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  2. 2. CPP-1 In your Chemical Engineering job you may use the skills acquired from this course more than the skills acquired from any other Chemical Engineering course
  3. 3. Sr. No. Topics 1 General Information, Grading policy, Syllabus, Learning outcomes 2 Introduction to engineering calculations : units and dimensions, conversion of units, system of units, force and weight 3 Process and Process variables: mass and volume, flow rate, chemical composition, pressure, temperature 4 Material balance: process classifications, balances on nonreactive systems, balances on reactive processes, balance on multiple units, recycle and bypass.
  4. 4. About Instructor  Name :Muhammad Ahsan Waseem  Designation: Lecture  Qualification: Master in chemical engineering, Karlstad University, Sweden  Area of research, rheology, viscoelastic polymers, Nano cellulose  Office : E 010 DCE ;  Phone : 0331-6508286  Email : ahsanwaseem@uog.edu.pk  Contact hours: 8:30 a.m. till 4:30 p.m.
  5. 5. Assignments (2) = 10% Quizzes (2) = 5% Presentation (1)=10% Midterm Exam = 25% Final Exam = 50% GRADING POLICY CH 120 Chemical Process Principles-I
  6. 6. Schedule for Sessional Sr. NO Activity Week 1 Quiz 1 6 2 Assignment 1 7 3 Quiz 2 14 4 Assignment 2 15 5 Presentations 15
  7. 7. Course Objectives  Obtain a feel for what Chemical Engineers do  Learn about basic chemical process units and their conversion.  Learn about the process and process variables used in chemical engineering.  Learn the practical application of law of conservation of mass while solving material balance problems both for without and with reaction.
  8. 8. Course Outlines  Units, dimensions, conversion of units, system of units, force and weight, dimensional homogeneity, mass and volume, flow rate, chemical composition, pressure, temperature  Mass balances for items of plant, Choice of basis/datum for balances. Overall and component balances, Limiting and excess reactants. Balances for systems with recycle, purge and bypass streams, Mass balances for unit operations. Tie components, Balances for batch and continuous plants.  Principles of Stochiometric Combination  Nature of balances; concept of a valance, Input-output relationships, steady state considerations, Sub-systems and interconnections. Familiarization with flow sheets, Mass and energy balance diagrams and tables
  9. 9. Course Learning outcomes By the end of the course the students should be able to: 1. Apply the principles of unit conversion, dimensional homogeneity and dimensionless numbers used in chemical engineering. 2. Apply the concept of process variable in solving chemical engineering problems. 3. Apply the law of conversation of mass to solve chemical engineering problems without reaction. 4. Apply the law of conversation of mass to solve chemical engineering problems with reaction.
  10. 10. Text books R. M. Felder and Rousseau, Elementary principles of chemical processes. J. Wiley & Sons, 3rd Update Edition, 2005. D. M. Himmelblau and J. B. Riggs, Basic principles and calculations in chemical engineering, Prentice Hall Professional Technical Reference, 8th Edition, 2004.
  11. 11. Chemical Engineering (Definition)  Chemical engineering is the profession in which a knowledge of  mathematics,  chemistry,  biology and  other natural sciences gained by  study,  experience, and  practice  is applied with judgment to develop economic ways of using materials and energy for the benefit of mankind.  The profession encompasses the spectrum from the products, to the processes and equipment for making them, and to their applications.
  12. 12. Chemical engineering (an introduction) Role of a chemical engineer in industry A chemist Discovers that if two reactants are mixed in a certain proportion at an elevated temperature A product will be attained which is More valuable than both reactants From this very point it is an engineering problem, or more precisely hundred’s of engineering problems
  13. 13. Chemical engineering (an introduction)  Based on the work performed by the chemist an engineer or more precisely a chemical engineer has to perform numerous tasks such as  Type of reactor?  A long pipe  A large tank  Several smaller tanks  An extremely large tube  Made of what?  Does it have to be heated?  How much and how?
  14. 14. Chemical engineering (an introduction) Type of reaction? exothermic or endothermic? Where should the reactants be obtained from? buy them or make them? What about the reactor effluent? Should it be wasted or sold? Reactants? Recycled? How to separate product from reactants? Add another substance that extracts the product? Or should it be heated to get the volatile product and recycled?
  15. 15. Chemical engineering (an introduction) Are controls needed to keep the operation of the process with in the rigid limits? What kind of controls? Automatic? Manual? Can the laboratory data be used directly to the industrial plant? Pilot plant? what can easily go wrong during the process? What pre cautions be taken?
  16. 16. Chemical engineering (an introduction) What about waste products if they are produced as a result of the process? Should they be wasted as it is or they have to be treated in order to keep environment clean? How much of the process should be automated? What about economics? Capital cost? Running cost? Man power? Profit? Is it worthwhile?
  17. 17. Chemical engineering (an introduction) What would be the procedure for start- up? What about safety? Setting up laboratories in order to ensure smooth running of the process? What to do if process specifications are change? How to do it without changing the whole layout of the plant?
  18. 18. Chapter 1 Introduction to Engineering Calculations Units and dimensions Conversion of units System of units Force and weight Dimensional homogeneity and dimensionless Quantities
  19. 19. Introduction to engineering calculations  Units and dimensions  A dimension is a property that can be measured such as length, time, mass, pressure or temperature.  It can also be calculated by multiplying or dividing other dimensions, such as length/time (velocity), length3 (volume) or mass/volume (density), Newton (mass x acceleration)  Units are the means of expressing the dimension such as feet or centimeters for length and seconds or hours for time.
  20. 20. Introduction to engineering calculations  Units can be treated by algebraic variables when quantities are added, subtracted, multiplied, or divided. The numerical values of two quantities may be added or subtracted only if the units are same.  3 cm- 1 cm = 2 cm  3 cm – 1 second = ?  On the other hand, numerical values and their corresponding units may always be combined by multiplication or division  3 N x 4m = 12 N.m  3 m x 4m = 12m2  5.0 km x 4 hr = 20 km/hr
  21. 21. Introduction to engineering calculations Conversion of units Converting a quantity expressed in terms of one unit to its equivalent in terms of another unit is known as to be conversion of units. A measured quantity can be expressed in terms of any units having the appropriate dimension. For instance velocity may be expressed in ft/sec, miles/hr, cm/year, or any other ratio of a length unit to a time unit. The numerical value of the velocity will depend entirely on what unit has been chosen.
  22. 22. Introduction to engineering calculations While converting units a conversion factor is always required that has some numerical value. What conversion factor you will use in order to convert Miles to km? Kg to g? Meter to centimetre? Atmospheric pressure to mm of Hg? Kelvin to Celsuius scale?
  23. 23. Introduction to engineering calculations Method of carrying out conversion A vertical line can be used to multiply two quantitates and then cancelling out common units. Convert an acceleration of 1 cm/s2 to its equivalent in km/year2 System of units A system of units has the following components
  24. 24. Base unit  Mass, length, time, temperature, electric current and intensity. Multiple units  Multiples of fractions of base unit such as minutes, hours and milliseconds.  It is easier to refer to 3 year than 94,608,000 seconds. Derived units (are obtained in one or two ways)  By multiplying and dividing base or multiple unit (cm2, ft/min, kg.m/s2)  As defined equivalents of compound units (e.g., 1 N = 1 kg.m/s2, 1 lbf = 32.174 lbm.ft/s2)
  25. 25. System of units used in engineering calculations There are more than one system of units that are being used across the globe such as SI (system of international units) CGS system American Engineering system
  26. 26. Quantity (Dimension) Base unit Symbol Length Meter (SI) m Centimetre (CGS) cm Foot (AE) ft Mass Kilogram (SI) kg Gram (CGS) g Pound (AE) lbm Time Second (SI) s Second (CGS) s Second (AE) s Temperature Kelvin (SI) k Degree Rankine or degree fahrenheit R, F Force Newton (SI) N Dyne (CGS) dyn Pound (AE) lbf Energy Joule (SI) J British thermal unit (AE) BTU
  27. 27. Units and their conversions Convert 30 g/s to its equivalent in kg/h. 30 lbm/s to its equivalent in kg/min. Convert a volume of 5 ft3 to its equivalent in m3. Convert 23 lbm.ft/min2 to its equivalent in kg.cm/s2.
  28. 28. Units and their conversions Carry out the following conversions 3 wk to seconds 554 m4/(day.kg) to cm4/(min.g) 38.1 ft/s to miles/h 0.04 g/(min)(m3) to lbm/(hr)(ft3) 2 L/s to ft3/day 60 mil/hr to ft/sec 6.2 cm/hr2 to nm/sec2
  29. 29. Force and Weight  Force and weight  Force is proportional to the product of mass and acceleration i.e. F = ma  1 N = 1 kg.m/s2  In an American engineering system, the derived force unit called a pound force lbf is defined as the product of a unit mass and the acceleration of gravity at sea level and 45O altitude which is 32.174 ft/s2.  So 1 lbf = 32.174 lbm.ft/s2
  30. 30. Force and Weight  The weight of an object is the force exerted on the object by the gravitational attraction. It is denoted by W.  Suppose that an object of mass m is subjected to a gravitational force W and if this object is freely falling its acceleration would be g.  W = mg  Conversion Factor gc (g sub c)  gc is used to denote the conversion factor from natural to derive units. It is simply a conversion factor and not to be confused with the gravitational acceleration which is usually denoted by g.  Value of gc is 32.174 lbm.ft/lbf.s2 or kg.m/N.s2
  31. 31. Value of g in different system of units SI CGS AE 9.8 m/s2 980.66 cm/s2 32.174 ft/s2
  32. 32. Conversion involving both lbm and lbf What is the potential energy in (ft)(lbf) of a 100 lb drum hanging 10 ft above the surface of the earth with reference to the surface of the earth?
  33. 33. Weight and mass Calculate Weight in lbf of a 25 lbm object The mass in kg of an object that weighs 25 newton. Water has a density of 62.4 lbm/ft3. How much does 2 ft3 of water weigh at sea level and 450 latitude and in Denver where the altitude is 5374 ft and the gravitational acceleration is 32.2 ft/s2?
  34. 34. Five hundred pounds of nitrogen is to be charged into a small metal cylinder at 25OC, at a pressure such that the gas density is 11.5 kg/m3. Estimate the required cylinder volume in American Engineering System of units?
  35. 35. Units and their conversions In American engineering system of units, the viscosity can have the units of (lbf)(hr)/ft2, while in a handbook the units are (g)/(m)(s). Convert a viscosity of 20 (g)/(m)(s) to the given American engineering units. 1 lbm water is flowing through a 2-inch diameter pipe with a velocity of 3 ft/s. What is the kinetic energy of the water in (ft)(lbf)? What is the flow rate in gal/min?
  36. 36. Convert the following from AE to SI Units 4 lbm/ft to kg/m 1 lbm/(ft3)(s) to kg/(m3)(s) Convert 1 J/sec to its equivalent (ft)lbf)/sec
  37. 37.  Dimensional homogeneity  Every valid equation must be dimensionally homogeneous, that is, all additive terms on both sides of the equation must have the same dimensions.  If two quantities can be expressed in terms of grams/second both must have the dimension (mass/time).  Consider the equation  u (m/s) = u0 (m/s) + g (m/s2) t (s)  The above equation is dimensionally homogenous, since each terms u, u0, and gt has the same dimensions (length/time).
  38. 38. If an equation is dimensionally homogeneous but its additive terms have inconsistent units, the terms (hence the equation) may be made consistent simply by applying the appropriate conversion factors. It is desired to express the time in minutes and the other quantities in the units given below u (m/s) = u0 (m/s) + g (m/s2) t (min)(60s/min) = u0 + 60gt
  39. 39.  The following equation is proposed to calculate the pressure drop (∆p) across the length of pipe (L) due to flow through the pipe. Determine the dimensional consistency of this equation  (∆p) = 1/2v2(L/D)f  The drag force is mathematically expressed  Fdrag= Vfluid. ρfluid.Across-sectional area  Where V (velocity), ρ (density)  Check if the above mentioned equation is dimensionally correct?
  40. 40. Bernoulli’s Equation  𝑣 2 + 𝑔𝑧 + 𝑃 𝜌 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Where V= fluid flow speed g = acceleration due to gravity z= elevation P= Pressure drop ρ= density of the fluid
  41. 41. Dimensional Homogeneity  The following equation is proposed to calculate the thermal conductivity in heat transfer 𝑘 = 𝑄𝐿 𝐴∆𝑇  Where  k= thermal conductivity in W/m.K  Q = the amount of heat transfer through the material in J/s.  A = area of the body.  ∆𝑇= difference in temperature.  Check if the above mentioned equation is dimensionally correct?
  42. 42. Dimensionless quantity A dimensionless quantity can be pure number (2,1.3,5/2) or a multiplicative combination of variables with no net dimensions. For example, consider the following equation M (g) = D (cm) µ(cm/s)ρ(g/cm3) Mo (g) µ[g/(cm.s)]
  43. 43. Reynolds number In fluid mechanics, the Reynolds number (Re) is a dimensionless quantity that is used to help predict flow patterns in different flow situations. It is defined as the ratio of momentum forces to viscous forces and consequently quantifies the relative importance of these two type of forces for given flow conditions.
  44. 44. Dimensionless group  A example of dimensionless quantities is the Reynolds number (Re)  Re = ρvD/µ  Where ρ=fluid density  v=fluid velocity  D=pipe diameter  µ=fluid viscosity  After substitution of units in the equation, it becomes clear that the Reynolds number is dimensionless group  Re=(kg/m3)(m/s)(m) kg/m.s
  45. 45. Calculating Reynolds Number  A useful dimensionless number called the Reynold’s number is DUρ/µ where  D= diameter or length  U= velocity  ρ= Fluid density  µ= fluid viscosity  Calculate the Reynold’s number for the following case  D= 2 inch  U= 10ft/s  ρ= 62.4 lb/ft3  µ= 0.3 lbm/(hr)(ft)
  46. 46. Calculating Reynolds Number Calculate the Reynold’s number for the following case D= 20 ft U= 10 miles/hr ρ= 1 lbm/ft3 µ= 0.14x10-4 lbm/(hr)(ft)
  47. 47. Prandtl number  The Prandtl number, NPr, is a dimensionless group important in heat transfer calculations. It is defined as Cpµ/k, where Cp is the heat capacity of the fluid, µ is the fluid viscosity, and k is the thermal conductivity.  For a particular fluid,  Cp=0.583 J/(g.oC),  k=0.286 W/(m.oC), and  µ=1936 lbm/(ft.h).  Estimate the value of NPr
  48. 48. Weber number Weber number is a dimensionless number in fluid mechanics that is often useful in analysing fluid flows where there is an interface between two fluids. Mathematically it is expressed as We = ρv2L/σ Where ρ = density of the fluid v = velocity of the fluid l = characteristic length σ = surface tension (N/m)
  49. 49. Weber number Calculate Weber number from the given data Where ρ = 0.5 kg/m3 v = 10 m/sec l = 7 m σ = 25 (N/m)
  50. 50. Chapter 3 Process and process variables
  51. 51. Chapter 3  Process and Process Variables  Mass and volume  Density  Specific volume  Specific gravity  Flow rate  Mass flow rate  Volumetric flow rate  Chemical Composition  Moles and molecular weight  Mass and mole fractions and average molecular weight  Pressure  Atmospheric pressure, absolute pressure and guage pressure  Pressure measuring devices  Differential pressure  Temperature  Temperature conversion
  52. 52. Process and Process Variables Processes and process variables A process is any operation or series of operations by which particular objective is accomplished. The materials that enters a process is referred to as the input or feed and that which leaves is the output or product.
  53. 53. Process Flow Diagram for ammonia solvey
  54. 54. Process and Process Variables Processes consist of multiple steps, each of which is carried out in a process unit, and each process unit has associated with it a set of input and output process streams. As a chemical engineer you would be called upon to design or operate a process. Design includes formation of process flow sheets as well as specification of individual process units. Operation involves day to day running of the process.
  55. 55. Mass and volume The density of a substance is the mass per unit volume of the substance (kg/m3, g/cm3, lbm/ft3). The specific volume of substance is the volume occupied by a unit mass of the substance and is the inverse of density. (m3/kg,cm3/g,ft3/lbm) Densities of pure solids and liquids are independent of pressure and slightly vary with temperature. Density of water increases from 0.999868 g/cm3 at OoC to 1.00000 g/cm3 at 3.98 oC.
  56. 56. Mass and volume The density of a substance can be used as a conversion factor to relate the mass and the volume of a quantity of the substance. ρ = m/v Specific Gravity The specific gravity of a substance is the ratio of the density ρ of the substance to the density ρref of a substance as a specific condition. SG = ρ/ ρref
  57. 57. Specific gravity  The reference most commonly used for solids and liquids is water at 4.0oC, which has the following density  ρH2O (4oC) = 1.000 g/cm3 = 1000 kg/m3 = 62.43 lbm/ft3  If you are given the specific gravity of a substance, multiply it by the reference density in any units to get the density of the substance in the same units. For example if the specific gravity of a liquid is 2.00, its density is 2.00 x 103 kg/m3 or 2.00 g/cm3.  Special density units called degree Baume (oBe), degrees API (0API) and degrees Twaddell (oTw) are used in the petroleum industry.
  58. 58. Calculations based on specific gravity A liquid has a specific gravity of 0.50. what is its density in g/cm3? what is its specific volume in cm3/g? what is its density in lbm/ft3? What is the mass of 3.0cm3 of this liquid? What volume is occupied by 18 g of this liquid? The specific gravity of gasoline is approximately 0.70 Determine the mass (kg) of 50 litres of gasoline.
  59. 59. Specific gravity At 25oC, an aqueous solution containing 35 wt% H2SO4 has a specific gravity of 1.2563. a quantity of the 35% solution is needed that contains 195.5 kg H2SO4. Calculate the required volume (L) of the solution using the specific gravity.
  60. 60. Flow rate Mass and volumetric flow rate Most processes involve the movement of material from one point to another sometimes between process units, sometimes between a production facility and a transportation depot. The rate at which a material is transported through a process line is the flow rate of that material.
  61. 61. Flow rate The flow rate of a process stream may be expressed as a mass flow rate (mass/time) or as a volumetric flow rate (volume/time). Suppose a fluid (gas or a liquid) flows in the cylindrical pipe as shown below.
  62. 62. Flow rate The shaded areas represent a section perpendicular to the direction of flow. If the mass flow rate of the fluid is m (kg/s), then every second m kilograms of the fluid passes through the cross section. If the volumetric flow rate of the fluid at the given cross section is V (m3/s), then every second V cubic meters of the fluid pass through the cross section.
  63. 63. Flow rate However, the mass m and the volume of a fluid, in this case, the fluid passes through the cross section each second. Are not independent quantities but are related though the fluid density, ρ ρ = m/V = m./V. Thus the density of the fluid can be used to convert a known volumetric flow rate of a process stream to the mass flow rate of that stream or vice versa.
  64. 64. Mass and volumetric flow rate?  The mass flow rate of n-hexane (ρ = 0.659 g/cm3) in a pipe is 6.59 g/sec. What is the volumetric flow rate of the hexane?  The volumetric flow rate of CCl4 (ρ = 1.595 g/cm3) in a pipe is 100.0 cm3/min. What is the mass flow rate of the CCl4?  Liquid benzene (sp.gravity = 0.879) and liquid n-hexane (sp.gravity = 0.659) are blended to form a stream flowing at a rate of 700 lbm/h. Stream has a Specific Gravity of 0.850. Estimate the mass and volumetric feed rates of the two hydrocarbons to the mixing vessel (in American Engineering units).
  65. 65. Flow rate A flow meter is a device mounted in a process line that provides a continuous reading of the flow rate in the line. Two commonly used flow meters are Rotameter Orifice meter
  66. 66. Flow measuring devices • Tapered vertical tube • Contains a float • Larger flow rate will result in high rise of the rotameter.
  67. 67. Flow measuring devices The orifice meter is an obstruction in the flow channel with a narrow opening through which the fluid passes. The fluid pressure drop (decreases) from the upstream side of the orifice to the downstream side.
  68. 68. Moles and the molecular weight The atomic weight of an element is the mass of an atom on a scale that assigns 12C (the isotope of carbon whose nucleus contains six protons and six neutrons) a mass of exactly 12. The molecular weight of a compound is the sum of the atomic weights of the atoms that constitute a molecule of a compound. For instance, atomic oxygen has atomic weight of approximately 16, therefore molecular oxygen has a molecular weight of 32.
  69. 69. Moles and the molecular weight A gram mole of a species is the amount of that species whose mass in grams is numerically equal to its molecular weight. Other type of moles e.g. kg moles, lb moles are similarly defined. Carbon monoxide (CO) has a molecular weight of 28 so 1 mol of CO therefore contains 28 grams, 1 lb mole consists of 28 lbm.
  70. 70. Moles and the molecular weight How much mole 34 gram of ammonia contain? And 4.0 lb moles of ammonia? One gram mole of any species contains approximately 6.02 x 1023 (Avogadro’s number) molecules of that species.
  71. 71. Moles  How many of the following are found in 15.0 kmol of Benzene (C6H6)?  Kg C6H6  mol C6H6  lb mol C6H6  mol C  mol H  gram C  gram H  Molecules of C6H6
  72. 72. Conversion between mass and moles How many of each of the following are contained in 100.0 g of CO2 mol CO2 lb moles CO2 mol C mol O mole O2 molecules of CO2
  73. 73. Conversion between mass and moles The molecular weight of a species can be used to relate the mass flow rate of a continuous stream of this species to the corresponding molar flow rate. For example if CO2 flows through a pipe line at a rate of 100 kg/hr, the molar flow rate of the CO2 is If the output stream from a chemical reactor contains CO2 flowing at a rate of 850 lb-moles/min, the corresponding mass flow rate is
  74. 74. Mass and mole fractions and average molecular weight Process streams occasionally contains one substance, but more often they consists of mixtures of liquids or gases, or solutions of one or more than one solute in a liquid solvent. The following terms may be used to define the composition of a mixture of substances, including a species A. Mass fraction: xA = mass of A/total mass Mole fraction: yA = moles of A/total moles The percent by mass of A is 100xA, and the mole percent of A is 100yA
  75. 75. Conversion using mass and mole fractions  Conversion using mass and mole fractions  A solution contains 15% A by mass (xA = 0.15) and 20% mole B (yB = 0.20)  Calculate the mass of A in 175 kg of the solution  Calculate the mass flow rate of A in a stream of solution flowing at a rate of 53 lbm/h.  Calculate the molar flow rate of B in a stream flowing at a rate of 1000 mol/min.  Calculate the total solution flow rate that corresponds to a molar flow rate of 28 kmol B/s.  Calculate the mass of solution that contains 300 lbm A.
  76. 76. Calculating molar composition O2 16% CO 4.0% CO2 17% N2 63%  What is the molar composition?
  77. 77. Conversion using mass and mole fractions The numerical value of a mass or a mole fraction does not depend on the mass units in the numerator and denominator as long as these units are same. If the mass fraction of benzene (C6H6) in a mixture is 0.25, then xC6H6 equals 0.25 kg C6H6/kg total, 0.25 g C6H6/g total, 0.25 lbm C6H6/lbm total, and so on.
  78. 78. Average molecular weight The average molecular weight of a mixture, M is the ratio of the mass of a sample of the mixture (mt) to the number of moles of all species (nt) in the sample. If yi is the mole fraction of the ith component of the mixture and Mi is the molecular weight of this component, then M = y1M1 + y2M2 + y3M3 +……..
  79. 79. Average molecular weight Calculate the average molecular weight of air From its approximate molar composition of 79% N2, 21% O2 and From its approximate composition by mass of 76.7% N2, 23.3 % O2.
  80. 80. Test Yourself If 100 lbm/min of A (MWA=2) and 300 lbm/min of B (MWB=3) flow through a pipe, what are the Mass fractions of A and B? Mole fractions of A and B? The molar flow rate of A? The molar flow rate of B? The total mass and molar flow rate of the mixture?
  81. 81. Mass concentration The mass concentration of a component of a mixture or solution is the mass of this component per unit volume of the mixture (g/cm3, lbm/ft3, kg/in3). The molar concentration of a component is the number of moles of the component per unit volume of the mixture (kmol/m3, lb-mol/ft3). The molarity of the solution is the value of the molar concentration of the solute expressed in grams- moles solute/litre solution (e.g., a 2 molar solution of A contains 2 mol A/litre solution).
  82. 82. Mass concentration The concentration of a substance in a mixture or solution can be used as a conversion factor to relate the mass (or moles) of a component in a sample of the mixture to the sample volume, or to relate the mass (or molar) flow rate of a component of a continuous stream to the total volumetric flow rate of the stream.
  83. 83. Conversion between mass, molar, and volumetric flow rates of a solution 0.50 molar aqueous solution of sulphuric acid flows into process unit at a rate 0f 1.25 m3/min. the specific gravity of the solution is 1.03. Calculate The mass concentration of H2SO4 in kg/m3 The mass flow rate of H2SO4 in kg/s and The mass fraction of H2SO4.
  84. 84. Problem 3.28 Page 71 Felder A 5.00 wt% aqueous sulphuric aid solution (ρ=1.03 g/mL) flows through a 45 m long pipe with a 6.0 cm diameter at a rate of 87 L/min. What is the molarity of sulphuric acid in the solution? How long (in seconds) would it take to fill a 55 gallon drum?
  85. 85. Pressure fluid and hydrostatic pressure A pressure is the ratio of a force to the area on which force acts. Accordingly, pressure units are force units divided by area units (e.g. N/m2, dynes/cm2 and lbf/in2). The SI unit of pressure is called a Pascal (Pa)
  86. 86. Different units for measuring pressure 1 atm = 1.01325 x 105 N/m2 (Pa) = 101.325kPa = 1.013 bar = 1.01325 x 106 dynes/cm2 = 14.7 psi (lbf/in2) = 760 mmHg = 29.92 inHg = 33.1 ft H2O = 10.33 m H2O
  87. 87. Units of Pressure in Different System of Units SI CGS AE Pascals 1 Pa= 1 N(kg.m /sec2)/ m2 Dynes 1 Dyne = 1 g.cm/sec2. Psi (Pound per Square inch) 1 psi=1 lbf/in2
  88. 88. Problem 3.32 Page 72 Felder Perform the following pressure conversions, assuming atmospheric pressure is 1 atm. 2600 mm Hg to psi. 275 ft H20 to kPa 280 cm Hg to dyne/m2 300 psi to ft H20 1000 bar to psi 101 kPa to atm
  89. 89. Pressure fluid and hydrostatic pressure  Let us introduce an additional definition of fluid pressure to explain the concept of atmospheric pressure and guage pressure.  Suppose a vertical column of fluid is h (m) high having a uniform cross sectional area A (m2).  Fluid has a density of ρ (kg/m3).  There would be two kinds of pressure being exerted on the column. One at the top on the surface of the column (Po) and the other at the base of the column.  Pressure at the base of the column is known as hydrostatic pressure.  Hence total pressure being exerted on the column will be  P = Po + ρgh
  90. 90. Pressure fluid and hydrostatic pressure
  91. 91. Pressure fluid and hydrostatic pressure  Pressure may also be expressed as a head of a particular fluid that is the height of a hypothetical column of the fluid that would exert the given pressure at its base if the pressure at the top were zero.  Mathematically  P = ρfluid g Ph  (head of the fluid)  Calculation of a pressure as a head of fluid  Express a pressure of 2.00 x 105 Pa in terms of mmHg.
  92. 92. Atmospheric pressure, absolute pressure and guage pressure  The pressure of the atmosphere can be thought of as the pressure at the base of a column of fluid (air) located at the point of measurement i.e. at sea level.  A typical value of the atmospheric pressure at sea level is 760.0 mm Hg has been designated as a standard pressure of 1 atmosphere.  The relationship for converting between absolute pressure and guage pressure is Pabsolute = Pguage + Patmospheric  A guage pressure of zero indicates that the absolute pressure of the fluid is equal to atmospheric pressure.  The abbreviations psia and psig are commonly used to denote absolute pressure and guage pressure lbf/in2.
  93. 93. Fluid pressure measurement The most common mechanical device used for pressure measurement is Bourden guage. Hollow tube closed at one end. Bent into a C configuration. Open end of the tube is exposed to the fluid whose pressure is to be measured. As the pressure increases the tube tends to straighten, causing a pointer attached to the tube to rotate.
  94. 94. Pressure Measuring Devices Measuring range: nearly perfect vacuum to about 7000 atm.
  95. 95. Manometer U shape tube partially filled with a fluid of known density (manometer fluid). Exposed to different pressure, the field level drop in the high pressure arm and rises in low pressure arm. The difference in pressure can be calculated by measuring the difference between the liquid level in each arm.
  96. 96. Manometer
  97. 97. Manometer Figure A shows and open end manometer whose one end is exposed to the atmosphere and the other one to a fluid whose pressure is to be measured. Figure B shows a differential manometer which is used to measure the pressure difference between two points in a process line. Figure C shows a seal end manometer which has a near vacuum enclosed at one end.
  98. 98. Manometer
  99. 99. Differential Pressure The formula that relates the pressure difference P1– P2 to the difference in manometer fluid level is based on the principle that the fluid pressure must be same at any two points at the same height in a continuous fluid. General manometer equation P1 + ρ1gd1 = P2 + ρ2gd2 + ρfgh Differential manometer equation P1 – P2 = (ρf – ρ) gh
  100. 100. Differential Pressure If either fluid 1 or 2 is a gas at moderate pressure the density of this fluid is 100 to 1000 times lower than the density of the manometer fluid, hence the term would be P1–P2 = ρf gh Manometer formula for gases P1 – P2 = h
  101. 101. Differential pressure measurement
  102. 102. Pressure measurement with manometers A differential manometer is used to measure the drop in pressure between two points in a process line containing water. The specific gravity of the manometer fluid is 1.05. The measured levels in each arm are shown below. Calculate the pressure drop between points 1 and 2 in dynes/cm2?
  103. 103. Pressure measurement with manometers
  104. 104. Calculation of Pressure Differences In measuring the flow of fluid in a pipeline as shown in the figure, a differential manometer was used to determine the pressure difference across the orifice plate. The flow rate was to be calibrated with observed pressure drop (difference). Calculate the pressure drop P1-P2 in pascals for the manometer reading in figure.
  105. 105. Calculation of pressure differences
  106. 106. Calculation of pressure (Problem 3.44 Page 75 Felder)  An open end manometer is connected to a low pressure pipeline that supplies a gas to a laboratory. Because paint was spilled on the arms connected to the line during a laboratory renovation, it is impossible to see the level of the manometer fluid in this arm. During a period when the gas supply is connected to the line but there is no gas flow, a Bourden guage is connected to the line downstream from the manometer gives a reading of 7.5 psig. The level of the mercury in the open arm is 900 mm above the lowest part of the manometer.  When the gas is not flowing, the pressure is same everywhere in the pipe. How high above the bottom of the manometer would the mercury be in the arm connected to the pipe?  When gas is flowing, the mercury level in the visible arm drops by 25 mm. what is the gas pressure (psig) at this moment?
  107. 107. Calculation of pressure
  108. 108. Temperature The temperature of a substance is the measure of average kinetic energy possessed by the substance molecule. Such energy cannot be measured. Hence indirect methods are employed by measuring some physical property of the substance whose value depend on temperature in a known manner.
  109. 109. Temperature Resistant thermometer, thermocouple, pyrometer and thermometer are commonly used devices for measuring the temperature of a substance. A defined temperature scale is obtained by assigning numerical values to reproducibly measureable temperatures, for example, assign a value of zero to the freezing point of water and a value of 100 to the boiling of water at 1 atm.
  110. 110. Temperature measuring devices Resistance Thermometer Thermocouple Pyrometer
  111. 111. Temperature  Celsius scale or centigrade scale: Tf is assigned a value of OoC and Tb is assigned a value of 100oC. Absolute zero is the lowest theoretically attainable temperature which falls at -273.15o C.  Fahrenheit scale: Tf is assigned a value of 32oF, and Tb is assigned a value of 212oF. Absolute zero falls at -459.67oF.  The Kelvin and Rankine scale are defined such that absolute zero has a value of 0 and the size of a degree is the same as a Celsius (Kelvin scale) or a Fahrenheit degree (Rankine scale).
  112. 112. Temperature  The following relationship may be used to convert a temperature expressed in one defined scale unit to its equivalent in another:  T (K) = T (oC) + 273.15  T (oR) = T (oF) + 459.67  T (oR) = 1.8T (K)  T (oF) = 1.8T (oC) + 32  The conversion factors refer to temperature intervals, not temperatures.  Temperature and temperature intervals.
  113. 113. Conversions oC oF K oR -40 77.0 698 69.8
  114. 114. Chapter 4 Material Balance
  115. 115. Topics on material balance Material Balance on non-reactive systems. Material Balance on multiple unit processes. Recycle and bypass Material balance on reactive systems.
  116. 116. What to learn in this chapter? Learn about some basic definitions to be used while solving material balance problems Solve material balance problems for continuous/transient and steady state/unsteady systems.
  117. 117. Material Balance A material balance is the application of law of conservation of mass.
  118. 118. What is system? Any arbitrary portion of or a whole process which is under consideration for analysis. A system can be defined as a reactor Or a section of pipe or an entire refinery by stating in words what the system is.
  119. 119. Material balance Process classification Chemical processes may be classified as Open/close system Continuous Batch Semi batch
  120. 120. Open and closed system  Open system  When material crosses the boundary of a system it is said to be a closed system.
  121. 121. Open and closed system Closed system  When no material crosses the boundary of a system it is said to be a closed system.
  122. 122. Steady and unsteady state system  If the values of all variables in a process such as temperature, pressure, volume and flow does not change with time the process is said to be operating at steady state.  If any of the process variables change with time the system is said to be transient or un steady state operation.  By nature batch and semi batch processes are unsteady state operations.  Continuous processes may be either steady state or transient.
  123. 123. Open steady state system
  124. 124. Process classification Continuous process The inputs and the outputs flow continuously throughout the duration of the process. Pump a mixture of liquids into a distillation column at a constant rate and steadily withdraw product streams from the top and the bottom of the column.
  125. 125. Process classification Batch process The feed is charge into the vessel at the beginning of the process. Vessel contents are removed sometimes later. No mass crosses the system boundary between the time the feed is charged and at the time the product is removed. Rapidly add reactants to a tank and remove the products sometime later.
  126. 126. Process classification Semi batch process Any process that is neither batch nor continuous Allow the contents of a pressurized gas container to escape to the atmosphere and slowly blend several liquids in a tank from which nothing is being withdrawn.
  127. 127. Process classification Classify the following processes as batch, continuous, or semi batch, and transient or steady state. A balloon is filled with air at a steady state of 2 g/min. A bottle of milk is taken from the refrigerator and left on the kitchen table. Water is boiled in a open flask.
  128. 128. Balances The general mass balance equation A balance on a conserved quantity (total mass, mass of a particular species, energy, momentum) in a system may be written in the following general way Input + generation – output – consumption = accumulation Input (enters through the system boundaries) Output (leaves through the system boundaries)
  129. 129. The general energy balance equation Generation (produced within the system) Consumption (consumed within the system) Simplification of general mass balance equation
  130. 130. Balances on continuous steady state process For continuous processes at a steady state, the accumulation term in the general mass balance equation will be Input + generation = output + consumption Furthermore If process is being carried out in the absence of any chemical reaction the above equation further reduces to Input = output
  131. 131. Differential and integral balance  Differential balance  Indicates what is happening in a system at an instant in time.  Each term of the balance equation is rate (rate of input kg/hr, rate of output, rate of general generation)  Usually applied to a continuous process.  Integral balance  Describes what happens between two instants of time.  This type of balance is usually applied to a batch process.
  132. 132. Material balance calculations  The catalytic dehydrogenation of propane is carried out in a continuous packed bed reactor. One thousand kilograms per hour of pure propane is preheated to a temperature of 670OC before it passes into the reactor. The reactor effluent gas, which includes propane, propylene, methane, and hydrogen is cooled from 800 OC to 110 OC and fed to an absorption tower, where the propylene and propane are dissolved in oil. The oil then goes to a stripping tower in which it is heated, releasing the dissolved gases, these gases are recompressed and sent to a distillation column in which the propane and propylene are separated. The propane stream is recycled back to join the feed to the reactor preheater. The product stream from the distillation column contains 98% propylene, and the recycle stream is 97% propane. The stripped oil is recycled to the absorption tower.
  133. 133. Flow charts When you are given process information like this and ask to determine something about the process, it is essential to organize the information in a way that is convenient for subsequent calculations.
  134. 134. A General strategy For Solving Material Balance Problems  Read and understand the problem statement.  Draw a sketch of the process and specify the system boundary.  Place labels (symbols, numbers, and units) on the diagram for all of the known flows, materials, and compositions.  Obtain any data you know are needed to solve the problem but are missing.  Choose a basis.  Determine the number of variables whose values are unknown.  Determine the number of independent equations and carry out degree of freedom of analysis.  Write down the equations to be solved in terms of the known and unknowns.  Solve the equations and calculate the quantities asked for in the problem.  Check your answers.
  135. 135. Material Balance on a continuous distillation column A continuous mixer mixes NaOH with H2O to produce an aqueous solution of NaOH. The problem is to determine the composition and flow rate of the product if the flow rate of NaOH is 1000 kg/hr and the ratio of the flow rate of the H2O to the product solution is 0.9. Draw a sketch of the process and put the data and unknown variables on the sketch with appropriate labels.
  136. 136. Flow chart  Draw flow chart and fill in all known variables, including the basis of calculation. Then label unknown stream variables.  A flow chart is a type of diagram that represents a process.  The total mass or mass flow rate and the mass fraction of all stream components.  The total moles or molar flow rate and the mole fractions of all stream components.  For each stream component, the mass, mass flow rate, moles, or molar flow rate.  Mole balance, mass balance and volume balance.
  137. 137. Material Balance on a continuous distillation column  One thousand kilograms per hour of a mixture of Benzene (B) and Toluene (T) containing 50% benzene by mass is separated in a distillation column into two fractions.  The mass flow rate of the benzene in the top stream is 450 kg B/hr and that of toluene in bottom stream is 475 kg T/hr.  The operation is steady state.  Write balances on benzene and toluene to calculate the unknown component flow rates in the output streams.
  138. 138. Flow Chart for the process
  139. 139. Marking of unknown variables Express what the problem statement asks you to determine in terms of the labelled variables. Mark the unknowns. If you are given mixed mass and mole units for a stream covert all quantities to one basis or the other.
  140. 140. Choosing a basis for the process  A basis of calculation is an amount (mass or moles) or flow rate (mass or molar) of one stream or stream component in a process.  Choose as a basis of calculation an amount or flow rate of one of the process streams.  If an amount or flow rate of a stream is given it can be conveniently used as a basis.  If several stream amounts or flow rates are given, use them collectively as a basis.  If no stream amount or flow rate is specified in the problem statement, take as a basis an arbitrary amount or flow rate with known composition.  If data related to product stream is given and you are asked to analyse the feed stream then you must take the flow rate of product stream as basis.
  141. 141. Degree of Freedom of Analysis Degree of freedom of analysis D.O.F = Number of unknowns – Number of independent equations. Write equations in an efficient order and circle the variables for which you will solve. Solve the equations.
  142. 142. Degree of freedom of analysis  In order to solve a problem the number of unknown variables must be equal to the number of independent equations
  143. 143. Degree of freedom of analysis Concept of independent equations! 3x1 + 4x2 = 0 6x1 + 8x2 = 0 Are the above equations independent of each other? If yes, find the value of x1 and x2 !!! How many independent equations we can formulate from the given problem?
  144. 144. Degree of freedom analysis Degree of freedom = number of unknowns – number of independent equations ndf = nunknowns – nindep eqns If ndf = O the problem can be solved. Ndf > O is greater than NE underspecified more independent equations are required. If ndf < O, there are more independent equations than unknown.
  145. 145. DOF  Sources of equation relating to unknown process stream variables include the following  Material balances  For a non reactive process no more than nms independent material balances may be written, where nms is the number of molecular species.  If Water and NaCl are the species in the stream entering and leaving the mixing unit you could write balances on Water, NaCl, total mass, atomic Sodium, atomic hydrogen, atomic chlorine but only two of them will be independent.
  146. 146. DOF Process specifications (Explicit Relation) The problem statement may specify how several process variables are related. For example you have been told that the ratio of the flow rate of the H2O to the product solution is 0.9. Hence an independent equation will be then W/P=0.9 Physical constraints (Implicit Relation) If the mole fractions of the three components of a stream are labelled as xA,xB,and xC, then the relation among these variables is xA+xB+xC =1
  147. 147. Writing down of equations What would be the material balance equations for? Water NaCl
  148. 148. Writing down of equations What would be the material balance equations for? Benzene Toluene
  149. 149. Summary  Read and understand the problem statement  Draw a sketch and specify the system boundary.  Place labels for unknown variables and values for known variables on the sketch.  Choose a basis  Determine the number of unknowns  Determine the number of independent equations and carry out a degree of freedom analysis  Write down the equations to be solved  Check your answers……
  150. 150. Material balance over a condensation process  (From Elementary Principles of chemical Processes 3rd Ed. Chapter 4 Page 100-101)  A stream of humid air enters a condenser in which 95% of the water vapour in the air is condensed.  The flow rate of the condensate (the liquid leaving the condenser) is measured and found to be 225 L/hr.  Dry air may be taken to contain 21 mole% oxygen, with the balance nitrogen.  Calculate the flow rate of the gas stream leaving the condenser and the mole fractions of oxygen, nitrogen, and water in this stream.
  151. 151. Material balance over a condensation process
  152. 152. Material balance over a condensation process Suppose now we have been given an additional piece of information that the entering air contains 10.0 mole% water. The flow chart would then appear as
  153. 153. Material balance over a condensation process
  154. 154. Mass balance on a distillation column  (From Elementary Principles of chemical Processes 3rd Ed. Chapter 4 Page 156 Problem 4.3)  A liquid mixture of benzene and toluene contains 55% benzene by mass.  The mixture is to be partially evaporated to yield a vapour containing 85% benzene and  residual liquid 10.6% benzene by mass.  Suppose the process is to be carried out continuously and at steady state, with a feed rate of 100.0 kg/hr of the 55% mixture.  Let mv (kg/hr) and mL (kg/hr) be the mass flow rates of the vapour and liquid product streams, respectively.  Draw and label a process flow chart, then write and solve balances on total mass and on benzene to determine the expected values of mv and mL.  For each balance state which terms of the general balance equation (accumulation = input + generation – output –consumption) you discarded and why you discarded them?
  155. 155. Flow Chart for the process
  156. 156. Mixing process  (From Elementary Principles of chemical Processes 3rd Ed. Chapter 4 Page 158 problem 4.10)  Three hundred gallons of a mixture containing 75 wt% ethanol (ethyl alcohol) and 25% water (mixture specific gravity = 0.877) and  a quantity of a 40 wt% ethanol 60% water mixture (SG = 0.952) are  blended to produce a mixture containing 60 wt% ethanol.  The object of this problem is to determine V40, the required volume of the 40% mixture.  Draw and label a flow chart of the mixing process and do the degree of freedom of analysis.  Calculate V40
  157. 157. Flow Chart for the process
  158. 158. Material balance on unsteady state system  (From Basic Principles and Calculations in Chemical Engineering 7th Ed. Chapter 8 Page 205 Example 8.4)  You are asked to prepare a batch of 18.63% battery acid as follows.  A tank of old week battery acid (H2SO4) solution contains 12.43% H2SO4 (the remainder is pure water).  If 200 kg of 77.7% H2SO4 is added to the tank, and the final solution is to be 18.63% H2SO4, how many kilograms of battery acid have been used.
  159. 159. Balance on a batch process  (From Basic Principles and Calculations in Chemical Engineering 7th Ed. Chapter 8 Page 222 Problem 8.18)  The solubility of barium nitrate at 100oC is 34 g/100 g of H2O and at OoC is 5.0/100 gram of H2O.  If you start with 100 g of Ba(NO3)2 and make a saturated solution in water at 100oC, how much water is required? If the saturated solution is cooled to 0oC, how much Ba(NO3)2 is precipitated out of the solution?  The precipitated crystals carry along with them on their surface 4g of H2O per 100 g of crystals.
  160. 160. Balance on a batch mixing process Two methanol water mixtures are contained in separate flasks. The first mixture contains 40.0 wt% methanol, and the second contains 70.0% methanol. If 200 g of the first mixture is combined with 150 g of the second, what is the mass and composition of the product?
  161. 161. Material balance on a distillation column (4.3.5)  (From Elementary Principles of chemical Processes 3rd Ed. Chapter 4 Page 102- 104)  A liquid mixture containing 45% benzene (B) and 55% Toluene (T) by mass is fed into a distillation column.  A product stream leaving the top of the column (the overhead product) contains 95 mole% B, and a bottom product stream contains 8% of the benzene fed to the column (meaning that 92% of the benzene leaves with the overhead product).  The volumetric flow rate of the feed stream is 2000 L/h and the specific gravity of the mixture is 0.872.  Determine the mass flow rate of the overhead product stream and the mass flow rate and composition (mass fractions) of the bottom product streams.
  162. 162. Balance on a distillation column  A distillation column is a process unit in which a feed mixture is separated by multiple partial vaporizations and condensations to form two or more product streams.  The overhead product stream is rich in volatile components of the feed mixture (the ones that vaporize most readily), and the bottom product stream is rich in the least volatile components.  The following flowchart shows a distillation column with two feeds and three product streams.
  163. 163. Balance on a distillation column How many independent material balances may be written for this system? How many of the unknown flow rates and/or mole fractions must be added before the others may be calculated? You may choose values of m1 and x2 at your own will. Once you have selected the values you may solve the problem following the steps you have provided with.
  164. 164. Balances on multiple unit processes Industrial chemical processes rarely involve just one unit process. One or more chemical reactors are often present as are units for mixing reactants. blending products. heating and cooling process streams.
  165. 165. Ammonia Solvey Process
  166. 166. Multi unit processes  The procedure for solving material balance problems for multiple systems is same as used earlier.  With multi unit processes we have to isolate and write balances for several subsystems.  Degree of freedom of analysis is not only done on the whole system but also on subsystems.  Carry out degree of freedom of analysis on every system.
  167. 167. Balances on multiple unit processes
  168. 168. Balances on multiple unit processes  Elementary Principles of Chemical Processes chapter 4 page 105 Example 4.4.1  A labelled flowchart of a continuous steady state process is shown in the figure.  Each stream contains two components, A and B, in different proportions.  Three streams whose flow rates and/or compositions are not known are labelled as 1,2 and 3.  Calculate the unknown flow rates and compositions of streams 1,2 and 3.
  169. 169. Balances on multiple unit processes
  170. 170. Material Balance on multi stage distillation column  Elementary Principles of Chemical Engineering page 165 problem 4.29  A liquid mixture containing 30.0 mole% benzene (B), 25.0% Toluene (T), and the balance Xylene (X) is fed to a distillation column.  The bottom product contains 98.0 mole% X and no B, and 96.0% of the X in the feed is recovered in this stream.  The overhead product is fed to the second column.  The overhead product of the second column contains 97.0 % of the B in the feed to this column. The composition of this stream is 94.0 mole % B and the balance T.
  171. 171. Material Balance on multi stage distillation column  Draw and label a flowchart of this process and do the degree of freedom of analysis to prove that for an assumed basis of calculation, molar flow rates and composition of all process streams can be calculated from the given information.  Write in order the equations you would solve to calculate unknown process variables.  Calculate the  percentage of the benzene in the process feed (i.e. the feed to the first column) that emerges in the overhead product from the second column and  the percentage of toluene in the process feed that emerges in the bottom product from the second column.
  172. 172. Material balance for multiple units in which no reaction occurs  Himmelblau chapter 11 page 315 example 11.2 Acetone is used in the manufacturing of many chemicals and also as a solvent. In its latter role, many restrictions are placed on the release of acetone vapour to the environment. You are asked to design an acetone recovery system having the flow sheet illustrated in the figure. All the concentrations are in weight percent. Calculate A,F,W,B and D per hour.
  173. 173. Recycle and Bypass  It is rare that a chemical reaction A to B proceeds to completion in a reactor.  Hence some of the reactant appears with the product.  We can find a way to separate most or all of the unconsumed reactant from the product stream and recycle the unconsumed reactant back to the reactor.  Although we have to pay for the recycling equipment but that can be compensated by purchasing less fresh reactant and being able to sell the purified product at a higher price.
  174. 174. Recycle • A labelled flowchart of a chemical process involving reaction, product separation and recycle is shown in the figure below.
  175. 175. Advantages of recycling ➢Recovery of catalyst ➢Recovering catalyst from product stream. ➢Dilution of a process stream ➢Control of a process variable ➢Circulation of a working fluid
  176. 176. Reference to the problem  Elementary principles of chemical processes by felder page 166 problem no. 4.31
  177. 177. Material balance on a recycling distillation column Problem 4.31 Page 166  An equimolar liquid mixture of benzene and toluene is separated into two product streams by distillation.  A process flowchart and a somewhat oversimplified description of what happens in the process follow  Inside the column a liquid stream flows downward and a vapour stream rises.  At each point in the column some of the liquid vaporizes and some of the vapour condenses.  The vapour leaving the top of the column, which contains 97 mole% benzene is completely condensed and split into two equal fractions one is taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column.  The overhead product stream contains 89.2% of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which 45% of it is vaporized.  The vapour generated in the reboiler (the boil up) is recycled to become the rising vapour stream in the column, and the residual reboiler is taken off as the bottom product stream. The composition of the streams leaving the reboiler are governed by the relation.
  178. 178. Material balance on a recycling distillation column
  179. 179. Material balance on a recycling distillation column  yb/ (1-yb) = 2.25 xb/(1-xb)  Where yb and xb are the mole fractions of benzene in the vapour and liquid streams, respectively.  Take a basis of 100 mol fed to the column, draw and completely label a flow chart, and for each of four systems (overall process, column, condenser and reboiler), do the degree of freedom of analysis and identify a system with which the process analysis might appropriately begin (one with zero degree of freedom).  Write in order the equations you would solve to determine all unknown variables on the flow chart, circling the variable for which you would solve in each equation.  Calculate the molar amounts of the overhead product and bottom products, the mole fraction of benzene in the bottom product, and the percentage recovery of toluene in the bottoms product (100 x moles toluene in bottoms/mole of toluene in feed).
  180. 180. An evaporative crystallization process  Elementary principles of chemical engineering 3rd edition chapter 4 topic 4.5 recycle and bypass problem 4.5.2 (example) page 112  The flow chart of a steady state process to recover crystalline potassium chromate (K2CrO4) from an aqueous solution of this salt is shown in the figure.  Forty five hundred kilograms per hour of a solution that is one third K2CrO7 by mass is joined by a recycle stream containing 36.4% K2CrO7 and the combined stream is fed into an evaporator.  The concentrated stream leaving the evaporator contains 49.4% potassium chromate, and this stream is fed into a crystallizer in which it is cooled (causing crystals of potassium chromate to come out of solution) and then filtered.  The filter cake consist of K2CrO7 crystals and a solution that contains 36.4% K2CrO7 by mass; the crystals account for 95% of the total mass of the filter cake.  The solution that passes through the filter, also 36.4% K2CrO7 is the recycle stream.  Calculate the rate of evaporation, the rate of production of crystalline K2CrO7, the feed rates that the evaporator and the crystallizer must be designed to handle, and the recycle ratio (mass of recycle/mass of fresh feed).  Suppose that the filtrate were discarded instead of being recycled. Calculate the production rate of crystals. What are the benefits and costs of recycling?
  181. 181. Bypass Stream A stream that skips one or more stages of the process and goes directly to another downstream stage is known as bypass stream. Can be rendered as opposite of recycle stream. Control the composition of a final exit stream.
  182. 182. Bypass stream
  183. 183. Material balance on a bypass stream  Reference to the problem (elementary principles of chemical processes) by felder page 167 problem no. 4.32  Fresh orange juices contains 12 wt% solids and the balance water, and concentrated orange juice contains 42 wt% solids.  Initially a single evaporation process was used for the concentration, but volatile constituents of the juice escaped with water, leaving the concentrate with flat taste, the current process overcomes this by passing the evaporator with a fraction of the fresh juice.  The juice that enters the evaporator is concentrated to 58% weight solids and the evaporator product stream is mixed with the bypassed fresh juice to achieve the desired final concentration.  Draw and label a flowchart for this process, neglecting the vaporization of everything in the juice but water.
  184. 184. Material balance on a bypass stream  First prove that the subsystem containing the point where the bypass splits off from the evaporator feed has one degree of freedom.  Then perform degree of freedom of analysis for the whole system, the evaporator and the bypass evaporator product mixing point, and write in order the equations you would solve to determine all unknown stream variables.  Calculate the amount of product (42% concentrate) produced per 100 kg fresh juice fed to the process and the fraction of the feed that bypass the evaporator.
  185. 185. Chemical reaction stoichiometry  The occurrence of a chemical reaction in a process brings several complications into the material balance calculations.  Stoichiometry is the theory of the proportions in which chemical species combine with one another.  The stoichiometric equation of a chemical reaction is a statement of the relative number of molecules or moles of reactants and products that participate in the reaction.
  186. 186. Chemical reaction stoichiometry  For example the stoichiometric equation  2SO2 + O2 2SO3  Indicates that for every 2 molecules of SO2 that react with one molecule of O2 to produce two molecules of SO3.  The numbers that precedes the chemical formulas are known as stoichiometric co-efficients.  A valid stoichiometric equation must be balanced, that is the number of atoms of each atomic species must be same on both sides of the equation since atoms can neither be created nor destroyed.
  187. 187. Chemical reaction stoichiometry  The stoichiometric ratio of two molecular species participating in a reaction is the ratio of their stoichiometric coefficients in the balanced reaction equation.  This ratio can be used to calculate the amount of a particular reactant (or product) that was consumed (or produced).  For the reaction the stoichiometric equations can be written as
  188. 188. Chemical reaction stoichiometry  2 moles of SO3 generated 1 mol of O2 consumed Or  2 lb-moles SO2 consumed 2 lb-mol SO3 generated  and so on  If we know for example that 1600 kg/h of SO3 is to be produced, we can calculate the amount of oxygen required as
  189. 189. Chemical reaction stoichiometry  Consider the reaction  C4H8 + 6O2 4CO2 + 4H2O  Is the stoichiometric equation balanced?  What is the stoichiometric co-efficient of CO2?  How many lb-moles of O2 react to form 400 lb-moles of CO2?  One hundred mol/min of C4H8 is fed into the reactor, and 50% reacts. At what rate is water formed?
  190. 190. Limiting and excess reactants  Two reactants A and B, are said to be present in stoichiometric proportion if the ratio (moles of A present)/(moles of B) equals the stoichiometric ratio obtained from the balanced reaction.  For the reactants in the reaction  2SO2 + O2 2SO3  to be present in stoichiometric proportion, there must be 2 moles of SO2 for every mole of O2 present in the feed to the reactor.
  191. 191. Limiting and excess reactants  If reactants are fed to a chemical reactor in a stoichiometric proportion and the reaction proceeds to completion, all of the reactants are consumed.  In the above reaction, if 200 mol of SO2 and 100 mole of O2 are initially present and the reaction proceeds to completion, the SO2 and O2 would disappear at the same instant.  However if we start wit 100 mol of O2 and less than 200 mol of SO2, SO2 would run out first.
  192. 192. Limiting and excess reactant  The reactant that would run out if a reaction proceeded to completion is called as limiting reactant and the other reactants are termed as excess reactants.  A reactant is limiting if it is present in less than its stoichiometric proportion relative to other reactant.  Suppose (nA)feed is the number of moles of an excess reactant, A, present in the feed to a reactor and that (nA)stioch is the stoichiometric requirement of A, or the amount needed to react completely with the limiting reactant.
  193. 193. Limiting and excess reactant  Then (nA)feed - (nA)stioch is the amount by which the A in the feed exceeds the amount needed to react completely if the reaction goes to completion.  The fractional excess of the reactant is the ratio of the excess to the stoichiometric requirement.  Fraction excess of A = (nA)feed -(nA)stoich (nA)stioch
  194. 194. Limiting and excess reactant  Consider the following reaction in which hydrogenation of acetylene is being carried out  C2H2 + 2H2 C2H6  Suppose 20.0 kmol/h of acetylene and 50.0 kmol/h of hydrogen is fed to the reactor.  The stoichiometric ratio of hydrogen to acetylene is 2:1.  Based on the amount of reactants fed to the reactor hydrogen is the excess reactant! how?
  195. 195. Limiting and excess reactant  Fractional excess of H2 = (50.0-40.0) kmol/h = 0.25 40.0 kmol/h  We say that there is 25% excess hydrogen in the feed.
  196. 196. Fractional conversion  The fractional conversion of a reactant is the ratio f = moles reacted/moles fed  If 100 moles of reactant are fed and 90 moles react, the fractional conversion is 0.90 (90%)  The unreacted fraction would be (1-f) which is this case is 0.10
  197. 197. Fractional conversion Considering the reaction discussed above C2H2 + 2H2 C2H6 Suppose 20 kmol of acetylene, 50 kmol of hydrogen, and 50 kmol of ethane are charge into the batch reactor. After some time 30.0 kmol of hydrogen has reacted, how much of each species will be present in the reactor at that moment?
  198. 198. Extent of reaction  The stoichiometric coefficient is the number that precedes the atoms, ions or molecules in a chemical equation to balance it.  It can be denoted as vi where stands for the ith species in the chemical reaction.  Its value is always negative for reactants and positive for products!!! why?  For a batch or continuous steady state process that involved chemical reaction the equation relating number of moles, stoichiometric co-efficient and extent of reaction is niout = niin + viE (Extent of Reaction)
  199. 199. Extent of reaction  Consider the ammonia formation reaction  N2 + 3H2 2NH3  Suppose the feed rate to a continuous reactor consists of 100 mol/sec of nitrogen, 300 mol/sec of hydrogen and 1 mol/sec of argon.  If we know the exit flow rate of any component or the fractional conversion of nitrogen or hydrogen, we can calculate  Extent of reaction  Two other unknown flow rates  Outflow rate of nitrogen is 40 moles,  Outlet flow rate of hydrogen,  Outflow rate of ammonia
  200. 200. Reaction stoichiometry  Elementary principles of chemical processes chapter 4, section 4.6 page 120 example 4.6.1  Acrylonitrile is produced in the reaction of propylene, ammonia and oxygen as follows C3H6 + NH3 + 3/2O2 C3H3N + 3H2O  The feed contains 10.0 mole% propylene, 12% ammonia and 78% air.  A fractional conversion of 30% of the limiting reactant is achieved.  Taking 100 mol of feed as basis, determine which reactant is limiting, the percentage by which each of the other reactants is in excess, and the molar amounts of all product gases constituents for a 30% conversion of the limiting reactant.
  201. 201. Reaction in which the fraction conversion is to be calculated  Basic Principles and Calculations in chemical engineering by Himmelblau chapter 10 page 266, example 10.2 Mercaptans, hydrogen sulphide, and other sulphur compounds are removed from natural gas by various so called sweetening processes that make available other wise useless sour gas. As you know H2S is toxic in very small quantities and is quite corrosive to process equipment. A proposed process to remove H2S is by reaction with SO2. 2H2S + SO2 3S + 2H2O
  202. 202. Reaction in which the fraction conversion is to be calculated In a test of the process, a gas stream containing 20% H2S and 80% CH4 was combined with a stream of pure SO2. The process produced 5000 lb. of S, and in the product gas the ratio of SO2 to H2S was equal to 3, and the ratio of H2O to H2S was 10. You are asked to determine the fractional conversion of the limiting reactant, and the feed rates of the H2S and SO2 streams.
  203. 203. Chemical reaction stoichiometry (Felder Problem 4.40 Page 171)  Ammonia is burned to form nitric oxide in the following reaction  4NH3 + 5O2 4NO + 6H2O  Calculate the ratio (lb mole O2 react/lb-mole NO formed).  If ammonia is fed to a continuous reactor at a rate of 100.0 kmol NH3/h, what oxygen feed rate (kmol/hr= would correspond to 40.0% excess O2?  If 50.0 kg of ammonia and 100.0 kg of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction (mol) and mass of NO produced (kg) if the reaction proceeds to completion.
  204. 204. Multiple reactions, yield and selectivity  In most chemical processes, reactants are brought together with the object of producing a desired product in a single stream.  Unfortunately the product once formed may react to yield something less desirable.  The result of these side reactions is an economic loss.  For example ethylene can be produced by the dehydrogenation of ethane C2H6 C2H4 + H2
  205. 205. Multiple reactions, yield and selectivity  Once some hydrogen is produced, it can react to produce methane C2H6 + H2 2CH4  More over ethylene can react with ethane to form propylene and methane C2H4 + C2H6 C3H6 + CH4  Since the object of the process is to produce ethylene, only the first of these reactions may be desirable.  Second one consumes the reactant without yielding the desired product.  The third consumes both the reactant and the desired product.
  206. 206. Multiple reactions, yield and selectivity The terms yield and selectivity are used to describe the degree to which a desired reaction predominates over competing side reactions. Yield = moles of desired product formed/moles that would have been formed if there were no side reactions and limiting reactant had react completely Selectivity = moles of the desired product formed/moles of undesired product formed
  207. 207. Multiple reactions, yield and selectivity If A is the desired product and B is an undesired product, one then refers to the selectivity of A relative to B. High values of yield and selectivity signify that the undesired side reactions have been successfully suppressed relative to the desired reaction.
  208. 208. Yield and selectivity Yield is also sometimes defined as the moles of desired product divided by either moles of reactant fed or moles of reactant consumed in the reactor.
  209. 209. Extent of reaction The concept of extent of reaction can be extended to multiple reactions as follows ni out = ni in + ∑ vijEj Considering the following multiple reaction C2H4 + ½ O2 C2H4O C2H4 + 3O2 2CO2 + 2H2O By using above mentioned equation we can calculate the molar flow rates of each of the five species involved in theses reactions.
  210. 210. Yield and selectivity in a dehydrogenation reactor  Elementary Principles of Chemical Processes 3rd edition by felder & Rousseau chapter 4 example 4.6-3 page 124  The reactions  C2H6 C2H4 + H2  C2H6 + H2 2CH4  Takes place in a continuous reactor at a steady state.  The feed contains 85 mole % ethane (C2H6) and the balance inerts.  The fractional conversion of the ethane is 0.501, and the fractional yield of ethylene is 0.471.  Calculate the molar composition of the product gas and the selectivity of ethylene to methane production?
  211. 211. Yield and selectivity  Elementary Principles of Chemical Processes chapter 4 problem 4.49 page 174  Methane and oxygen react in the presence of a catalyst to form formaldehyde. In parallel reaction, methane is oxidized to carbon dioxide and water  CH4 + O2 HCHO + H2O  CH4 + 2O2 CO2 + 2H2O  The feed to the reactor contains equimolar amounts of methane and oxygen. Assume a basis of 100 mole feed/sec.  Draw and label a flowchart. Use a degree of freedom of analysis based on extents of reaction to determine how many process variables must be specified for remaining variable values.  Use equation to derive expressions for the product stream component flow rates in terms of two extents of reaction, E1 and E2.  The fractional conversion of methane is 0.900 and the fractional yield of formaldehyde is 0.855.  Calculate the molar compositions of the reactor output stream and the selectivity of formaldehyde production relative to carbon dioxide production.
  212. 212. Balance on reactive process Systems that involve chemical reactions may be analysed using Molecular species balances Atomic species balances Extents of reaction
  213. 213. Balances on reactive processes Molecular and atomic balances Consider the following reaction C2H6 C2H4 + H2 One hundred kmol/min of ethane is fed to the reactor. The molar flow rate of hydrogen in the product stream is 40 kmol/min.
  214. 214. Molecular Species Balance
  215. 215. Atomic Species Balance
  216. 216. Degree of freedom of analysis  If two molecular species are in the same ratio to each other where ever they appear in the same process and this ratio is incorporated in the flow chart labelling, balances on those species will not be independent equations. Similarly if two atomic species occur in the same ratio where ever they appear in the process, balances on those species will not be independent equations.  Chemical reactions are independent if the stoichiometric equation of any one of them cannot be obtained by adding and subtracting multiples of stoichiometric equations of the others.
  217. 217. Independent reactions  Since nitrogen and oxygen are shown as being in the figure have same ratio wherever they appear on the flow chart (3.76 molN2/mol O2), we cannot count them as two independent species and so you may count only two independent molecular species balances in a degree of freedom of analysis one for either O2 or N2 and one for CCl4.  Similarly atomic nitrogen and atomic oxygen are always in the same proportions to each other in the process (3.76:1) as are atomic chlorine and atomic carbon (4 mol Cl/Cl).  Hence even though four atomic species are involved in the process, we may only count only two independent atomic species balances in the degree of freedom of analysis one for either O and N and one for either C or Cl.
  218. 218. Independent equation and reactions  Chemical equations are independent if the stoichiometric equation of any one of them cannot be obtained by adding and subtracting multiples of the stoichiometric equations of the others.  For example  A 2B  B C  A 2C  Are not independent since [3] = [1] + 2 x [2]
  219. 219. Molecular species balance Molecular species balance If molecular species balances are used to determine unknown stream variables for a reactive process, the balances on reactive species must contain generation and/or consumption terms. The degree of freedom of analysis is as follows. No. of unknown labelled variables + no. of independent reactions – no. of independent molecular species balances – no. other equations relating unknown variables.
  220. 220. Atomic species balance All balances on atomic species take the form input= output, since atomic species can neither be generated nor consumed. The number of degree of freedom is determined directly by subtracting equations from labelled unknowns. No. of unknown variables – no. independent atomic species balances – no. of molecular balances on independent nonreactive species – no. other equations relating unknown variables =no. degree of freedom
  221. 221. Extent of Reaction  The third way to determine unknown molar flow rates for a reactive process is to write expressions for each product species flow rate in terms of extents of reaction. The degree of freedom of analysis is as follows  No. of unknown labelled variables + no. of independent reactions (one extent of reaction for each) – no. of independent reactive species – no. of independent nonreactive species- no. of other equations relating unknown variables no. of degrees of freedom
  222. 222. Degree of freedom of analysis Consider the following reaction that shows the dehydrogenation of ethane to illustrate required procedures. The flow chart is shown again here for ease of reference.  degree for freedom of analysis 2 unknown labelled variables (n1,n2) + 1 independent chemical reaction - 3 independent molecular species balance - 0 other equations relating unknown variables.
  223. 223. Use of atomic, molecular and extent of reaction balance Atomic species balances generally lead to the most straightforward solution procedure, especially when more than one reaction is involved. Extents of reaction are convenient for chemical equilibrium problems and when equation solving software is to be used. Molecular species balances require more complex calculations than either of the two approaches and should be used only for simple systems involving one reaction.
  224. 224. Chemical reaction stoichiometry  The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor.  The product stream is analysed and found to contains 51.7 mol% C2H2Br and 17.3% HBr.  The feed to the reactor contains only ethylene and hydrogen bromide.  Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess.  If the molar rate of the feed stream is 165 mol/s, what is the extent of reaction?
  225. 225. Incomplete combustion of methane  Methane is burned with air in a continuous steady state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide, and water. The reactions taking place are  CH4 + 3/2 O2 CO + 2H2O  CH4 + 2O2 CO2 + 2H2O  The feed to the reactor contains 7.80 mole% CH4, 19.4 mole % O2, and 72.8% N2.  The percentage conversion of methane is 90% and the gas leaving the reactor contains 8 mol CO2/mol CO.  Carry out a degree of freedom of analysis on the process.  Then calculate the molar composition of the product stream using molecular and atomic specie balance as well as extents of reaction.
  226. 226. Chlorination of ethane (Problem 4.50 page 174, 4th chapter, felder and Rousseau)  Ethane is chlorinated in a continuous reactor  C2H6 + Cl2 C2H5Cl + HCl  Some of the product monochloro ethane is further chlorinated in an undesired side reaction  C2H5Cl + Cl2 C2H4Cl2 + HCl  Suppose your principal objective is to maximize the selectivity of monochloro ethane production relative to dicholoro ethane production. Would you design a reactor for high or low conversion of ethane? Explain your answer.  Take a basis of 100 mol C2H5Cl produced. Assume that the feed contains only ethane and chlorine and that all of the chlorine is consumed and carry out a degree of freedom of analysis based on atomic species balance.  The reactor is designed to yield a 15% conversion of ethane and a selectivity of 14 mol C2H5Cl/mol C2H4Cl2, with a negligible amount of chlorine in the product gas. Calculate the feed ratio (mol Cl2/mol C2H6) and the fractional yield of the monochloro ethane.
  227. 227. Production of ethanol(Problem 4.51 page 175, 4th chapter, felder and Rousseau)  Ethanol is produced commercially by the hydration of ethylene:  C2H4 + H2O C2H5OH  Some of the product is converted to diethyl ether in the side reaction  2 C2H5OH (C2H5)2O + H2O  The feed to the reactor contains ethylene, steam, and an inert gas. A sample of the reactor effluent gas is analysed and found to contain 43.3 mole% ethylene, 2.5% ethanol, 0.14% ether, 9.3% inerts, and the balance water.  Take a basis of 100 mol of effluent gas, draw and label a flowchart, and do a degree of freedom of analysis based on atomic species to prove that the system has zero degrees of freedom.  Calculate the molar composition of the reactor feed, the percentage conversion of ethylene, the fractional yield of ethanol, and the selectivity of ethanol production relative to ether production.
  228. 228. Example 5.8 Basic Principles by Himmelbalu Page 270  Formaldehyde (CH2O) is produced industrially by the catalytic oxidation of methanol (CH3OH) by the following reaction: CH3OH + ½ O2 CH2O+H2O  Unfortunately, under the conditions used to produce formaldehyde at a profitable rate, a significant portion of the formaldehyde can react with oxygen to produce CO and H2O: CH2O+1/2 O2 CO+H2O  Assume that methanol and twice the stoichiometric amount of air needed for complete oxidation of the CH3OH are fed to the reactor, that 90% conversion of the methanol results, and that a 75% yield of formaldehyde occurs (based on the theoretical production of CH2O by Reaction (1).  Determine the composition of the product gas leaving the reactor.
  229. 229. Product separation and recycle Two definitions of reactant conversion are used in the analysis of chemical reactors with product separation and recycle of unconsumed reactants. Over all conversion = reactant input to process- reactant output from process/reactant input to process. Single-pass conversion = reactant input to reactor – reactant output from reactor/reactant input to reactor
  230. 230. Product separation and recycle  For example consider the following labelled flowchart for a simple chemical process based on the reaction A B
  231. 231. Product separation and Recycling
  232. 232. Dehydrogenation of propane Example 4.7-2 page 136, 4th chapter, felder and Rousseau)  Propane is dehydrogenated to form propylene in a catalytic reactor  C3H8 C3H6 + H2  The process is designed for a 95% overall conversion of propane.  The reaction products are separated into two streams.  The first which contains H2, C3H6 and 0.555% of the propane that leaves the reactor is taken off as product.  The second stream which contains the balance of the unreacted propane and 5% of the propylene in the first stream, is recycled to the reactor.  Calculate the composition of the product, the ratio (moles recycled)/(mole fresh feed), and single pass conversion.
  233. 233. Advantages of recycle stream  Consider what's happening in the process  Only 10% conversion has been achieved in the reactor where as overall conversion of the process is 95%.  How come we were able to achieve such higher conversion???  Higher overall conversion can be achieved in two ways.  Design a reactor to yield a high single pass conversion OR  Design a reactor to yield a lower single pass conversion.  Then, use a separator to recover the product and allow it to react again hence increasing the overall conversion of the process.
  234. 234. Production of formaldehyde Problem 4.56 Page 177 3rd Edition Felder  A catalytic reactor is used to produce formaldehyde from methanol in the reaction.  A single pass conversion of 60.0% is achieved in the reactor. The methanol in the reactor product is separated from the formaldehyde and hydrogen in a multiple unit process. The production rate of formaldehyde is 900.0 kg/hr.  Calculate the required feed rate of methanol to the process (kmol/hr) if there is no recycle.  Suppose the recovered methanol is recycled to the reactor and the single pass conversion remains 60%. Without doing any calculations prove that you have enough information to determine the required fresh feed rate of methanol (kmol/hr) and the rates (kmol/hr) at which methanol enters and leaves the reactor. Then perform calculations.
  235. 235. Purging  Suppose a material that enters with the fresh feed or is produced in a reaction remains entirely in a recycle stream, rather than being carried out in a process stream.  If not removed that particular substance would continuously enter the process and would have no way of leaving.  Hence it would steadily accumulate, making the attainment of steady state process impossible.  In order to prevent this a portion of recycle stream is must be withdrawn as a purge stream to get rid of the substance.
  236. 236. Recycle and purge in synthesis of methanol  Methanol is produced in the reaction of carbon dioxide and hydrogen  CO2 + 3H2 CH3OH + H2O  The fresh feed to the process contains hydrogen, carbon dioxide, and 0.4 mole % inerts. The reactor effluent passes to a condenser that removes essentially all of the methanol and water formed and none of the reactants or inerts.  The latter substance are recycled to the reactor.  To avoid build up of the inerts in the system, a purge stream is withdrawn from the recycle.  The feed to the reactor (not the fresh feed to the process) contains 28 mole% CO2, 70 mole % H2, and 2 mole% inerts.  The single pass conversion of hydrogen is 60 %.
  237. 237. Recycle and purge in synthesis of methanol Calculate the molar flow rates and molar compositions of the fresh feed The total feed to the reactor And recycle stream, and the purge stream.
  238. 238. Combustion Reactions  Combustion the rapid reaction of a fuel gas with oxygen is more important than any other class of industrial chemical reactions.  The combustion products such as CO2, H2O and CO along with SO2 are worth much less than the fuels burned to obtain them.  The significance of these reactions lies in the fact that tremendous amount of energy they release is then use to drive the turbines that generate most of the world’s electric power.
  239. 239. Combustion Reactions The job of designing power generation equipment usually falls to mechanical engineers, but The analysis of combustion reactions and reactors and the abatement and control of environmental pollution caused by combustion products such as CO, CO2, and SO2 are problems with which chemical engineers are heavily involved. In preceding sections different terminologies commonly used in the analysis of combustion reactions will be discussed.
  240. 240. Combustion Chemistry Most of the fuel used in power plant combustion furnaces is either are Coal (carbon, some hydrogen and sulphur, and various non combustible materials). Fuel Oil (mostly high molecular weight hydrocarbons, some sulphur) Gaseous fuel (such as natural gas, which is primarily methane) Or LPG (Liquified Petroleum Gas) which is usually propane and or butane.
  241. 241. Combustion Chemistry When a fuel is burned, carbon in the fuel reacts to form either CO2 or CO. Hydrogen forms H2O and sulphur forms SO2. It is interesting to note that at temperatures greater than 1800OC, some nitrogen in the air reacts to form nitric acid (NO). A combustion reaction in which CO is formed from a hydrocarbon is referred to as partial combustion or incomplete combustion.
  242. 242. Combustion Chemistry
  243. 243. Combustion Chemistry For obvious economic reasons, air is the source of oxygen in most combustion reactors. Dry air has the following average molar composition.
  244. 244. Combustion Chemistry In most combustion calculations, it is acceptable to simplify this composition to 79% N2, 21% O2. 79 moles N2/21 moles O2 or 3.76 moles of N2/mol O2. The term composition on a wet basis is commonly used to denote the component mole fraction of a gas that contains water and Composition on a dry basis signifies the component mole fraction of the same gas without water.
  245. 245. Combustion Chemistry For example, a gas that contains 33.3 mole% CO2, 33.3% N2, and 33.3% H2O (wet basis), Contains 50% CO2 and 50% N2 on a dry basis. The product that leaves a combustion furnace is referred to as the stack gas or flue gas. When the flow rate of a gas in a stack is measured, it is the flow rate of the gas including water.
  246. 246. Combustion Chemistry You must be able to convert a composition on dry basis to its corresponding composition on a wet basis before writing material balances on the combustion reactor. The procedure for converting from wet basis to dry basis is similar to the one used to convert mass fractions to mole fractions and vice versa.
  247. 247. Example 4.8-1 Composition on Wet and Dry Basis A stack gas contains 60.0 mole% N2, 15.0% CO2, 10.0% O2, and the balance H2O. Calculate the molar composition of the gas on a dry basis.
  248. 248. Theoretical and Excess Air If two reactants participate in a reaction and one is considerably more expansive than the other, the usual practice is to feed the less expansive reactant in excess of the valuable one. As a result the conversion of the valuable reactant is increased at the expense of the cost of the excess reactant and additional pumping cost. The extreme case of an expensive reactant is air, which is free.
  249. 249. Theoretical and Excess Air Combustion reactions are therefore invariably run with more air than is needed to supply oxygen in stoichiometric proportion to the fuel. Theocratical oxygen The moles or molar flow rate of O2 needed for complete combustion of all the fuel fed to the reactor, assuming that all carbon in the fuel is oxidized to CO2 and all the hydrogen is oxidized to H2O. Excess Air The quantity of air that contains the theoretical oxygen.
  250. 250. Theoretical and Excess Air Percent Excess Air If you know the fuel feed rate and the stoichiometric equation for complete combustion of the fuel, you can calculate the theoretical O2 and air feed rates.
  251. 251. Example 4.8-2 Theoretical and Excess Air One hundred mol/hr of butane (C4H10) and 5000 mol/hr of air are fed into a combustion reactor. Calculate the percent excess air.
  252. 252. Material Balance on Combustion Reactors The procedure of writing and solving material balances for a combustion reactor is the same as that for any other reactive system. Bear in mind these points, however, When you draw and label the flow chart, be sure the outlet stream (stack gas) includes Un reacted fuel unless you are told that all the fuel is consumed. Unreacted oxygen Water and carbon dioxide, and Nitrogen if the fuel is burned with air and not pure oxygen.
  253. 253. Material Balance on Combustion Reactors To calculate the oxygen feed rate from a specified percent excess oxygen or percent excess air (both percentages have the same value, so it doesn’t matter which one is stated). First calculate the theoretical O2 from the fuel feed rate and the reaction stoichiometry for complete combustion, then calculate the oxygen feed rate by multiplying the theoretical oxygen by (1 + fractional excess oxygen).
  254. 254. Material Balance on Combustion Reactors If only one reaction is involved, all three balance methods (molecular species balances, atomic species balances, extent of reaction) are equally convenient. If several reactions occur simultaneously, however such as combustion of a fuel to form both CO and CO2 atomic species balances are usually most convenient.
  255. 255. Example 4.8-3 Combustion of Ethane Ethane is burned with 50% excess air. The percentage conversion of the ethane is 90%. Of the ethane burned, 25% reacts to form CO and the balance reacts to form CO2. Calculate the molar composition of the stack gas on a dry basis and the mole ratio of water to dry stack.
  256. 256. Example 4.8-4 Combustion of a Hydrocarbon Fuel of Unknown Composition A hydrocarbon gas is burned with air. The dry basis product gas composition is 1.5 mole% CO, 6.0% CO2, 8.2% O2, and 84.3% N2. There is no atomic oxygen in the fuel. Calculate the ratio of hydrogen to carbon in the fuel gas and speculate on what the fuel might be. Then calculate the percent excess air fed to the reactor.