AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri. Fundamentals for Chemical Engineering Calculations

Transcription

1 AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri Fundamentals for Chemical Engineering Calculations

2 AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri Dimensions, Units, and Unit Conversion

3 Dimensions Dimensions are concepts of measurement in engineering works. The basic dimensions we are familiar with are length, mass, temperature and time. Other dimensions are called derived dimensions, because they are derived from the basic dimensions. The common derived dimensions are force, pressure, energy, concentration, etc.

4 Relation between basic and derived dimensions Density Volume Volume Flow Rate Area Time Mass Flow Rate Mass Length Velocity Acceleration Force

5 Units Units are the means of expressing the dimensions such as metre(m) for length, kilogram(kg) for mass, degree Celcius( C) for temperature and second(s) for time. Derived units are those that can be developed in terms of fundamental units such as Newton(N) for force, Pascal(Pa) for pressure, Joules(J) for energy and Molar(M) for concentration.

6 Common Dimensions and Units (SI) Dimensions Units Symbols for units Length metre m Mass kilogram kg Time second s Temperature Kelvin K Force Newton N Molar amount gmole gmol (or mol) Energy Joule J Power Watt W Density kilogram per cubic metre Kg/m 3 Velocity metre per second m/s Acceleration metre per second squared m/s 2 Pressure Pascal Pa Heat Capacity Joule per kilogram Kelvin J/kg K

7 Common Dimensions and Units (AE) Dimensions Units Symbols for units Length foot ft Mass pound mass lb m Time second, minute, hour, day s, min, hr, day Temperature degree Rankine or degree Fharenheit R or F Force pound force lb f Molar amount pound mole lb mol Energy British thermal unit Btu Power horsepower hp Density pound mass per cubic foot lb m /ft 3 Velocity feet per second ft/s Acceleration feet per second squared ft/s 2 Pressure pound force per square inch psi Heat Capacity Btu per pound mass per Btu/lb m F

8 Examples of Derived Dimension 1. Force (F) In AE system, 1 lb f is a force required to accelerate a mass of lb m at a rate of 1 ft/s 2 or 1 lbf = lbm ft/s 2 In SI, 1 N is a force required to accelerate a mass of 1 kg at a rate of 1 m/s 2 or 1 N = 1 kg m/s 2

9 Example 1: Unit of Force From the definition F = ma When F = force, m = mass, and a = acceleration Then, F = m(kg) x a(m/s 2 ) F(kg m/s 2 ) While the definition of 1N is the movement of 1 kg-mass with the acceleration of 1 m/s 2 Hence, 1 N = 1 kg m/s 2

10 Examples of Derived Dimension 2. Pressure (P) Pressure is a force exerted by fluid per unit area Or P = F/A SI; Unit of pressure is Pascal (1 Pa =N/m 2 ) AE; Unit of pressure is psi, (1 psi = 1lb f /in 2 )

11 Example 2: Unit of Pressure From the definition P = F/A When P = pressure, F = force, and A= cross-sectional area Therefore, P = F(N)/A(m 2 )= F(kg m/s 2 )/(A (m 2 ) P (kg/m s 2 ) Or P (Pascal) since 1 Pa = 1 kg/m s 2 Pascal is usually used as a common unit for pressure.

12 Units Conversion Conversion of units using conversion factors Conversion of units using prefix factors

13 Conversion for Fundamental Units Mass (m) 1 lb m = kg Length (s) 1 ft = m Temperature (T) T( F) = 1.8 T( C) + 32 T(R) = 1.8T(K)

14 Common Conversion Factors Dimensions Length Mass Time Temperature Force Molar amount Energy Power Pressure Conversion Factors 1 m = in = ft 1 kg = lb m Same unit T(K) = T( C) , T(R) = 1.8 T(K) T ( F) = 1.8 T( C) N = lb f 1 lb f = lb m ft/s 2 1 mol = 1 gmol = lb mol 1 kj = Btu 1W = 1 J/s 1 kw = hp 1 Pa = 1N/m 2 = x 10-4 psi 1 atm = kpa = bars = 760 mm.hg

15 Example 3A: Convert 2 km to miles. Conversion Factor is 1 mile = 1.61 km Example 3B: Convert 400 in 3 /day to cm 3 /min Conversion Factor is 1 inch = 2.54 cm 1 day = 24 hours 1 hour = 60 minutes

16 Example 4: Unit conversion for biochemical engineering Glucoamylase is an enzyme that aids in the conversion of starch to glucose. Experiments show that 1 g mol of glucoamylase in a 4% starch solution results in a production rate of glucose of 0.6 g mol/ml min. Determine the production rate of glucose for this system in the units of lb mol/ft 3 day.

17 Standard prefixes in SI units Multiple Prefix Multiple Prefix Tera (T) 10 9 Giga (G) 10 6 Mega (M) 10 3 Kilo (k) 10 2 Hecto (h) 10 1 Deka (da) Pico (p) 10-9 Nano (n) 10-6 Micro ( ) 10-3 Milli (m) 10-2 Centi (c) 10-1 Deci (d)

18 Example 5: Conversion of particle size A semiconductor is ZnS with a particle diameter of 1.8 nm. 5A: Convert this value to dm 5B: Convert this value to inches

19 Dimension Consistency (Homogeneity) Basic principle states that equations must be dimensionally consistent. Each term in an equation must have the same net dimensions and units as every other term to which it is added, subtracted or equated. Dimensional considerations can be used to help identify the dimensions and units of or quantities in an equation.

20 Example 6: Conversion of particle size The handbook shows that microchip etching roughly follows the relation; d = e t Where d = depth of the etch in microns t = time of the etch in seconds (A) What are the unit associated with the numbers 16.2 and 0.021? (B) Convert the relation so that d is expressed in inches and t is in minutes.

21 Example 7: Homogeneity of dimension Leaking oil tanks have become an environmental problems that the government has implemented a number of rules to reduce the problem. A leak from a small hole in a tank can be predicted from the following relation: Q = 0.61 S 2 P/ρ Where Q = leakage rate S = cross-sectional area of the leak P = pressure drop = fluid density To test the tank, the vapor space is pressurized with N 2 to a pressure of 23 psig. If the tank is filled with 73 inches of gasoline (sp.gr ) and the hole is 0.25 inch in diameter, what is the value of Q in ft 3 /hr?

22 AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri Temperature and Pressure

23 Temperature? The zero th law of Thermodynamics Two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact. Freezing point (ice point) : a mixture of ice and water in equilibrium with saturated air at 1 atm Boiling point (steam point) : a mixture of liquid water and water vapor in equilibrium at 1 atm

24 Temperature Scales Boiling point of water at 760 mm Hg scales 100 scales Freezing point of water at 760 mm Hg F = C [ F] [R] Absolute zero [K] [ C]

25 Thermodynamics Temperature Scales S.I. System; AE System; Kelvin scale (K) T (K) = T (C) T (K) = T ( C) Rankin scale (R) T (R) = T (F) T (R) = T ( F) Conversion factors; T ( F) = 1.8 T ( C) + 32

26 Relative and Absolute Temperature 1. Relative temperature; based on relative scales and expressed in degree Fahrenheit [ F] and degree Celsius [ C] 2. Absolute temperature; based on absolute temperature scales and expressed in Rankine [R] and Kelvin [K].

27 Example 8: Conversion of temperature unit The heat capacity of sulfuric acid has the unit of J/(gmol)( C), and is given by the relation as shown below Heat capacity = (1.56 x 10-1 ) T Where T is expressed in [ C]. Modify the formula so that the resulting expression has the associated units of Btu/(lbmol)(R), with T expressed in [R].

28 Pressure Pressure (P) Pressure is a force exerted by fluid per unit area Or P = F/A SI; Unit of pressure is Pascal (1 Pa =N/m 2 ) AE; Unit of pressure is psi, (1 psi = 1lb f /in 2 )

29 Pressure in a fluid does not change in horizontal direction P 1 Assuming fluid density ( ) is constant x F z ma z 0 z W P 2 When, F z P2 x P1 x m x z a z g P P x 2 1 P P x P g x z g z 2 1 gage 0 P

30 If fluid density ( ) is a function of elevation, =fn(z) P 1 x From P x P x g x z z W Divide by x z, then taking the limit z --> 0 dz P 2 dw= xdz 1 2 dp dz g P gdz 2 1

31 Definition of pressure Absolute pressure (P a ) actual pressure at a given position Gauge pressure (P gage ) pressure read from measuring device Atmospheric pressure (P atm ) local atmospheric pressure P abs = P gage + P atm

32 The Manometer measure P g Assume; P gas Pgas P gas1 Constant at any point GAS h 1 2 and P P gas1 liq2 therefore P gas Pliq2 liq gh

33 Atmospheric pressure (Barometric pressure) = weight of air above the location per unit surface area The Barometer Patm P B P B W A 1 1 gha A 1 1 gh A 1 A 2 A 3 B B B h W P B P B W A 2 2 W A 3 3 gha A 2 2 gha A 3 3 gh gh

34 Standard atmosphere at 1 atm Pressure of 760 mm. height of mercury column ( Hg = 13,595 kg/m 3 ) Temperature of 0 C Gravitational acceleration of m/s 2 Standard atmospheric pressure: 1 atm = 760 torr = 1.013x10 5 Pa = bars = psi

35 Example 9: Calculation of pressure Small animals such as mice can live at reduce air pressure down to 20 kpa absolute. In a test, a mercury manometer attached to a tank reads 64.5 cm. Hg and the barometer reads 100 kpa. Will the mice survive?

36 Differential pressure measurement For a certain type of liquid (with constant density), the pressure is the same at any point lied on the same height. For gas at a constant temperature, its density can be considered constant. The gas pressure at every points in the same container is equal.

37 Example 10: Calculation of pressure difference A differential manometer is used to determine the pressure difference across the orifice plate. The flow rate was to be calibrated with the observed pressure drop. Calculate the pressure drop (P1 P2) in Pascals for the manometer reading in the following figure.

38 Example 11: Calculation of gas pressure Air flows through a duct under a draft of 4.0 cm H 2 O. The barometer indicated the atmospheric pressure of 730 mm Hg. What is the absolute pressure of the air in the unit of inches of mercury?

39 AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri Molar, Density, and Concentration

40 Molar (n) S.I. unit 1 g mol = mass in g Molecule Weight 1 g mol = x molecules AE unit 1 lb mol = mass in lb Molecule Weight 1 lb mol = x x molecules Conversion Factor: 1 lb mol = g mol

41 Molecular Weight (MW) The atomic weight of an element is the mass of an atom based on the scale that assigned a mass of exactly 12 to the carbon isotope 12 C. The molecular weight of a compound is the sum of the weights of atoms of which the compound composed. In SI, MW is presented in a unit of g/gmol. In AE system, MW is presented in a unit of lb m /lbmol.

42 Example 12: Calculation of Molecular Weight Recently, a superconductivity material in a form of YBa 2 Cu 16 O 24 was invented. What is the molecular weight of the superconductor? Given: The Periodical Table in the next slide

43 Periodical Table of Elements

44 Example 13: Use molecular weight to convert moles to mass How many pounds of NaOH are in 7.5 gmol of NaOH? Given: MW of NaOH = 40.0 g/gmol

45 Density ( ) Density is a ratio of mass per unit volume, in kg/m 3 or lb m /ft 3. Density of Incompressible substance ( solid & liquid) does not change significantly, but density of compressible substance (gas & vapor) do. The inverse of density is called specific volume (v^). ρ = m V and V^ = V m

46 Density of solution A homogeneous mixture of two or more components is called solution. V solution = n i=1 V i m solution = n i=1 m i ρ solution = i=1 n mi i=1 n V i

47 Specific gravity (sp. gr.) Specific gravity is a dimensionless ratio. It is a ratio of two densities, i.e. that of substance of interest to that of a reference substance. sp. gr. of A = ρ ref Common reference substance is usually water at 4 C. ρ A

48 Example 14: Calculation of solution density If you add 50 g of sugar to 500 ml of water, how do you calculate the density of the sugar solution? Given: Density of water = 1 g/ml

49 Flow Rate Mass flow rate (m ): m = m t Volumetric flow rate (F): F = V t Molar flow rate (n ): n = n t

50 Concentration 1. Mass (or weight) fraction & Mole fraction n i 1 x i 1 and n i 1 y i 1

51 Example 15: Calculation of the mass fraction The microstructure of nanosized particles has proved to be important in nanotechnology in developing economic magnetic performance of nanocomposites. In a ternary alloy such as Nd 4.5 Fe 77 B 18.5,the average grain size is about 30 nm. By replacing 0.2 atoms of Fe with atoms of Cu, the grain size can be reduce to 17 nm. (a) What is the molecular formula of the alloy after adding Cu to replace Fe? (b) What is the mass fraction of each atomic species Given information; Neodymium (Nd) has a MW of Iron (Fe) has a MW of Boron (B) has a MW of Copper (Cu) has a MW of 63.55

52 Example 16: Conversion between mass fraction and mole fraction An industrial-strength drain cleaner contains 5.00 kg water and 5.00 kg of NaOH. What are the mass fraction and mole fraction of each component in the cleaner?

53 Example 17: Changing basis for calculation A medium grade bituminous coal analyzes as follows: Component % by mole S 2 N 1 O 6 Ash 11 Water 3 Residuum (C & H) 77 Total 100 The mole ratio of C and H in the residuum is 9. Calculate the mass fraction of each component of the coal with ash and moisture-free basis.

54 2. ppm & ppb: Concentration The unit of ppm (parts per million) and ppb (parts per billion) are used to express concentrations of trace species. The definition may refer to mass ratio (for liquid and solid) or mole ratio (for gas). For liquid & solid; For gas; 1g solute ppm 10 g solution 1 6 1gmol solute ppm 10 gmol solution 1 6 1g solute ppb 10 g solution 1 9 1gmol solute ppb 10 gmol solution 1 9

55 Example 18: Use of ppm The current OSHA 8-hour limit for HCN in air is 10.0 ppm. A lethal dose of HCN in air is 300 mg/kg air at room temperature. 5.1) How many mg HCN / kg air is 10.0 ppm? 5.2) What fraction of the lethal dose is 10.0 ppm? * OSHA = Occupational Safety and Health Agency

56 Concentration 3. Other ways to express concentration - Mass per unit volume; kg/m 3, lb/ft 3, g/l - Mole per unit volume; gmol/cm 3, gmol/l, lbmol/ft 3 - Molarity - Molality - Normality - etc.

57 Example 19: Calculation of concentration and flow rate In the production of a drug having MW = 192, the exit stream from the reactor flows at a rate of 10.5 L/min. The drug concentration is 41.2% (in water), and the specific gravity of the solution is Calculate the concentration of the drug (in kg/l) in the exit stream, and the flow rate of the drug in kg mol/min.

58 AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri Terms, Definitions, and Concepts of Material Balance

59 Process Process Path State A State B Process is a pathway to obtain the system is desirable state. Examples are heat transfer process, fermentation process, production process, etc. Chemical processes can be classified as batch, semibatch, or continuous and operated as steady state or unsteady-state processes.

60 Types of Processes

61 System and Surroundings surroundings system boundary System is a process or part of process to be considered. System and surroundings are divided by boundary.

62 Closed and Open Systems Closed System: no mass entering nor exiting the closed system. However, there might be energy transferred across the boundary. Opened System: Mass and energy are transfer across the boundary. They can be considered as Flow systems

63 Closed system Open system (control volume) Mass in Mass out Energy movement Energy movement e.g., storage tank gas in piston cylinder e.g., surge tank liquid flow past valve

64 Steady-state Process There is no accumulation (or change) of materials and energy in the system throughout the process. Mass (or molar) and energy of the system do not change with time.

65 Unsteady-state Process Mass (or molar) and energy of the system change with time. There will be accumulation term existed in the balance equation.

66 Common Unit Operations in Chemical Processes

67 1. Splitter Splitter are used to divide the flow rate in a certain stream into two or more streams with different flow rates, although the compositions of all streams are the same. Y-shape pipe connector

68 2. Mixing Tank There are two or more entering streams, and only one exit stream resulting form the blending of the incoming streams. The streams can be in any phase; gas, liquid, or solid. Feed 1 Feed 2 Product

69 3. Reactor A chemical reactor carries out chemical reaction that convert molecular species in the input to different molecular species in the output.

70 4. Evaporator Evaporation process is used to concentrate the solutions by supply energy to the solution so that the solvent (usually water) is removed be mean of evaporation. Vapor solution

71 Moisture 5. Dryer Drying is a process resulting in the removal of moisture from solid, semi-solid, or liquid to produce a solid state product. Hot Air

72 6. Distillation Column Distillation is a method of separating chemical substances based on differences in their volatilities. More volatile components are in the overhead product, while less volatile components are in the bottom. Separation is accomplished by boiling of the feed mixture.

73 7. Filter Filtration is a method used to remove impurities from liquid or to isolate solid from fluid mixture.

74 8. Crystallizer Crystallizer is used to achieve liquid-solid separation as solid crystals are formed by a change of temperature, and filtered out of the original solution.

75 9. Extractor Extractor is used to separate out the solute from original solution via additional solution, in which the solubility of solute is higher. The two solvents are immiscible so that they can be separated easily.

76 10. Absorber Absorber is a similar unit to extractor, usually used to de-contaminate gas. Common name of gas absorber column is gas scrubber, or stripper.

77 11. Adsorption Column Adsorption is used when the solute is collected on the surface of adsorbent. It is one of the separation method commonly used in chemical industries.

78 12. Heat Exchanger Heat exchanger is a device that heat transfer is performed between two types of substances that are not touching each other. Heat is transfer from the higher temperature substance to the lower temperature substance. At the end of the process the exit temperature of the hotter feed will be lower, while the exit temperature of the cooller fees will be raised.

79 13. Humidifier and Dehumidifier Humidifier is a device that increases the amount of moisture in the air, by allowing water to evaporate into air stream. Dehumidifier is a device to reduce the moisture in the air, by condensing the water out of the air. The use of humidifier and dehumidifier relies on humidity reported in the psychometric chart,

80 Concepts of Material Balance Material balance is an application of the law of the conservation of mass: Matter is neither created nor destroyed. Material balance for a single-component process is [Accumulation of material within the system] = [Total input] [Total output] Material balance for a multi-component process is [Accumulation of material within the system] = [Total input] [Total output] [Generation within system] - [Consumption within system]

81 Material Balance Equation Let m 0 = initial mass of the system m t = mass of the system at time t m i = mass of inlet stream(s) m e = mass of exit stream(s) m gen = mass of generated material(s) m con = mass of consumed material(s) m t m 0 = m i m e + m gen m con

82 Differential and Integral Balance

83 Example 1: Differential balance Ex 1: 1000 kg/hr of a mixture containing benzene(b) and toluene(t), with 50% by mass of B. The two components are separated by distillation process, so that the top product contains 450 kg B/hr and the bottom product contains 475 kg T/hr. What are the unknown components in top and bottom streams?

84 Example 2: Batch-integral balance Ex 2: Two methanol-water solutions are contained in two flasks. The first mixture contains 40%wt of methanol, and the second mixture contains 70%wt of methanol. If 200 g of the first mixture is combined with 150g of the second mixture, What are the mass and composition of the product?

85 Example 3: Semibatch-integral balance Ex 3: Air is bubbled through the drum of liquid hexane at a rate of kmol/min. The gas stream leaving the drum contains 10.0%mole hexane vapor. Air is not soluble in hexane. Estimate the time required to vaporize 10.0 m 3 of the liquid.

86 The strategy for problem solving 1. Understanding the problem 2. Drawing a diagram of process and specify the system boundary 3. Place labels for all given information on the diagram 4. Obtain the derived data needed for problem solving 5. Choose basis 6. Determine the unknowns 7. Perform degree of freedom analysis 8. Write down equations to be solved 9. Do the calculation 10. Check your answer(s)

87 What is Degree of Freedom? Degree of freedom (N D ) means the number of variables whose values are unknown (N U ) minus the number of independent equation (N E ). N D = N U - N E Case (1):N D = 0 when N U = N E a solution exists with exactly specified Case (2): N D > 0 when N U > N E underspecified, more equations needed Case (3): N D < 0 when N U < N E over-specified, additional unknowns needed

88 Example 4: Extraction of Streptomycin from a fermentation broth Streptomycin is produced by the fermentation of bacterium in a biological reactor with a nutrient of glucose and amino acids. After the fermentation process, streptomycin is recovered by contacting with an organic solvent in an extraction process. The diagram of extraction process is shown below. Fresh organic solvent (S) 10 L/min Aqueous solution 200 L/min 10g Strep/L Extraction Process Aqueous solution 0.2 g Strep/L organic solvent (S) and extracted medicine Determine the mass fraction of streptomycin in the exit organic solvent assuming that no water exits with solvent and no solvent exits with an aqueous solution. Density of the aqueous solution is 1 g/cc, and density of organic solvent is 0.6 g/cc.

89 Example 5: Separation of gases using a membrane Membrane separation process is used to separate nitrogen and oxygen from air. According to the given diagram and information, what is the composition of the waste stream if the waste stream amounts to 80% of the input stream? High-pressure side Membrane Low-pressure side Air stream Air contains 21%mol O 2 and 79%mol N 2 Product O 2 25%mol and N 2 75%mol Waste Stream

90 Example 6: Overall analysis for a continuous distillation column The demand for ethyl alcohol (EtOH) is increasing due to the gasohol production. Most difficult part of the manufacturing is to separate water from EtOH. It appears that too much alcohol is lost in the bottom product when distillation process is applied. If 1000 kg of water-etoh mixture (10%wt water and 90%wt EtOH) is fed into the distillation column, the distillate contains 60% EtOH. The distillate (top product) is amount to 0.1 of the feed. Calculate the composition of the bottoms and the mass of EtOH lost in the bottoms. Theses information were collected within 1 hour.

91 Example 7: Mixing of Sulfuric Acid You are asked to prepare a batch of 18.63% battery acid (H 2 SO 4 ) as follows. A tank of old-weak battery acid solution contains 12.43% H 2 SO 4 (the remainder is pure water). If 200 kg of 77.7% is added to the tank, and the final solution is to be 18.63% H 2 SO 4. How many kilograms of battery acid have been made? Added 200 kg solution of 77.7% H 2 SO % H 2 SO % H 2 SO 4 Note: all % H 2 SO 4 are by mass

92 Hemodialysis Hemodialysis is the most common method used to treat advanced and permanent kidney failure. When your kidneys fails, harmful wastes build up in your body, your blood pressure may rise, and your body may retain excess fluid and may not make enough red blood cells. In hemodialysis, your blood flows through a device with a special filter that removes wastes. The dialyzer is a large canister containing thousands of small fibers through which the blood passes. Dialyzer solution is pumped around these fibers. The fibers allow wastes and extra fluids to pass from your blood to dialyzer solution that carries them away. The plasma components are usually water, uric acid, area, creatinine, phosphorus, potassium, and sodium.

93 Hemodialysis canister (Dialyzer)

94 Example 8: Application of hemodialysis Calculate the concentration (in g/l) of each component of the plasma in the outlet solution. The flow rate of water in inlet blood is 1,100 ml/min, and that in returning blood is 1,200 ml/min. Component Concentration in inlet blood (g/l) Concentration in returning blood (mg/l) Uric acid Creatinine Urea 18 1,510 Phosphorus Potassium Sodium 13 5,210

95 Crystallization Principles

96 Example 9: Crystallization of Na 2 CO 3 A tank holds 10,000 kg of a saturated solution of Na 2 CO 3 at 30 C. If you want to crystallize from this solution 3,000 kg of Na 2 CO 3 10H 2 O without any accompanying water. At which temperature must the solution be cooled. Solubility data of Na 2 CO 3 are in the following table: Temp ( C) Solubility (g Na 2 CO 3 / 100 g H 2 O)

97 AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri Stoichiometry and MB with Reactions

98 Stoichiometry Stoichiometry provides a quantitative means of relating the amount of products produced by chemical reactions to the amount of reactants. The stoichiometric equation of a chemical reaction is a statement of a relative numbers of molecules or moles of reactants and products that participate in the reaction. The stoichiometric coefficients of a balanced equation tell us the mole ratios among substances that reacted and produced by the reaction. The stoichiometric ratio is the ratio of stoichiometric coefficients in the balanced reaction equation.

99 Sulfur Trioxide (SO 3 ) Production During combustion of sulfur bearing fuels, such as coal, sulfur oxides are produced. While most of the sulfur forms sulfur dioxide (SO 2 ), a small amount is further oxidized to sulfur trioxide (SO 3 ) due to the oxidation of SO 2. The stoichiometric equation for this reaction is written as: 2SO 2 + O 2 2SO 3 Let i = stoichiometric coefficient of substance i in the reaction, then SO2 = (-) 2 O2 = (-) 1 SO3 = (+) 2 The stoichiometric ratios from this reaction can be; 2 moles of SO 3 produced 2 mole ofso 2 reacted or 2 moles of SO 3 produced 1 mole ofo 2 reacted or 2 moles of SO 3 reacted 1 mole ofo 2 reacted

100 Example 1: Combustion of heptane (C 7 H 16 ) In the combustion of heptane, carbon dioxide is produced. Assume you want to produce 500 kg of dry ice per hour, and that 50% of the CO 2 can be convert into dry ice. How many kilograms of heptane must be burned per hour? Other gases C 7 H 16 O 2 Combustion Chamber CO 2

101 Example 2: Application of Stoichiometry when more than one reaction occurs By heating limestone we can recover oxides known as Lime. A limestone composition is analyzed as Component % wt CaCO MgCO Inert 1.70 (1) How many pounds of CaO can be made from 1 ton of limestone? (2) How many pounds of CO 2 can be recovered per pound of limestone? (3) How many pounds of limestone are needed to produce 1 ton of lime?

102 Limiting and Excess Reactants In industrial reactors, it is very rare to find exact stoichiometric amounts of materials used. Some reactants are costly that we need to use it up. However, there will always be excess materials come out of the reactor together with the products. The limiting reactant is the specie in a chemical reaction that would theoretically run out first if the reaction proceeds to completion according to the chemical reaction. The reactant is limiting since it presents in less than its stoichiometric proportion relative to every other reactants. The other species of reactants are, then, called excess reactants. % excess reactant = 100[ amount of excess reactant fed stoichiometric amount stoichiometric amount ]

103 Example 3: Hydrogenation of Acetylene The hydrogenation of acetylene will form ethane as shown below: C 2 H 2 + 2H 2 C 2 H 6 Suppose that 20 kmol/h of acetylene and 50kmol/h of hydrogen are fed to a reactor. Therefore, the ratio of H 2 to C 2 H 2 in the reactor is 50:20, or 2.5:1. However, the stoichiometric ratio of H 2 to C 2 H 2 is 2:1. Hydrogen is fed in a greater than stoichiometric proportion to acetylene. Therefore, acetylene is the limiting reactant. With 20 kmol/h of C 2 H 2 fed, we will need 40 kmol/h of H 2 to do the reaction. %excess of H 2 = (50 40) kmol/h 40 kmol/h x 100 = 25%

104 Extent of Reaction ( ) Extent of reaction denotes how much reaction occurs, mostly reflecting by the consumption of limiting reactants. The unit of extent of reaction is presented as mole reacting. When, Or, = n i n io i n i and n io are moles of specie i present in the system after reaction and when reaction starts, respectively. i = stoichiometric coefficient of specie I n i = n io + i

105 Example 4: Calculation of the Extent of Reaction Determine the extent of reaction for the following chemical reaction N 2 + 3H 2 2NH 3 Given the following analysis of feed and product: Specie Amount in Feed (g) Amount in Product (g) N na. H 2 50 na. NH

106 Maximum Extent of Reaction ( Max ) The maximum extent of reaction is the of each reactant, based on the complete reaction. The amount of products produced would be controlled by the amount of limiting reactant. Max = 0 n io i The amount of n i will always be zero for the complete reaction. The reactant with the smallest Max is the limiting reactant.

107 Example 5: Determination of limiting reactant using Max Consider the combustion of heptane; C 7 H O 2 7CO 2 + 8H 2 O If 1 gmol of heptane and 12 gmol of oxygen are mixed, which specie will be considered a limiting reactant? Using maximum extent of reaction to answer this problem.

108 Example 6: Calculation of the Ammonia Production If we feed 10 g of N 2 and 10 g of H 2 into a reactor; (A) What is the maximum amount of NH3 that can be produced? (B) Which specie is a limiting reactant? (C) What is the %excess of an excess reactant? Knowing that MW. of N 2 =28 MW. of H 2 = 2 MW. of NH 3 = 17 N 2 + 3H 2 2NH 3

109 Conversion Conversion is the fraction of the feed that is converted into products; amount of feed that reacted %conversion = 100 x [ amount of feed introduced ] Or, Conversion = extent of reaction that occurs extent of reaction for complete reaction = Max = n i n io 0 n io

110 Yield 1. Yield based on feed: Yield 1 = amount of desired product obtained amount of the limiting reactant fed 2. Yield based on reactant consumed: Yield 2 = amount of desired product obtained amount of the limiting reactant consumed 1. Yield based on theoretical consumption of the limiting reactant: Yield 3 = amount of desired product obtained amount of that product that would be obtained theoretically

111 Selectivity Selectivity is the ratio of desired product produced to the amount of other products co-produced. In most chemical processes, reactants are bought together with the object of producing a desired production, Unfortunately, reactants can usually combine in more than one way, and the product once formed may react to yield something less desirable. The result of these side reactions is an economic loss. Take an example on ethylene production by dehydrogenation of ethane; C 2 H 6 H 2 + C 2 H 4 Once hydrogen is produced, it can react with ethane and form methane; C 2 H 6 + H 2 2CH 4 Ethylene can also react with ethane to form propylene and methane; C 2 H 6 + C 2 H 4 C 3 H 6 + CH 4

112 Example 7: Production of Ethanol from Sugar Yeasts are living organisms that consume sugars and produce a variety of products. For example, yeasts are used to convert malt to beer, and convert corn to ethanol. The growth of S. cerevisiae on glucose under anaerobic conditions proceeds by the following reaction to produce biomass, glycerol, and ethanol. C 6 H 12 O NH CH 1.74 N 0.2 O C 3 H 8 O CO C 2 H 5 OH H 2 O Calculate the theoretical yield of biomass (in g. biomass per g. glucose) and yield of ethanol (in g. ethanol per g. glucose).

113 Example 8: Selectivity in the Production of Nanotubes A carbon nanotube may consist of a single wall tube or a number of concentric tubes. A single wall tube may be produced as unaligned structures or bundles of ropes packed together in an orderly manner. The structure of the carbon nanotubes influences its properties, such as conductance. In nanotechnology, numerous methods exist to produce nanotubes. For example, large amount of single wall carbon nanotubes can be produced by the catalytic decomposition of ethane over Co and Fe catalysts supported on silica. C 2 H 6 2C + 3H 2 C 2 H 6 C 2 H 4 + H 2 (a) (b) If you collect 3 gmol of H 2 and 0.5 gmol of C 2 H 4 in the products, what is the selectivity of C relative to C 2 H 4?

114 Carbon Nanotubes Structures Conceptual diagram of single-walled carbon nanotube (SWCNT) (A) and multi-walled carbon nanotube (MWCNT) (B) delivery systems showing typical dimensions of length, width, and separation distance between graphene layers in MWCNTs. THE JOURNAL OF NUCLEAR MEDICINE Vol. 48 No. 7 July 2007

115 Example 9: Determination of limiting reactant when there are more than two reactants involved Acrylonitrile (C 3 H 3 N) is produced in the reaction of propylene (C 3 H 6 ), ammonia (NH 3 ), and oxygen (O 2 ): C 3 H 6 + NH O 2 C 3 H 3 N + 3H 2 O The feed contains 10%mol propylene, 12%mol ammonia and 78%mol air. A fractional conversion of 30% of the limiting reactant is achieved. Taking 100 gmol of feed as a basis, determine (A) Which reactant is limiting? (B) %excess of the excess reactants (C) Molar amount of all products

116 Review of Mass Balance Equation Let m 0 = initial mass of the system m t = mass of the system at time t m i = mass of inlet stream(s) m e = mass of exit stream(s) m gen = mass of generated material(s) m con = mass of consumed material(s) m t m 0 = m i m e + m gen m con

117 Species Mole Balance When reaction occurs, equation of mass balance can be modified for each specie, replacing mass by mole. Let n A,0 = initial mass of specie A in the system n A,t = mass of specie A in the system at time t n A,i = mass of specie A in inlet stream(s) n e = mass of specie A in exit stream(s) n gen = mass of specie A generated n con = mass of specie A consumed n A,t n A, 0 = n A, i n A,e + n A,gen n A,con

118 The use of For an open and steady-state process n A,t na, 0 = 0 From the definition of extent of the reaction ( ), = n A,out n A,in A For reacting species (reactants); n A,con = n A,out n A,in = A For producing species (products); n A,gen = n A,out n A,in = A

119 Example of ammonia production (open, steady state process): N 2 + 3H 2 2NH 3 6 gmol NH 3 18 gmol H 2 15 gmol N 2 Reactor 9 gmol H 2 12 gmol N 2 Species mole balance can be written as follow; For H 2 ; n H2,con = n H2,out n H2,in = 9 18 = 9 gmol reacted For N 2 ; n N2,con = n N2,out n N2,in = = 3 gmol reacted For NH 3 ; n NH3,gen = n NH3,out n NH3,in = 6 0 = 6 gmol produced

120 Example 10: Reaction in which the fraction conversion is specified The chlorination of methane occurs by the following reaction: CH 4 + Cl 2 CH 3 Cl + HCl Determine the product composition if the conversion of the limiting reactant is 67%, and the feed composition in mole percent is 40% methane, 50% chlorine gas, and 10% nitrogen gas.

121 Processes involving multiple reactions For open, steady-state processes with multiple reactions, R n A,e n A,i = ij j Where ij = j = R = j=1 Stoichiometric coefficient of specie i in the reaction j in the minimal reaction set extent of reaction for the j th reaction, in which component i is present in the minimal set number of independent chemical reactions

122 The Minimal Set The minimal set is the set of independent chemical reactions among all the multiple chemical reactions occur in the process of interest. Consider the carbon dioxide (CO 2 ) generation; C + O 2 CO 2 (1) C + ½ O 2 CO (2) CO + ½ O 2 CO 2 (3) The independent reactions are (1) and (2), since reaction (3) is the operation of (1) and (2).

123 Example 11: Material balances involving two ongoing reactions Formaldehyde is produced by catalytic oxidation of methanol by the following reaction, CH 3 OH + ½ O 2 CH 2 O + H 2 O (A) Unfortunately, under the conditions used, a significant portion of formaldehyde can react with oxygen to produce carbon monoxide. CH 2 O + ½ O 2 CO + H 2 O (B) Methanol and twice the stoichiometric amount of air needed for complete oxidation of methanol are fed to the reactor. This results 90% conversion of methanol, and 75% yield of formaldehyde (based on reaction A). Determine the composition of product leaving the reactor.

124 Analysis of Bioreactor (unsteady-state process) In the anaerobic fermentation of grain, the yeast Saccharomyces cerevisiae digests glucose from plants to form ethanol (C 2 H 5 OH) and propenoic acid (C 2 H 3 CO 2 H) by the following reactions: Reaction 1: C 6 H 12 O 6 C 2 H 5 OH + 2CO 2 Reaction 2: C 6 H 12 O 6 C 2 H 3 CO 2 H + 2H 2 O In a process, a tank is initially charged with 4,000 kg of a 12% glucose solution (in water). After fermentation, 120 kg of CO 2 have been produced and 90 kg of unreacted glucose remain the the broth. What are the weight percent of ethanol and propenoic acic in the broth at the end of the process?

125 Combustion Process Combustion is a reaction of substance with oxygen, with the association of energy released and generation of product gases such as H 2 O, CO 2, CO and SO 2. Most combustion uses air as a source of oxygen. We usually approximate the composition of air as it contains 79%mol N 2 and 21%mol O 2, with MW =29 O entering - O required O required 2 2 %Excess Air = x 100 2

126 Example 12: Excess Air Calculation Fuels other than gasoline are being used for motor vehicles because the create lower levels of pollutants than gasoline does. Compressed propane (C 3 H 8 ) is one such proposed fuel. Suppose that 20 kg of C 3 H 8 is burned with 400 kg of air to produce 44 kg of CO 2 and 12 kg of CO. What is the percent excess air? C 3 H 8 + 5O 2 3 CO 2 + 4H 2 O

127 Element Balance Element balance is useful when the exact stoichiometric equations are not known. For element balance, there will be no generation nor consumption terms in the material balance equations. All elements are conserved through out the reactions. Example 13: Octane cracking process produces the cracked products with the following composition. C 3 H %mol C 4 H %mol C 5 H %mol Determine the molar ratio of hydrogen to octane reacted for this process.

128 Example 14: Combustion of Fuel Component of coal (Fuel) %wt C H 4.45 O 3.36 N 1.08 S 0.70 Ash 7.36 Total Component of Stack gas (Product) %mol CO2 + SO O2 4.0 N Total A local utility burns coal and report the gas product as tabulated. Moisture in the fuel was 3.9%wt and the air contains lb water/lb dry air. The refuse showed 14% unburned coal, with remainder being ash. You are asked to check the consistency of the report, and find the percent excess of air used.

129 AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri Material Balance for Multi-Unit Operations

130 Process Flowsheet (Flow Chart) 1. Process Information: key properties data, market conditions, technical information, etc., are collected. 2. Input/Output Diagram: All major inputs and outputs streams and their stoichiometric balances are illustrated, according to the potential chemical pathways. 3. Functions Diagram: All the major functions of the process and materials flow to and from those functions are specified. The functions include reaction boxes, separation boxes, finishing boxes, etc. 4. Operations Diagram: Detail technologies used such as types of reactor, operating conditions, etc. are specified. 5. Process Flowsheet: Details for each function are described, ready to use.

131 Process Flowsheet Development

132 Sodium Dodecylbenzene Sulfonate Surfactant Manufacturing Input/Output Diagram

133 Function Diagram

134 Operation Diagram

135 Process Flowsheet

136 MB for Multiple Units Process

137 Example 1: on MB for Multi-units Process

138 Example 2: on MB for Multi-units Process

139

140 Recycle System The stream containing the recycled material is known as a recycle stream. Because of relatively high cost of industrial feeds, recycle of unused reactants/feed is a common practice. Recovery of catalyst, circulation of working fluids, dilution of process streams, control of process variables, etc., are all reasons for applications of recycle system. Recycle ratio = mass of recycle/mass of fresh feed

141 Example 3: Evaporative Crystallizer Forty-five hundred kilograms per hour of a solution that is one-third K 2 CrO 4 by mass is joined by a recycle stream containing 36.4% K 2 CrO 4, and the combined stream is fed into an evaporator. The concentrated stream leaving the evaporator contains 49.4% : this stream is fed into a crystallizer in which it is cooled (causing crystals of K 2 CrO 4 to come out of solution) and then filtered. The filter cake consists of crystal and a solution that contains 36.4% K 2 CrO 4 by mass; the crystals account for 95% of total mass of filter cake. The solution that passes through the filter, also 36.4% K 2 CrO 4, is the recycled stream. Calculate the rate of evaporation, the rate of production of crystalline K 2 CrO 4, the feed rates that the evaporator and the crystallizer must be designed, and the recycle ratio

142 The flowchart of steady state process to recover crystalline potassium chromate (K 2 CrO 4 ) from an aqueous solution of this salt is shown below.

143 Recycle with Chemical Reaction Overall Conversion: based on streams that enter and leave the overall process Overall Conversion = moles of reactants in fresh feed moles in the output of the overall process moles of reactant in fresh feed Single-pass Conversion: based on streams that enter and leave the reactor Overall Conversion = moles of reactants fed into reactor moles that exiting the reactor moles of reactant fed into reactor

144

145 Example 4: Cyclohexane Production Cyclohexane can be made from the reaction of benzene and hydrogen gas. C 6 H 6 + 3H 2 C 2 H 12 After the reaction in the reactor, the gross product is separated to spit the net product and the recycle stream. Determine the ratio of the recycle stream to the fresh feed stream if the overall conversion through the reactor is 95% and the single-pass conversion through the reactor is 20%. Assume that 20% excess of hydrogen is used in the fresh feed, and that the composition of recycle stream is 22.74mol% benzene and 78.6mol% hydrogen.

146 Example 5: Fructose Production from Glucose Immobilized glucose isomerase is used as a catalyst in producing fructose from glucose in a fixed-bed reactor (water is a solvent). For the system shown below, what percent conversion of glucose results on one pass through the reactor when a ratio of the exit stream to the recycle stream in mass units is equal to 8.33? Recycle Feed 40% glucose in water 4% fructose Fixed-bed Reactor Product

147 Purging A problem may arise in processes that involve recycling. 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 product. If nothing were done about this situation, the substance would continuously enter the process and would have no way of leaving; it would therefore steadily accumulate, making the attainment of steady state impossible. To prevent this buildup. A portion of the recycle stream must be withdrawn as a purge stream to rid the process of the substance in question.

148 Example 6: on Purging in Ethylene Oxide Production

149 Example 7: Recycle and Purge in the synthesis of Methanol Methanol is produced in the reaction of carbon dioxide and hydrogen: CO 2 + 3H 2 CH 3 OH + H 2 O 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 unused reactants and inerts are recycled to the reactor. To avoid buildup of the inerts in the system, a purge stream is withdrawn from the recycle. The feed to the reactor contains 28.0 mole% CO 2, 70.0mole% H 2, and 2.0 mole% inerts. The single-pass e conversion of hydrogen is 60%. Calculate the molar flow rates and molar compositions of the fresh feed, the total feed to the reactor, the recycle stream, and the purge stream for a production rate of 155 kmol CH 3 OH /hr.

150 Recycle and Purge in the synthesis of Methanol

151 Bypass Bypass is a fraction of feed stream that is diverted around the unit and combined with the output streams of such a unit. By varying the fraction of the feed that is bypassed, we can vary the composition and properties of the product.

152 A stream containing 5.15wt% chromium is contained in the wastewater from a metal finishing plant. The wastewater stream is fed to a treatment unit that removes 95% of the chromium in the feed and recycles it to the plant. The residual liquid stream leaving the treatment unit is sent to a waste lagoon. The treatment unit has a maximum capacity of 4500 kg wastewater/hr. If the wastewater leaves the finishing plant at a rate higher than the capacity of the treatment unit, the excess by passes the unit and combines with the residual liquid leaving the unit, and the combined stream goes to the waste lagoon. Wastewater leaves the finishing plant at a rate of 6,000 kg/hr. Calculate the flow rate of liquid to the waste lagoon, and the mass fraction of Cr in this liquid.

153 Example 8: Bypassing wastewater to waste Lagoon

154 AE 205 Materials and Energy Balances Asst. Prof. Dr. Tippabust Eksangsri Energy Balances on Chemical Processes

155 Thermodynamics system surroundings system boundary

156 Forms of Energy 1. Energy Possessed within the system 1.1 Internal Energy 1.2 Kinetic Energy 1.3 Potential Energy 2. Energy transferred across the boundary 2.1 Heat 2.2 Work

157 Forms of energy that possessed within the system Define E = total energy (kj), e = E/m (kj/kg) In an absence of exerted energy, E = U + KE + PE when, or e = u + ke + pe U = internal energy (kj), u = U/m (kj/kg) KE = kinetics energy (kj), ke = KE/m (kj/kg) PE = potential energy (kj), pe = PE/m (kj/kg)

158 Internal energy (U) Kinetics energy (KE) Potential energy (PE) He v He He He He He He He He He He He He He KE mv 2 2 PE mgz z Related to molecular structure and activities in a system When m = mass of a car v = velocity of a car When m = mass of an airplane g = gravitational acceleration z = elevation of an airplane

159 Point Function (State Function) Properties such as T, P, v are point function (or state function), they have exact differentials designated by d For example; Volume change by process A is dv A ; dv A = V 2 V 1 = ΔV A Volume change by process B is dv B ; dv B = V 2 V 1 = ΔV B It means the volume change by process A and B are the same although the system takes different path.

160 Energy Transfer by Work Work (W) is an energy transfer associated with a force (F) acting through a displacement (s), when both are measured in the same direction. It is a path function. δw = F ds (1) Integration of Eq. (1) from state 1 to state 2 gives amount of work done 2 2 W = F ds (2) 1 1

161 Work has magnitude and direction! Formal sign convention of work: Work done by the system is positive (W > 0) Work done on the system is negative (W < 0) surroundings Work done by system system Work done on system

162 The time rate at which work is done by or on the system is called power P = δw /dt (Watt, hp) (3) Efficiency is defined as E = actual work maximum work

163 Energy Transfer by Heat (Q) A form of energy transferred solely by a temperature difference between system and surroundings A path function A positive sign of heat refers to heat transfer into a system (from surroundings) A negative sign of heat refers to heat transfer from a system (to surroundings)

164 Define: Q = heat amount (kj or Btu) q = heat transfer per unit mass q Q m (kj/kg or Btu/lb m ) Q = rate of heat transfer (Watt or Hp) Q Q dt (4)

165 A Process with heat transfer surroundings system Q out (-) Q in (+) The system is under an adiabatic process when Q = 0 Adiabatic process Isothermal process

166 Modes of heat transfer Heat transfer by conduction Heat transfer by convection Heat transfer by radiation

167 Enthalpy (H,h) Enthalpy is a combined property of internal energy (U) and pressure-volume product (PV, or Flow Work). H = U + PV (kj) or h = u + Pv (kj/kg)

168 Flow work Control volume involves mass flow across their boundary and some work is required to push the mass into or out of the control volume. This work is called flow work or flow energy. If Then, P = fluid pressure A = cross-sectional area of the fluid L = distance of fluid movement F = PA W flow = PAL = PV (5) W flow is a combination property for flowing system only!

169 Specific heat (C v and C p ) Specific heat is the energy required to raise the temperature of unit mass of substance by 1 degree Specific heat at constant volume, C v, occurred when the system s volume is constant C v u T v (kj/kg K) or (kj/kmol K) Specific heat at constant pressure, C p, occurred when the system s pressure is constant (6) C p h T p (kj/kg K) or (kj/kmol K) (7)

170 Internal energy, enthalpy and specific heats of ideal gas From ideal gas EOS: Pv = RT (8) Enthalpy: h = u +Pv (9) Combine (1) and (2) h = u +RT (10) Since u = u(t), then h = h(t)

171 Internal energy, enthalpy and specific heats of ideal gas (cont.) Then specific heats are function of T only: du = C v (T) dt (11) And dh = C p (T) dt (12) Regarding Eq.(23) h = u +RT (13) dh = du +RdT (14) Replacing Eq.(12) and (13) into (14) C p = C v +R (15)

172 Internal energy, enthalpy and specific heats of solid and liquid Solid and liquid are incompressible substance; dv = 0 and C p = C v = C (16) Internal energy change (du): du = C av (T 2 T 1 ) (17) Enthalpy change (dh): dh = du + vdp = C av (dt) + vdp (18)

173 The 1 st law of Thermodynamics (the law of energy balance) E E in E out (19) when and E = Energy change of the system Contributed by the change of internal energy, kinetic energy and potential energy E in E out = Energy transfer across the system boundary Contributed by work, heat and mass flow across the boundary

174 Application of Energy Balances for Non-reactive System Simplified Energy Balance Equation: (U + PE + KE) system = Q - W + (H + PE + KE) flows (20) When U = internal energy KE = kinetic energy PE = potential energy Q = heat transfered to the system W = work done on the system H = enthalpy = U + PV

175 The three common systems found in engineering applications are 1. Closed system 2. Open system with steady-state flow 3. Open system with unsteady-state operation

176 Energy changes on Non-reactive systems Changes in pressure at constant temperature Changes in temperature at constant pressure Phase change at constant pressure and temperature Mixing or dissolving at constant pressure and temperature

177 Application of Energy Balances for Non-reactive System Closed (and stationary) Systems: There will be no PE nor KE for the system. There is no material flow across the boundary. U system = Q - W (21)

178 Example 1: reaction Energy Balance on a closed system without 10-lbm of CO 2 at 80F are stored in a fire extinguisher that has a volume of 4.0 ft 3. How much heat must be removed from the extinguisher so that 40% of CO 2 becomes liquid? CO 2 CO 2 Initial state: CO 2 gas T 1 = 80 F m 1 = 10 lbm V 1 = 4.0 ft 3 Final state: 40% liquid m 2 = 10 lbm V 2 = 4.0 ft 3

179

180 Application of Energy Balances for Non-reactive System Open system with steady-state flow: There is no change of system s energy Total energy flows in is equal to total energy flows out E in = E out (22) (H+PE+KE) incoming flows + Q in + W done on system = (H+PE+KE) outgoing flows + Q out + W done by system

181 Example 2: Mixing 2 streams of sucrose solutions with different concentrations; Feed 1 with mass fraction of sucrose of 0.1 and Feed 2 with mass fraction of sucrose of 0.5, will give the 100 kg sucrose solution product with 0.3 mass fraction of sucrose, as specified by customer s demand. Given that the specific enthalpy for Feed 1, Feed2 and Product are 200 kj/kg, 300 kj/kg, and 275 kj/kg, respectively. Find (1) ratio of Feed 1 and Feed 2 to obtain the product specified, and (2) the amount of heat transfer between the system and surroundings.

182 Solution: Basis: 100 kg Product Assuming that this is a steady-state flow system, Total material balance is F 1 + F 2 = P Sucrose balance is 0.1 F F 2 = 0.3(100 kg) Therefore, F 1 = 50 kg and F 2 = 50 kg Ratio of Feed1 to Feed 2 is 1 ANS Furthermore, energy balance shall be used to find hear transfer amount. Total energy balance is h 1 F 1 + h 2 F 2 + Q = -h p P (200 kj/kg)(50 kg) + (300 kj/kg)(50 kg) + Q = (275 kj/kg)(100 kg) Q = 2,500 kj ANS

183 Example 3: Energy Balance on an open, steady-state system without reaction Hot gas 100 kmol/min at 500 C is being cooled to 300 C by transferring heat to the liquid water that enters at 20 C and exits at 213 C, in an insulated heat exchanger. The cooling water does not mix with the gas. Calculate the mass flow rate of water stream. m W,1 T W,1 m G,3 1 m 3 4 G,4 T G,3 2 T G,4 gas %mol CO 2 20 N 2 10 CH 4 30 H 2 O 40 m W,2 T W,2

184 Given information for Example 3 Data retrieved from reference CD Component in hot gas h from 3 to 4 (kj/kmol) CO 2-9,333 N 2-6,215 CH 4-11,307 H 2 O -7,441 Data retrieved from steam table T ( C) H f (kj/kg)

185 Application of Energy Balances for Non-reactive System Open system with unsteady-state operation: Mass (or mole) of the system change with time. Energy possessed by system changes with time. (U + PE + KE) system = Q - W + (H + PE + KE) flows Neglect the changes of KE and PE, U final U initial = Q W + H incoming flows - H outgoing flows (23)

186 Sensible Heat Sensible heat is a specific enthalpy change associated with temperature change of pure substance due to the process in a single phase system. Where

187 Heat Capacity of a Mixture For a mixture of gases or liquids, calculate the total enthalpy changes as the sum of the enthalpy changes for the pure mixture components. (24)

188 Latent Heat Latent heat is a specific enthalpy change associated to phase transition at constant pressure and temperature. The specific enthalpy change when substance changes from liquid to vapor phase, or oppositely, is a Latent heat of vaporization; h vap. The specific enthalpy change when substance changes from solid to liquid phase, or oppositely, is a Latent heat of fusion; h fus or Heat of melting; h melt

189 Example 4: Energy Balance on a condenser

190 Data required for Example 4 Substance Cp (kj/mol C) Acetone (liq) x10-5 T Acetone (vap) x10-5 T 12.78x10-8 T 2 Nitrogen (gas) x10-5 T x10-8 T 2 h vap,ac = kj/mol

191 Heat of Solution (or Mixing) The heat of solution; h sol, is defined as enthalpy change for a process in which 1 mole of solute is dissolved in r moles of a liquid solvent at constant temperature. As r becomes large (solution is dilute), h sol approaches a limiting value. The heat of mixing has the same meaning as the heat of solution when the process involves mixing two fluids rather than dissolving solid or gas in a liquid.

192

193 Example 5: Concentration of Sulfuric acid A 5.0 wt% H 2 SO 4 solution at 60 F is to be concentrated to 40.0 wt% by evaporation of water. The concentrated solution and water vapor emerge from the evaporator at 180 F and 1 atm. Calculate the rate at which heat must be transferred to the evaporator to process 1,000 lb m /h of the feed solution.

194 Application of Energy Balances for Systems with Chemical Reactions Concerning the chemical reactions, more terms of energy involved to the reactions are applied; The Standard Heat (Enthalpy) of Formation The Heat (Enthalpy) of Reaction The Sensible Heat (Enthalpy Change) The Heat (Enthalpy) of Combustion

195 The Standard Heat of Formation ( H F) The standard heat of formation is the enthalpy change associated with the formation of 1 mol of a compound from its constituent elements and products in their standard state of 25 C and 1 atm. Example: The standard heat of formation for CO 2 is obtained from the reaction of C-element (C) and O-element (O 2 ), C(s) + O 2 (g) CO 2 (g) [1] H F,CO2 is heat transferred from a steady-state reaction [1]. At standard state, Q = kj/kmol CO 2 = H F,CO2

196 Quiz: Calculate the standard heat of reaction for the following reaction. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Given that; Compound Standard specific heat of formation (kj/gmol) NH O 2 0 NO H 2 O

197 The Standard Heat of Reaction ( H RXN ) The standard heat of reaction is the enthalpy change when stoichiometric quantities of reactants in the standard state (25 C and 1 atm) react completely to produce the products. For example: Consider the steady-state flow process for Benzene (C 6 H 6 ) production from Cyclohexane (C 6 H 12 ), at 25 C and 1 atm. C 6 H 6 (g) + 3H 2 (g) C 6 H 12 (g) [2] The energy balance is reduced to be Q = H RXN = (1) H F,C6H12 + (-1) H F,C6H6 + (-3) H F,H2

198 At 25 C and 1 atm C 6 H 6 C 6 H 12 Reactor H 2 C 6 H 6 (g) + 3H 2 (g) C 6 H 12 (g) [2] Compound Standard specific heat of formation (kj/gmol) C 6 H H 2 0 C 6 H

199 When the reaction is not completed. Regarding the process to produce cyclohexane from benzene, if the extent of reaction is given as, n i,out = n i,in + i H RXN = [(1) H F,C6H12 + (-1) H F,C6H6 + (-3) H F,H2 ] Suppose the fractional conversion is given as equal to 0.80, and benzene is a limiting reactant, fed to the process for 1 mole. Find the heat of reaction actually transferred.

200 Example 6: What can we do with H RXN? For the Bhopal process to produce carbaryl (C 12 H 11 O 2 N), methyl isocyanate (C 2 H 3 NO) which is a very toxic gas is generated during the process. C 3 H 2 NH 2 + COCl 2 C 2 H 3 NO + 2HCl (a) C 2 H 3 NO + C 10 H 8 O C 12 H 11 O 2 N (b) To eliminate the generation of methyl isocyanate, an alternative process for green chemistry is proposed. C 10 H 8 O + COCl 2 C 11 H 7 O 2 Cl C 11 H 7 O 2 Cl + CH 3 NH 2 C 12 H 11 O 2 N + HCl (d) (c) Show the alternative process is safer than the Bhopal process.

201 Example 7: What if the reaction does not take place at standard conditions? An inventor believes that he has developed a new catalyst that can make the gas phase reaction as shown here; CO 2 + 4H 2 2H 2 O + CH 4 The conversion of CO 2 is 40%. Assume that 1.5 mole of CO 2 enters the reactor at 700 C together with 4 moles of H 2 at 100 C. Determine the heat of reaction if the exit gas leaves at 1 atm and 500 C.

202 Heat of Reaction and Sensible Heat CO 2 at 700 C H 2 at 100 C Sensible Heats Sensible Heats Actual Reaction Product(s) at 500 C (H T H 25 ) Reactants Reaction at 25 C and 1 atm (H 25 H T ) Products H RXN

203 Example 8: Batch and reactive process For the commercial production of citric acid (C 6 H 8 O 7 ) in a batch process, three different phases occur for which the stoichiometry are slightly different. Initial phase, between hr; 1 gmol glucose gmol O 2 (g) 3.81 gmole biomass gmole citric acid gmol CO 2 (g) gmole polyols Medium phase, between hr; 1 gmol glucose gmol O 2 (g) 1.54 gmole biomass gmole citric acid gmol CO 2 (g) gmole polyols Late phase, between hr; 1 gmol glucose gmol O 2 (g) gmole polyols 0.86 gmole citric acid gmol CO 2 (g)

204 In anaerobic batch process, 30% glucose solution at 25 C is introduced into a fermentation vessel. Citric acid is to be produced by using the fungus Aspergillus niger. Stoichiometric sterile air is mixed with a culture solution by a 100 hp aerator. Only 60% overall of the glucose supplied is eventually converted to citric acid. The initial phase is run at 32 C, the medium phase is run at 35 C, and the late phase is run at 25 C. Base on the given data, how much net heat has to be added or removed from the fermenter during the production of a batch of 10,000 kg of citric acid. Aerator 100 hp 30% glucose 25 C 32 C 35 C 25 C 10,000 kg Citric acid At the beginning Initial phase Medium phase Late phase At the End

205 Example 9: M& E Balances for Multiple Units Process Air 60 F 450 F Heat Exchanger Reactor 1 CO(g) 60 F CO(g) CO 2 (g) N 2 (g) 1400 F Reactor F 800 F Steam Gen Products 1000 F SO 2 2,200 lbmol/hr 77 F Sat. steam At 450 psig H 2 O 60 F

206 Example 9: (Cont.) CO is burned with 80% of the theoretical air in reactor 1. The combustion gases are used to generate steam, and also to transfer heat to the reactants in reactor 2. A portion of the combustion gases that are used to heat the reactants in reactor 2 are recycled. SO 2 is oxidized in reactor 2. Calculate the moles of CO burned per hour in reactor 1. The gas involved in the SO 2 oxidation do not come in direct contact with the combustion gas used to heat the SO 2 reactants and products. Data for reactor 2 are: Reactants Mole fraction T ( F) Products Mole fraction T ( F) SO O Total SO O SO Total 1.000

207 The Heat of Combustion ( H C) Standard heat of combustion was commonly used to calculate the enthalpy changes when chemical reactions occur. The conventional used are as follows: 1. The compound is oxidized with oxygen or some other substances to the products CO 2 (g), H 2 O(l), HCl(aq), and so on. 2. The reference conditions are still 25 C and 1 atm. 3. Zero values of H C are assigned to certain substances such as CO 2 (g), H 2 O(l), HCl(aq) and O 2 (g). 4. Stoichiometric quantities react completely. H C (25 C) i Products n H C, i i Reactan ts n H C, i

208 Example 10: Conversion of H C to the corresponding H F The heat of combustion of liquid lactic acid (C 3 H 6 O 3 ) is -1,361 kj/gmol [Electronic J. of Biotech., 7 No. 2, 2004]. What is the corresponding standard heat of formation? The formation of lactic acid can be considered as; 3C(s) + 3H 2 (g) + 1.5O 2 (g) C 3 H 6 O 3

209 Psychrometric Charts and Applications Psychrometric charts, or Humidity charts give information on humidity of mixture of airwater system The applications of psychrometric charts are humidification, dehumidification, drying and Cooling tower.

210 Humidity (H) Humidity means the ratio of mass of water vapor(moisture) per unit mass of dry air H = m water m dry air = 18P water 29(P total P water )

211 Relative Humidity (RH) For any gas mixture, When, RH = P vapor P P vapor = partial pressure of vapor in gas mixture P* = vapor pressure of the vapor component Note: Vapor pressure is a pressure of vapor component when it is saturated in gas mixture at a given temperature and mixture.

212 %Relative Humidity (%RH) Consider the moisture (water) in air, When, %RH = P water P water P water = partial pressure of moisture in air mixture P water * = vapor pressure of the moisture at the given temperature

213 Wet-bulb Temperature (T wb ) Wet-bulb temperature is the equilibrium temperature of water vapor in the air. Dry-bulb Temperature (T db ) Dry-bulb temperature is system temperature.

214 Dew Point Temperature (T dp ) Dew point temperature is the temperature at which the water vapor in the air will first condense at constant pressure and composition.

215 Psychrometric Charts

216 Psychrometric Charts Reading

217 Psychrometric Charts Reading

218 Psychrometric Charts Reading