steadystate condition the first term in Eq 1 is zero and we have J pvndA 0 2 Ac Here the symbol Ae represents all the entrances and exits for the control volume shown in Figure 35b One of these entrances is at the single stream entering the lake Another entrance is at the bottom of the lake where the ground water seeps into the lake The surface of the lake is both an entrance for rain and an exit for evaporation When the lake has reached its nal steady state con guration we can use the nomenclature illustrated in Figure 35b to express Eq 2 as rn1rn2 rn3 rn40 3 We must keep in mind that these various mass ow rates are average flow rates and that they are averaged over a period of many years Each term in Eq 3 has the units of mass per unit time and each term represents a portion of the area integral given in Eq 2 For example the mass ow associated with the single stream owing into the lake is given by JpvnalA 7M3 4 A3 where A3 represents the crosssectional area of the stream It is convenient to arrange Eq 3 in a form that equates the loss owing to evaporation to the three sources of water and this leads to the steadystate mass balance in the form m2 m4 steady state after 1941 5 loss dueto flow into the lake evaporatlon The rate of evaporation from the lake depends on a number of factors such as water temperature salt concentration humidity and wind velocity and it varies considerably throughout the year It appears that the average rate of evaporation from Mono Lake is on the order of 36 inches per year Note that this rate is given in terms of a Single Component Systems 9 convenience unit and in order to extract the actual mass ux from this measure we need to multiply by the density of water Thus we express the mass ux as W r2 lt6gt where B is the convenience unit of inches per year and r is the radius of the lake as indicated in Figure 35a The parameter 3 should be thought of as an average value for the entire lake over a suitably long period of time say 10 years thus we have in mind that B is a constant for our calculations In Problem 315 the transient behavior of Mono Lake is examined a more reliable value of l is obtained from the available experimental data If we let r1 be the radius of the surface of the lake in the nal steadystate configuration the mass ow rate owing to evaporation is given by 2 m2 W n m If we use this result in Eq 3 we obtain 2 BPHZOTE i m1 7713 m4 8 and it becomes clear that we need to know the mass ow rates for the three sources that supply the lake with water While the average evaporation rate and the average rainfall can be measured in the Mono Basin the ground water ow into the lake is difficult to determine However our knowledge of the state of the lake in 1941 prior to the beginning of shipments of water to Los Angeles will allow us to determine the ground water ow In the primeval state the water diverted to Los Angeles 1 in Figure 35b was actually owing into the lake Under those conditions the steadystate mass balance would yield m1 m2 m3 m4 m5 0 priorto 1941 9 rather than Eq 3 The mass ow rate owing to evaporation is now given by Eq 5 where Va is the radius of the lake in 1941 Use of Eq 6 in Eq 9 allows us to obtain 39 39 4 m5 steady state prior to 1941 10 W ow into the lake evaporat1on We assume the evaporation rate as indicated by B and the mass ow rates indicated by m1 m3 and m4 have remained essentially constant over the past hundred years and will continue to remain so during the next hundred years This allows us to eliminate the term m1 m3 m4 from Eq 8 and solve for the final steadystate radius 7515 r1 r0 17 2 szo Wo 11 10 Chapter3 This is known as a solution to the problem and in order to obtain an answer to the problem it will be convenient to express the mass ow rate of water to Los Angeles as 7515 pHZOQLA 12 Use of Eq 12 in Eq 11 leads to a simpler expression for the radius r1 r0 1 542 13 On the basis of the model pro le shown in Figure 35a we can express the depth of the lake at the nal steadystate condition as h1 r0 1 9M2 tanB 14 WU Since the surface area of the lake in 1941 was 56000 acres we determine r0 and ho to be r o 55 meters The volumetric ow rate to Los Angeles is 479gtlt109 ft3 year and we can use that gure in Eq 13 to obtain 2 12 048 m 1 Carrying out the indicated calculations yields r1 4990 m and the new depth is I39ll 325 m Here we see that the diversions of water to Los Angeles will lead to a 65 decrease in the surface area and if the model pro le shown in Figure 35a is reasonable the maximum depth will decrease by 41 Section 33 36 A cylindrical tank having a diameter of 100 ft and a height of 20 ft is used to store water for distribution to a suburban neighborhood The average water consumption stream 2 in Figure 32 during predawn hours midnight to 6 ANT is about 100 m3hr From 6 AM to 10 AM the average water consumption increases to 500 m3hr and then diminishes to 300 m3hr from 10 AM to 5 PM During the night hours from 5 PM to midnight the average consumption falls even lower to 200 m3hr The tank is replenished using a line stream 1 that delivers water steadily into the tank at a rate of 1120 galmin Assuming that the level of the tank at midnight is 3 m plot the level of the tank for a 24 hour period Single Component Systems 1 l CD 7 m lZo 14 gm w oI ow I l 0 In H u I y s Q K Figure 3 6 Water storage tank for distribution 36 The control volume to be used in this analysis is illustrated in Figure 36 and in terms of this control volume the principle of conservation of mass is expressed as d pdV pvndA 0 1 dt V A This expression can be integrated to constant density and constant cross section area to obtain dh pXACX pXQm prmt 2 dt Rearranging dh Q i Q 2 m 011 3 dt A Integrating hhrorero 4 where to is the initial time and ht0 is the height at t 2 to Two parameters are going to be constant through the different intervals One is the cross section area and the other is the entering volumetric rate The cross section area can be calculated 2 7r100 03048 2 A m 730m And the entering volumetric ow in the right units 3 39 3 Q 3 007065 mm ga 60 s s 12 Chapter 3 a During prodown hours In3 1 hr In3 Q7 100 002778 hr 3600 s s t0 0 t1 6hr or II 216003 ht0 h0 3m Apply now equation 4 3 007065 002778 h 6AM S 21600 0 730ml h6AM 42685m b 6 AMto 10 AM 7723 1 hr 7723 QM 500 013888 hr 3600 s 3 t1 216003 2 10m or 2 36000s ml h21600 42685m Apply now equation 4 3 007065 013888quot hIOAM 3 36000 21600 730ml c h10AM 29225m c 10 AM to 5 PM In3 1 hr 7723 QM 300 008333 hr 3600 s s Single Component Systems 13 t2 360006 t3 17hr orta 61200s 140 036000 29225m Apply now equation 4 3 007065 008333 hSPM S 61200 36000 mmz lt s c h5PM 24852m d 5 PM to 12 midnight m3 1 hr m3 QM 200 005556 hr 3600 s 3 t3 61200s t4 24hr or t 864003 hm h61200 24852m Apply now equation 4 3 007065 005556quoti h12 d S 86400 61200 m 730m2 S c h12md 30069m 3m Plotting the data Chapter 3 Problem 36Evolution of h with time hm 10 timehr 37 The 7th Edition of Perry s Chemical Engineering Handbook 1984 gives the following formula to compute the volume of liquid inside a partially filled horizontal cylinder 2 V TELRziLf aisinoc Where L is the length of the cylinder and R is the radius of the cylinder The angle a is measured in degrees as indicated in Figure 37 The distances L R a can be measured in any units as long as they are the same units Compute the level of the tank as a function of time if the net flow rate into the tank is equal to Q m3s 5 Figure 3 7 Definition of geometric variables in horizontal cylindrical tank The distances L R and H can be measured in any units as long as they are the same units Compute the level of the tank as a function of time if the net flow rate into the tank is equal to Q 3 m s 37 According to the problem the volume of the horizontal cylindrical tank is given by LR2 oc smot 1 V TELRZ where on is a function of time Single Component Systems 15 If we try to calculate the value of h high with this de nition of variables we have h R R cos 2 Then I h0whenot3600 I hRorH2whenoc1800 h2Roerhen0c00 To calculate the variation of the x with time dV 7 p dt Q 3 Now plugging 3 into 1 2 i 39LRZ LR a sina Q dt 2 4 or 2 doc 1 cosor dt Q 5 or doc 2 1 Q 6 dz 7 LRZ coscx l Now to calculate the variation of h with time the equation 2 must be used gmwmm 7 dt 7 2 2 2 dt Plugging equation 6 into equation 7 dh Q sin3 8 dt 2 2 LR2 cos 1 LR cos 1 38 A cylindrical tank of diameter D is filled to a depth 110 as illustrated in Figure 38 At t 0 a Chapter 3 0 We begin the A0 W 11 control A0 a z 433 Man x x xxxx m7xxx xmxxo m xWNxxxxxxnxxxxxxx xxxx x area of ori ce area of orifice xxx aHoQB C4140 2A179 Ap Figure 38 Draining tank Vn wn 9 4 9 f e w 72272 2 x wawawvwwwe Q x x x xx xx x 04 mxxxxxoxxxxxxwxxxxxxuxxx 37494060 4U r I xxx x x xxxn xxxuxx 7 h 4 5 x 9 xxxw xxx xxxxxxxx xxxxx xxxx xxxxxxxx xxxxxxxx xxx xxxx xxxxxx xxxxxx x xx xx x xxx xx xx x uvivw 3 crosssectional area of the tank is large compared to the area of the orifice the pressure in the tank is essent1ally hydrostatlc and Ap 1s g1ven by diameter of 20 cm how long will it take to lower the depth to 1 cm if the diameter of the orifice is plug is pulled from the bottom of the tank and the volumetric flow rate through the orifice is given Here Cd is a discharge coefficient having a value of 06 and A0 is the area of the ori ce If the pgh where h is the depth of the uid in the tank Use this information to derive an equation for the 3 mm by what 1s somet1mes known as Torricelli s law Rouse and lnce 1957 38 The control volume to be used in the analysis ofthis process is shown in Figure 38a The construction of this control volume begins with a cut at the eXit of the tank where information is given and a cut at the gasliquid interface where information is required These two cuts arejoined where V n is known ie Vn Single Component Systems 17 Figure 38a Moving control volume for a draining tank analysis with the macroscopic mass balance for a moving control volume and we express that result as d ndA 0 1 dt I V t A I Here we have used V t to represent the moving control volume and A t to represent the boundary of the control volume We assume that the density can be treated as a constant so that Eq 1 simpli es to dV t dt v wndA 0 2 A t The normal component of the relative velocity v 7 wn is zero everywhere except at the exit of the tank thus Eq 2 simpli es to M JvndA 0 3 dt A0 and this result can be expressed as 61 TC 2 D ht 0 4 61th Q We are given that the volumetric ow rate leaving the tank can be expressed as Q CdAosIZgh 5 and use of this expression in Eq 4 leads to dh 4C A 2 gh 7 7 51 6 I TED We can express the initial condition in the form 1C h ho t 0 7 and this leads to the solution for the depth of uid in the tank as a function of time WMW2g 8 TE Chapter 3 In this problem we are asked to determine the time required to lower the uid depth from 16 m to 1 cm and for this type of problem we rearrange Eq 8 in the form JZ w 702 t 9 2CdA0 2g 1 Here t1 represents the time required to lower the uid depth from h0 to hl and for the parameters associated with this problem the time is given by t1 64 minutes and 58 seconds 10 In subsequent courses there should be an opportunity to compare this result with experiment and this will provide an indication of the validity of the representation of the volumetric ow rate given by Eq 5 39 In Figure 39 we have shown a tank into which water enters at a volumetric rate Q1 and leaves at a rate Q0 that is given by Q0 06 1401 gh Here A0 is the area of the orifice in the bottom of the tank If the tank is initially empty when 91 2 064M 3 9 3 Figure 39 Tank filling process water begins to ow into the tank we have an initial condition of the form IC h 0 t 0 In this problem you are asked to derive an equation that can be used to predict the height of the water at any time For the special case given by Q1 10th A0 05 cm2 D 15m H 278m you are asked to determine if the water will over ow the tank i 445414 if amp ny ewxwy 0 7400 42 9 a QB Ix 5 1 xx zooN Zed5V x i 574 0 j lil 1 2 3 4 ndA 0 W Vjeza Single Component Systems 39 The control volume used to analyze this process is illustrated in Figure 39a where we have shown a moving control volume with cuts at the entrance and exit which are Q 9 x ng h x to eveHZwa intJeni x l lux Figure 39a Tank lling and over ow process joined by a surface at which v wn is zero When the water over ows a second exit will be created however we are only concerned with the process that occurs prior to over ow Our analysis is moved on the macroscopic mass balance for a moving control volume A z and the assumption that the density is constant leads to v 7 wndA dV t dt The normal component of the relative velocity v wn is zero everywhere except at the entrance and exit thus Eq 2 simplifies to tgtQoQl dV dt The volume of the control volume is given by 71D2 4 V I 20 Chapter 3 and use of this expression in Eq 3 provides TED2 deet 4 dt dt Q Q1 0 5 If the area of the jet is much much smaller than the cross sectional area of the tank we can impose the following constraint on Eq 5 dee lt 702 dt 4 dt 6 however one must keep in mind that there are problems for which this inequality might not be valid When Eq 6 is valid we can simplify Eq 5 and arrange it in the compact form a BJZ 7 dt where or and B are constants given by 4Q1 060J2g 2 p 2 8 TED TED The initial condition for this process is given by 1C h 0 t 0 9 and we need to integrate Eq 7 and impose this initial condition in order to determine the uid depth as a function of time Separating variables in Eq 7 leads to 611 d and the integral of this result is given by The constant of integration C can be evaluated by means of the initial condition given by Eq 9 and the result is 2 C 2 otiocanL 12 52 This allows us to express Eq 11 in the form Eli NZ wan Ema r 13 Single Component Systems 21 and this is an equation that can be used to predict the height of the water at any time Since Eq 13 represents an implicit expression for ht some computation is necessary in order to produce a curve of the uid depth versus time In order to answer the second part of this question ie will the tank over ow we note that Eq 13 provides the result kBJZoc 0 too 14 and this can be used to determine whether ht is greater than H as time tends to in nity One can also obtain Eq 14 directly from Eq 7 by imposing the steadystate condition that a qdt 0 We can arrange Eq 14 in the form hm aB zoo 15 and in terms of Eqs 8 we obtain hm Q106XA0J2 g2 I from the values given in the problem statement we nd that hm 0566m at steady state thus the tank will not over ow 310 A bathtub is filled with water from a faucet at a ow rate of 10 liters per minute The volume of the bathtub is 25 gallons when the depth of the water is one foot How long will it take to fill the bathtub Suppose the bathtub plug has a hole and water drains at a rate of qlmk 01 JE where the leak ow rate is in liters per second and the water height inside the tank is in meters Assume that the cross section of the tub is constant How long will it take to fill the bathtub with a leak 310 The macroscopic mass balance for a moving control volume d 11 CM 0 l d w v z A z and with the assumption of constant density leads to dV 1 dt v wndA 0 2 A z The normal component of the relative velocity v wn is zero everywhere except at the entrance and exit thus Eq 2 simplifies to dVt 7 Q i Q 0 3 where Q0 is the outing volume ow rate and Q is the entering volume ow rate 22 Chapter 3 a Considering that there is no outing volume ow rate equation 3 becomes W E Qi 7 0 4 This equation can be integrated to calculate the time needed to ll the bathtub The limits of the integrals are given by 0 the initial conditions V 0 t 0 I the nal conditions V 25 gal or 95 l t t Then 95 t IdVZ J Qidt 5 0 0 Then being Q constant we have 95 7 0 10l min t 7 0 6 or t 95 min b Now we have a hole and water drains at a rate of qlmk 01 h Equation 3 becomes here d A h 0142 Q 0 7 dt To arrive to this equation the cross section area A is considered constant and the Q0 has been substituted by the above equation of q leak We can simplify again equation 7 and arrange it in the compact form a NZ 8 dt where or and B are constants given by g 2 A B A 9 0c The initial condition for this process is given again by h0 t 0 and we need to integrate Eq 8 and impose this initial condition in order to determine the uid depth as a function of time Separating variables in Eq 8 leads to dh dt 10 a IN Single Component Systems 23 and the integral of this result is given by The constant of integration C can be evaluated by means of the initial condition and the result is l on on In on 12 52 This allows us to express Eq 11 in the form Bil NZ ocln l BxZOL t 13 Now we have to calculate the values of A on and B to apply equation 13 The high of the bathtub can be calculated 3 V 9510001772 AZ 0 3048 0312m2 l 397 1ft 0L and B are given by the equation 9 as follows in 10 a U2534103d7m s 0312m2Mj 1m 3 01 d J 2 3 3232110 3 117 0312m2M m 2 5 lm Then apply equation 13 The resulting time is t 7405 or t 1233 min 31 1 The ow of blood in the veins and arteries is a transient process in which the elastic conduits expand and contract As a simplified example consider the artery shown in Figure 31 1 At some 24 Chapter 3 instant in time the inner radius has a radial velocity of 0012 cms The length of the artery is 13 cm and the volumetric ow rate at the entrance of the artery is 03 cm3s 1f the inner radius of the artery is 015 cm at the particular instant of time what is the volumetric ow rate at the exit of the artery F L d Figure 311 Expanding artery 311 In this problem we consider the expanding artery having an inner radius of 015 cm a length of 13 cm and a uniform radial velocity of 0012 cms We begin our analysis with the control volume illustrated in Figure 311a along with the appropriate macroscopic mass balance given by d E pdV pv7wndl 0 1 t V t A I At the wall of the tube we have the condition v 7w n 0 at the tube wall 2 while at the entrances and exits the control volume illustrated in Figure 1 requires that w n 0 at the entrance and exit 3 Under these circumstances Eq 1 takes the form d pdV pvndA 0 4 dt V t At where At represents the area of entrances and exits Ifwe assume that the density is 70 Figure 31111 Control volume for an expanding artery constant Eq 4 can be expressed in terms of the volumetric ow rates according to Single Component Systems 25 W 7 Qexit Qentrance 0 5 Since we are interested in the volumetric ow rate at the exit we arrange this result in the form 6 Qex1t Qentrance dt and we need only compute the time rate of change of the control volume in order to determine Qexit This can be done directly however it is of some interest to make use of the general transport theorem given by g deV Jil JqundA 7 Vat Va 1 A 11 For the special case indicated by vac ltr 1 8 and the particular control volume shown in Figure 3 1 la we see that Eq 7 takes the form W t dt windl wr21triL 9 A10 Here we have used Ait to represent the area of the interface between the uid and the wall of the artery r to represent the inner radius of the artery and w to represent radial component of the velocity of the uidsolid interface Use of Eq 9 in Eq 6 leads to Qexit Qantrance wr2 riL and substitution of the appropriate numbers provides the volumetric ow rate at the exit given by Qexit n3s 0153cm3s One should note that if the radial velocity w is large enough the volumetric ow rate at the exit could be negative 312 A variety of devices such as ram pumps hydraulic jacks and shock absorbers make use of moving solid cylinders to generate a desired uid motion In Figure 312 we have illustrated a cylindrical rod entering a cylindrical cavity in order to force the uid out of that cavity In order to determine the force acting on the cylindrical rod we must know the velocity of the uid in the annular region If the density of the uid can be treated as a constant the velocity can be determine by application of the macroscopic mass balance and in this problem you are asked to develop a general representation for the uid velocity Chapter 3 o Figure 312 Flow in an hydraulic ram 312 In this problem the density is assumed to be constant so that we can express Eq 344 as M v wncL4 0 1 all A 1 Here we have replaced Vat and A at with V t and A t since the volume and area are no longer arbitrary but are determined by the process illustrated in Figure 312 The moving control volume V t is illustrated in Figure 312a where we have identi ed the diameter of the ram as D0 and the diameter of the cavity as D1 A little thought will indicate that v w n 0 on the face of the ram and that everywhere else HH 1 7 zLt Single Component Systems 27 Figure 31261 Flow in an hydraulic ram on the surface of the control volume except over the area of the area of the annulus at z 0 At that surface the outwardly directed unit normal vector for the control volume is given by n k atthe exit 2 and since Wn 0 at that surface we have v vz at 0 3 Substitution of this result into Eq 1 leads to the following form W t dt VzdA 0 4 Aexit This result can be expressed in terms of the area of the annulus and the average velocity to obtain W t dt ltvzgt BIZ Di 0 5 and use of the obvious expression for V t leads to 2 From this we see that the average velocity at the exit of the annular region is given by u ltvzgt 2 7 D1 D0 7 l A uid mechanical analysis will allow one to express the force acting on the ram in terms of vz thus one can determine the force required to move the ram at the velocity u0 313 In Figure 313 we have illustrated a capillary tube that has just been immersed in a pool of water The water is rising in the capillary so that the height of liquid in the tube is a function of time Later in a course on uid mechanics you will learn that the average velocity of the liquid ltvz gt can be represented by the equation 8pltv gth 26r0 7 pgh 22 1 gmvitational 0 ca orgy force Visc OHS force in which v2 is the average velocity in the capillary tube The surface tension a capillary radius ro and uid viscosity p can all be treated as constants in addition to the uid density p and the gravitational constant g From Eq 1 it is easy to deduce that the final height of the liquid is given by has 2SDgro In this problem you are asked to determine the height h as a function of time for the initial condition given by TC h 0 20 You may find it convenient to work with a dimensionless height defined by hthm and you may wish to verify your analysis by doing a simple experiment This would require that you measure the height as a function of time thus you should design the experiment so that ha is sufficiently large and the entire process is sufficiently slow Keep in mind that G can be determined experimentally by Eq 2 If the liquid does not completely wet the walls of the capillary tube one must replace 5 in Eqs 1 and 2 with s cosq where G is the contact angle Under these circumstances the product 8 cosq can be determined experimentally by Eq 2 H Figure 313 Transient capillary rise 313 The control volume to be used in this problem is shown in Figure 313a and we begin our analysis with the macroscopic mass balance expressed as d dt v z pdV pv7wndA 0 A t Chapter 3 2 3 4 1 Single Component Systems 29 00 moving control volume Figure 31311 Transient capillary rise At the moving gasliquid interface we have the condition v n w n at the gasliquid interface 2 while at the walls of the capillary tube we can write v n w n 0 at the walls ofthe capillary tube 3 At the entrance to the capillary tube the control volume is xed in space and this leads to vn ivz wn 0 at 20 4 When Eqs 2 3 and 4 are used in Eq 1 and we take the density to be constant Eq 1 takes the form W t dt Jvzz0 dA 0 5 A in whichA represents the cross sectional area of the capillary tube 7V3 We can express this form of the transient macroscopic mass balance as 2 601 2 Tub E re ltVzgtz0 in which v2 represents the areaaveraged velocity de ned explicitly by 1 V2 Z J Vz dA A The cross sectional area in Eq 6 can be cancelled to obtain dh E ltvZgt0 8 30 Chapter 3 and this equation must be solved subject to the initial condition given by LC h 0 t 0 9 If we apply the xed control volume form of Eq 1 pdV JpvndA 0 10 V A to the xed control volume shown in Figure 313b one concludes that the area averaged 7 xed contml volume Figure 3 I 3 b Fixed control volume velocity is independent of z ie ltvZgtz ltvzgtz0 0 s 2 she 11 Since v2 is independent ofz we can express Eq 8 according to dh V 12 dt lt z and make use of the uid mechanical result given in the problem statement to express the transient macroscopic mass balance as 2 dh 7r0 7 pgr0 13 E 4w 8 Our mathematical problem can now be stated in the form h 0L 3h dt 1 14a IC h 0 10 14b Single Component Systems 31 in which the dimensionless parameters are given by 2 a 6quot B 0 15 4p 8p The solution to Eqs 14 is given as an implicit equation for the uid depth that takes the form on 0L h t 7 TZIII70LBZ B 16 As time tends to in nity we must have a Bhao astaoo 17 and from this we conclude that the nal height is given by 26 Pg 0 11 aB 18 This indicates that the capillary rise is inversely proportional to the radius of the capillary tube and very large values of haC can be achieved for very small values of r0 We can use Eq 18 with Eq 16 to express the timeheight relation in the following dimensionless form 7 4 km 7 Int Hm Ht 19 in which the dimensionless height is given by HG hthw 20 Equation 19 can be used to determine the liquid height in a capillary tube as a function of time and empirical versions of this result are used to determine the rate of imbibition of water into dry porous media 314 In Figure 314 we have illustrated a crosssectional view of a barge loaded with stones The barge has sprung a lead as indicated and the volumetric ow rate of the leak is given by leak flow rate Cd Aoquotgh hz Here Cd is a discharge coefficient equal to 06 A0 is the area of the hole through which the water Chapter 3 xxx wa x a Z 9 7 r 7 WWW 7 4 4004quot 0 39 t r 1 of r 4 lM Iz 9 39 39 M aw6 IoWw w 64001 Oi3 4574340Z 022 och754022 690232M223640200 Maggie2 5902275 7 7 9570034 Z0745 amp0464 43 leak 7065 wM 4W awa Wz awx WOW7 x W MWIWMWX wm Figure 314 Leaking barge is leaking h is the height of the external water surface above the bottom of the barge and h is the internal height of the water above the bottom of the barge The initial conditions for this problem are TC hho hl0 t0 and you are asked to determine when the barge will sink The length of the barge is L and the space available for water inside the barge is eHwL Here 8 is usually referred to as the void fraction and for this particular load of stones 8 035 In order to solve this problem you will need to make use of the fact that the buoyancy force acting on the barge is buoyancy force pgh wL where p is the density of water This buoyancy force is equal and opposite to the gravitational force acting on the barge and this is given by gravitational force mg Here m represents the mass of the barge the stones and the water that has leaked into the barge The amount of water that has leaked into the barge is given by volume of water in the barge shlwL Given the following parameters w h 8ft H 12ft determine how long it will take before the barge sinks You can compare your solution to this problem with an experiment done in your bathtub Fill a coffee can with rocks and weigh it39 then add water and weigh it again in order to determine the void fraction Remove the water but not the rocks and drill a small hole in the bottom Measure the diameter of the hole it should be about 01 cm so that you know the area of the leak and place the can in a bathtub filled with water Measure the time required for the can to sink 314 In Figure 314a we have illustrated the control volume to be used in the analysis of this problem The control volume contains only the water that has leaked into the barge and the macroscopic mass balance for this case is given by Single Component Systems 1 IpdV Jpv7wndA 0 1 dt A t V 0 Even though the density could be treated as a constant it is convenient to work directly with Eq 1 in order to obtain 2 prs d dt rate of accumulation mass ow rate of water into the barge of mass of water in the barge in which prepresents the density of water The initial condition for this process is given by 1C 3 0 represents the time at which the leak began Since we have two unknowns where I ha and ht we need more information in order to solve Eq 2 This additional x x x x xxxxxxxx xx x xxx xx xxxx x xxa xxxxxxxxxxa xxxxxxxxxxa xxxxxxxxx x xxxxxxxxxxxxxxxxxxx x x x x x xxxxxxxxxxxxx xxxxxxxxx xxaxxxxxxxxxxxa xxxxxxxxxaxxxxxxxxx xxxxxxxxxxoxvxxxxx xpxxxx xxx xxx xxUxxx xx quotft contml volume x 5 x x xxxx xxx x xxxx xxxxx xxxxx xx xNx xxnxxxxxux xxvxxxx x xxu xxx xx xxxx xx xxx x xx xx x x axxxxxxxa axxxxxx x x x xxx xx xx xxxx xx x x x xx x xxxxx x xxxxx x xxxx xxx xnxxxxxxw xeOxxnxmxow xxxxmxxxxxvx Figure 31411 Control volume for leaking barge information is provided by the buoyancy condition which can be stated as 4 prhg mg body force buoyancy force Here m represents the mass of the barge the stones in the barge and the water that has leaked into the barge We can express the mass according to 5 MD p sthi in which m0 represents the mass of the empty barge and the stones in the barge Although m0 is not given we can use Eqs 3 and 4 to deduce that 6 p th0 mo 34 Chapter 3 and when Eqs 5 and 6 are used in Eq 4 we nd that the buoyancy condition leads to h0 ghi h 7 In order to use this result with Eq 2 we rst arrange it in the form ha 7 17 S h h 7 h 8 8 and we also take the time derivative to obtain d dh g 9 dt 39 dt Use of these two results allows us to express our macroscopic mass balance as dh CdA quot2 2 7 O g lh07178 h 10 dt wL If we de ne the constants aand according to CdAm Zg 1 2 s B 2 ha 11 wL 1 our boundary value problem takes the form dh z a 7h 12 dt 5 1C h ho t0 13 One should note that if h 3 we have reached an equilibrium condition for which h barge oats in a new h 14 equilibrium condition On the other hand if h gtH water will ow over the sides of the barge and it will sink For this particular problem 3 123 ft and H 12 ft thus a new equilibrium condition is not possible Integration of Eq 12 and imposition of the initial condition given by Eq 13 leads to r xBho JBh 15 and the time required for the barge to sink is given by rm lls ho J H 16 Single Component Systems 35 Use of the numbers provided in the problem statement gives I mquotk 155 hours 17 and this does not leave much time to reach a safe harbor 315 The solution to Problem 35 indicates that the diversion of water from Mono Lake to Los Angeles would cause the level of the lake to drop 19 meters A key parameter in this prediction is the evaporation rate of 36 inyear and the steadystate analysis gave no indication of the time required for this reduction to occur In this problem you are asked to develop the unsteady analysis of the Mono Lake water balance Use available experimental data to predict the evaporation rate and then use your solution and the new value of the evaporation rate to predict the final values of the radius and the depth of the lake You are also asked to predict the number of years required for the maximum depth of the lake to come within 10 cm of its final value The following information is available Year Surface Area Maximum Depth prior to 1941 56000 acres 1815 ft 1970 45700 acres 1640 ft During these years between 1941 and 1970 the diversion of water from Mono Lake Basin was 56000 acreft per year and in 1970 this was increased to 110000 acreft per year The additional water was obtained from wells in the Mono Lake Basin and as an approximation you can assume that this caused a decrease in m see Figure 35b by an amount equal to 54000 acreft per year Your analysis will lead to an implicit equation for B and a trial and error calculation is required 315 Our objective in this problem is to develop an equation that can be used to predict the depth of Mono Lake as a function of time This depth ht is illustrated in Figure 35a Once we have developed an equation for ht we can use the experimental data for the depth of the lake between 1941 and 1970 to predict the rate of evaporation from the lake Knowing this parameter we can predict how the depth of the lake will change with time after 1970 Our mass balance for the lake can be expressed as 4 m1 7 m2 1 Here the mass of the lake for our particular model is given by 75 3 m p h 2 3tan2 0 in which p is the density of water ml is the total mass ow rate entering the lake form all sources and m2 is the mass ow rate leaving the lake owing to evaporation This can be expressed as m2 2 3pan Bpntan20 h2 3 where b has units of inches per year or some equivalent Concerning the depth h we have the following information 36 Chapter 3 h 170 1815ft priorto 1941 4 h h1 1640 ft in 1970 We assume that prior to 1941 the lake was in a steadystate condition in the time averaged sense and we express Eq 1 as 0 rho 7 m2 prior to 1941 5 Here m0 represents the primeval mass ow rate into the lake and on the basis of Eqs 3 and 5 we can express MD as m spatanze hi 6 After 1941 shipments of water to Los Angeles began and we will assume that any water owing to Los Angeles diminishes the ow into the lake by an equal amount If mm is used to represent the mass ow rate to Los Angeles the general form of Eq 1 is given by dm E mo mLA 7739120 7 in which we have explicitly noted that m2 is a function oftime Use of Eqs 2 3 and 6 along with the representation mm pQLA 1941 1970 8 allows us to write Eq 7 as h QM hanz 2 9 This result can be arranged in the form hz x H12 10 ch in which the constant on is given by 2 X pkg M 11 TI The initial condition associated with Eq 9 can be expressed as IC h ho t0 12 thus our origin for the time is taken to be 1941 We can separate variables in Eq 10 to obtain Wk 0L 7 M22 dz 13 Single Component Systems 37 and integration leads to l3 tC 14 13 13 j The constant of integration can be determined by the initial condition given by Eq 12 and after some algebraic manipulation the result can be expressed as H r 5 2 H g 1 ho Here the new dimensionless variables are given by H hho JoeBk 16 A little thought will indicate that t h0 can be thought of as a dimensionless time and that the dimensionless height is bounded by g s H s 1 17 At this point we would like to use Eq 15 and the experimental data given by Eq 4 to evaluate 3 We express our data as h h0 1815 ft 10 18 h h1 1640 ft t 29 years and use these values in Eq 15 to obtain 00964 i 19 2 J 1815 ft This represents an implicit equation for B since depends on B as indicated by the second of Eqs 16 and a trialanderror calculation leads to B 405 ftyear 486 inches 20 This is in reasonable agreement with the previous estimate of 36 inches per year In 1970 ground water pumping began in the Mono Basin and an additional 54000 acrefeet per year were diverted to Los Angeles Our solution for ht from 1970 onward is identical in form to that given by Eq 15 however in this case the initial condition is given by LC h 13921 164 ft t01970 21 and the volumetric ow rate to Los Angeles is given by QLA 1970 22 38 Chapter 3 The solution for the depth as a function of time is given by 23 24 The nal depth is given by H39 a ast gtoo 25 and this can be expressed as he him1 QiAtanzeBnhi 26 Carrying out the numerical calculation leads to hm 395 m 1295 ft 27 If we wish to determine the time required for the lake to reach a depth of h 396 m 1299 ft 28 we return to Eq 23 and use H39 942 486 inchesyear 29 in order to nd a time of 81 years One must remember that these results are based on a very crude model of the topology of the lake and a more accurate model could lead to signi cantly different results 316 During the winter months on many campuses across the country students can be observed huddled in doorways contemplating an unexpected downpour In order not to be accused of idle ways engineering students will often devote this time to the problem of estimating the speed at which they should run to their next class in order to minimize the unavoidable soaking This problem has such importance in the general scheme of things that in March of 1973 it became the subject of one of Ann Landers syndicated columns entitled What Way is Wetter Following typical Aristotelian logic Ms Landers sided with the common sense solution the faster you run the quicker you get there and the drier you will be Clearly a rational analysis is in order and this can be accomplished by means of the macroscopic mass balance for an arbitrary moving control volume In order to keep the analysis relatively simple the running student should be modeled as a cylinder of height h and diameter D as illustrated in Figure 316 The rain should be treated as a continuum with the mass flux of water represented by pv Here the density p will be equal to the Single Component Systems 39 Ky Figure 316 Student running in the rain density of water multiplied by the volume fraction of the raindrops and the velocity v will be equal to the velocity of the raindrops The velocity of the student is given by w and both v and w should be treated as constants You should consider the special case in which v and w have no component in the ydirection and you should separate your analysis into two parts In the first part consider iv gt 0 as indicated in Figure 316 and in the second part consider the case indicated by irv lt 0 In the first part one needs to consider both iv gt iw and iv lt iw When iv gt iw the runner gets wet on the top and back side On the other hand when the runner is moving faster than the horizontal component of the rain the runner gets when on the top and the front side When iv lt 0 the runner only gets wet on the top and the front side and the analysis is somewhat easier In your search for an extremum it may be convenient to represent the accumulated mass in a dimensionless form according to m0 mo F parameters pLDh in which t is equal to the distance run divided by i w u and m0 is the initial mass of the runner 316 In this problem we wish to determine the amount of rain that is accumulated by our model student body shown running in the rain in Figure 316 We begin our analysis by considering the case for which iV gt 0 and vx is always positive P I We rst recall that the macroscopic mass balance for an arbitrary moving control volume that is given by d pdV pv7wndA 0 1 t Vat A 10 and for this particular problem we choose a control volume that encloses the body Our macroscopic mass balance then takes the form 40 Chapter 3 d1 pdV pv7wncZl 0 2 t v z A z in which we have identi ed the control volume as V t since it is no longer arbitrary but moves with a speci c velocity given by w in0 3 The mass in the control volume can be expressed as m pdV 4 V t and this allows us to write Eq 2 in the form 2 pv iun 4 5 A t The integrand in this result will be zero over the bottom of the cylinder shown in Figure 316 and it will be nonzero over the top of the cylinder The value of the integral over the top of the cylinder is easy to obtain however evaluation of the integral over the side of the cylinder requires some thought Sometimes it is convenient to think about this process in terms of the relative velocity de ned by v v 7 in0 6 and the relative velocity eld is illustrated in Figure 316a where we can see that the runner will get wet on the back side when ivr gt 0 and on the front side when ivr lt 0 The dividing line between the region that gets wet and the region that remains dry is given by nvr 0 and this dividing line is illustrated in the three dimension representation shown in Figure 316b There we have illustrated the rain shadow that occurs when the running is moving more slowly than the horizontal component of the rain Because both V and w are constant vectors exactly one half of the cylindrical surface will get wet for either i vr gt0 or for ivr lt0 and we can solve the mass transfer problem in terms of the two obvious special cases In both case we will express the velocity of our runner as and we will be interested in learning whether the accumulated moisture is minimized by letting u0 tend to in nity CASE 1 ivgtiW We begin our analysis by noting that Eq 5 can always be expressed as dm E w MA 7 Assz Amp Single Component Systems 41 in which we have used 11 k on the top surface For the case in which iV gtiw the iv gt 0 iv lt O a b Figure 31611 Relative velocity of the rain runner will accumulate rain on the top and the back as illustrated in of Figure 316aa From the projected area theorem we know that area integral over the side Figure 316b Wet and dry regions of the cylinder can be expressed as JlndA hD 8 AI while the area integral over the top surface takes the form 42 Chapter 3 J vde van24 9 Amp Use of Eqs 8 and 9 in Eq 7 leads to dm 2 t pox mm pvan 4 10 which must be integrated subject to the initial condition IC m m0 t0 11 Integration provides a solution for the mass as a function of time ma t v D24 t 12 and if the distance that the runner must cover is denoted by L this result can be arranged in the form mt 7 m0 pLDh M W back top vx MO 13 o This represents a two parameter problem in which vx u0 represents one parameter and the second term on the right hand side represents the other parameter There are two special cases that provide some insight and these are given by m gtoo Lin gt0 14a pLDh t V TED m m M7 u0vx 14b pLDh uo 4h top and it becomes apparent that if the rain is coming from behind one would want to run at least as fast as the horizontal component of the rain To determine if one would want to run faster that the horizontal component we need to consider the second special case associated with Part I of this problem CASE II ivltiW In this case we can again express Eq 5 as dm E w v MA 15 Assz Amp Single Component Systems 43 however when no gtvx it is the front side of our runner that will get wet and the projected area theorem gives us JlmdA hD 16 AI We can use this result along with Eq 9 in the mass balance given by Eq 15 to obtain dm 2 E D 4 Mo v 17 and we can follow the analysis from Eq 10 to Eq 13 in order to express the integrated mass balance as 7711 mo pLDh no EH front top no vx 18 Once again there are two special cases that can be expressed as m Lghmo Mow 19a 9 Mo top a1 unaoo 19b p and here we see that the common sense solution that you should run as fast as you can appears attractive However there are two terms in Eq 18 and the rst increases with increasing u0 while the second decreases with increasing u0 In order to clarify the two solutions given by Eqs 13 and 18 we express them in the form m 0 Vx 20a pLDh m MO VX 20b pLDh front l OP in which the parameter Q can be thought of as a constant for any given runner and any given situation This parameter takes the form 44 Chapter 3 Q if 21 and from the representation of Eqs 20 illustrated in Figure 3l6c we see that it is Q that controls the speed of our runner The results can be expressed as 1 Q gt 1 Run as fast as you can 2 Q 1 Run at least as fast as the horizontal component of the velocity of the rain but it will do no good to run faster 3 Q lt 1 Run at least as fast as the horizontal component of the velocity of the rain but do not run faster One should remember that these results are based on the special case for which V has no component in the y direction however it is a routine matter to include this effect in the mass balance given by Eq 2 Part 11 Our previous analysis dealt with the case represented by iv gt 0 and we now wish to consider the situation in which the horizontal component of the rain is negative ie ivlt0 In this case the runner will always be wet on the front side as indicated in Figure 316ab and we can make use of Eq 18 to represent the accumulated moisture as mt 7 m LDh uo vx 22 P a front top Since vx is negative is convenient to express this result in the form 7710 mo pLDh 23 front mp Here it becomes obvious that the minimum value for the accumulated moisture is given by mt m0 LDh l u gt00 24 P min and one would want to run as fast as possible One can express Eq 23 in a more compact form according to mt mquot w l 25 pLDh uo where the parameter Q is now defined by Single Component Systems 45 26 Some results from Eq 25 are shown in Figure 316d where one can see that the accumulation of moisture is reduced by running faster regardless of the value of Q as de ned by Eq 26 30 i 20 i quot10 mo pLDh 10 7 l l 0 0 05 1 0 Figure 3160 Accumulated moisture for Vx gt 0 uoVx 46 Chapter 3 MO mo pLDh unlel Figure 31651 Accumulated moisture for Vx lt 0 Miscellaneous 317 The steadystate average residence time of a liquid inside a holding tank is computed by the ratio of the volume of the tank to the volume flow rate of liquid in and out of the tank I VQ A cylindrical tank with volume V 3 m3 and the input mass flow rate of water is 250 kgminute At steadystate the output flow rate is equal to the input flow rate What is the average residence time of water in the tank What would be the average residence time if the mass flow rate of water is increased to 300 kgminute What would be the average residence time if a load of 12 m3 of stones is dropped into the holding tank 317 The volume ow rate of liquid in and out the tank can be calculated as Q 1 In the rst case applying equation 1 we have Single Component Systems 47 2503 d 3 mm 250 quotl kg J mm 1 3 dm Then the steadystate average residence time is T 7 V 7 3000dm3 7 7 3 Q 250 mln 12min In the second case we have that the ow rate of liquid is 3003 d 3 mln m 1 g min aim3 And the steadystate average residence time is T V 3000dm3 3 Q 300 mln 10min If a load of 12 m3 of stones is dropped into the holding tank the free volume of the tank will be V M mG 44W 218 Now the steadystate average residence time is in the rst case 3 239 72 min 250 quot7 mm And in the second case V 1800alm3 6mm Q aim3 300 mm 48 Chapter 3 318 If the delivery charge for the propane tank described in Example 33 is 3750 and the cost of the next largest available tank is 2500 for a 22 cubic meter tank how long will it take to recover the cost of a larger tank 318 The volume of the new tank in gal is 581 which is almost twice the volume of the initial tank 250 gal These means that if the first tank has to be lled twice a month the second one will have to be lled once a month Then the already existent tank tank 1 will give the following monthly expenses C1272375n775n 1 Where no initial investment has been taken into account and with n being the number of months The minus sign indicates that this is spent money Now the new tank tank 2 will give one half of the monthly expenses of the rst tank In addition we have in this case an initial investment of 2500 So the capital C2 2500 37539n 2 In order to recover the cost of the new tank C1 and C2 have to be equal 775n725007375n 3 From equation 3 the time required to recover the cost is n 667m0nths Chapter 4 September 12 2009 Multicomponent Systems Section 41 We need a problem any problem for this section Need help from Brian and Ramon Section 42 41 Determine the mass density p for the mixing process illustrated in Figure 41 4 1 The mixing process illustrated in Figure 4 1 is a batch process in which the mass of each species is conserved We express this idea as Pix VA PAV Pi VB PBV Pi VC PCV 1 and we can solve for the three species densities in the final mixture to obtain PA PXVAV P3 P EVBV PC ngCV 2 The numerical results are given by p A 0283 gcm3 p B 0317 gcm3 pC 0356 gcm3 3 This type of calculation is intuitively obvious however it is worthwhile to understand the framework of the calculation in terms of the concepts presented in Chapter 4 The macroscopic mass balance for species A can be expressed as dt v t A t V t i J pAdV IpAvA wndA IrAdV 4 where V t represents a control volume that properly represents the mixing process during which species B and C are added to species A Since there is no chemical reaction involving species A and since no species A enters or leaves the system at the control surface A t Eq 4 takes the form d 7 dV 0 5 tilIDA Vt Integration from the initial time when all three species are in separate containers to the final time when all three species are mixed lead to p2 dV p A dV V I1 V 02 Since the species density is uniform this result simplifies to p AV t1 pAV Q The initial and final volumes are given by V01 VA V02 V and when these representations are used in Eq 7 we obtain PillA PAV Chapter 4 6 7 8 9 This is the first of Eqs 1 indicating that those equations are based on an application of Eq 4 for each of the three species involved in the mixing process 42 A liquid hydrocarbon mixture was made by adding 295 kg of benzene 289 kg of toluene and 287 kg of pxylene Assume there is no change of volume upon mixing ie to determine 1 The species density of each species in the mixture 2 The total mass density 3 The mass fraction of each species AVmix O in order 4 2 This mixing process is analogous to the process illustrated in Figure 4 1 except for the fact that Alex 0 If we let species A represent benzene species B represent toluene and species C represent p xylene the three species densities are given by PA PjiVAVa P3 P iBVBVa PC in which the total volume Vis given by V VA VB VC We can also write Eq 1 in terms of the masses of the three components mA PillA m3 DllBVBa mC to obtain PA mAV P3 mBV PC 1 2 3 4 Since we are given the masses we can determine the individual volumes by Eqs 3 provided that we know the pure species densities Once we know the individual volumes we can determine the total volume by means of Eq 2 and then the individual species densities by Eq 4 The total density is then determined by Multicomponent Systems 3 P PA PB PC 5 and the mass fractions are determined by DA pAp DB 939 me Deli 6 The key to the solution to this problem is knowledge of the pure species densities 43 A gas mixture contains the following quantities per cubic meter of carbon monoxide carbon dioxide and hydrogen carbon monoxide 0 5 molesm3 carbon dioxide 05 molesm3 and hydrogen 0 6 molesm5 Determine the species mass density and mass fraction of each of the components in the mixture 4 3 With the given data the species mass density can be calculated as follows Notice the problem of units here ie meters to the sixth power on the right hand side pvltmi lxMWlt3 pltk g m3 101 1 m3 Vm3 Then for each species mol Kg 05 x0 02801 7 p 32Mzo 014 CO m3 m3 mol Kg 05 x0044 7 p 1113 mol 002 Kg C02 C02 m3 1m3 m3 0i6 m21x0i0023 p k g 1 101 0i0012 Kg H2 H2 m3 1m3 m3 The total density is determined by p pco pcoz sz 2 Then p 00372kgm3 and the mass fractions are determined by Chapter 4 mco pcop mcoz pcozlp DHZ pHZp Then mco 00140037203763 n O 00220037205914 2 C 03C 00120037200323 Something is wrong with the last equation 44 The species mass densities of a three component liquid mixture are acetone pA 3264 kgm3 acetic acid rAA 3264kgm3 and ethanol pa 2176 kgm3 Determine the following for this mixture 1 The mass fraction of each species in the mixture 2 The mole fraction of each species in the mixture 3 The mass of each component required to make one cubic meter of mixture 4 4 The mass fractions are determined by DA PAPa 0AA pAAp DEA PEAP 1 The total density is determined by p pApAApEA 2 and use of the appropriate values leads to p 8704 kgm3 3 On the basis of Eqs 1 we have 03A 2 32648704 2 0375 mAA 2 32648704 2 0375 4 DEA 21768704 0250 2 The molar fraction can be calculated considering 100 kg of mixture Then the amount of mass of each component will be Multicomponent Systems 5 mA 375kg mAA 375kg mEA 250kg Now the number of moles of each component can be calculated using the molecular weight as n i iAAAEA 3 MW Then nA 6457kmol nAA 6245kmol mm 5425kmol The total number of moles 3 nm 2 nl i A AA EA 4 Then HM 18127kmol Finally the molar fraction can be calculated as xi 1 iAAAEA 5 quotmt Then xA 0356 xAA 0345 xEA 0299 3 The amount of mass of each component can be calculated as miin iAAAAE 6 Since the total volume of mixture is 1 m3 the mass of each component Will be Chapter4 mA 3264kg mAA 3264kg mm 2176kg 45 A mixture of gases contains one kilogram of each of the following species methane ethane propane carbon dioxide nitrogen Calculate the following 1 The mole fraction of each species in the mixture 2 The average molecular mass of the mixture 4 5 Let assume 1 kg of each species Then the total amount of mass Will be 5 kg The number of moles of each species can be calculated as m v n zCH4EPRC02N2 1 Then CH4 1 00625kmol kmol n 1k g 00333kmol kmol nPR M g 00227kmol kg 441 kmol nCO 1L 0022714an 2 44 kmol nNZ 1k lfg 00357kmol kmol The total number of moles 5 quotm Z quoti 1 z i CH4 ET PR C0 N2 2 Then nm 01770kmol Finally to calculate the molar fraction Multicomponent Systems x i i CH4 E PR C0N2 3 quotW Then xCHA 03532 xE 01884 xPR 01281 xcoz 01284 xNZ 02018 2 The average molecular weight can be calculated as 5 MWinMWi iCH4EPRC02N2 4 i1 Applying equation 4 MW 2826 kg kmol 46 Two gas streams having the ow rates and properties indicated in Table 461 are mixed in a pipeline Assume perfect mixing ie no change of volume upon mixing and determine the composition of the mixed stream in molesm3 Table 461 Composition of gas streams 4 6 The total mass density of each stream can be calculated as pjZpi iMEP j12 1 i1 Then applying equation 1 we have 8 Chapter 4 p1 22655 m 02 098k m Now the volumetric ow rate of each stream is given by m J Q J p 12 J Then 3 Q1 0100 S 3 Q2 0302 Then the total volumetric ow rate 3 QM 0402 3 S Now the total amount of moles of the each species can be calculated as Q1quot n 131W 2 Then nM 601910 3km 016019m ol S S nE 9040103 I m 01 9040 S S n 3501103kL0l3501m quotl S S Finally the concentration of each species is calculated as Cii iMEP 3 th By applying equation 3 the concentrations are Multicomponent Systems 9 CM 14973m l m CE 2249m il m CF 8709m il m 47iDevelop a representation for the mole fraction of species A in an N component system in terms of the mass fractions and molecular masses of the species Use the result to prove that the mass fractions and mole fractions in a binary system are equal when the two molecular masses are equal 4 7 We begin this problem with the definition of the mole fraction of species A expressed as xA LA 1 in which 0 represents the total molar concentration given by c cAcBcCcN 2 We can also use compact mixed mode notation and express the total molar concentration in the form GN c E as 3 01 Returning to Eq 1 we express the molar concentration in terms of the species density and the molecular weight CA PA MW A 4 and then make use of the definition of the mass fraction to obtain CA 03A 9 MW A 5 This representation can be used in Eq 3 so that the total molar concentration takes the form GN c 2 p MWG 6 01 Substitution of this result and Eq 5 into Eq 1 leads to x 0JAPMWA A GN 2 00 PMWG 01 and we can cancel the total density to obtain muMWA XA GN ZwGMWG Gl Chapter 4 7 8 If mass fraction data is given for a system and it is more convenient to work in terms of mole fractions this result provides the transformation from one measure of concentration to another For a binary system Eq 8 reduces to x 0A M WA A DAMWA wBMWB For the special case in which MWA MWB this result reduces to 0 M A 0A binary system 0A DB 9 10 This suggests that the mole fractions and mass fractions in a mixture tend to follow the same general trend Section 43 48 The species velocities in a binary system can be decomposed according to VAVllA VBVllB 1 in which V represents the mass average velocity defined by Eq 436 Use this result along with the definition of the mass average velocity to prove that l v v in A B 1wA A This means that the approximation V A 6 VB requires the restriction uA ltlt 1 03A VA 2 3 Since l oA is always less than one we can always satisfy this inequality whenever the mass diffusion velocity is small compared to the species velocity ie iuAi ltlt VA 4 Mnlticomponent Systems 11 For the sulfur dioxide mass transfer process illustrated in Figure 44 this means that the approximation quot502 k Vair k 5 is valid whenever the mass diffusion velocity is restricted by lust k ltlt lvsoz k 6 In many practical cases this restriction is satisfied and all species velocities can be approximated by the mass average velocity Now you are asked to direct your attention to the mass transfer process at the gasliquid interface illustrated in Figure 44 Assume that there is no mass transfer of air into or out of the liquid phase and prove that VSOZ n jusoz n at the gasliquid interface 7 A Under these circumstances the mass diffusion velocity is never small compared to the species velocity and the type of approximation indicated by Eq 444 is never valid for the component of the velocity normal to the gasliquid interface As a simplification treat the sulfur dioxideair system as a binary system with species A representing the sulfur dioxide and species B representing the air In addition assume that the normal component of the velocity of the gas liquid interface is zero ie w n 0 4 8 In this problem we assume that there is no mass transfer of species B air normal to the gas liquid interface and this leads to the following constraint on the species velocity V B n 0 at the gas liquid interface 1 For a binary system the mass average velocity is given by V OAVA DBVB 2 and at the gas liquid interface we can use Eq 1 to obtain V n OAVA n at the gas liquid interface 3 We now decompose the species velocity in terms of the mass average velocity and the mass diffusion velocity VA v HA 4 and use this to eliminate the mass average velocity from Eq 3 in order to obtain l coA VA n uA n at the gas liquid interface 5 Provided that 0A is not equal to one this result indicates that the diffusion velocity is not small compared to the species velocity One can also use Eq 4 in Eq 3 to eliminate the species velocity V A rather than the mass average velocity V and this leads to m V n uA n at the gas liquid interface 6 03A 12 Chapter 4 If the mass fraction of species A is small compared to one this result indicates that the normal component of the mass diffusion velocity for species A is large compared to the normal component of the mass average velocity It is important to keep in mind that Eqs 5 and 6 are based on Eq 1 Which indicates that there is no mass transfer of species B between the gas and liquid phases Section 44 49 A three component liquid mixture flows in a pipe with a mass averaged velocity of 09 ms The density of the mixture is p 850 Kgm3 The components of the mixture and their mixture and their molar fractions are npentane xp 2 02 benzene x3 03 and naftalene xN 05 The components diffusion fluxes are pentane pp up 1564 10396 Kgm2 s benzene pB uB 1563 10396 Kgm2 s and naftalene pN uN 1603 1039 Kgm2 s Determine the diffusion velocities and the species velocities of the three components Use this result to determine the molar averaged velocity v Notice that you must carry numbers with eight digits in your computations 4 9 In order to calculate the diffusion velocities the species mass densities should be calculated These are given by piwip iPBN 1 The molar fractions are given To calculate the mass fractions MWznxi 3 ZMWyxi i1 w z 2 Then W 0142 w 0230 wN 0628 Apply equation 1 now 0 1207k m p 1955k m pN 5338k m Using the data given in the problem the diffusion velocities are given by Multicomponent Systems 13 1564406 up 1310 8 0P 5 1563406 m MB quot 0810 8 0P 5 1603406 m MN quot 0310 3 0P 5 Now the species velocities are given by v v u 3 Then m VP 0900000013 s VB 0900000008 S vN 0900000003 S Finally the molar averaged velocity vquot is given by Then v 09000000065 E S Section 45 410 Sometimes heterogeneous chemical reactions take place at the walls of tubes in which reactive mixtures are flowing lf species A is being consumed at a tube wall because of a chemical reaction the concentration profile may be of the form em a kWref 1 Here r is the radial position and r0 is the tube radius The parameter 1 depends on the reaction rate at the wall and the molecular diffusivity and it is bounded by O S 131 If I is zero the concentration across the tube is uniform at the value c2 If the flow in the tube is laminar the velocity profile is given by 14 Chapter 4 vzr 2ltvzgtl rro2 2 and the volumetric flow rate is Q ltvzgtnr3 3 For this process determine the molar flow rate of species A in terms c2 1 and ltvzgtr When 1 05 determine the bulk concentration CA 1 and the areaaveraged concentration ltCAgt r Use these results to determine the difference between MA and ltCAgtQ r 4 10 The molar ow rate of species A at some arbitrary exit is given by MA I VA mum 1 Asem and if the exit is fixed in space this simplifies to MA JcAVAndA 2 Aexit If Ae represents the cross section of a tube and the diffusion velocity is negligible this expression simplifies to MA cAvndA cszdA 3 A A exit exit To be explicit we write this result as rro 927 MA J Jc21 rn22ltvzgt1 rn2lrdrde lt4 r0 90 and since the integrand is independent of 6 we obtain rro MA 4nc2ltvzgtJ1 rn21 rn2rdr lt5 r0 With some algebraic effort this integral can be evaluated to obtain MA TEr0262Vzl 6 Multicomponent Systems 15 The volumetric ow rate is given by the area averaged velocity times the cross sectional area Q ltnm n and this can be used with Eq 6 to express the molar ux of species A in the form MA 620Q amp The area averaged concentration is defined in precisely the same way that the area averaged velocity is defined and we list the representation as ramp 62n 1 CA AleCAdA 72quot IcArdrde 9 AS 7 r0 60 For the concentration profile given for this problem we can evaluate the integrals to obtain CA 021 10 An alternative averaged concentration is the bulk or cup mixed concentration that is defined by J CAV n dA A i cpbeiigggi GD VndA Aa and this can be expressed as 1 M em 7J1 dA 7A 12 Q z Q Aa The advantage of the bulk concentration is that it provides a simple representation for the molar ux MA CAbQ 13 that is valid under all circumstances We can compare Eqs 8 and 13 to demonstrate that the bulk concentration is given explicitly by cob d0 no 16 Chapter 4 and here we see that the bulk concentration Eq 14 and the area averaged concentration Eq 10 are different when l is non zero From Eqs 13 and 14 we have the exact representation for the molar ow rate given by MA cwg 6130 Q 15 and when l is non zero the molar ow rate is not given by the area averaged concentration times the volumetric ow rate ie MA lt6AgtQ 0 16 When the molar ow rate of some species is described as the concentration times the volumetric ow rate one should ask the question What concentration 4lli A ash unit is used to separate vapor and liquid strwms from a liquid stream by lowering its pressure before it enters the ash unit The feed stream is pure liquid water and its mass ow rate is 1000 kghri Twenty percent by weight of the feed stream leaves the ash unit with a density p 10 kgmsi The remainder of the feed stream leaves the ash unit as liquid water with a density p 1000 kgm3 Determine the following 1 The mass ow rates of the exit strme in kgsi 2i volumetric ow rates of exit streams in m3si 4 11 Let denote the feed stream as stream 1 the top stream as stream 2 and the bottom stream as stream 3 Considering that the 20 by weight of the feed stream leaves the ash unit with a density p 10 kgm3 the top stream will be m13i6106k g S m2 02m1 702105k g S Then m3n41 m2 2i88106kTg 2 The volumetric ow rates of exit streams in m3 s can be calculated using the densities 3 Q1 721041 01 S 3 Q2 288103m 02 S Multicomponent Systems 17 Section 46 412 Show that Eq 477 results from Eq 476 when either cVn or xA is constant over the area of the exit 4 12 We begin this problem with Eq 4 77 MA JxAcvndA 1 A exit and consider the case for which CAV n is a constant For this case Eq 1 takes the form MA JxAcvndA LIxAdA Aaxncvn 2 Aexit exit Aexit A The constant quantity 0 vn represents the total molar ux molesarea time and when multiplied by the area it gives the total molar ow molestime M Under these circumstances Eq 2 takes the form MA ALJxAdA M xAM constantcvn 3 exit A exit Returning to Eq 1 we consider the case for which xA is a constant Under this circumstances we have MA JxAcvndA xAvandA 4 Aein Aein Since the area integral of c v n represents the total molar ow this result takes the form MA xAvandA mill 5 Am When xA is a constant we have xA xA and we see that this result can be expressed as MA xAvandA xAM constantxA Am 18 Chapter 4 When neither cAvn nor xA is a constant a more detailed calculation is required in order to determine M A 4l3i Use Eqi 477 to prove Eqi478i 4 13 Given Eq 4 77 MA XAgtM 1 we sum over all N species to obtain AN ZMA ltXAgtM ltXBgtM ltXCgtM ltXNgtM A1 2 3 x3 WC ltXNgtM At any point the mole fractions are constrained by xAxBxCxNl 3 and the area average of this result is xx 963 Xe ltXNgt 1 4 Use of this result in Eq 2 leads to AN M A M 5 A1 which is Eq 4 79 4 14 Derive Eqi 481 given that either pv n or 03A is constant over the area of the exit 4 14 It was suggested in the text that Eq 4 81 BN lt03Agt MA Zmn 1 31 was valid when either pvn or 0A is constant over the area of an exit or an entrance To explore this matter we express the mass ow rate in detailed form according to MA IcoApVhdA 2 A exit Multicomponent Systems 19 and note that the sum of all the species mass ow rates is given by BN BN m Em E JprvndA 31 31 Am 3 BN J Em pvndA JpvndA Aexit 31 Aexit Returning to Eq 2 we consider the two special cases to obtain 1 pvnconstant m J coApvndA J wAdA Aexnpvn 4 A exit A exit exit 2 03Aconstant m IwApvndA 03A J pvndA 5 A A exit exit For these two special cases the total mass ow rate is given by 1 p Vn constant m A6in p V n 6 2 0A constant m p VndA 7 Aexit and we can use these results in Eqs 4 and 5 to obtain 1 pvnconstant MA oAn391 8 2 0A constant MA 0A m 9 Both of these representations for the mass ow rate of species A can be expressed in terms of the single equation rearranged in the form BN MA 1amp2 coA 7 mA mg 10 m 31 This is Eq 4 81 which is also listed above as Eq 1 415 Prove Eq484 20 Chapter 4 4 15 In order to prove the relation 20 1 1 we begin with the definition of the bulk mole fraction at an exit which is given by J xAVndA ltxAgtb Aequoti 2 VndA Aexn Since the volumetric ow rate is given by Q J v ndA 3 Aexit we can express the bulk mole fraction according to 1 xAb i xAvndA 4 Q Aerdt We now sum over all mole fractions to obtain AN AN ZltxAb 5 J Zn VndA 5 A1 Aexit A1 in which we have interchanged the integration and summation operations Given the constraint on the mole fractions AN ZxA 1 6 A1 we see that Eq 5 leads to the desired result given by AN gum jvndA g 1 7 Aexit Multicomponent Systems 21 This result was given as Eq 4 85 in the text Section 47 416 Determine M3 and the mole fractions for species B and C for the distillation process described in Sec 47 subject to the following conditions 39 1200 moleshr 39 250 moleshr 03 08 02 4 16 In this problem we consider the distillation column illustrated in Figure 416 where we have constructed a control volume that cuts the single inlet stream and the two outlet streams Since we are dealing with a three component system N 3 containing three 0502 2 05292 39 Web I I I I I I control xA1 I If volume xB1 I xch I I I I I I I I I g 0503 3 05193 XC3 Figure 416 Three component distillation process streams M 3 we can quickly construct the generic portion of our degrees of freedom table as indicated in Table 416 and this is followed with the completion of the particular specifications and constraints Chapter 4 Table 4 I 6 Degrees of Freedom Table Stream Variables compositions N XM 9 ow rates M 3 ngMM12 Generic Degrees of Freedom Number of Y Balance Equations massmole balance equations N 3 Number of Constraints for C 39 39 M 3 Generic Specifications and Constraints N M 6 The information given in the problem statement allows us to complete the particular portion of the degreesoffreedom table to obtain Specified Stream Variables compositions 4 ow rates 2 Constraints for Compositions 0 Auxiliary Constraints 0 Particular Specifications and Constraints 6 This leads to the degrees of freedom given by Degrees ofFreedom 12 6 6 0 This degrees of freedom analysis indicates that we have a solvable problem and we can proceed to determine the single unknown ow rate and the ve unknown compositions The three macroscopic mole balances for species A B and C are given by Species A Species B Species C ltXAgt1M1 ltXAgt2M2 ltXAgt3M3 0 la ltXBgt1M1 ltXBgt2M2 XBgt3M3 0 1b ltxcgt1M1 ltxcgt2M2 ltxcgt3M3 0 1C while the constraints on the mole fractions in streams l 2 and 3 take the form Multicomponent Systems 23 Stream 1 xA1 xB1 xC1 1 2a Stream 2 xA2 x32 xC2 1 2b Stream 3 xA3 xB3 xC3 1 2c We now make use of the data provided to obtain six equations and six unknowns expressed as SpeciesA 031200 moleshr 08250 moleshr xA3M3 0 3 Species B 021200 moleshr xB2250 moleshr 025M3 0 4 Species C xC11200 moleshr xC2 250 moleshr xC3M3 0 5 Stream 1 03 02 xC1 1 6 Stream 2 08 x32 xC2 1 7 Stream 3 xA3 025 xC3 1 8 There are several possible routes to a solution for the six unknowns and one route begins With Eq 3 for species A to obtain Species A xA3M3 160 moleshr 9 While the mole balance for species B can be expressed as x192 250 moleshr 025M3 240 moleshr 10 The mole fraction constraints can be arranged in the form Stream 1 xC1 05 11a Stream 2 xC2 02 x32 11b Stream 3 xC3 075 xA3 11c Use of Eq 11a With Eq 5 leads to Species C xC2 250 moleshr xC3M3 600 moleshr 12 and use of Eqs 11b and 11c provides x3 2 250 moleshr xA3M3 075M3 550 moleshr 13 We can add Eqs 10 and 13 in order to eliminate xB2 leading to M3 xA3M3 790 moleshr 14 and substitution of Eq 9 provides the solution for the molar ow rate of stream 3 24 Chapter 4 M3 950 moleshr 15 With this value for the molar ow rate in stream 3 we can use Eq 10 to determine the mole fraction of species B in stream 2 240 moleshr 025950 moleshr 250 moleshr ltXBgt2 0010 16 At this point we can return to Eq 9 to determine the mole fraction of species A in stream 3 as 160 moleshr 160 moleshr 0168 17 M3 950m01eshr XA gt3 With this value for the mole fraction of species A we can use Eq 11c to determine xC3 075 0168 0582 18 and with the value of xB2 given by Eq 16 we can use Eq 11b to compute the mole fraction of species C in stream 2 according to ltXCgt2 02 ltXBgt2 0190 19 The sequence of algebraic steps in this solution is arbitrary and the key to a solution is the fact that Eqs 1 and 2 and the specified parameters provide the necessary six independent equations given by Eqs 3 through 8 to determine the six unknowns 417 A continuous filter is used to separate a clear filtrate from alumina particles in a slurry The slurry has 30 by weight of alumina specific gravity of alumina 45 The cake retains 5 by weight of water For a feed stream of 1000 kghr determine the following 1 The mass flow rate of particles and water in the input stream 2 The volumetric flow rate of the inlet stream in m s 3 The mass flow rate of filtrate and cake in kgs 4 17 Let denote the inlet stream as 1 the ltrate stream as 2 and the cake as 3 The mass ow rate of particles in the input stream can be calculated as m 1 WP 1 m 1 And that of water will be mw1m m 1 2 Since the weight fractions of particles in the feed stream and the mass ow rate of the feed stream are known equations 1 and 2 can be applied to get k mp1300h Multicomponent Systems 25 kg 700 mw 1 hr 1 The volumetric ow rate of the inlet stream can be calculated as follows Q1 Qw1Qp1 3 where m Q 1 1 4 PW my Qpl pp 1 s and pp specific gravitypw 6 Applying equation 6 kg pp 4500 The volumetric ow rates of water and particles are calculated using equations 4 and 5 respectively ltle 0M Qp 1 00667m3 And the total ow rate of the input stream Will be equation 3 3 Q1 07667m hr In m3s m3 Q 213 10 4 s 3 Making a balance of particles 0300k g 095m3 7 hr And the balance of water Chapter 4 kg 027007 005m3 m2 8 From equation 7 we can get the mass ow rate of cake kg 3158 mg hr In kgs m3 00878k g s Plugging this value back into equation 8 we can get m2 6842k g hr m2 0190k g S 418 A BTX unit associated with a refinery produces benzene toluene and xylenes as illustrated below Stream 1 leaving the reactorreforming unit with a volumetric flow rate of 10 m3hr is a mixture of benzene toluene and xylenes with the following composition CB1 6kmolm3 or 1 2kmolm3 CX 1 2kmolm3 Stream 1 is the feed to a distillation unit where the separation takes place according to the following specifications 1 98 of the benzene leaves with the distillate stream stream 2 2 99 of the toluene in the feed leaves with the bottoms strwm stream 3 3 100 of the xylenes in the feed eaves with the bottoms stream Assuming that the volumes of components are additive and using the densities of pure components from Table l in the Appendix compute the concentration and volumetric flow rate of the distillate stream 2 and bottoms stream 3 stream leaving the distillation unit A39distilation column Multicomponent Systems 27 Figure 418 Distillation unit 4 18 The molar ow rate of each component can be calculated as i1Q1Ci1 iBTX 1 Considering the data given in the problem these values can be calculated MB 1 60 kmol hr kmol 39 20 nr 1 hr kmol 39 20 nX 1 hr Now we know that 98 of the benzene leaves with the distillate stream stream 2 99 of the toluene in the feed leaves with the bottoms stream stream 3 and 100 of the xylenes in the feed eaves with the bottoms stream Then it 098n311 099 r1 2 n3 1 0981 B1099 T1 X1 3 with n82 098MB1 4 m 1 099gt r1 s n3 1 0981 B 1 6 ltmgt3099ltnrgt1 lt7 1 Ml 8 Applying equations from 2 to 8 we get kmol h r kmol h r i559 ig41 with kmol kmol hr quotT 2 0 2 mg 5ss Chapter 4 kmol kmol nB312h r nT3198h r kmol it 20 lt X 3 hr Now considering the following data of densities pB 885k g3 pr 867k g3 pX 885k g3 m m m The volumetric ow rate of each component can be calculated as it M W Q J12 9 1 Pi Then applying equation 9 to each component in the distillate stream Q 2 5190 Q 2 0021 And the total distillate stream Q2 521 121 The bottom stream QB 3 0106 QT3 21042173 Q 3 2446 And the total bottom stream 3 m 4656 Q3 hr The concentration can be calculated as C 10 I Q Multicomponent Systems 29 Then kmol C 0258 c 11284 kmquotl BL m3 B 2 3 kmol m C 4253 C 0038km01 T3 m3 T 2 3 kmol m CX 3 42967 419 A standard practice in refineries is to use a holding tank in order to mix the light naphta output of the refinery for quality control During the first six hours of operation of the refinery the stream feeding the holding tank at 200 kgmin had 30 by weight of npentane 40 by weight of nhexane 30 by weight of nheptane During the next 12 hours of operation the mass flow rate of the feed stream was 210 kgmin and the composition changed to 40 by weight of n pentane 40 by weight of nhexane and 20 by weight of nheptane Determine the following The average density of the feed streams The concentration of the feed streams in molesm3 After 12 hours of operation and assuming the tank was empty at the beginning determine The volume of liquid in the tank in ms The concentration of the liquid in the tank in molsm3 The partial density of the species in the tank 4 19 a The average density of the feed streams can be calculated as pj J 112 1 where 1 denotes the first six hours stream and 2 denotes the second twelve hours stream The values of m are given in the problem The volumetric ow rates of each stream are given by w ri1 3 QJZI p iPHHP 2 z Being the densities of the components k k k p 626 pH 659i pH 6844 m m m The volumetric ow rates are calculated using equation 2 30 Chapter 4 3 Q1 03050 quot H1111 3 Q2 03231 quot H1111 Now applying equation 1 p1 6558k i m k 02 6500 a The concentration of the feed streams in rnolesrn3 is given by qt i lt3 Where all the values are known and the calculations can be done With the following results CP1 27263quot17 l CP2 36039 17 l mol mol CH130440W CH2 3017l mol mol CHP119633W CHP 2 12974 The volume of liquid in the tank in In3 after 12 hours of operation Will be Vliq660 660 4 1 2 With m in kgmin Then Vliq 2261m3 The concentration of the liquid in the tank in rnolsrn3 can be calculated as M1mi606wi2ma606 MW MW Cith s liq Multicomponent Systems 31 By applying this equation the concentrations are mol C1mnk 1778 mol CH tank 30300 mol CHP tmk 16210 The partial density of the species in the tank Will be pitank Citank k mm 2293 k ma zmgi k pHP mk 2162 4mga 420 A distillation column is used to separate a mixture of methanol ethanol and isopropyl alcohol The feed stream with a mass ow rate of 300 kghr has the following composition Separation of this mixture of alcohol takes place according to the following specifications a 90 of the methanol in the feed leaves with the distillate stream b 5 of the ethanol in the feed leaves with the distillate stream c 3 of the isopropyl alcohol in the feed leaves with the distillate stream Assuming that the volumes of the components are additive compute the concentration and volumetric ow rates of the distillate and bottom streams 4 20 In order to make a balance of components let calculate the total mass density of the feed stream stream 1 It is given by pZpi iMaEJ 1 i1 And the mass fraction can be calculated as 32 Chapter 4 w 2 Then by applying equations 1 and 2 we get k WM10501 p17893 g WE10250 W110249 Now the balance of components considering the conditions given in the problem is given by 00501300 090501300 wM3ri13 3 00250300 0050250300 wE 3ri13 4 00249300 0030249300 wl3ri13 5 Where distillate stream is stream 2 and bottom stream is stream 3 After performing computations WM 3 m 150323 w53m 712410 w 3 n3913 724458 3 By adding these equations and considering that Z w 1 1 0 of stream we have i1 m3 158719k g h Equivalent calculations can be done to stream 2 to get WM2ri12 1352908 WE2Vi12 37495 14524 512 22406 And by adding these equations m 1412809k7 g Now the volumetric ow rates of distillate and bottom streams are give by Multicomponent Systems 33 Q Z Z 6 Being the densities of the pure components pM 791k g3 pE 789k g3 p1 786k g3 m m m The volumetric ow rates of distillate and bottom streams can be calculated using equation 6 as 3 m 01786 Qz h m3 02015 Q h Finally the concentration of the distillate and bottom streams are given as 7 Then after calculations we get CM 2 236357 CM 3 2386 kmol kmol C12 C13 m m 421 A mixture of ethanol and water is separated in a distillation column The volumetric ow rate of the feed stream is 5 m3hr The concentration of ethanol in the feed is CE 28kmolm3 The distillate leaves the column with a concentration of ethanol CE l3kmolm3 The volumetric ow rate of distillate is one cubic meter per hour How much ethanol is lost through the bottoms of the column in kilograms of ethanol per hour 4 21 Let make the balance of ethanol 0CE1Q1 CEQ2 CE3Q3 1 Chapter 4 Where stream 1 denotes the feed stream stream 2 denotes the distillate stream and stream 3 denotes the bottom stream Substitution of the data given by the problem leads to 0285 131 r153 2 Then from equation 2 we have This is the ethanol lost through the bottom stream This amount can be transformed in kilograms of ethanol per hour by multiplying by its molecular weight kg mE3 4609h r 422 A ternary mixture of benzene ethylbenzene and toluene is fed to a distillation column at a rate of 100 kmolhr The composition of the mixture in moles is 74 benzene 20 toluene and 6 ethylbenzene The distillate ows at a rate of 75 kmolhr The composition of the distillate in moles is 9733 benzene 2 toluene and the rest is ethylbenzene Find the molar ow rate of the bottoms stream and the mass fractions of the three components in the distillate arid bottoms stream 4 22 Let denote feed stream as stream 1 distillate stream as stream 2 and bottom stream as stream 3 Making a balance to the components 02XB1 7hxBz i 2XB3 n3 1 02XEB1 L1XEB2 7 2XEB3 n3 2 0XT1iaXT2XT3 n3 3 Considering the data given in the problem kmol kmol 00741007 09733757 x834g 4 kmol kmol 0020100h r 00275h r xT3 n3 5 kmol kmol 0 006100 hr 1 09733 00275 hr xEB3 n3 6 Solving equations 45 and 6 Multicomponent Systems kmol x8 3 n3 10025h r kmol JCT 3 n3 185h r kmol 39 54975 XEB3 quot3 hr And the total bottom stream can be calculated by adding the above three equations to get kmol hr Now to nd the mass fractions of the three components in the distillate and bottoms stream the mass ow rates should be calculated by using the molecular weight In the distillate steam we have ig25 mixiquot39l MW iBTEB j12 7 Applying equation 7 kg mB2 5702127 kg 138212 mr 2 hr kg m 53349 EB Iquot And the total mass ow rate of distillate 3 kg m2 Zm2 5893688h r i1 Now the mass fractions in the distillate stream are given by w 39 8 Applying equation 8 36 Chapter 4 In the bottom stream we can calculate de mass ow rates of each component by applying equation 7 kg m 78309 B h kg m 3 1704609 hr 39 583 659kg mEB 3 E And the total mass ow rate of bottom stream 3 m3 205193 2366577k g i1 hr Finally the mass fractions in the bottom stream are calculated using equation 8 w 3 00331 w 3 07203 WEB 3 02466 423 A complex mixture of aromatic compounds leaves a chemical reactor and is fed to a distillation column The mass fractions and ow rates of distillate arid bottoms streams are given in Table 423 Compute the molar ow rate and composition in molar fractions of the feed stream Table 423 Flow rate and composition of distillate arid bottoms streams I In kghr I CDBenzene I CDToluene I CDBenzaldehide I CDBenzoicAcid I m I Distillate 125 01 085 003 00 002 Bottoms 76 00 005 012 08 003 4 23 Let denote the feed stream as stream 1 the distillate stream as stream 2 and the bottom stream as stream 3 The component balances in the distillation column are 0wB1n39t wB a wg3m 1 0 wTLM wnzm wngm 2 0wBA1m WBA2 nrpm mg 3 0wBA1nt WBAC2m wgttm 4 0wMB1 quotvi Wm nitWmtm 5 Multicomponent Systems 37 In order to calculate the molar ow rate of each component in the feed stream the individual amounts of the components have to be divided by the molecular weight ni1 wi1ri11MWi iBTBABAcMB 6 After computations we get kmol 01600 quotB 1 hr kmol 11944 nr 1 hr kmol nBA1 nBAC 1 04979M hr kmol nMB1 003517 Then the molar ow rate of the feed stream is given by 5 11 2M1 7 The molar fractions of the components in the feed stream can be calculated as 8 By applying equations 7 and 8 the following results are got 771 20087km 01 hr With compositions x3 1 00797 XI 1 05946 x3191 00797 Jam 1 02479 ng 1 00175 424 A hydrocarbon feedstock is available at a rate of 1000 kmolehr and consists of propane XI 02 nbutane x3 03 npentane xPZ 002 and nhexane xH 03 0 The distillate contains all of the propane in the feed to the unit and 80 of the pentane fed to the unit The mole fraction of butane in the distillate is yB 04 The bottom stream contains all of the hexane fed to the unit Calculate the distillate and bottoms streams ow rate and composition in terms of mole fractions 38 Chapter 4 4 24 Let denote the feed stream as stream 1 the distillate stream as stream 2 and the bottom stream as stream 3 The component balances in the distillation column are 0XP1 ihyP2quotl2XP3quotL3 1 0XB1 ihy32quotl2x33 i 2 02XP21 L1YP22 XP23 7i3 3 0XH1 7L1YH2 iL2xH3 iL3 4 Considering the data given in the problem we have 0021000 yP2ri2 0 5 0031000 04i12 x83ri3 6 0 021000 08021000 x3 43913 7 0031000 0 xH3rl3 8 In the distillate stream we have propane n butane and n pentane Solving equations 5 and 7 we can know the molar ow rate of propane and n pentane kmol nP2 ZOOh r kmol nPE 2 l60h r And we know that the mole fraction of butane in the distillate is yB 04 This means that 04 9 Since rip2 and r532 are known rig2 can be calculated using equation 9 After calculations kmol n8 2 2407 Then the distillate ow rate is given by kmol r12 600 h r Multicomponent Systems 39 Substitution of r22 into equation 6 7 and 8 leads to kmol x8 3 n3 60 hr kmol xPe3 n3 40 hr m 7 300Wl h By adding these equations we get fa 400 kmol hr Finally the molar fractions of the components in the distillate and bottom streams can be calculated as yi 10 And xi3 11 After computations yP2 033 x8 2 015 a o4o a o1o m 027 ltth o7s Section 4 8 425 It is possible that the process illustrated in Figure 411 could be analyzed beginning with Control Volume lrather than beginning with Control Volume II Begin the problem with Control Volume I and carry out a degreeof freedom analysis to see what difficulties might be encountered 4 25 In this problem we examine the process studied in Sec 48 and illustrated in Figure 425 In the text we began our analysis with Control Volume II and the degrees of freedom analysis indicated that the material balance problem for Control Volume II 40 Chapter 4 4 x w I 2 Control I Volumel I I I I II I l I an u l 5 I I I I I I l l 1 I l l I x Figure 425 Control volumes for two column distillation unit was solvable Thus our procedure consisted of solving the problem associated with Control Volume II and then solving the problem associated with Control Volume I In this problem we consider the inverse ie solving the problem associated with Control Volume I and then moving on to the solution of the problem associated with Control Volume II To explore this strategy we focus our attention on Control Volume I illustrated in Figure 425 and note the following N 3 species A B and C l M 3 streams l 2 and 5 2 0 05 o 03 o 0045 Specified stream variables Ah B1 Ah 3 03 2 0091 m1 10001bmhr Our degrees of freedom analysis for Control Volume I is given as follows Multicomponent Systems 41 Table 425 Degrees of Freedom Table Stream Variables compositions N X M 9 ow rates M 3 Generic Degrees of Freedom N X M 12 Number ofl Balance Pauations massmole balance equations N 3 Number of Constraints for Compositions M 3 Generic Specifications and Constraints N M 6 Specified Stream Variables compositions 4 ow rates 1 Constraints for Compositions 0 Auxiliary Constraints 0 Particular Specifications and Constraints 5 Degrees of Freedom 12 6 1 1 Here we are left with one degree of freedom and we cannot solve the material balance problem associated with Control Volume I in order to determine the seven unknowns that we list as m2 ms LOC1 002 oA5 LOB 5 and oC5 However we could solve for six unknowns in terms of the seventh unknowns and sometimes this is a suitable option In this case we could solve for m2 oCl 002 coA5 LOB 5 and 005 in terms of ms however this is not a useful solution strategy Instead we follow the development in the text and solve the problem associated with Control Volume II which provides a value for m2 This additional specified ow rate increases the Particular Specifications and Constraints in Table 425 to six and the problem for Control Volume I becomes solvable 426 In a glycerol plant a 10 mass basis aqueous glycerin solution containing 3 NaCl is treated with butyl alcohol as illustrated in Figure 426 The alcohol fed to the tower contains 2 42 Chapter 4 water on a mass basis The refinate leaving the tower contains all the original salt 10 glycerin and 10 alcohol The extract from the tower is sent to a distillation column The distillate from this column is the alcohol containing 5 water 10 glycerin extract 95 alcohol 1000 lbmhr 98 alcohol moo lbwhr Figure 426 Solvent extraction process The bottoms from the distillation column are 25 glycerine and 75 water The two feed streams to the extraction tower have equal mass ow rates of 1000 lbIn per hour Determine the output of glycerin in pounds per hour from the distillation column 4 26 The process under consideration in this problem represents a clean up operation in the extraction tower where salt is separated from the glycerin This occurs because the glycerin is soluble in the butyl alcohol while the NaCl is not thus glycerin is transferred to the alcohol stream while the salt remains in the aqueous stream For convenience we identify the molecular species according to A glycerin B alcohol C water D salt and for each species we will make use of a steady state mass balance in the absence of chemical reaction For species A this takes the form JwApvndA 0 1 A and we must now think about the control volume or control volumes that need to be constructed Since we want to determine the mass ow rate of glycerin in the bottom stream leaving the distillation column we must cut that stream where information is required In addition we must cut the glycerin input stream to the extraction tower Multicomponent Systems 43 where information is given This indicates that at least we need to make use of the control volume shown in Figure 426a It is good practice at this point to construct a 5 39 diEtiEtl h column product Figure 4 26a Control volume for the extraction distillation process degree of freedom table to verify that we have a solvable problem however this is a particularly simple problem that we can immediately explore in terms of Eq 1 for all the species and in terms of the single control volume shown in Figure 425a The mass balances can be represented as SpeciesAi mom wA3m3 mum COA5quot3915 wA1m1 2a species B wghma coB3n 13 coB4m4 ohms comma 2b species C 021412 mama wc4m4 wasms cochml 2c species D commZ mD3n 18 coD4m4 60D5ms wD1m1 2d and in each stream we have a constraint on the mass fractions that we could express in the form co4 cog OJCj LODj 1 j12345 3 This certainly has the appearance of a complex problem one which is well suited to the use of a computer routine provided that a degree of freedom analysis indicates that we have a solvable problem Often problems of this type can be simplified by using N l species balances and the total mass balance because of the simplicity of the latter If we think about replacing one species balance with the total mass balance it seems logical 44 Chapter 4 that we should eliminate the most complex species balance and this suggests that we should eliminate the water balance since water appears in all five streams that are cut by the control volume illustrated in Figure 426a This leads to species A 0A 2 12 0A 3 quot3913 60 4 m4 0A 5 m5 01 1 ml 43 species B 03 2 m2 03 3 quot3913 03 4 m4 03 5 m5 03 1 m1 4b total mass 12 m3 W quot3915 11 4C species D 0D2n3912 LOD3n3913 60D4 m4 C0D5 15 COD 1 m1 4d This still has the appearance of a very complex problem however if we recognize that the salt appears only in streams 1 and 2 we can simplify this set of equations to the form SPeCieS A1 01 2 m2 03A 3 3 03A 4 m4 01 5 m5 01 1 m1 5 21 SPeCieS B 1 033 2 m2 033 3 3 033 4 m4 03 5 m5 033 1 1 5b total mass 12 13 m4 ms m1 5 c species D 012n3912 0 0 0 0D1n3911 5d The mass balance for species D salt is quite simple because of the zeros associated with those streams that contain no salt If we identify the zeros that exist in the mass balances for species A and B we obtain further simplification indicated by species A coA2n3912 0 coA4n3914 0 oA1n3911 6a species B 03 2 m2 03 3n3913 0 03 5n3915 0 6b total mass 12 13 m4 m5 m1 6c species D 012n3912 0 0 0 0D1n3911 6d If we now identify the mass fractions and mass ow rates that are known by enclosing them in brackets we find an interesting simplification illustrated in the following four equations speciesA coA2m2 0 A4n 14 0 nA1n 11 7a speciesB coB2n3912 mB3m3 0 DB5n3915 0 7b total mass ma m3 m4 m5 m1 7c species D coD2n391Z 0 0 0 wDlml 7d Multicomponent Systems 45 Here we are confronted with four unknowns 12 m4 ms and oD2 and four equations thus a solution is possible In particular we see that the first three equations contain only the three unknowns 12 m4 and ms thus we need only solve three equations for three unknowns This is much simpler than solving four equations for four unknowns Since we are asked to determine the output of glycerin in pounds per hour we make use of Eqs 7a through 7c to solve for W and the solution takes the form 1 ROBE 03 1 0320A1 CO1 033 5 03A 2 03A2 m4 1 8 1 03 2 MM CO1 1 03 2 MM CO1 033 5 wA2 01 2 033 5 03A 2 03A 2 Making use of the specified mass ow rates and mass fractions leads to 1 376 lbm hr 9 and the mass ow rate of glycerin leaving the distillation column is given by mass ow rate LOA4m4 376lbmhr 10 of glycer1n The algebra associated with Eq 8 is non trivial when done by hand and even for a simple mass balance problem of this type the use of computer software has some very important advantages Section 49 427 In Sec 48 the solution to the distillation problems was shown to reduce to solving the matrix equation A m b in which 1 l l 1000 A 0045 0069 0955 u b 500 0091 0901 0041 quot3914 300 Here it is understood that the mass ow rates have been made dimensionless by dividing by lbm hr The augmented matrix aug A for this problem is l l l 1000 augA 0045 0069 0955 500 0091 0901 0041 300 Define the following lists in Mathematica corresponding to the rows of the augmented matrix aug A Chapter 4 R1 1 1 1 1000 R2 01045 01069 01955 500 R3 01091 01901 01041 300 Write a sequence of Mathematica expressions that correspond to the elementary row operations for solving this ystemr The first elementary row operation that given Eqr Error Reference source not found is R2 01045R2 R1 Show that you obtain an augmented matrix that defines Eqr Error Reference source not found 4281 Rework Problem 427 using row operations that reduce the matrix problem A m b to the result m A71 b 1 Hint Define the lists R1 R2 R3 for the matrix A not the augmented matrix and then another set of lists B1 B2 B3 for the augmented matrix b Do we write this as aug lb or as lb b9 irer R1 111 B1 1001000 R2 0041006900955 32 010500etc i Show that your answer corresponds to the one found in Sec 481 ii Show that the inverse found satisfies A IA 1 iii Use Mathematica s built in function Inverse to find the inverse of A iv Use Mathematica s Reduce function on Ab and show that you obtain the same result as in 1 Chapter 5 January 13 2010 GasLiquid Systems Section 51 51 Show that the mole fraction in an ideal gas mixture can be expressed as yA pAp 51 The equation of state for an ideal gas mixture is given by pAV nART A12N 1 for the individual species and by pV nRT 2 If we divide Eq 1 by Eq 2 we obtain M A12N 3 pV nRT which leads to the relation between the mole fraction the partial pressure and the total pressure expressed as p A yA A12N 4 p The simplicity of this result often leads to its use for gas mixtures that are not ideal and one must be very careful to use Eq 4 judiciously 52 Assuming ideal gas behavior determine the average molecular weight of a mixture made of equal amounts of mass of chlorine argon and ammonia 52 Let assume 3 kg of mixture Since the mixture is made of equal amounts of mass of chlorine argon and ammonia we have that the mass fractions w are 333 NH 0333 and the amount of mass are m w1m202 1 Applying equation 1 to each of the components Chapter 5 mm 03333lkg mczz W012 mm WA mm 03333 lkg mm wNHK mm 0333 3 1kg The amount of moles of each of the components can be calculated m n 2 1 MW Then m 1701 1kg 7 00141107201 012 7090 mm Mp 1kg 7 00250107201 4 3995 m nNH H3 1kg 00587107201 3 MW NHK 17031 And the total amount of moles m ncl2 mm nNHK 00978kmol 3 Then molar fraction can be calculated as n y n 4 202 Then n ya I 0144 quot202 yA quot 1 0256 quot202 n yNHK 0600 quot202 Finally the average molecular weight is given by A3 MW 2y Mm A ClzArNH3 5 A1 GasLi qui d Systems 3 MW 30656 kg kmol Section 52 53 A liquid mixture of hydrocarbons has 40 by weight of cyclohexane 40 of benzene and 20 toluene Assuming that volumes are additive compute the following a partial densities of components in the mixture overall density of the mixture concentration of components in molesm3 d molar fractions of components in the mixture A 33 53 Let assume 100 kg of mixture The amount of mass of each of these components can be calculated as mi Wi mtot 1 Applying equation 1 to each component mCH wCH 4720 040 100 40kg m3 wB mm 040100 40kg mT lemm 020 100 20kg Considering now the densities of the components m V 7 2 P Then the volume of each component will be VCH MUCH 40kkg 00513m3 PCH 779i3j m VB m3 403 00452m3 PB 885i3 VT 7 quotLT 20k 7 00225m3 P 864 m Then the total volume 4 Chapter 5 Vm VCH V VB 0119m3 a Finally the partial densities of components in the mixture are given by m p t 3 t Vtot Then applying equation 3 pCH LCH 336131Lg3 Vtot m 7723 kg p3 73361373 tot quot2 pT 16807l 3 tot quot1 b The overall density of the mixture is given by A3 p 2 pl ACHBT 4 Al Then p 84034k g3 m c The concentration of components in molesm3 is given by PA C ACHTB 5 A MWA GasLi qui d Systems 5 33613 103i3 CCH 399 103l 8416i m mol 16807 103i3 CB 430103quotLo3l 78114i m mol 33613103 CT 182103quotLo3l 92141i m mo e Let consider 1 m3 ofmixture The amount ofmoles of each species is 17ch 39910720 nT 43010720 n3 l82kmol And the total amount of moles mm 1011kmol The molar fraction is given by n t 6 quottot Then yCH 0395 yB 0425 yT 0180 Section 53 54 Determine the vapor pressure in Pascal of ethyl ether at 25 C and at 30 C Estimate the heat of vaporization of ethyl ether using these two vapor pressures and the ClausiusClapeyron equation 54 The vapor pressure is given by the Antoine s equation 6 Chapter 5 B 1 A 0g10pAvap 1 in which p Ayvap is determined in mm Hg and T is speci ed in C The coefficients for ethyl ether are A 698467 B 109064 0 23121 Then the vapor pressure at 25 C can be calculated using the equation 1 pee vap 25 C 5344rman And at 30 C peewp 30 C 64466ng The ClausiusClapeyron equation can be expressed as pAvap pziyap 39 exp i 2 Or taking 25 C as reference temperature AH 1n gt 1 LLl T30 T25 3 Applying equation 3 J AHWIP 2783011 55 Determine the vapor pressure of methanol at 25 C and compare it to that of ethanol at the same temperature Consider the ethanolmethanol system to be an ideal solution in the liquid phase and an ideal gas mixture in the vapor phase Determine the mole fraction of methanol in the vapor phase when the liquid phase mole fraction is 050 If the liquid phase is allowed to slowly evaporate will it become richer in methanol or ethanol Here you are asked to provide an GasLi qui d Systems intuitive answer concerning the composition of the liquid phase during the process of distillation In Chapter 10 a precise analysis of the process will be presented 55 The vapor pressure is given by the Antoine s equation B l A 0g10pAvap e T 1 in which p Ayvap is determined in mm Hg and T is speci ed in C The coefficients for ethanol are A 81629 B 162322 6 22898 And the coefficients for methanol are A 807246 B 157499 6 23886 Then the vapor pressure of ethanol can be calculated using the equation 1 pew 59124mmHg And of methanol pmwp 127989mmHg The total pressure will be Pt Fe pm 2 Then p 93557mmHg The mole fraction of methanol in the vapor phase is given by yi 39pi xi 39piyap 8 Chapter 5 M0316 Pt Ye xm 39 pmvap 0 Pt J m Ifthe liquid is allowed to evaporate it will become richer in ethanol 56 Determine the vaporliquid equilibrium curve of a binary mixture of glycerol and acetone Plot the data as molar fraction of glycerol in the vapor and liquid phases as a function of temperature 56 In an ideal vaporliquid multicomponent system the partial pressure of species g glycerol and a acetone in the gas phase is given by Pg xg39Pgwap Pa xa39Pavap 12 where the vapor pressures are given by the Antoine s equation 19487 10 P 748689 g10 gvap 13296 T 3 127703 10 P 7231577 g10 mu 23723 T 4 with the temperature in 0C and the pressure in mm Hg Now the total pressure can be calculated as P101 2 Pg Pu 5 In order to nd the molar fraction of glycerol in the vapor and liquid phases as a function of temperature the following procedure is used 1 Assume a temperature T and a composition of glycerol in the liquid phase xa 2 Calculate the composition of acetone as x 1 x8 3 Using equation 3 y 4 calculate the vapor pressures GasLi qui d Systems 9 4 Calculate the partial pressures using equation 1 and 2 5 Calculate the total pressure 6 It the total pressure is 760 mm Hg the initial T is correct If it is not go back to step 1 7 Calculate the composition of glycerol in the vapor phase using Dalton s law y Pg g Prat 6 8 Finally plot xg step 1 and y8 step 7 as a function of the temperature step 1 By doing this procedure the results are the following And the plot Figure 56Vapor liquid equilibrium curve of glycerol 35000 30000 25000 20000 15000 10000 5000 000 T 0C Xg yg 57 Use Eqs 527 and 528 order to derive 529 57 We begin this problem with Eq 527 for both species A and species B VA xB pBvap p and divide the first by the second to obtain yA 7 xA pAvap 7 xA MB 963 pBvap 963 in which 1A3 is the relative volatility defined by pAvap OLAB pBvap Given the constraints on the liquid and vapor phase mole fractions xAxBL yAyB1 we can eliminate yB and x3 to obtain Here we could compact our result in the form yA B l yA and solve for yA to obtain x 0L AB M ill 1 A 1B 1 xA 0c l xA quot3quot quot ofthe andJ 39 bylixAleadsto x on M A AB I XA t xAOLAB and this can easily be arranged in the classic form given by OLAB xA M 2 1 xAxAB l Chapter 5 1 2 3 4 5 6 7 8 9 This is an extremely useful result for binary systems however when three or more species are present computational methods are more appropriate for the determination of vaporliquid equilibrium relations GasLiquid Systems 1 1 58 Demonstrate that Eq 532 is valid for an ideal system containing three components and think about replacing the constraint x A ltltl with something more appropriate 58 We begin this problem with Eq 527 for species A B and C m p yc xc PavapP 1 In an effort to develop a representation for species A we add the two results for species B and C to obtain x x yB yC BPBvap p CPCvap 2 The left hand side of this result is 17 y A leading to prBvap xCpCvap 1 M 3 P and we can use this in the first of Eqs 1 to obtain yA xApAvap 4 liyA xBPByap xCpCvap We now de ne two relative volatilities according to pBvap pCvap XBA 0ltc4 5 pAvap pAvap so that Eq 4 takes the form yA xA 6 1 yA xBOLBA xCaCA In general this is not a very useful relation however if we impose the constraint yA ltlt 1 7 Eq 6 simpli es to xA yA xBO BA f xCOLCA Typically this result is expressed in the form yA KA xA where K A is a strong function of the temperature through OLBA and OLCA and a function of the composition through the dependence on x3 and xC Chapter 5 59 Consider air to consist of nitrogen and oxygen and indicate under what circumstances the mole balances for these two components can be added to obtain the special form Icmvmdl 0 A that was used in Example 55 Here the molar concentration of air is defined by can 0N2 002 59 We begin this problem with the macroscopic mole balances for nitrogen species A and oxygen species B d E 1 i 2 dt Addition of these two macroscopic balances leads to d E RA RB av a V A V and if we make use of the decomposition of the two species velocities vAvuA vBvuB 4 we can express the sum of the two equations in the form d c v ndA dt 3 5 JcAuAcBuBndl IRAFR dV If species A and B are not involved in a chemical reaction No Chemical Reaction R A RB 0 6 and the diffusion velocities are negligible compared to the mass average velocity Negligible Diffusion uA 113 ltlt v 7 GasLi qui d Systems 13 we can simplify Eq 5 to g MA 0 s V A This can be expressed as iJ cmrdV Icairvndl 0 9 V A where the molar concentration of air has been de ned as calr 0N2 002 10 The use of lumped macroscopic balances such as Eq 9 is often very useful however one must be careful to note that it can only be done when both diffusion and reaction make negligible contributions to the balance equation 510 In Example 55 we used a control volume located outside of the scrubber This control volume is illustrated in Figure 55c and in this problem you are asked to construct a suitable control volume that lies insider the scrubber 510 To construct of control volume that lies inside the scrubber we first cut the entrances and exits inside the scrubber and then join the cuts with a surface at which the ux is zero Thus wejoin the cuts where cAvn 0 as indicated in Figure 510 Figure 510 Construction of a control volume located inside the scrubber 14 Chapter 5 511 In Example 55 we used a macroscopic mole balance to determine the mole fraction of SO2 in the exit water stream In actual fact the exit water stream contains two molecular species that involve sulfur dioxide ie SO2 and H2503 This occurs because of the reaction SO2 H20 J HZSO3 A more precise analysis of this problem would make use of a total macroscopic mole balance for 02 and H2503 Repeat the analysis given in Example 55 for the molar concentration of species C where cc CA c3 Here C A represents the concentration of 02 and CB represents the concentration of H2503 Be careful to consider the reaction rate term in the species mole balance and comment on your treatment of the species velocities VA and VB 511 In this problem we wish to take into account the reaction between 802 species A and H20 to produce H2803 species B For steadystate conditions and the xed control volume shown in Figure 511 the macroscopic mole balances for species A and B take the form SOZ J cAvndA J RAdV 1 A V H2SO3 IchndA J RB dV 2 A V We assume that species B is not present in either the entering liquid and gas streams and GasLi qui d Systems 1 5 Figure 51 Control volume for the analysis of gas adsorption we will assume that the concentration of H2303 species B is negligible in the gas stream leaving the scrubber Under these circumstances we can evaluate the area integrals in Eqs 1 and 2 to obtain 802 cA4 Q4 IRA dV 3 V H2303 03 2 Q2 J RB 51V 4 V The stoichiometric constraint on the molar rates of reaction is given by RA RB 5 and this means that we can add Eqs 3 and 4 to obtain 02H2803 CA4Q4 0 6 In terms of mole fractions this result for the combined mole balance for sulfur dioxide and sulfurous acid takes the form 802H2803 yA4c4Q4 0 lt7 16 Chapter 5 Following Example 55 we assume that the absorption of nitrogen oxygen etc in the water is negligible so that the mole balance for air see Problem 59 provides Air 01 Vair 1 Q1 0404 Q4 2 0 8 A little thought will indicate that this mole balance for air can be expressed as 01 1 Vs02 1Q1 C4 1 Vs024Q4 9 and if the mole fraction of sulfur dioxide is small compared to one 02302 ltlt 1 10 Eq 8 takes the form 01Q1 04 Q4 11 Since the temperature and pressure are assumed to be constant the total molar concentration in the air stream can be treated as a constant 01 c4 pRT 12 and this condition can be used in Eq 11 to obtain Q1 Q4 13 Use of this constraint on the volumetric ow rates for the gas entering and leaving the scrubber allows us to solve Eq 7 for the liquid mole fractions according to SO2 H2SO31 9502 0902 VA1 0 A4 14 2 2 This result has the same form as Eq 11 in Example 55 and it indicates that the mole fraction of 02 in Example 55 must be interpreted as the sum of the mole fractions of 02 and H2303 512 Derive the form ofthe solid phase mass balance given by Eq 8 in Example 56 512 Imagine that the solid phase in the drying process consists of chemical species A B and C For a fixed control volume the mass balances for these three species are given by d dV la dt rA This is o en referred to as the dilute solution condition GasLi qui d Systems 17 dijdeV Jva JrBdV 1b t V A V d dV lc dt rC V provided that diffusion is negligible Addition of these three macroscopic mass balances leads to d dA dV 2 dt rA rB rC V If any chemical reactions occurring in the solid phase involve only those chemical species that make up the solid phase this result simplifies to d vndA 0 3 dt B C Here it is convenient to de ne the solid phase density as 0mm 9A 93 9c 4 so that the macroscopic mass balance for the solid phase takes the form d E psolid dV psolidv 0 5 V A For steadystate conditions Eq 5 simpli es to J pmhdv ndA 0 6 A and when this result is applied to the drier described in Example 56 we obtain Solidi Pm1h Q1 Pmdb Q2 8 The total areaaveraged density in the wet solid stream is de ned by ltpgt ltpsozidgt ltszogt ltpmgt lt9 and the associated mass fractions are given by 18 Chapter 5 The constraint on the mass fractions in the wet solid stream takes the form wsolidgt wHZOgt mat 1 11 however the contribution of the air to this constraint will be negligible allowing us to express the mass fraction of the solid as msolidgt 1 CDHZOgt 10 This result is used to obtain Eq 8 of Example 56 513 Complete the analysis in Example 56 in order to determine the total molar flow rate of fresh air entering the drier 513 The macroscopic mole balance for air can be written as see Problem 59 d E calrdV calrv ncZl 0 l and for steady state this takes the form JcmrvndA 0 2 A The gas streams in the drier described in Example 56 contain both air and water vapor thus it is convenient to express Eq 2 as J ymr cv ndl 0 3 A and assume that cvn is constant across the area of the entrance and exit in order to obtain 6sz mum 5 Here we have made a plausible and very subtle assumption that the air entering and leaving with the wet solid streams can be ignored in the mole balance for air Given that yair 1 szo 6 we can alrange Eq 5 in the form l M4 yHZO 3 M3 7 1 yHZO 4 Substitution of this result into Eq 12 of Example 56 leads to GasLi qui d Systems 19 M3 H2041 VHZO3 l 8 H203 L1VHZO4 j and the molar ow rate of the incoming air is given by M3 259 lbm hr 9 514 A gas mixture leaves a solvent recovery unit as illustrated in Figure 514 The partial pressure of benzene in this stream is 80 mm Hg and the total pressure is 750 mm Hg The volumetric analysis of the gas on a benzenefree basis is 15 CO2 4 O2 and the remainder is nitrogen This gas is compressed to 5 atm and cooled to 100 F Calculate the percentage of benzene condensed in the process Assume that CO2 O2 and N2 are insoluble in benzene thus the liquid phase is pure benzene Figure 514 Recoverycondenser system 514 Since the inlet stream to the compressorheat exchanger is completely de ned in terms of the composition we need only consider a bole balance for the control volume illustrated in Figure 514a For steadystate conditions and no chemical reaction the 20 Chapter 5 Figure 51411 Control volume for recoverycondenser system macroscopic mole balance for benzene can be written as Benzene JcAvndA 0 1 AC provided we assume that VA n vn at the entrances and exits If we further assume that the concentration of benzene is uniform at the entrances and exits the benzene mole balance takes the form Benzene 0A1Q1 0A2Q2 CA3Q3 2 The percentage of benzene that is condensed in this process can be expressed as percentage ofbenzene x 100 3 CA 1 Q1 condensed and with the aid of Eq 2 this takes the form jpercentagel of benzene g 100 4 lcondensed J J If we are willing to use the ideal gas law the molar concentrations of benzene in the two gas streams can be expressed in the form on pA2RT2 5 GasLi qui d Systems 2 1 Here p A 1 represents the partial pressure of benzene in the stream entering the compressorheat exchanger unit and p A2 represents the partial pressure of benzene in the gas stream leaving this unit Use of the ideal gas relations in Eq 4 provides percentage ofbenzene m 100 6 condensed A 1 2 1J and all that remains to be done is to determine the ratio of the volumetric ow rates In order to determine the ratio of the volumetric ow rates w return to the mole balance given by Eq 1 to obtain the balance equations for carbon dioxide species B oxygen species C and nitrogen species D These are given by Carbon dioxide 03 1Q1 CB ZQZ 7a Oxygen Cc 1 Q1 Cc2Q2 7b Nitrogen 0D 1Q1 0D 2Q2 7 We can use the sum of these three equations and use ideal gas representations analogous to Eqs 5 to obtain Q2 8 PB PB PC PD ZRTZ This result can be expressed in terms of the total pressures in streams 1 and 2 p1 and p2 in order to obtain pl 7 m p2 7 pm 3 9 From this we obtain the desired ratio of volumetric ow rates that is given by g P2 PA2 A 10 Q2 P1 PA1 T2 and substitution of this result into Eq 6 leads to percentage condensed 2 From Antoine s equation we can calculate the vapor pressure of benzene at 100 F to be 167 mm Hg thus all of the pressures on the right hand side of Eq 11 are known and given by 22 Chapter 5 pl 1 80760 atm p2 5atm pA2 167760 atm and this leads to the result percentage of benzene 62 condensed In thinking about a result of this type one must always remember that the ideal gas law is an approximation as is Antoine s equation thus there is always some error associated with the calculated result 515 Small amounts of inorganic salts contained in an organic uid stream can be removed by contacting the stream with water as illustrated in Figure 515 The process requires that the organic and aqueous streams be contacted in a mixer that provides a large surface area for mass transfer and then separated in a settler 1f the mixer is efficient the two phases will be in equilibrium as they leave the settler and you are to assume that this is the case for this problem You are given the following information a Organic stream flow rate 1000 lbmmin b Specific gravity of the organic fluid pmg pH20 087 c Salt concentration in the organic stream entering the mixer CA 0g 00005 molesliter d Equilibrium relation for the inorganic salt cam Keg CAmg where Keg 60 Here CA W represents the salt concentration in the aqueous phase that is in equilibrium with the salt concentration in the organic phase CAUg In this problem you are asked to determine the mass flow rates of the water stream that will reduce the salt concentration in the organic stream to 01 001 and 0001 times the original salt concentration The aqueous and organic phases are to be considered completely immiscible ie only salt is transferred between the two phases In addition the amount of material transferred is so small that the volumetric flow rates of the two streams can be considered constant C 1 Figure 515 Liquidliquid extraction Gnqumd ysfm 23 5715 In thrs problem no rnformataon rs grven or reqmred about the stream between the murer and the settler thus the appropnate eontro volume rs the one ruustrated tn on Frgure 5 15a We assume thatthe system operates at steadyrstate and that no reaetr gure 5 15a Cuntqu vulume fur mxa39rsettla39 1norgamesa1t J s ndA 0 1 Ar Here we have used vn tn p1aee of v wrth the rdea that dAffusne effects are When Eq 1 volume ruustratedrn Frgure 5 15a we obtatn Inorganre sa1t 494 0 2 We are grven that EA or CA KqrA4 3 andthrs allows us to express Eq 2m the form KaqEAAQz 90292 i 0494 4 We are a1so grven that the volumetne ow rates ean be eonstratned by the apprommatrons Q er Q 94 5 and thrs allows us to express the volumemc ow rate ofthe stream entenng the mixer settler system as Q m e we 92 6 Chapter 5 In terms of the mass ow rates of water and the organle phase entenng the system we have a solution to our problem glven by mm ml HQ 7 quot2qu anolthls soluuon ean also be expressed as mu mm 7 3 8 pegpup Klq To eompute the answers requesteol ln thls problem we make use of the appropnate numbers to obtaln r nl llgte 10 m r 18961bmmln bmu e 100 llgte 1000 Thls problem was partleularly slmple beeause anol organle hases were approlrlmateol as eo the volumetrle ow rates of the aqueous nstan aqueous stream entenng the system was zero ts and the salt eoneentrauon ln the If these slmpllfleauons are not made 546 ln thls problem we ellmlne the probess bf xecavenng nsson mltenlls mm spem nuelelr fuel mus Ths lsusua y referred to ls tsptqesrrmg quhe fuel to mauve plulamum pu ml the letlve lsotope bf ulnum um Reproeesslng em be ume by seplrnlon orthe soluble lsotope t t c soluble nucleur ruels ls done by ebuntereurrent operltlun af munyllqululllqulu sepnvmow mg These sepurltlon stages eonast uf wellrmlxed eontletlng s where 1020032 ls exchangzd ere sepurlteu A sehemltle af l sepurltlun stlge ls shawanlgu e 5 lbu r phb hrgl l pllnl nge 515 quulullqulu sepnltlon stlge funepmcessng GasLi qui d Systems 2 5 In this process an aqueous solution of uranil nitrate UO2 N03 2 is one of the feed streams to the separation stage and the mass flow rate of the aqueous feed phase is 400kg hr The second feed stream is an organic solution of TBP in dodecane which we assume to be a single component The organic and inorganic phases are assumed to be completely immiscible thus only the uranil nitrate is transferred from one stream to the other The process specifications are indicated in Figure 516b where species A represents water species B represents uranil nitrate and species C represents the organic solution Since all mass fractions are specified it is the mass flow rates that we wish to determine in this problem Figure 5161 Specified stream variables 516 In this problem the information about the composition of the streams is given in terms of mass fractions thus we make use of the steadystate mass balance equation given by JmApvndl 0 A123 1 AC Application of this result to the system illustrated in Figures 512 leads to the following three mass balance equations species1 3mg A 0 2a speciesB 3mg 3 0 2b species c mg C 0 2c In addition to these three mass balance equations there are four constraints associated with the mass fractions at the entrances and exits These take the form Stream 1 coA 1 03 1 00 1 1 3a Stream 2 03A 2 03 2 DC 2 1 3b Stream 3 03A 3 033 3 00 3 1 3c Stream 4 03A 03 DC 1 3d 4 26 Chapter 5 The degreeoffreedom table for this process is given by Table 516 DegreesofFreedom Stream Variables compositions NXM 12 ow rates M 4 Generic Degrees of Freedom N X M M 16 Number ofY J J Balance F quot massmole balance equations N 3 Number 0f I for nmnn itinn A Generic Speci cations and Constraints N M 7 Specified Stream Variables compositions 7 ow rates 1 39 for F quot39 0 Auxiliary Constraints 0 Particular Speci cations and Constraints 15 Degrees of Freedom 16 7 8 1 Here we see that there is one degree of freedom thus it would appear that we are missing one equation or we are missing a variable that must be speci ed The missing specified variable is in stream 2 where we have indicated that this stream contains only the organic material Since mass fractions are always positive we need to specify that either 03A 2 or 033 2 is zero This type of constraint is obvious when a solution is developed by hand however if computer software is used to solve this set of equations one must explicitly identify the variables that are specified Thus in addition to those variables that are specified in Figure 516b we add the condition 03 2 0 4 Since all the quantities in the first column of Eqs 2 are known we move them to the right hand side of the equations and express our three mass balances in the form GasLi qui d Systems 27 species1 sa speciesB sh speciesc so At this point we can make use of Eqs 5 to determine all the mass fractions of the streams illustrated in Figure 512b and this leads to the situation illustrated in Figure 512c Using the information concerning the mass fractions allow us to simplify Eqs 5 to the form speciesA 0 002 23 02 24 40 cghr 6a species B 0 098 m3 0 360 kg hr 6b species C 7 7522 0 08 24 0 6c This set of three equations and three unknowns is especially easy to solve because of the EDA LOB 04 Figure 5160 Mass fractions numerous zeros in the may of coefficients on the le hand side To develop a solution we first use Eq 6b to determine m3 according to m3 360098kghr 367kghr 7 This result allows us to eliminate m3 from Eq 6a to obtain m 163kghr 8 and we are now in a position to eliminate m from Eq 6c leading to the solution for m2 given by 22 l30kghr 9 For more complex problems in which there are no zeros in Eqs 6 one needs to make use of computer routines to avoid the extensive algebraic effort required to solve a threeby three matrix of equations 28 Chapter 5 517 The concept of an equilibrium stage is a very useful tool for the design of multicomponent separations and a typical equilibrium stage for a distillation column is shown in Figure 517 A liquid stream 51 owing downward encounters a vapor stream 52 owing upward We assume that the vapor 2 Figure 517 Sketch of an equilibrium stage process exchange mass inside the equilibrium stage until they are in equilibrium with each other Equilibrium is determined by a ratio of the molar fractions of each component in the liquid and vapor streams according to KA yA4 A12 xA3 The streams leaving the stage S3 liquid and S4 vapor are in equilibrium with each other that is the molar fractions of the individual components satisfy the above relation The ratio of the molar flow rates of the output streams is a function of the energy balance within the stage For our example we will assume that the ratio of the liquid output molar flow rate to the vapor output molar flow rate R M3 3914 is given Assuming that the compositions ie the mole fractions of the components and the molar flow rates of the input streams 1 and 52 are known and the equilibrium constant for one of the components is given develop the mass balances for a two component vaporliquid equilibrium stage 517 The macroscopic mole balance to the control volume shown in the gure 1517 is given by Species A J CAV ndA 0 1 AC In terms of mole fractions we express Eq 1 as Species1 JxAcvmdA 0 2 AC and note that the total mole balance is given by GasLi qui d Systems 29 Total mole balance J c V ndA 0 3 Ac Use of Eqs 2 and 3 in the control volume given by the gure 1517 xA1M1 2M2 xA3M3 4 M1M4 M2 M3 5 where the mole fractions of the components Xi and yi and the molar ow rates Mi of the input streams SI and 2 are known In addition the relation R 2M3 M4 6 is known The equation 6 can be written as M3 R M 4 7 Plugging equation 7 into equation 5 M1M4 M2 R M4 8 or M M 4 71 2 9 R 1 Then the value of the molar ow rate of the stream 4 can be calculated by means of the equation 9 And plugging the value of M4 back into equation 7 10 Now the considering that the equilibrium constant of the component A is known we have KA yA4 9843 11 01 yA4 KA 39xA3 12 Plugging equations 9 10 and 12 into equation 4 we have 30 Chapter 5 xA1MlA2M2 9603 R II 13 01 xA1M1yA2M2 14 M1 szR KA xA 3 71 Then Xa3 can be calculated using the equation 14 Plugging this equation into equation 12 MM KA39 15 Now 953 3 1 xA3 16 and yA4 1 yB4 17 Then the equations 9 10 14 15 16 and 17 de ne the unknown variables as functions of the known variables In this way the system can be solved 518 A single stage binary distillation process is illustrated in Figure 518 The total molar flow rate entering the unit is M1 and the mole fraction of speciesA in this liquid stream is x A 1 Heat is supplied in order to generate a vapor stream and the ratio 11121113 3 depends on the rate at which heat is supplied The liquid phase can be assumed to be wellmixed and the vaporliquid equilibrium relation is given by 01AB xA yA lxA0cAB 71 in which a is the relative volatility Derive an implicit expression for yA 2 in terms of xA 1 and examine three special cases 1 chB gt 0 2 3 a 0 3 ocAB 1 GasLiquid Systems 3 l v Q gt Figure 518 Single stage binary distillation 518 The appropriate control volume for the analysis of the singlestage binary species A and B distillation process illustrated in Figure 518 is shown in Figure 518a and the appropriate macroscopic mole balance takes the form SpeciesA JcAvnM 0 1 Ae Here we have used v n in place of V A n with the idea that diffusive effects are negligible at the entrances and exits of the system While this should be a satisfactory Figure 51811 Control volume for singlestage binary distillation simpli cation one should remember see Example 42 that a separation will not occur unless somewhere in the system the species velocities are different and diffusive effects are not negligible Because we are provided information in terms of mole fractions we express Eq 1 as SpeciesA JxAcvndl 0 2 AC and note that the total mole balance is given by 32 Chapter 5 Total mole balance J c V ndA 0 3 Ac Use of Eqs 2 and 3 with the control volume shown in Figure 518a leads to xA1M1 yA 2M2 XA3M3 4 M1 M2 M3 5 These two results can be used to obtain xA3 xA1 13 301202 6 where B is the ratio of molar ow rates de ned by M2 M3 5 7 In order to derive an implicit expression for y A 2 in terms of x A1 we substitute the equilibrium relation 0 AB 9903 8 WA 1xA3ocAB 71 into Eq 6 to obtain AB yA2 yA2 9 1 AB 1 yA2 Given this result we can examine three special cases 1 For the special case ocAB a0 we obtain yA2 v0 XAB v0 10 This indicates that no species A appears in the vapor stream when the relative volatility of A to B tends to zero 2 For the special case 3 a 0 we obtain yA2 m 0 11 xA 3 XABi This indicates that the vapor leaving the still is in equilibrium with the liquid entering the still This represents a maximum in terms of separation e iciency however when B gt0 the vapor ow rate also tends to zero and there is no overhead product This suggests that there is an optimum in terms of yA2 xA1 versus B that must be determined by an economic analysis Gmrlngmd system 3 For the speera1 ease ME 1 we obtarh m a lter VAE 1 Thrs mdmates that the vapor stream has the same eomposrhor as the ehterrhg hqurd stream p the analysrs ofduullauon processes 49 A satunted salmm hr talcum hydraxxde enters a tender as Shawn m Frgure 5 w and a freeth a whim water emermg the bmlerxsvapunzed Under these ermstmees rpamah hr snlubdxty s e garcaomzghmzo and that the umpexmxe hr the hqund entermg andluvmg the tender 15 a cahstmt Develap a murms hat was Fax e n m n 21 and n75 deurmmz themessrreetwharsuspehaeasmra whehs2s m3 srlm Egue 519 Precxpxtauan uf calcan hydraxxde ma barter 5719 In thrs problem we are deahhg wrth two eherhrea1 speeres ea1erurh hydmxxde mole fraetaor rs r5 andthe vapor phase where the mole fraetror rs yp The ea1erurh hydmxxde exrsts rh both the hqurd phase where the mole fraetror rs m and the sohd phase where we have he mform eorwehtror forthe mole fraetror The nature of the ruustratedrh Frgure 5 19a shouldbe used For a steadyrstat ehe e process m the absehee of rhrea1 reaetron the two speeres mass balances ear be expressed as Ca1erurh hydmxxde I v mu 7 0 1 4 ChuptevS Water IpEVndA 0 2 promoted that dAffusne effects are neghgrble at the entranees and ernts s own m It h rgure 5 19a rs eonvenrent to express these results m terrns of the mass fraetrons to leadmg Calciumhydroxxde mApv 11114 7 0 3 Ae Water ntppvvntu 0 4 Ae fraehon r Mum Tum h n mhw Hwy 5 19a Cuntxul vulume furprempltahun uf eelerurn hydmxxde m abmler ofsorneestabhshedeonventronweuse rmquot and Wood and apply Eq 3 to the eontrol volume rllustxatedrn Frgure 5 19ato obtam Calerurn hydroxrde m2 Anna M 5 nssnhen nssnhen sepennsn snhn wanes wanes wanes Fouowrng the same proeedure for the mass balance for water we obtam Water 5W quota 5 hnnrn hqmd vopnr phase phase phase andxfwe assume that there rs no calmum hydroxxde m the vaporphase thrs srrnph es to Water 9501 m 051W 2 quot 3 7 GasLi qui d Systems 3 5 The solubility was expressed as Solubility S g ofCaOH2g of H20 8 and what is really meant by grams of calcium hydroxide is grams of dissolved calcium hydroxide Because of this the solubility takes the form AJiq Solubility S m 9 Bliq and this relation applies to both in incoming and outgoing liquid streams This leads to two important relations 03Aiiq1 03Aiiq2 Solubility 7 Waugh Waugh 5 10 The mole fraction constraints in the incoming and outgoing liquid streams are given by Constraint 03AM OJBJlq1 l 11 Constraint 03Aiiq2 03Asoiid2 DBliq2 1 12 In addition to the solubility being a speci ed parameter the fraction of water that is vaporized p is also a speci ed parameter that we express as 13 31121 1 7711 fraction of water m3 P vaporized At this point we note that there are eight unknowns wAJlqh oAJlq 2 wAJohdb 0131in ooBJlq 2 ml r512 and m3 and only seven equations Eqs 5 7 10two ll 12 and 13 however this is a solvable problem since the mass balance equations are homogeneous in the mass ow rates and we need only determine the ratio of these ow rates This means that the three unknowns m1 m2 and M can be replaced with the two unknowns m2 m1 and m3 m1 and a little algebra leads to the solution given by pS 0Asolid2 m 14 Given that S 25x10 3 we nd oi1M 249x10 3 q 050 oi1M 662x10 4 p 021 15 oi1M 202x104 p0075 36 Chapter 5 Section 54 520 An equimolar mixture of ethanol and ethyl ether is kept in a closed container at 103 KPa and 95 C The temperature of the container is slowly reduced to the dew point of the mixture Determine a What is the dew point temperature of the mixture b What is the pressure of the container at the dew point temperature of the mixture c What is the composition of the first drop of liquid at the dew point 520 In the container we have an equimolar mixture of ethanol e and ethyl ether ee at at 103 KPa and 95 C Applying the ideal gas law P V n R T 1 We can get 103000Pa ml 00337quotl 2 VI 95273 K I In the dew point we will have the following composition of the gas 34 ygg05 o In an ideal vaporliquid multicomponent system the partial pressure of species e and ee in the gas phase is given by P x P 39 45 Pg 9622 Pgmp where the vapor pressures are given by the Antoine s equation 162322 10 P 81629 6 gm W 22898 109064 10 P 2 698467 7 gm map 23121 T with the temperature in 0C and the pressure in mm Hg 0 According to the Dalton s law GasLi qui d Systems 3 7 Pg ye 3980 89 Pee yee quotPm In order to nd the dew point temperature of the mixture a trial 7 error method has to be used 9 Assume a temperature T 10 Calculate the total pressure using equation 1 and 2 11 Using equation 34 and 89 calculate the partial pressures 12 Use equations 6 and 7 to calculate the vapor pressures 13 With equation 4 and 5 calculate the liquid molar fractions 14 Sum x6 and xee If the summation is equal to l the assumed temperature is the dew point temperature If it is not assume a new temperature By doing this procedure the results are the following a What is the dew point temperature of the mixture Tdew point 6465 0C b What is the pressure of the container at the dew point temperature of the mixture Pm 093 atm c What is the composition of the rst drop of liquid at the dew point xe 0822 xee 0178 Chapter 5 Chapter 6 Stoichiometry Section 61 6l By counting atoms provide at least one version of a balanced chemical equation based on 7C2H6 702 gt vco 7C2H4O 7H20 co2 61 One possibility different from the two versions given in the text is 302 H6 02 300 C2H4O 7H20 CO2 Section 62 62 Construct the chemical composition matrix N M for the following set of components Sodium hydroxide Na OH methyl bromide CHgBr methanol CHgOH and sodium bromide NaBr 62 A Visual illustration of the chemical composition matrix for this group of molecular species is illustrated by Molecular Species NaOH CH3Br CH3OH NaBr Sodium 1 0 0 1 Hydrogen 1 3 4 0 1 Oxygen 1 0 1 0 Carbon 0 1 1 0 Bromine 0 1 0 1 and an explicit representation of the chemical composition matrix takes the form 1 0 0 1 1 3 4 0 N M 1 0 1 0 2 0 1 1 0 0 1 0 1 2 Solutions for Chapter 6 63 Construct a chemical composition matrix for a system containing the following molecular species 63 Construct a chemical composition matrix for a system containing the following molecular species NH3 O2 NO N2 H20 and N02 Find the rank of this matrix 63 We begin this problem with the chemical composition matrix illustrated by Molecular Species NH3 02 N0 N2 H20 NO2 nitrogen l 0 l 2 0 l 1 hydrogen 3 0 0 0 2 0 oxygen 0 2 l 0 l 2 If we divide row 2 by three and row 3 by two we obtain 1 1 2 0 1 NJA1000 0 2 1 1 0 1 5 0 5 1 A cyclic permutation of the rows leads to 2 0 0 0 3 0 NJA20101 3 0 1 2 0 1 and subtraction of row 1 from row 3 provides the following simpli ed form 1 0 0 0 g 0 NJA 0 1 0 lt4 2 0 0 1 2 73 1 Multiplying row 3 by one half and subtracting that result form row 2 gives 0 NM 5 D I N H and no additional zeros can be produced in row 3 indicating that the rank of the matrix is three r rank 3 Ifwe make use ofthis matrix in Eq 621 we obtain Staichiametry 0 6 and one can solve for the reaction rates to obtain RNH3 RHZO 7a R02 RN2 gRHzo RNo2 7b 7C RNO ZRNZ gRHzo RN02 64 Verify that Eq 630 can be obtained from Eq 629 65 Show how Eq 633 can be obtained from Eq 632 and use the former to solve for x1 x2 and x3 in terms of x4 66 Demonstrate that the chemical composition matrix in Example 62 has a rank of three In the course of your analysis identify a row equivalent form a row echelon form and the row reduced echelon form 66 Given Eq 2 of Example 62 2 0 0 1 NM 6 0 2 0 1 0 2 1 2 we divide row 1 by two row2 by six and row 3 by two in order to obtain 4 Solutions for Chapter 6 2 1 lt3le 10 NJA10 01 N H le O This is a row equivalent form of Eq 1 and progress toward the nal form can be achieved by interchanging rows 2 and 3 leading to 1 l 0 0 5 NM 0 1 1 3 1 l 0 3 0 We now subtract row 1 from row 3 in search of a row of zeros This provides 1 l 0 0 5 NM 0 1 1 4 1 1 0 0 3 5 This can be considered a row reduced form in that no additional zeros can be created in row 3 thus the rank of the matrix is three r rank 3 The row echelon form is now created by multiplication of row 3 by three to obtain 1 l 0 0 5 N M 0 1 1 5 3 0 0 1 Multiplication of row 3 by one half and subtracting the result from row 2 leads to 1 l 0 0 5 N M 0 1 0 g 6 3 0 3 This is the row reduced echelon form that this so convenient for use with Eq 621 and other elementary row operations can be used to achieve this result from Eq 1 67 Begin with the statement that mass is neither created nor destroyed by chemical reaction and use it to derive Eq 621 Be careful to state any restrictions that might be necessary in order to complete the derivation Staichiametry 5 67 We begin this development with AN VA 0 1 A21 and express the mass rate of production in terms of the molar rate of production to obtain AN ZRAMWA 0 2 A1 Given that the molecular mass of species A can be expressed as JT MWA ZMWJNJA 3 J1 we see that Eq 2 takes the form AN JT RA ZMWJNJA 0 4 A1 J1 Since R A is independent of the sum over J we can express this result as AN JT MWJNJARA 0 5 A1 J1 and the order of the summation process can be inverted to obtain JT AN ZMWJNJARA 0 6 J21 A21 Here we note that M WJ is independent of the sum over A so that this result can be JT AN E MWJ E NJARA 0 7 J1 A1 This can also be written in the expanded form expressed as A MW1 Al NJARA Solutions for Chapter 6 AN Al 8 Ifthe sum of NJARA from A l to A N is independent ofthe molecular masses of the atomic species J l23 we conclude that the sum must be zero and we have derived Eq 621 given by AN E N JARA 0 A21 68 For the partial oxidation carbon described in Example 63 explore the use of pivot species other than CO and CO2 This will lead to five more possibilities in addition to those given by Eqs 7 in Example 63 68 All six possibilities that were mentioned in Example 63 are given by 111 VI RC Rco Rcoz R02 RCO RCOZ RC0 2RC 2R02 Rcoz RC 2R02 R02 RC t Rco Rcoz Rc RC0 RC 2R02 Rcoz RC0 2R02 2Rcoz RC R02 Rco Rcoz Roz Rco RC0 RC RC02 R02 RC Rco2 1 2 3 4 5 6 69 Demonstrate that the chemical composition matrix in Example 64 has a rank of three In the course of your analysis identify a row reduced form and the row echelon form 9 Staichiametry 7 69 We begin with the chemical composition matrix represented by Molecular Species 12m 02 H20 C0 C02 C2H4O carbon F2 0 0 1 1 2 1 1 hydrogen 6 0 2 0 2 4 oxygen 0 2 l l 0 l and express the matrix as 2 1 NM 4 2 1 l We can begin to simplify this matrix by dividing the rst and third rows by two and the second row by six in order to obtain 1 2 NJA j I 3 1 3 J If we subtract row 1 from row2 we have 1 NJA l i 4 l 2 and interchanging rows 2 and 3 leads to a row echelon form given by 1 NJA I 5 1 3 J Multiplication of the last row by three provides 1 NJA i 6 1 l If we multiply row 3 by one half and subtract that result from row 2 we obtain the desired row reduced echelon form 8 Solutions for Chapter 6 N 1 7 JA 1 J This indicates that the rank of the matrix is three r rank 3 Use of this result with Eq 4 of Example 64 leads to 1001 0 0101 0 8 33 001 5 5 1 0 and carrying out the multiplication produces Eqs 5 of Example 64 Rch I Rco ERCOZ Rc2H4o 9a R02 ZRC Rco2 RC2H4O 9b RHZO Rco Rco2 RC2H4O 9C 610 For the molecular species listed in Problem 62 determine the ratio of reaction rates given by RNaOH RCHKBr RCHKOH RNaBr RNaBr RNaBr 610 The visual representation of the chemical composition matrix for this system of four molecular species and ve atomic species is given by Molecular Species NaOH CH3Br CH3OH NaBr Sodium l 0 0 Hydrogen l 3 4 0 1 Oxygen l 0 l 0 Carbon 0 l l 0 Bromine 0 l 0 l The matrix itself is represented by Stoichiometry 391 0 0 1 1 3 4 0 N M 1 0 1 0 0 1 1 0 70 1 0 1 while the rowechelon reduced form is 391 0 0 1 0 1 0 1 NM z 0 0 1 71 0 0 0 0 70 0 0 0 In terms ofthis reduced form Eq 621 can be expressed as A24 2 NjARA 0 A21 and the set of equations determining the reaction rates is given by RNaOH RNaBr 0 0 RNaBr 0 0 RNaBr 0 2 3 4 5a 5b 50 Here we are confronted with three equations and four unknowns thus we can solve for three of the reaction rates in terms of the rate of reaction of the fourth or pivot species If we choose sodium bromide to be the pivot species we can represent the ratio of reaction rates according to RCH3OH RNaBr RNaOH RNaBr r 6 611 The production of alumina NaAlO2 from bauxite Al OH 3 requires sodium hydroxide as a reactant and yields water as product For this system determine the ratio of reaction rates given by RA10H3 RNaOH RHz 0 RNaAlO 2 RNaAlOZ RNaAlOZ Solutions for Chapter 6 611 The visual representation of the chemical composition matrix for this system of four molecular species and four atomic species is given by Molecular Species AlOH3 NaOH H20 NaAIO2 Aluminium 39 l 0 Oxygen 3 1 Hydrogen 3 1 Sodium 0 l The matrix itself is given by l 0 0 1 NJA 3 l l 2 3 l 2 0 0 l 0 l and the rowechelon reduced form is l 0 0 1 NM 2 0 l 0 l 0 0 l 2 0 0 0 0 In terms ofthis reduced form Eq 621 can be expressed as A4 E NjARA 0 Al and the set of equations determining the reaction rates is given by more aA102 0 RNaAle 0 RNaAlez 0 l 2 0 1T 2 0 1 1 2 4 5a 5b 50 Here we are confronted with three equations and four unknowns thus we can solve for three of the reaction rates in terms of the rate of reaction of the fourth or pivot species If we choose alumina to be the pivot species we can represent the ratio of reaction rates according to Slaichiametry ll RA1ltOHgt3 RH20 2 lt6 RNaAlOZ NaAlOZ RNaAlOZ 612 At T 20C the rate of disappearance of methyl bromide in the reaction between methyl bromide and sodium hydroxide Problems 62 and 610 is 7RCH3BK 02molm3s and the stoichiometric schema takes the form NaOH CH3Br a CH3OH NaBr a Determine the reaction rates of sodium hydroxide methanol and sodium bromide in molm3s b Determine the reaction for all components of the reaction in kgm s c Show that mass is neither created nor destroyed by this chemical reaction 612a From the stoichiometric schema and from the solution of Problem 610 we have RNaOH RCH3OH 1 1 RNaBr NaBr RNaBr This allows us to express all the reaction rates in terms of RCH3Br RNaBr OH RCH3Br 2 and the values are given by 3 RNaBr aOH Rem 02 molesm s 3 612b In order to determine the mass rates of production of the four molecular species we make use of the general relation rA R AMTV A 4 in order to obtain rNaBr m8 10 3kgm3s rNaOH 7997x10 3 kgm3s rCHaBr 1898x10 2 kgm3s 5 612c To show that mass is neither created nor destroyed by chemical reaction we consider the sum of the mass rates of reaction to obtain A24 rA 103 1898 102 LE 0 6 ms A21 12 Solu onsfor Chapter 6 613 Use the chemical composition matrix determined in Problem 62 and the reaction rates found in Problem 612a to prove that atomic species are neither created nor destroyed by this chemical reaction 613 A Visual illustration of the chemical composition matrix for this group of molecular species is illustrated by Molecular Species NaOH CH3Br CH3OH NaBr Sodium 1 0 0 Hydrogen 1 3 4 0 1 Oxygen 1 0 1 0 Carbon 0 1 1 0 Bromine 0 1 0 1 and this can be used to express Eq 621 as 1 0 0 1 02 1 3 4 0 2 3 1 0 1 0 2 molesm s 0 1 1 0 02 L0 1 0 1J Carrying out the matrix multiplication leads to 0 A24 0 E NJARA 3s 0 molesm3s A1 0 M which indicates that the ve atomic species are neither created nor destroyed in this chemical reaction 614 Methanol can be synthesized by reacting carbon monoxide and hydrogen over a catalyst The rate of production of methanol at 400 K is of rCHKOH 0035kgm35 Determine the reaction rates in molm3s for all components of the synthesis reaction 614 The molecular species involved in this reaction are CO H2 and CH3OH and the Visual representation of the chemical composition matrix takes the form Stoichiometry Molecular Species gt CO H2 CH3OH Carbon Oxygen Hydrogen The chemical composition matrix is given by l 0 1 NM l 0 0 2 4 and the rowechelon reduced form is 0 NM 0 1 0 In terms ofthis reduced form Eq 621 can be expressed as A4 E NjARA 0 J123 Al and the set of equations determining the reaction rates is given by Rco0 RCH3OH 0 0 RHZ 2RCH3OH 0 1 2 3 4 5a 5b Given three reaction rates and two equations we can determine two of the reaction rates in terms of the reaction rate for the pivot species Ifwe choose methanol to be the pivot species we can solve Eqs 5 to obtain RC0 12 ZRCH3OH The molar reaction rate for methanol can be calculated according to rCH OH 3 RCH OH 3 109 molm s 3 MWCH3OH and use of this result in Eqs 6 provides RC0 RH2 218 molm3s 6 8 9 Solutions for Chapter 6 Section 63 615 Given a matrix equation of the form c Ab having an explicit representation of the form 01 1 1 12 13 14 b 1 c2 2 1 22 23 24 b 7 2 1 c3 3 1 32 33 34 b 3 c4 4 1 42 43 44 b 4 c5 5 1 5 2 5 3 54 develop a partition that will lead to an equation for the column vector represented by If the elements of C are related to the elements of b according to 2 Lil lt3 What are the elements of the matrix normally identified as A21 7 V 616 Construct the complete columnrow partition of the matrix equation given by 11 12 13 14 b 1 C1 21 22 23 24 b2 c2 1 31 32 33 34 b3 03 41 42 43 44 74 c4 and show how it can be represented in the form b1 b2 b3 3 2 Section 64 6l7 Given a reacting system containing C3H6 NH3 02 C3H3N and H20 construct the chemical composition matrix and determine the rank of that matrix Use the development presented in Sec 64 to develop the stoichiometric schema for the single independent reaction involving propylene ammonia oxygen acrylonitrile and water Stoichiometry 15 618 Construct the stoichiometric schema for the reaction described in Problem 67 The stoichiometric coefficients determined on the basis of Problem 67 are given by RNaDH 30H 1 RNaBr 1 RNaBr NaBr NaBr RNaBr 618 To develop the stoichiometric schema associated with this reaction we make use of Eqs 630 and 631 which take the form 1 2 This valid mathematical equation can be expressed as 3 and we can use the given stoichiometric coef cients to obtain 4 Since we are provided with no information about the sign of the reaction rate for the pivot species the transformations represented by Eqs 638 lead to the picture given by NaOH H NaBr 5 Section 65 619 Fogler1 has proposed the following gas phase reactions involving the oxidation of ammonia and the reduction of nitric oxide 1 4NH3 RNH chNchozY H 2NH3 RNH3 kchcho2 III 2N0 02 gt 2N02 R02 7k3cNO2cOZ Iv 4NH3 RNo kicmwmgm Analyze a system containing these six molecular species and these three atomic species to determine the number of independent reactions Are there any restrictions on the choice of pivot and nonpivot species Do you have any ideas about how one would measure the rate of 1 Fogler SH Elements of Chemical Reaction Engineering Second Edition Prentice Hall Englewood Cliffs New 16 Solu onsfor Chapter 6 consumption of ammonia in Reactionl independently from the rate of consumption of ammonia in Reaction H in order to determine the rate constants C1 and k2 7 619 We begin this problem with the chemical composition matrix illustrated by Molecular Species NH3 02 N0 N2 H20 N02 nitrogen l 0 l 2 0 l 1 hydrogen 3 0 0 0 2 0 oxygen 0 2 l 0 l 2 and carry out a series of elementary row operations to obtain 0 NJA i 5 1 This indicates that the rank of the matrix is three and if we make use of this matrix in Eq 621 we obtain 0 0 6 0 From this we nd that the reaction rates are related by RNH3 gRHzo 73 R02 RN2 gRHzo RNo2 7b RNO ZRNZ gRHzo RN02 7C Slaichiametry 17 620 When methane is partially combusted with oxygen one finds the following molecular species CH4 02 C0 C02 H20 and H2 Determine the number of independent reactions and comment on the restrictions concerning the choice of pivot and nonpivot species 621 Develop the stoichiometric schemata for the system described in Problem 620 Solutions for Chapter 6 Chapter 7 Multicomponent Systems with Chemical Reactions Section 71 71 A ue gas stream 1 composed of carbon monoxide carbon dioxide and nitrogen can undergo reaction with water gas stream 2 and steam to produce the synthesis gas stream 3 for an ammonia converter The carbon dioxide in the synthesis gas must be removed before the gas is used as feed for an ammonia converter This process is illustrated in Figure 71 and the Figure 71 Synthesis gas reactor product gas in stream 3 is required to contain hydrogen and nitrogen in a 3to1 molar ratio In this problem you are asked to determine the ratio of the molar ow rate of the ue gas to the molar ow rate of the water gas ie M1 M2 that is required in order to meet the specification that yH23 3yN2339 Answer M1 12 0467 71 Our rst step in the solution of this problem is to construct a control volume and in this case there is only one possibility that we show in Figure 71a Next we want to construct a table of speci ed and unknown data as indicated in Table 7 la where we have been careful not to over specz the mole fractions in the four streams Solutions for Chapter 7 We begin the degree of freedom analysis by noting that there are five molecular species N 5 four Figure 71a Control volume for the synthesis gas reactor atomic species T 4 and four streams M24 We need to know how many of the atomic species balances are independent thus we begin with a chemical composition Table 7 la Specified and unknown conditions Stream 1 Stream 2 Stream 3 Stream 4 M12 M22 M3 M42 yCO 020 yCO 050 yCO 00 yCO 00 yCO2 002 yco2 00 yco2 00 yH2 050 yH2 00 yHZO 00 yHZO 00 yHZO 00 yHZO l0 yN2 078 matrix that is convenient to represent in the following form Multicomponent Systems with Chemical Reactions 3 Molecular species HZO CO H2 N2 C02 hydrogen 0 carbon 1 I 1 oxygen 2 I nitrogen 0 Use of this matrix with Eq 621 leads to RHZO 0 O l 0 H2 0 2 N2 0 OZ and a series of elementary row operations leads to the reduce row echelon form given by RHZO 3 0000 This indicates that the rank of the matrix is four rank 4 and the reaction rates are constrained by RHzo 2 R002 RNZ 0 4 In terms of the entries in Table 7lb this leads to the generic degrees of freedom and the generic constraints given by Generic Degrees of Freedom A N X M M N 29 Generic Constraints B N M r 13 To determine the particular speci cations and constraints we examine the control volume streambystream using Table 7 la to obtain Stream 1 Four mole fractions are speci ed Stream 2 Four mole fractions are specified Stream 3 Two mole fractions are specified Stream 4 Four mole fractions are specified Solutions for Chapter 7 This provides 14 speci ed compositions and with the ratio of hydrogen to nitrogen speci ed in stream 3 we have 15 particular speci cations and constraints as indicated in the degreesoffreedom table This leads to one degree of freedom which means that nine of the unknowns listed in Table 7 la can be determined in terms of one of the unknowns Since we want to determine the ratio of molar ow rates M1 2 we need only develop a solution in terms of M2 Table 71b Degreesof Freedom Stream Variables and Reactions compositions ow rates chemical reactions Generic Degrees of Freedom A Number ofY J J Balance F massmole balance equations Number 0f I for nmnn itinn NXM20 M4 N5 1NXMMN29 Number of C 39 for Reaction Rates N5 M4 r4 Generic Constraints B Speci ed Stream Variables NMr13 compositions l4 ow rates none Constraints for Compositions 1 Auxiliary Constraints none Speci ed Reaction Rates none Particular Speci cations and Constraints C 15 Degrees of Freedom A B C 1 We begin to develop the solution to this problem with the molar form of the rst aXiom Multicampanent Systems with Chemical Reactions 5 d c dV V A l2N 5 dt IA V A V in which we have replaced the species velocity with the mass average velocity based on the assumption that diffusion effects are negligible at the entrances and exits We further assume that the process operates at steady state so that Eq 5 takes the form A H20 co H2 N2 c02 6 Ac Here we have used A6 to represent the entrances where vn is nonzero and we have noted that the number of molecular species is five In addition we have identified the total rates of reaction according to RA H20 co H2 N2 c02 7 V Since information is given in terms of mole fractions we express Eq 6 in the form yAcv A H20 co H2 N2 02 8 Ae and represent the five molecular species balances according to H201 szo 4M4 RH20 9 CO3 yc01M1 yco 2M2 RC0 10 H21 J H2 2M2 H2 3M3 RH2 11 N21 J N2 1M1 N2 3M3 0 12 C021 yco2 1M1 yco2 3M3 Rco2 13 Because of the simplicity of the mole balance for water it is reasonable to express all the reaction rates in terms of R H20 according to RH2 Rcoz RHzo 14 This allows us to express Eqs 9 through 13 in the form 6 Solutions for Chapter 7 H203 szo 4M4 RH20 15 CO3 yco 1M1 yco 2M2 RH20 16 H21 J H2 2M2 H2 3M3 R H20 17 N21 J N2 1M1 N2 3M3 0 18 C023 yco2 1M1 yco2 3M3 R H20 19 Since we seek to determine M1 2 we de ne the ratio of molar ow rates as a MM2 139124 20 so that Eqs 15 through 19 simplify to H201 szo 4014 RHZOMZ 21 CO3 yco 1 11 yco 2 RH20 M2 22 H23 H2 2 H2 3013 R H20 M2 23 N21 N2 1011 N2 3013 I 0 24 C021 J Co2 1 11 yco2 3 13 R H20 M2 25 At this point it is helpful to list the seven unknown quantities as Unknowns RHZO M2 3 H2 3 N2 3 yco2 3 In addition to the five equations given by Eqs 21 through 25 we also have the conditions given by 1123 mm 1 26 thus we can solve for Oil as indicated by the degree of freedom analysis One way to begin is to recognize that Stream 4 is pure water and this allows us to use Eq 15 to obtain the following species balance equations Multicampanent Systems with Chemical Reactions 7 H20 14 RHZOM2 27 CO3 yco 10 1 yco 2 0 4 28 H21 H2 2 H2 3W3 014 29 N21 J N2 10 1 N2 30 3 0 30 C023 yco2 10h yco2 30 a4 31 From Eqs 26 we have sz3 1 4m 32 and this allows further simpli cation of Eqs 27 through 31 to H20 oc4 RHZOM2 33 CO3 yco 10 1 yco 2 0 4 34 H2 yH2 2 3B0L3 L4 35 N21 J N2 10h 3013 I 0 36 C02 yco2 log 1 46 x3 oc4 37 in which the unknown yN2 3 has been replaced with 3 so that the unknowns are isolated as Greek letters In Eqs 33 through 37 we see four unknowns and four equations thus a solution is available as indicated by the degreesof freedom analysis We can express Eq 36 as N21 360 s 3 N2 10 1 38 and use this result in Eq 35 to obtain H21 H2 2 3 N2 10h 014 39 Now recall Eq 34 as Solutions for Chapter 7 C03 yco 10 1 yco 2 0 4 40 Adding these two equations will eliminate X4 and provide the solution yH yco a1 22 2 41 3yN21 ycol which leads to the answer given by x1 0467 42 72 In a process for the production of formaldehyde CHZO by catalytic oxidation of methanol CH3OH an equimolar mixture of methanol vapor and air is sent to a reactor in which the catalyst is finely divided silver supported on alumina An undesirable side reaction occurs in which CHZO reacts with oxygen to produce carbon dioxide and water For a specific reactor 20 of the methanol in the input stream reacts to form form aldehyde and the selectivity for this reactor is given by 5 ed 85 m0 eso car 011 10x1 e pro ced Determine the mole fraction of all components in the stream leaving the reactor Treat the air as 79 nitrogen and 21 oxygen 72 The system under consideration is illustrated in Figure 72 where air and methanol are fed to a catalytic reactor in order to produce formaldehyde The control volume for D Figure 72 Catalytic reactor this analysis is obvious and we begin the analysis by preparing a table of specified and unknown conditions and this is information is given in Table 72a In addition to these speci cations we need to clearly identify the constraints that are identi ed in the Multicomponent Systems with Chemical Reactions 9 Table 72a Speci ed and unknown conditions M oxygen nitrogen water problem statement The comment that 20 of the methanol in the input stream reacts to form formaldehyde can be interpreted mathematically as 020 ycngoH IMJ R CHZO 1 and the statement concerning the selectivity takes the form given by R S 8 5 2 R coZ In order to develop the degreesoffreedom analysis we need to know the number of independent reactions We begin this part of the analysis with the chemical composition matrix represented by Molecular Species CH30H 02 N2 H20 C02 CHZO carbon F 1 0 0 0 1 1 1 hydrogen 4 0 0 2 0 2 3 oxygen 1 2 0 l 2 l nitrogen 0 0 2 0 0 0 Using a series of elementary row operations the chemical composition matrix can be expressed in reduced row echelon form according to NJA 4 Owl 1 I ll This indicates that the rank of the matrix is four rank 4 This allows us to identify the generic degrees of freedom as 20 and the generic constraints as 12 as indicated in Table 72b To determine the specified stream variables we examine Table 72a to Solutions for Chapter 7 conclude that ve compositions are speci ed From Eq 1 we have a single constraint on the compositions and from Eq 2 we have a single auxiliary constraint As indicated in Table 72b this leads to one degree of freedom and the problems statement implies that we should solve this problem in terms of the ratio of ow rates on Ml M 2 Table 72b Degreesof Freedom Stream Variables and Reactions compositions ow rates chemical reactions Generic Degrees of F reedom A NXM12 M2 N6 Number ofY J J Balance F quot massmole balance equations N 6 Number ofC 39 for nmnn itinn M 2 Number of C 39 for Reaction Rates r 4 Generic Constraints B N M r 12 Speci ed Stream Variables compositions 5 ow rates none Constraints for Compositions 1 Auxiliary Constraints 1 Speci ed Reaction Rates none Particular Speci cations anal Constraints C 7 Degrees of Freedom A B C 1 We are now ready to solve this problem making use of the six species balances that are expressed as Multicampanent Systems with Chemical Reactions J yAcvndA A RA A1236 In addition we make use of the constraints on the atomic species that are given by R coZ This leads to the following constraints on the molar rates of reaction R cw oz R CHZO 1 5R CHZO 0 col RCHZO IRCHZO I 0000 0 5 6 7 8 9 10 The six species balances can be expressed in terms of the uxes at the entrance and exit along with the reaction rate term to obtain CH3OH W2 R CH3OH 02 3 yoZ 1M1 yoZ 2M2 R 02 N21 JNZ 1M1 J NZ 2M2 R NZ H20 1 yHZO 1M1 yHZO 2M2 R H20 C02 3 ycoZ IMI J coZ 2M 2 R coZ CH203 yCHZO IMI l39 yCHZO 2M 2 R CHZO 11 12 13 14 15 16 12 Salu ansfar Chapter 7 As indicated by the degree of freedom analysis we can only determine the ratio of molar ow rates given by 01 M1 M 2 and this suggests that we arrange the set of equations as RCHgoH c1120 M2 17 R oz Mm CHZO M2 18 R N2 M2 0 19 R H20 c1120 M2 20 CH3OH 3 yCH30H 1 X yCH3OH 2 R CH3OH M 2 21 02 foam yoz2 R02 11 22 N21 yNZ 10 J NZ 2 RNZ M2 23 H20 1 yHZO 10 yHZO2 R H20 M 2 24 C02 3 JcoZ 1 0 JcoZ 2 R coZ M 2 25 CH201 yCHZO 1 a yCHZO 2 I R CHZO M 2 26 In addition to these equations we have the constraints on the mole fractions given by Stream 1 yCH3OHl oZ 1 yCHZOl 1 27 Stream 2 yCH3OH2 oZ 2 yCHZO2 1 28 along with the constraints given by Eqs 1 and 2 which we express as 020yCH3OH10L R CHZO M 2 29 R M S L 85 30 RCOZMz Multicampanent Systems with Chemical Reactions There are a variety of approaches to developing a solution to this set of equations One approach makes use of Eqs 17 through 20 along with Eq 30 to express all the reaction rates in terms of the reaction rate for formaldehyde according to RCI30H RCH20 S3911 M2 M2 R92 30 25 RNZ Z 0 M2 RH20 RCHZO 2S711 M2 M2 RcoZ RCHZOS4 M2 M2 This allows us to express the species balances as CH3OH1 yCH30H1OL yCH3OH2 l3 S 11 02 y021a yogz gs 1 N23 yNZlOL 39l 0sz 0 H20 7szo1oc yH202 B 2S 11 C023 ycozha t yc022 l3 S CH203 ycnzohq yCHZO2 l3 in which we have made use of l3 RCHZOMz 31 32 33 34 35 36 37 38 39 40 41 42 Here we should note that the mole fractions in stream 1 can be completely determined by the data in Table 72a and the constraint given by Eq 27 thus we are left to determine 14 Salu ansfar Chapter 7 the 6 mole fractions in stream 2 the parameter 01 and the parameter 3 To accomplish this we have the 6 species balances given by Eqs 36 through 41 the constraint on the mole fractions in stream 2 given by Eq 28 and the constraint on the molar ow of methanol and formaldehyde given by Eq 29 We can sum Eqs 36 through 41 to obtain 1 a1 BS 1 43 We know the mole fractions in stream 1 thus Eq 29 provides 01001 B 44 and we can use this result with Eq 43 to obtain R 1 1 45a M2 10 1 a 10 45b M2 10 S 11 We are now in a position to return to the species balances given by Eqs 36 through 41 in order to calculate the mole fractions in the exit stream of the reactor These are given by CH3OH yCH3OH2 0368 46 02 yolk 0035 47 N2 yNZ2 0374 48 H20 yHZo2 0117 49 C02 ycoz2 0011 50 CHZO yCHZO2 0095 51 73 Carbon is burned with air with all the carbon oxidized to C02 Calculate the ue gas composition when the percent of excess air is 0 50 and 100 The percent of excess air is defined as Multicampanent Systems with Chemical Reactions 15 percent of excess air molar rate of consumption of oxygen owing to reaction Take the composition of air to be 79 nitrogen and 21 oxygen Assume that no NOX is formed 73 74 A fuel composed entirely of methane and nitrogen is burned with excess air The dry ue gas composition in volume percent is CO2 75 02 7 and the remainder nitrogen Determine the composition of the fuel gas and the percentage of excess air as defined in Problem 73 74 75 A process yields 10000 ft3 per day of a mixture of hydrogen chloride and air The volume fraction of HCl is 0062 the temperature of the mixture is 550 F and the total pressure is represented by 292 inches of mercury Calculate the mass of limestone per day required to neutralize the HCl if the mass fraction of CaCO3 in the limestone is 092 Determine the cubic feet of gas liberated per day at 70 F if the partial pressure CO2 is 18 inches of mercury Assume that the reaction between HCl and CaCO3 to form CaCl2 CO2 H20 goes to completion 76 In a typical experimental study such as that described in Example 72 one would normally determine the complete composition of the entrance and exit streams If these compositions were given by yczn 1 2 050 0 02 1 2 050 gt szo 1 2 00 yco 1 00 gt yco2 1 00 gt yc2H40 1 00 yczH 2 yH202 043 yco2 005 gt yco2 2 003 gt yC2H402 027 how would you determine the six reaction rates associated with the partial oxidation of ethane 16 Solu onsfar Chapter 7 Section 72 77 An ammonia converter was analyzed in Sec 722 in order to determine the fraction of stream 6 that was purged in stream 3 For design purposes the molar ow rate in the recycle stream M 4 must be determined as a function of the molar ow rate of the feed stream M1 Determine the molar ow rate in stream 4 as a function of M1 77 Returning to Figure 75 we note that we must cut stream 4 and this will lead to a control volume around either the splitter of the mixer Since a mole balance around the splitter leads to a simpler relations see Sec 721 we place the control volume around the splitter as indicated in Figure 77 This leads to a total mole balance of the form M4 M6 M3 1 In the analysis of the ammonia reactor or the ammonia converter in Sec 722 the Figure 7 7 Construction of control volume to determine recycle ow rate solution is presented in terms of the ratio of ow rates de ned by Eq 738 which we repeat here as Multicampanent Systems with Chemical Reactions 17 OL i ll26 2 1 7 M6 We can use this result in Eq 1 to obtain 0L4 1 x3 3 where the solution for X3 is given in Sec 722 as 013 0062 4 This allows us to write Eq 3 in terms of the molar ow rates M4 093ng6 5 which can be arranged in the form L0938 11 M 6 This indicates that we need to determine Otl and from Eq 748 we have ltny 0002 7 In order to determine y Ar 6 we return to Eq 750 and make use of 3 010 yAr5 0005 8 in order to calculate the mole fraction of argon in stream 6 as yAI6 000555 9 Use of this result in Eq 7 determines x1 which can be used in Eq 6 to determine the molar ow rate in the recycle stream in terms of the molar ow rate in the feed stream M4 0161M1 78 A simple chemical reactor in which a reaction A gtproducts is shown in Figure 78 The reaction occurs in the liquid phase and the feed stream is pure species A The overall extent of reaction is designated by Q where z is defined by the relation CO1 1 E 01 2 Here we see that E 0 when no reaction takes place and when i l the reaction is complete and no species A leaves the reactor We require that the mass fraction of species A entering the reactor be constrained by COAX 80901 in which a is some number less than one and greater than zero Since the products of the reaction are not specified assume that they can be lumped into a single spec1es B Under these conditions the reaction can be expressed as A AB and the reaction rates for the two species 18 Salu ansfar Chapter 7 system must conform to rA irB The objective in this problem is to derive an expression for the ratio ofmass flow rates m rh4 in terms ofi and s Figure 78 Chemical reactor with recycle stream 78 We begin this problem by making some decisions about the primary cuts that will form the basis for the construction of the necessary control volumes The cuts are illustrated in Figure 78a and they are based on the following information that is given and required I Since the extent of reaction is given in terms of DA 2 and 1 A 4 we place primary cuts on streams 2 and 4 11 Since we are given information relating 03A2 to wA1 we place primary cuts on streams 1 and 2 111 Because we are asked to determine the ratio of mass ow rates m5 n3914 we place primary cuts on streams 4 and 5 Figure 78a Primary cuts One possibility for joining the primary cuts is to form control volumes around the reactor the mixer and the splitter This arrangement is illustrated in Figure 78b where we see two redundant cuts ie streams 2 and 5 are each out twice and we see two secondary cuts ie stream 3 is out twice and no information is given or required for this stream This situation is improved for the control volumes illustrated in Figure 78c where we nd only two redundant cuts ie streams 1 and 2 are each out twice and there is one Multicampanent Systems with Chemical Reactions 19 Figure 78b Control volumes joining primary cuts Figure 780 Improved control volumes secondary cut of stream 3 The number of redundant cuts can be reduced to one and the number of secondary cuts can be reduced to zero for the control volumes illustrated in Figure 78d thus this is the preferred set of control volumes to be used in the analysis of this reactor and recycle system 4 Figure 7861 Preferred control volumes Having selected a pair of control volumes we are ready to explore the matter of degrees of freedom The nature of the problem statement indicates that we can treat the system as a twocomponent system ie species A and the products that can be lumped together as 20 Solutions for Chapter 7 species B Given a two species system we can immediately complete the first two parts of the degreesoffreedom analysis as indicated in Table 78 As usual it is the particular speci cations and constraints that require careful consideration and the results given in Table 78 are explained as follows 1 The feed stream is pure species A and this provides one constraint for the compositions 2 We are given explicitly two constraints on the compositions that we list here as A Ah 1 3 We require that the splitter concentrations be constrained by 03A3 09AM 0305 2 and since there is no cut of stream 3 this provide only a single constraint This provides a total of four constraints in the category of particular specifications anal constraints as indicated in Table 78 This leads to one degree of freedom and we can Multicomponent Systems with Chemical Reactions 21 Table 78 DegreesofFreedom Stream Variables and Reactions compositions NXM 8 ow rates M 4 chemical reactions N 2 Generic Degrees of Freedom A N X M M N 14 Number ofY J J Balance F massmole balance equations N 4 Number ofC 39 for nmnn itinn M 4 Number of C 39 for Reaction Rates Generic Constraints B 9 Speci ed Stream Variables compositions 1 ow rates none Constraints for Compositions 3 Auxiliary Constraints none Speci ed Reaction Rates none Particular Speci cations and Constraints C 4 Degrees of Freedom A B C 1 develop a solution to the problem in terms of the mass ow rate m4 We begin our analysis of the system shown in Figure 78d with the xed control volume form ofthe species mass balance given by d dV 3 dt VA 22 Solutions for Chapter 7 We assume that the process is steady and that diffusional effects can be neglected so that Eq 3 reduces to J39mApv39ndl JrAclV 4 A V and the total mass balance is given by J p vndA 0 5 A For Control Volume I we have the two mass balances given by 1 Species A uA1rn1 uA5 m5 wA2 m2 6 1 Total mass 7521 m5 m2 7 while for Control Volume II we can express the total mass balance as 11 Total mass m1 m4 8 If we make use of the species A mass balance for Control Volume II we will encounter the rate of reaction term rA for which we have no explicit information This means that the mass balance for species A over Control Volume II will create one more equation and one more unknown Instead we summarize the constraints for this problem as C014 4 I 1 03A2 9a A2 809A 1 9b 01 4 03A 5 9C and remember that that the mass fraction in stream 1 is given by nA1 10 10 Here we see that we have seven equations and eight unknowns and this means that we can determine seven of the unknowns in terms of one of the eight Since we wish to determine the ratio m5 r514 we de ne the ratio of mass ow rates as x il2345 10 and make use of Eqs 9 so that our set of balance equations takes the form I SpeciesA AIA1oc1 s l E coAloc5 wA1x2 11 Multicampanent Systems with Chemical Reactions 23 1 Total mass 11 x5 12 12 II Total mass x1 1 13 Here we nd that our mass balance for species A in Control Volume I is homogeneous in 1A1 thus our problem takes the following form ocl 8 leg 015 0L2 14a 11 as 12 14b x1 1 14c which leads to a solution for x5 given by 17 as is 15 m4 8 This result is independent of the value of coA 1 indicating that it is true for any value of the mass fraction of species A in stream 1 This situation does not normally occur however in this case the constraints given by Eqs 9 lead to a form of Eq 6 that is homogeneous in 03A 1 79 Assume that the system described in Problem 78 contains N molecular species thus species A represents the single reactant and there are N 1 product species The reaction rates for the products can be expressed as III N 1 r5 rA m n Where the overall rate of reaction for species A is given by 7 I n m N rA 7 Ql rA rA rA egin your analysis with the axioms for the mass of an N component system and identify the conditions required in order that N 71 product species can be represented as a single spec1es 79 Solutions for Chapter 7 710 In an air drier part of the effluent air stream is to be recycled in an effort to control the inlet humidity The solids entering the drier stream 3 contain 20 water on a mass basis and the mass flow rate of the wet solids entering the drier is 1000 lbmhr The dried solids stream 4 are to contain a maximum of 5 water on a mass basis The partial pressure of water vapor in the fresh air entering the system stream 1 is equivalent to 10 mm of mercury and the partial pressure in the air leaving the drier stream 5 must not exceed 200 mm of mercury In this particular problem the recycle stream stream 6 is to be regulated so that the partial pressure of water vapor in the air entering the drier is equivalent to 50 mm Hg For this condition calculate the total molar flow rate of fresh air entering the system stream 1 and the total molar flow rate of the recycle stream stream 6 The system which operates at one atmosphere is illustrated in Figure 710 Figure 710 Air drier with recycle stream 710 7ll Metallic silver may be obtained from sulfide ores by roasting to sulfates leaching with water and precipitating the silver with copper It is this latter process involving the chemical reaction AgZSO4 Cu 2Ag Cuso4 that we wish to consider here In the system illustrated in Figure 711 the product leaving the second separator contains 90 by weight silver and 10 copper The percentage excess of copper in feed stream 1 is defined by lar rate of umption of l in the reactor j 100 molar rate of consumption of copper in the reactor percentage of excess copper For the conditions given what is the percentage excess of copper The percentage completion of reaction is defined y Multicampanent Systems with Chemical Reactions percent completion of reaction If the reaction is 75 complete What is m6 M15 7 Figure 711 Metallic silver production 711 x100 25 Solutions for Chapter 7 Chapter 8 Transient Mass Balances Section 81 81 A tank containing 200 gallons of saturated salt solution 3 lbm of salt per gallon is to be diluted by the addition of brine containing 1 lbm of salt per gallon If this solution enters the tank at a rate of 4 gallons per minute and the mixture leaves the tank at the same rate how long will it take for the concentration in the tank to be reduced to a concentration of 101 lbm per gallon 81 The process under consideration in this problem is illustrated in Figure 81 where we have shown a control volume with cuts at the entrance and exit that are joined by a Figure 81 Perfectly mixed stirred tank surface at which we assume that VA w n 0 This means that we have assumed that there is negligible transfer of species A the salt at the liquidair interface Since the process is described in terms of the species mass density we begin the analysis with the macroscopic mass balance in the form d dV 1 dt VA V t In the absence of any chemical reaction involving species A the term on the right hand side of this result is zero and the assumption of transport of species A at only the entrance and exit allow us to simplify Eq 1 to the form Solutions Chapter 8 dt v A6 1 pAdV IpAvndl 0 2 Here we have assumed that the control volume does not change with time and we have replaced the species velocity V A with the mass average velocity V on the basis that diffusion of species A can be ignored Evaluation of the terms in Eq 2 leads to d v MQ1 0 3 where p A and p A represent the bulk species densities at the inlet and outlet respectively The assumption associated with a perfectly mixed stirred tank is that the concentration of species A is uniform throughout the tank and this assumption leads to PAb2 2 PA 2 PA 4 One can use this result along with the simplification that the volumetric ow rate enter and leaving is a constant Q2Q1Q 5 in order to express Eq 3 in the form 61 rltmgt pm 6 Here we have used I to represent the mean residence time given by 1 V Q 7 The initial condition for this particular process can be expressed as 10 m m r 0 8 where p A 0 is given as 3 lbIn per gallon Following the development given in Sec 81 we solve for the concentration as a function of time in order to obtain ltpAgt a4 ML94 e 9 Since we are concerned with the time required to reach a certain concentration it is convenient to arrange Eq 8 in the form t TlnPAo 920171 9 PA PAM The mean residence time is determined as TransientMass Balances Solutions 5 VQ M 50min 10 4galm1n and the time required for p A to reach the value of 101 lbm gal is given by t 50minln 265 min 11 82 A salt solution in a perfectly stirred tank is washed out with fresh water at a rate such that the average residence time V Q is 10 minutes Calculate the following a The time in minutes required to remove 99 of the salt originally present b The percentage of the original salt removed after the addition of one full tank of fresh water 82 The concentration in this problem is given by Eq 821 CA 2 CA 01 1eit r 1 lt0Agt1 CA with CA 1 set equal to zero in order to obtain Q e tT 2 631 We can arrange this result in the form t Iln ems1 3 To solve the rst part of this problem we use I 10 min CA 00102 4 in order to obtain a t 10 min ln100 46 min 5 In order to solve Part b we note that we want to determine the percentage of the original salt removed during the time period from t 0 to t 1 and this is given by percentage of salt removed between i l 63 6 t0andt r CA 83 Two reactants are added to a stirred tank reactor as illustrated in Figure 83 In the inlet stream containing reactant 2 there is also a miscible liquid catalyst which is added at a level that yields a Solutions Chapter 8 concentration in the reactor of 0002 moles per liter The volumetric ow rates of the two reactant streams are equal and the total volumetric rate entering and leaving the stirred tank is 15 gallons per minute It has been decided to change the type of catalyst and the change will be made by a substitution of the new catalyst for the old as the reactant and catalyst are continuously pumped into the 10000 gallon tank The inlet concentration of the new catalyst is adjusted to provide a final concentration of 00030 moles per liter when the mixing process is operating at steady state Determine the time in minutes required for the concentration of the new catalyst to reach 00029 moles per liter Figure 83 Catalyst mixing process 83 If we assume that the tank is perfectly mixed the governing equation for the concentration of the catalyst is given by V 0AbiQi ltCAgt 2Q1 1 Here we have used CAb1 to represent the bulk concentration of the new catalyst in the inlet stream having a volumetric ow rate given by Q1 Since the volumetric ow rates of the two entering streams are equal the exit ow rate is 2Q1 as indicated in Eq 1 The initial condition associated with this process is given by IC CA 0 t0 2 and the solution takes the form ltcAgt cAb11 6W 3 Here I is the mean residence time given by r V Q 66667 min 4 and because we wish to determine the time required to reach a given value of the concentration we arrange Eq 3 in the form Salmons THZVLSZEVLZ Mass Balance 1 7 c t 3 1 2 Ab1 5 2 754bl CA From Eq 3 with t 00 and the problem statement We know that c A 1 0003 molesliter 6 and the time required for cA to reach 00029 molesliter is given by z Wm 18hrsand54rnin 7 f uent Initially 84 Three perfectly stirred tanks each of10000 gallon capacity are arranged so that the e en i an a ct Hwy ure vlmter is then fed to the rst tank at the rate of 50 the concentration in each tank is 5 gallons per rninute You are asked to determine a 10 b The concentrations in the second and third tanks at this time c A general equation for the concentration in the nth tank in the cascade system illustrated in Figure 84 Fzgmz 84 Sequence ofstirred tanks 1 In order to solve an ordinary ditferential equation of the form d gem ft explore the possibility that this rams Equatlm t can be transformed to the Slmplz Equatzm t given Y d 7 t bt 2 dt 0XEAgt 39 39 because it canL A 439 39 to obtain w e me Solutions Chapter 8 0a IltCAgt20 3 The new functions at and bt can be determined by noting that lda E 80 170 atft 4 Any initial condition for at will suffice since the solution for CA does not depend on the initial condition for at Use of dummy variables of integration is essential in order to avoid confusion 84 The system under consideration is illustrated in Figure 84 along with a simpli ed nomenclature used to identify the various concentrations Since the inlet concentration to the rst tank is zero we can use Eq 818 to express the concentration in the rst tank as wwg m Here c0 represents the original concentration in the sequence of stirred tanks In order to analyze the second tank in the sequence shown in Figure 84 we follow the analysis leading to Eq 812 and assume perfect mixing to obtain dc3 T t 2 dt 03 02 1C c3 c0 t0 3 This initial value problem is similar in form to Eqs 814 however in this case the non homogeneous term is a function of time and we cannot separate variables as indicated in Eq 815 Here we are confronted with a new mathematical problem and one of the classic methods for solving a new mathematical problem that we cannot solve is to transform it to an old mathematical problem that we can solve We begin this process by expressing Eq 2 in the classic form W m where the two new functions are de ned by Pm WHW Clearly Pt is a constant equal to the inverse of the mean residence time however there is no harm associated with a study of Eq 4 instead of Eq 2 We begin our search for an ordinary differential equation that we can solve by asking whether Eq 4 can be expressed as d Emam TransientMass Balances Solutions which is a separable rst order ordinary differential equation To explore this possibility we expand the derivative of the product and divide by at to obtain dt 1 dac3 btat 7 E In order to transform Eq 4 to the separable form given by Eq 6 we need to find the functions at and bt such that 1 W Q0 8 E Directing our attention to the first of these we integrate from zero to any arbitrary time according to TF1 71 1 da 761 P d 9 Iamdn n J n n n0 n0 This integration leads to TF1 lily J Hum 10 n0 in which 10 is the unknown initial condition represented by IC at a0 t0 ll Solving Eq 10 for the function at leads to 71 00 do exp 13096111 12 110 and this result can be used to express bt in the form 130 a0QteXP 13 We now return to Eq 6 and integrate leading to Solutions Chapter 8 dg jaw 14 C 51053 0 EFO rw Ca 53 Here we have been very careful to represent the integration in terms of the dummy variables of integration Q and i Carrying out the integration for the left hand side of Eq 14 provides atc3 aoc0 15 in which we have made use of the initial condition given by Eq 3 At this point we can make use of the representation for bt given by Eq 13 in order to express our solution for c3 t in the form 16 a103 Dividing this result by at and making use of Eq 12 leads to t 030 Q eXP P d d t t 0 L 0 J 17 Here we have found a solution the general equation represented by Eq 4 and in order to extract a solution to the specific equation represented by Eq 2 we need to make use of the definitions of Pt and Qt given by Eq 5 which allow us to express Eq 17 as c3t one tT equot Wig 18 The final step in this process requires the solution for 02t which is given by Eq 1 and the use of that result leads to 030 coe t r 1 19 It is certain that one could avoid the steps between Eq 2 and Eq 19 by nding the solution somewhere in a text on ordinary differential equations or by using the recipe TransientMass Balances Solutions known as the integrating factor The advantage of the analysis given here is that it illustrates the concept that one can often transform a new problem that one cannot solve into an old problem that one can solve We can now move on to the third stirred tank in the sequence illustrated in Figure 84 and describe the initial value problem as dC 4 1 t 20 dt 04 030 1C 04 co t0 21 The solution is obtain by the procedure illustrated above and the result is given by The solutions for 02 t c3t and 04t do not provide enough information to guess the general form for the nth tank however the solution for the concentration in the fourth tank allows us to conclude that the concentration in the nth tank is given by A convenient form of this result represents the solution to Part c of this problem which is given by mnil 1 Pan c Cn1t one W E 7 rr quot 25 m m0 We now return to Part a and express Eq 1 as Part a t T ln 00020 V In 00020 26 The time required for the concentration in the rst tank to diminish to 02 t 01 00 is given by Pama t 4230 460mm 27 ga mm The answers to Part b are available from Eqs 19 and 22 which provide Solutions Chapter 8 c3 00 1 23 e 03300 Part b 28 85 Develop a solution for Eqs 820 and 821 when ft is given by 0quot c1 c rm 0lttltAt A A A Ci r This represents a process in which the original steadystate concentration is c and the final steadystate concentration is CIA The response time for the mechanism that creates the change in the concentration of the incoming stream is At while the response time of the tank can be thought of as the mean residence time I The time required to approach within 1 of the new steady state is designated as t1 and we express this idea as cAgt all 001 cg c ttl For At 0 we know that t1 7 46 on the basis of Eq819 In this problem we want to know what value of II r is required to approach within 1 of the new steady state when At T 02 85 The initial value problem under consideration is given by dltc gt r A CA f0 1 dt IC CA 021 10 2 The function f t describes the manner in which the concentration at the entrance changes with respect to time In Sec 81 we considered an instantaneous change from c to 01 and in this problem we explore the in uence of a nite response time described by 0 0A 0 t t f0 3 0114 t t In this problem it is convenient to use a dimensionless concentration defined by 0 gt 0 CA H 4 0A 0A so that the problem statement takes the form 10 TransientMass Balances Solutions dCA T C Ft 5 dt A IC CA 0 t0 6 Ft A OS A tl t tlt t 7 tZAt The solution to this problem follows from the type of analysis described in Problem 84 and after some algebra we obtain lt24gt31 2 4 2J s 0A 0A As At 1 tends to zero this result tends to the solution given earlier by Eq 819 When the response time of the mechanism that creates the change in the inlet concentration is given as AlT 02 Eq 8 takes the form 0 Ufa CA 2 2J 1 11962 9 0 CA CA If we designate t1 as the time required to reach within 1 of the new steadystate condition Eq 9 yields 001 1196 651 10 and we can solve for t1 T to obtain tl r 478 At r 02 11 This result should be contrasted with that obtained for the instantaneous change of the inlet concentration which is given by Eq 819 In that case the time required to reach within 1 of the new steadystate is given by tl I 460 At r 0 12 Here we see that the model of an instantaneous change can be an excellent approximation provided that the details of the solution are not required at times that are small compared to the mean residence time 86 Volume 2 of the Guidelines for Incorporating Safety and Health into Engineering Curricula published by the Joint Council for Health Safety and Environmental Education of Professionals JCHSEEP introduces without derivation the following equation for the determination of concentration of contaminants inside a room G Qquot C W G where C concentration of contaminant at time t G rate of generation of contaminant Solutions Chapter 8 Q effective rate of ventilation Q QK V volume of room enclosure K design distribution ventilation constant I length of time to reach concentration C Making the appropriate assumptions derive the above equation from the material balance of contaminants What would be a proper set of units for the variables contained in this equation 86 The macroscopic mole balance for a xed control volume is given by d RdV 1 dt A v in which species A represents the contaminant If we assume that the rate of generation of contaminant is due to chemical reaction we can express this result as cAdV IcAvndl G 2 V A provided that the species velocity is equal to the mass average velocity Further assuming that there is mass transfer of contaminant only at an exit Eq 2 takes the form cAdV JcAvmdl G 3 V in which Ae represents the area of the exit where ventilation is taking place Assuming that the velocity pro le is at or that the concentration is uniform at the exit leads to d E J CAdV lt0Agtexitanit G The classic assumption associated with a perfectly mixed stirred tank is that the concentration of the contaminant is uniform throughout the tank This assumption leads to ltCAgtexit CA 5 in which the volume averaged concentration 0A is de ned by c gt i 0 61V 6 A V A V Use of Eqs 5 and 6 in Eq 4 leads to d v ltcAgtQex G 7 12 TransientMass Balances Solutions This differential equation is separable and we can write Eq 10 as dltCAgt V 7dr 8 ltCA gtQexit G We take the initial condition to be IC ltCAgt 0 t0 9 which leads to the solution given by W exp rQexv 10 In order for this result to correspond to that given in the Guidelines for Incorporating Safety and Health into Engineering Curricula we are forced to conclude that effective rate of ventilation Qexit 1 1 design distribution ventilation constant and that the concentration of contaminant at time I actually means the volume averaged concentration of the contaminant at time t In addition we conclude that one proper set of units for the variables in the equation contained in the Guidelines for Incorporating Safety and Health into Engineering Curricula is given by 3 big C467 QI ts Vm3 S m S Finally we conclude that vague and poorly defined equations are in fact unhealthy for any engineering curriculum 87 Often it is convenient to express the transient concentration in a perfectly mixed stirred tank in terms of the dimensionless concentration defined y ltCAgt 031 CA 1 0 CA CA Represent the solution given by Eq 819 in terms of this dimensionless concentration 87 In terms of a dimensionless concentration we can represent Eq 819 as 0 CA CA 2 1 ef 014 031 This dimensionless concentration has the characteristic that it satisfies the initial condition given by 1C CA 0 t0 2 and the asymptotic condition that takes the form Solutions Chapter 8 CA gt 1 t gtoo 3 This particular dimensionless concentration was used in the solution of Problem 85 88 A perfectly mixed stirred tank reactor is illustrated in Figure 88 A feed stream of reactants enters at a volumetric flow rate of Q0 and the volumetric flow rate leaving the reactor is also Qo Under steady state conditions the tank is half full and the volume of the reacting mixture is Vo At t 0 an inert species is added to the system at a concentration c and a volumetric flow rate Q1 Unfortunately someone forgot to change a downstream valve setting and the volumetric flow rate leaving the tank remains constant at the value Qo This means that the tank will over ow at t Vo Q1 While species A is inert in terms of the reaction taking place in the tank it is mildly toxic and you need to predict the concentration of species A in the fluid when the tank over ows Please derive a general expression for this concentration taking into account that the concentration of species A in the tank is zero at t 0 Figure 88 Accidental overflow from a stirred tank reactor 88 We begin this problem with the general macroscopic balance for species A d RdV 1 dt A v vac a a note that species A is inert and make use of the obvious control volume to obtain d dA 0 2 dt vn VLIraVLCE Amt The assumption of perfect mixing leads to d E V1lt0Agt CZQ1 ltCAgtQ0 0 3 which is subject to the initial condition given by 1C ltcAgt 0 t0 4 TransientMass Balances Solutions The volume of uid in the tank is given by Vt V0 Q1t OStSVOQl 5 V0 2V0 tgtV0Q1 and we need to solve for the concentration of species A when the tank over ows ie at the time t V0 Q We can use the rst of Eqs 5 in Eq 3 in order to obtain VOZQ1 CA Q0 0 6 and it is convenient to arrange this result in the form dltCAgt Q1 Qo Q1 021 dt V0 0 Q1 t In more compact notation we can express this ordinary differential equation as d c PrltcAgt Q0 8 where Pt and Qt are de ned in the obvious manner This equation is not separable however it can be transformed to a separable equation having the form d EatltcAgt 130 9 and the details are given in the solution to Problem 104 where we show that the integrated form of Eq 8 is given by t t t P d Q exp4 P d gt61 110 n0 i0 110 10 On the basis of the initial condition given by Eq 4 this general result simpli es to 11 Given that Pm is de ned by Solutions Chapter 8 Pm 12 V0 an we can evaluate the rst integral to obtain nzt exp 13mm 13 1120 where the parameter 3 is de ned by B Q1 Q0 14 Q1 Use of the result given by Eq 13 along with the fact that Q39 is de ned by Qi 15 V0 Q1 leads to the solution for CA that can be expressed in the form ltcAgt CZB I osrsVoQl 16 When the tank over ows at t V0 Q the concentration of the mildly toxic species A is given by ltcAgt aim 1 f tVoQ1 17 One might avoid this complex calculation by assuming that it would be suf cient to determine the concentration of the mildly toxic species A when the over owing tank has reached a steady state On the basis of Eq 6 this concentration is given by cAgt 0216 1 1900 18 Since this concentration is always greater than the concentration when the tank over ows it provides a conservative estimate of the concentration of the mildly toxic species that may be released to the environment Section 82 89 A perfectly mixed batch reactor is used to carry out the reversible reaction described by kl AEaB 16 TransientMass Balances Solutions k2 B AE and use of mass action kinetics provides the chemical kinetic rate equation given by RA 7 kchcE k2 CB secondorder rst order eactlon decomposition If species E is present in great excess the concentration CE will undergo negligible changes during the course of the process and we can define a pseudo first order rate coefficient by ki klcE Use this approach to determine the concentration of species A and B as a function of time when the initial conditions are LC CA 02 CB 05 10 810 When molecular species A and B combine to form a product one often adopts the chemical kinetic schema given by k A B 4 products Use of mass action kinetics then leads to a chemical kinetic rate equation of the form R A kcA 63 One must always look upon such rate expressions as hypotheses to be tested by experimental studies For a homogeneous liquidphase reaction this test can be carried out in a batch reactor which is subject to the initial conditions ICl CA c2 t0 IC2 CB cg t0 Use the macroscopic mole balances for both species A and B along with the stoichiometric constraint RA R8 in order to derive an expression for CA as a function of time In this case one is forced to assume perfect mixing so that CA CB can be replaced by cAgtltcB 811 A batch reactor illustrated in Figure 811 is used to study the irreversible decomposition reaction k A gt products The proposed chemical kinetic rate equation is R A 7 k CA Salmons Chapter 8 and this decomposition reaction is catalyzed by sulfuric acid To initiate the batch process asmall volume of catalyst is placed in the reactor as illustrated in Figure 811 At the time t0 the solution of species A is added at a Volumetric ow rate Q1 and a concentration 52 When the reactor is full the stream of speciesA is shut offand the system proceeds in the normal manner for abatch reactor During the startup time the volume of uid in the reactor can be expressed as V t K gt and the final Volume ofthe uid is given by Vl Vo Q 1 Here tl is the startup time In this problem you are asked to determine the concentration of speciesA during the startup time and all subsequent times The analysis for the startup time can be simplified by means ofthe transformation yt W t and use ofthe initial condition Ic y 0 t0 After you have determined y you can easily determine ltcAgt during the startup period The concentration at ttl then becomes the initial condition for the analysis of the system for all subsequent times O en one simplifies the analysis ofa batch reactor by assuming that the time required to fill the reactor is negllglble Use your solution to this problem to identify what is meant by negligible for this particular problem thm 81 Batch reactor startup process 812 In the perfectly mixed continuous stined tank reactor illustrated in Figure 812a species A undergoes an irreversible reaction to form products according to k A gt pmdhctr RA rkcA 1 The original 39 w L is VB 4 L 39 39 39 mm and out ufthe reactor is Q The concentration ofspeciesA emermg the reactor is xed at c and under 18 TransientMass Balances Solutions steady state operating conditions the concentration at the exit and therefore the concentration in the reactor is ltCA Determine the concentration ltCAgt under steady state operating conditions Figure 81211 Steady batch reactor Because of changes in the downstream processing it is necessary to reduce the concentration of species A leaving the reactor This is to be accomplished by increasing the volume of the reactor by adding pure liquid to the reactor as illustrated in Figure 812b at a volumetric ow rate Q1 until the desired reactor volume is achieved During the transient period when the volume is given by V t VO Q1t the exit ow rate is constant at Q In this problem you are asked to determine the exit concentration during this transient period Figure 8121 Transient process in a perfectly mixed stirred tank reactor Section 83 813 Consider the process studied in Example 81 subject to an initial condition of the form 0 0 LC CA CA CE CB i0 Solutions Chapter 8 and determine the concentration of species A as a function of time 814 For the process studied in Example 81 assume that the equilibrium coefficient and the first order rate coefficient have the values KM 101 k1 10min1 and determine the time II required for the concentration of species A to be given by ltcAgt c 099 ltcAgtzq c H1 Here 61 represents the equilibrium concentration 815 In this problem we consider the heated semibatch reactor shown in Figure 815 where we have identified the vapor phase as the y phase and the liquid phase as the Bphase This reactor has been designed to determine the chemical kinetics of the dehydration of tbutyl alcohol species A to produce isobutylene species B and water species C The system is initially charged with tbutyl alcohol39 a catalyst is then added which causes the dehydration of the alcohol to form isobutylene and water The isobutylene escapes through the top of the reactor while the water and tbutyl alcohol are condensed and remain in the reactor If one measures the concentration of the tbutyl alcohol in the liquid phase the rate of reaction can be determined and that is the objective of this particular experiment The analysis begins with the fixed control volume shown in Figure 815 and the general macroscopic balance given by g Wdv D AM 1 V A V The moles of species A tbutyl alcohol in the yphase can be neglected thus the macroscopic balance for this species takes the form tbutyl alcohol I CAB dV J R AB dV 2 um um and in terms of average values for the concentration and the reaction rate we have tbutyl alcohol chABW a RABWBO 3 dt If we also assume that the moles of species B isobutylene and species C water are negligible in the yphase the macroscopic balances for these species take the form d isobutylene EltCB3VB0 MB REWVBQ 4 d water Ekcqomo ltRC gtVBagt 5 20 TransientMass Balances Solutions This indicates that alcohol and water are retained in the system by the condenser while the isobutylene Figure 815 Semibatch reactor for determination of chemical kinetics leaves the system at a molar rate given by M 31 The initial conditions for the three molecular species are given by TC CA c I 0 6a 1C CB 0 t 0 6b 1c cc 0 t 0 6c Since the molar rates of reaction are related by Rep Rqs gt and Rm Rip 7 we need only be concerned with the rate of reaction of the t butyl alcohol If we treat the reactor as perfectly mixed the mole balance for tbutyl alcohol can be expressed as ch Qua dVEO dt V50 dz 8 RAB This indicates that we need to know both CA6 and VB as functions of time in order to obtain experimental values of R A The volume of uid in the reactor can be expressed as Solutions Chapter 8 V130 7va quotBBVB quotCIva 9 in which nA nB and no represent the moles of species A B and C in the 3phase and VA VB and VG represent the partial molar volumes To develop a useful expression for R AB assume that the liquid mixture is ideal so that the partial molar volumes are constant In addition assume that the moles of species B in the liquid phase are negligible On the basis of these assumptions show that Eq 8 can be expressed as dew RA at 1 10 This form is especially useful for the interpretation of initial rate data ie experimental data can be used to determine both CAB and chB dt at t 0 and this provides an experimental determination of R AI for the initial conditions associated with the experiment An alternate approach1 to the determination of R AI is to measure the molar flow rate of species B that leaves the reactor in the yphase and relate that quantity to the rate of reaction Section 84 816 When Eqs Error Reference source not found and Error Reference source not found are valid Eq Error Reference source not found represents a valid result for the chemostat shown in Figure 810 One can divide this equation by a constant my to obtain Eq Error Reference source not found39 however the average mass of a cell mm in the chemostat may not be the average mass of a cell in the incoming stream lf mm is different than mmll1 Eq Error Reference source not found is still correct but our interpretation of ltngt1 is not correct Consider the case mmll1 i my and develop a new version of Eq Error Reference source not found in which the incoming flux of cells is interpreted properly in terms of the number of cells per unit volume in the incoming stream 817 Develop a general solution for Eqs Error Reference source not found and consider the behavior of the system for D lt u D y and D gt it Section 85 818 Repeat the analysis of binary batch distillation when Raoult s law is applicable ie the process equilibrium relation is given by eff 54 y gt A 2 1 xA am 1 1 Gates BC and Sherman J D 1975 Experiments in heterogeneous catalysis Kinetics of alcohol dehydration reactions Chem Eng Ed Summer 124127 22 TransientMass Balances Solutions Here 012 is the effective relative volatility and is temperature dependent however in this problem you may assume that 0121 is constant Use your result to construct figures comparable to Figure 815 for x 01 and x 05 with values of de given by 14 12 l 2 and 4 Section 86 819 Use Eqs Error Reference source not found and Error Reference source not found to obtain a chemical kinetic rate equation for species E 820 Develop the global reaction rate equation for species E on the basis of Eqs Error Reference source not found and Error Reference source not found 821 Show how Eqs Error Reference source not found are obtained from Eqs Error Reference source not found through Error Reference source not found 822 Provided a detailed development of Eq Error Reference source not found 823 Derive Eq Error Reference source not found on the basis of Eqs Error Reference source not found through Error Reference source not found 824 It is difficult to find a real system containing three species for which the reactions can be described by A i B la B E c 1b however this represents a useful model for the exploration of the condition of local equilibrium The stoichiometric constraint for this series of first order reactions is given by RARBRC0 2 since the three molecular species must all contain the some atomic forms For a constant volume batch reactor and the initial conditions given by 10 cl 0 t 0 3 Solutions Chapter 8 one can determine the concentrations and reaction rates of all three species as a function of time If one thinks of species B as a reactive intermediate the condition of local equilibrium takes the form Local reaction equilibrium RB 0 4 In reality the reaction rate for species B cannot be exactly zero however we can have a situation in W ic RB ltlt RA 5 Often this condition is associated with a very large value of k2 and in this problem you are asked to develop and use the exact solution for this process to determine how large is very large You can also use the exact solution to see why the condition of local equilibrium might be referred to as the steadystate assumption for constant volume batch reactors 825 The global staichiametric schema associated with the decomposition of N205 to produce NO2 and 02 can be expressed as N205 gt 2N02 02 and experimental studies indicate that the reaction can be modeled as first order in N205 Show why the following chemical kinetic schemata give rise to a first order decomposition of N205 2 I N205 No2 NO3 k H No2 NO3 Zgt No2 02 NO k 111 NO N03 gt 2N02 k4 IV NO2 No3 N205 825 We begin this problem with J 39 for the 39 y 39 39 39 ic relations and the mass action kinetics for each chemical kinetic schema This leads to k1 Kinetic Schema I N205 gt N02 N03 1a 1 I I I RN02 RN205 RN03 RN205 Elementary Stoichiometry 1 1b 1 I R02 0 RNO 0 Rate Equation 1 RIIQZOS iklcNZOS 1c k2 Kinetic Schema 11 N03 N02 gt N02 02 NO 2a 24 TransientMass Balances Solutions 11 II II RN02 0 R02 RNO3 Elementary Stoichiometry II 2b II II II RNO RNO3 RN205 0 Rate Equation II R1303 7 k2 cNOacNo2 2c k3 Kinetic Schema III NO N03 gt 2N02 3a III III III III RN03 RNO RN02 RNO Elementary Stoichiometry III 3b III III R02 0 RNZOS 0 Rate Equation III R135 7 k3cNOcN03 3c k4 Kinetic Schema IV NO2 N03 gt N205 4a IV IV IV IV RN03 RN02 RN205 RNOZ Elementary Stoichiometry IV 4b IV IV RNO 0 R02 0 Rate Equation IV Rik32 k4CN02 CNO3 4c The local rates of reaction are given by III IV RN205 RN205 RN205 5a RN02 Rkoz R1302 R gz RIVsz 5b RN03 Rim Riio3 Riib3 RITE 5c R02 R z R32 R R33V2 5d RNO Rho R o RISE R136 59 and we can use the elementary stoichiometric conditions and the rate equations to express these local rates according to RNZOS kICNZOS k4cNocho3 6a RN02 No3 k4cNOZ CNO3 6b

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