Mass and Energy Balances
NEAR EAST UNIVERSITY ENGINEERING FACULTY FOOD ENGINEERING DEPARTMENT FDE 201 MATERIAL & ENERGY BALANCES LECTURE NOTES PREPARED BY : Filiz ALSHANABLEH NICOSIA – 2012
Table of Content Chapter 1 Dimensions, Units, and Unit Conversion 1 Chapter 2 Introduction to Process Variables and Basic Food Engineering Calculations 21 Chapter 3 Introduction to Material Balances 41 Chapter 5 Material Balances for Multiple Units 87 Chapter 6 Energy Balances 121 Chapter 7 Material & Energy Balances of Some Selected Unit Operations 157 APPENDIX 1 : SYMBOLS, UNITS AND DIMENSIONS 182 APPENDIX 2 : UNITS AND CONVERSION FACTORS 186
Chapter 1: Dimensions, Units, and Unit Conversion
Chapter 1 Dimensions, Units, and Unit Conversion Learning Objectives Upon completing this Chapter, you should be able to: • understand the definitions and physical meanings of dimensions and units • perform operations (i.e. addition, subtraction, multiplication, and division) of numbers accompanied by corresponding units • identify units commonly used in engineering and scientific calculations, including those in cgs, SI, and American Engineering (AE) systems • convert one set of units (or one unit) associated with numbers (or a number) or in an equation into another equivalent set of units (or another unit), using a given conversion factor • explain and utilize the concept of dimensional homogeneity (consistency) of equations to identify units of specific numbers in those equations • appreciate the importance and rationale of dimensionless groups (quantities) James Clark Maxwell, a Scottish Mathematician and Theoretical Physicist (1831– 1879) expressed the definition of unit that “Every physical quantity can be expressed as a product of a pure number and a unit, where the unit is a selected reference quantity in terms of which all quantities of the same kind can be expressed”. Physical Quantities • Fundamental quantities • Derived quantities Fundamental Quantities • Length • Mass • Time • Temperature • Amount of substance • Electric current • Luminous intensity
Examples of Derived Quantities • Area [length × length or (length)2] • Volume [(length)3] • Density [mass/volume or mass/(length)3] • Velocity [length/time] • Acceleration [velocity/time or length/(time)2] • Force [mass × acceleration or (mass × length)/(time)2] Dimension A property that can be measured directly (e.g., length, mass, temperature) or calculated, by multiplying or dividing with other dimensions (e.g., volume, velocity, force) Unit A specific numerical value of dimensions. Systems of Units (commonly used in engineering and scientific calculations) • cgs (centimetre, gram, second) • SI (Le Système International d’ Unitès) • American Engineering (AE) (or fps : foot, pound, second) Units & Dimensions of Fundamental Quantities UNIT SYSTEM QUANTITY DIMENSION cgs SI AE • Length cm m ft L • Mass g kg lb M m • Time s s s t or θ • Temperature oC K o F T • Amount of Substance mol n _ _ • Electric current A A A I • Luminous intensity can* _ _ * stands for “Candela”
Examples of the Dimensions of Derived Quantities • Area (A ) [ A = L×L = L2 ] • Volume ( V ) [ V = L×L×L = L3 ] • Density (ρ ) [ρ = M / V = M / L3 ] • Velocity ( ) [ = L / t ] • Acceleration (a) [ a= V / t = ( L / t) /( t ) = L / t2 ] Example 1.1 Sir Isaac Newton (English physicist, mathematician, astronomer, philosopher, and alchemist: 1643– 1727) established a second law of motion equation that the force (F ) is the product of mass (m) and its acceleration (m), which can be described in the equation form as follows: F=ma What is the dimension of F (force)? From a previous page, the dimensions of • mass (m) ≡ M • acceleration (a) ≡ L / t2 The dimension of force (F ) can, thus, be expressed by those of fundamental quantities as follows F = ma ≡ ( M ) ( L / t2 ) F ≡ ( M )( L ) / ( t2 ) Example From a Physics or Chemistry courses, pressure (P ) is defined as “the amount of force (F ) exerted onto the area (A) perpendicular to the force” What is the dimension of P ? • Dimension of force F ≡ ( M )( L ) / ( t2 ) • Dimension of area (A) ≡ L2 Hence, the dimension of pressure is Dimensions and Units The “dimension” is the property that can be measured experimentally or calculated, and in order to express the physical quantity of a dimension, we use a pure number and its corresponding unit
For example, a ruler has a dimension of “length” (L ), its physical quantity can be expressed as 1 foot (ft) or 12 inches (in) or 30.45 centimetres (cm) Another example, Americans express their normal freezing point of water as 32 o F, while Europeans say that the normal freezing point of water at 1atm is at 0 oC We can see that a physical property [e.g., length (L ) or temperature (T )] with the same dimension may be expressed in different numerical value if it is accompanied with different unit. Units of Derived Quantities and Alternative Units The units of fundamental quantities of different unit systems are summarized on Page 3. What are the units for derived quantities? We can assign the unit (in any unit system) to each individual derived quantity using its dimension For instances, • the unit of area (A ), in SI system, is m2 2 , since its dimension is L • the unit of volume (V ), in AE system, is ft3 Example 1.2. Determine the units of density, in cgs, SI, and AE systems? Since the dimension of density (ρ ) is its corresponding units in • cgs unit system is g / cm3 • SI unit system is kg/ m3 • AE unit system is lb 3 m / ft What are “alternative” units? From our previous example we have learned that the dimension of force (F ) is F ≡ ( M )( L ) / ( t2 )
Hence, its corresponding units in • SI system is (kg)(m) / (s2) • AE system is (lb 2 m)(ft) / (s ) Have you ever heard that the unit of force (F ) is (kg)(m) / (s2)? To honour Sir Isaac Newton (1643-1727), who established the 2nd law of motion, a community of scientists gave the name of the unit of force as “Newton (N)”, which is defined as Newton (N) is an example of an alternative unit. Accordingly, instead of expressing the unit of pressure (P), in SI system, as since the dimension of pressure is we can, alternatively, write the unit of pressure, in SI system, as This comes from the fact that and that the units of force (F) and area (A), in SI system, is N and m2, respectively. However, the unit of pressure in SI system is expressed as “Pascal (Pa)”, which is defined as 1 Pa ≡ 1 (try proving it yourself that 1 Pa ≡ 1 ) In AE system, the unit of pressure is expressed as or psi (note that lb is the unit of force in AE system, not a unit of mass, and that “psi” stands for pound f force per squared inches” 1 N ≡ 0.224809 lb f Example 1.3. Work ( W ) is defined as “force (F ) acting upon an object to cause a displacement (L )”. What are the dimension and the corresponding unit of work, in SI system? 2 • Dimension of force F ≡ ( M )( L ) / ( t )
• Dimension of a displacement = L Hence, the dimension of work is W = FL ≡ ( M )( L 2 2 ) / ( t ) Accordingly, the unit of work, in SI system, is Alternatively, the unit of work, in SI system, can be expressed as W = FL ≡ (N)(m) Commonly, the unit of work, in SI system is expressed as “Joule (J)”, in which 1 J ≡ 1 (N)(m) Units of Work, Energy, and Heat We have just learned that the unit of work, in SI system, is J or (N)(m) or and from Physics courses, we learned that work, energy, and heat are in the same unit Is it true? From the definition of work: “force acting upon an object to cause a displacement” the unit of work can be expressed as J or (N)(m) or ,as mentioned above. How about the unit of “Energy”? Energy: • Potential Energy; Ep =mgL where g is an acceleration (a) caused by gravitational force Thus, in SI system, potential energy has the unit of E =mgL p
• Kinetic Energy; E 2 k = ½(m)(V) 2 Hence, the unit of kinetic energy, in SI system, is E = ½(m)(V) k It is clear that E and E are in the same unit, i.e. P K , and we have already got the fact that 1 J ≡ 1 (N)(m) ≡ Accordingly, we can conclude that work and energy are in the same unit Since, from Physics or Chemistry courses, both work and heat are the form of energy transferring between a system and surroundings, the unit of heat is as same as that of work. Units of Temperature A unit of temperature used in any calculations must be an absolute temperature unit Absolute temperature unit • SI K (Kelvin) • AE R (Rankine) T(K)=T( oC)+273.15* (1.1) T(R)=T( o F)+459.67** (1.2) * For convenience, the value of 273 is used ** For convenience, the value of 460 is used The Conversion of the Temperature Units between o o C and F Principle where Tnb = normal boiling point of water Tnf = normal melting/freezing point of water Hence,
o o T( F)=1.8T( C)+32 (1.3) o Example 1.4. The specific gravity of liquid is normally reported at 60 F in AE system. What is the equivalent temperature in SI system? Employing Eq. 1.3 yields o o T( F)=1.8T( C)+32 o 15.6 C Temperature Difference (ΔT ) Consider the following example: ΔT ( o o o C) = 15 C – 10 C = 5 oC ΔT (K) = (273+15) K – (273+10) K = 5 K Thus, it can be concluded that ΔT ( oC) = ΔT (K) (1.4) When considering in the same manner for o F and R, we shall obtain the fact that ΔT ( o F) = ΔT (R) (1.5)
How about the relationship between ΔT ( o o C) and ΔT ( F) or between ΔT (K) and ΔT (R)? The temperatures of 10 and 15 oC are equivalent to the temperatures in AE system of (using Eq. 1.3) T ( o o F) =1.8( C) + 32 T ( o o F) =1.8(10) + 32 and T ( F) =1.8(15) + 32 respectively Hence, the temperature difference between 15 and 10 oC, in AE system, is [1.8(15) + 32] − [1.8(10) + 32] = [1.8(5)] o F Accordingly, the relationship between ΔT ( o o F) and ΔT ( C) can be written in a general form as follows o o (1.6) ΔT ( F) =1.8ΔT ( C) When doing the same for ΔT (R) and ΔT (K), we obtain the following relationship (1.7) ΔT (R) =1.8ΔT (K) Units of Pressure In addition to Pa and psi, pressure can be presented in the unit of, e.g., • atm • bar • mm Hg • in H2O As same as temperature, the unit of pressure used in any calculations must be an absolute pressure unit. Operations with Units What is the correct answer for 9+5? • Is 14 the only correct answer? • Could 9 + 5 be 2? It is certain that 9 apples + 5 apples = 14 apples 9 oranges + 5 oranges = 14 oranges but 9 oranges + 5 apples = ? A fruit salad !!
Can 1 + 1 be 13 or 1 – 1 = 99? These are some examples of confusion caused by writing numbers without units. Some other worse examples Consider the following news “A Chinese air-traffic controller at Shanghai international airport directed the pilots of a Korean Airlines plane to take the plane to the altitude of 1,500 meters (the plane was at the altitude of ~1,000 meters at the time), but the pilots thought that it was 1,500 feet, which is equivalent to 455 meters. So, instead of climbing up, the pilots lowered the altitude of the plane. This misunderstanding in “unit” caused the plane to crash, which killed all crewmembers and another five people on the ground” (Modified from Wall Street Journal, June 6th, 2001, Page A22) “In 1999, the Mars Climate Orbiter was crashed to the Martian surface, because engineers forgot to convert units in SI system to AE one. This damage cost ~US$ 125 million!” (Modified from Basic Principles and Calculations in Chemical Engineering (7th ed.), Page 15) By attaching units to all numbers when performing any calculations, you can get the following benefits: • reduce/diminish the possibility of errors in your calculations • a logical approach to solve the problem rather than remembering a formula • easy interpretation of the physical meaning of the number you are dealing with Example 1.5. You want to calculate the mass (m) of substance A, when you are given a volume ( V ) and a density (ρ ) of 0.2 m3 3 and 1,250 kg/m , respectively, but you totally forgot the relationship between m, V, and . How would you do? Addition & Subtraction You can add or subtract numerical quantities only when they are in the same dimension. On top of that, to obtain the correct answer, the units of those numerical values must be the same. For example: You cannot carry out the following addition/subtraction; • 5 kg – 3 N (mass vs force) 3 3 • 45 m /kg + 250 m (specific volume vs volume) as the numbers in operations are in different dimensions.
You, however, can do the following addition/subtraction • 5 km + 5 mi 3 3 • 50 m /kg – 12 ft /lbm BUT you have to do the UNIT CONVERSION before carrying out an addition/a subtraction (unit conversion will be discussed in the next section) Multiplication & Division You can multiply or divide any units, but you cannot cancel the units unless they are identical For example: You can do the followings • ( 15 kg) (9.81 ) = 147.2 or 147.2 N • = 600 or 600 Pa • = 30 but you cannot do the following • = 30 • = 4 • = 2 since the units are different. Unit Conversion In engineering calculations, there are TWO commonly used unit systems: SI and American Engineering (AE) As a prospective engineer, you must be careful of handling all sorts of unit systems and be able to convert a given unit to another competently.
The advantages of SI system over AE one Consider the unit increment in AE system Length: 12 inches = 1 foot 3 feet = 1 yard 1,760 yards = 1 mile Mass: 16 ounces (oz) = 1 pound (lbm) 14 pounds (lbm) = 1 stone It is evident that the increment in unit is NOT systematic, which usually leads to confusion and errors On the other hand, the unit increment in SI system is systematic For example Length: 100 cm = 1 m 1,000 m = 1 km Mass: 1,000 mg = 1 g 1,000 g = 1 kg You can see that the unit increment, in SI system, is in the power of 10, and the incremental patterns are the same for (almost) all kinds of quantities. Can you think of any quantity in SI system whose unit increment is NOT in the power of 10? The power of 10 can be expressed by prefixes, and some commo nly used prefixes are • centi- (c) = 10-2 • milli- (m) = 10-3 • micro- (micro- or μ ) = 10-6 • nano- (n) = 10-9 • deci- (d) = 10-1 • kilo- (k) = 103 • mega- (M) = 106 • giga- (G) = 109
Examples 9 1 GBytes = 10 Bytes 6 15 MW = 15 × 10 Watts 30 kN = 30 × 103 N -2 35 cm = 35 × 10 m Example 1.6. Find the length in ft that is equivalent to 47.25 cm 1 m ≡ 100 cm and 1 m ≡ 3.28084 ft Thus, Example 1.7. An example of nano-sized semiconductor is ZnS (in a semiconductor plant, chemical engineers produce this kind of semiconductor). If its size is 1.8 nanometres (nm), what is the size in inches (in.) o 3 Example 1.8 At 4 C, water has a density of 1 g/cm . Liquid A has a density at the same temperature of 60 lb 3 m/ft . When water is mixed with liquid A, which one is on the upper layer? Can you answer that which substance is on the upper layer? 2 2 Example 1.9. Convert the mass flux of 0.04 g / (min. m ) to that in the unit of lb /( h.ft ) m
Dimensional Consistency (Homogeneity) As stated previously, the answers of adding and/or subtracting numerical quantities can be obtained only when the unit of each quantity is identical This is a basic principle of “dimension homogeneity (consistency)” The basic principle states that, in order to add, subtract, or equate any terms, each term must be in the same dimension and unit. By employing this principle, it leads to a conclusion that the numerical values in any non-linear forms (e.g., log, exp) must be dimensionless (i.e. have no unit) Example 1.10 What is the unit of R in an ideal-gas equation of state (EoS)? An ideal-gas equation of state (EoS) can be written as follows PV = nRT (1.8) where P has a unit of Pa (or N/m2) V has a unit of m3 n has a unit of mol T has a unit of K From the principle of dimensional homogeneity, it is required that each side of equation (i.e. EoS in this case) must be in the same unit Hence, the unit of R can be calculated by substituting the unit of each quantity into Eq. 1.8 and, then, rearranging the equation, as follows PV = nRT 3 (Pa)(m ) = (mol) R (K) Performing further unit conversions yields
Example 1.11. One of the real-gas EoS is that proposed by van der Waals (called van der Waals equation of state), which can be written as follows: Use the principle of dimensional homogeneity to determine the unit of constants a and b (note that the units of each quantity is as described in the previous Example). The principle of dimensional consistency states that, to add, subtract, and equate the terms, each term must have the same unit. Hence, the term • must be in the same unit as P • b must have the same unit as V 3 Accordingly, b is in the unit of m . Since the term must have the unit of Pa (i.e. the unit of P ), the unit of a can be calculated as follows; Dimensionless Groups or Quantities Quantities or properties that have NO unit. Mostly, used in “process design” and/or “scaling up/down” An example of dimensionless groups Reynolds number (Re) (used to describe the flow of fluid), which is defined as where ρ = density (unit in SI ≡ )[ dimension ≡ ] D = diameter (unit in SI ≡ m)[dimension ≡ L] V = velocity (unit in SI ≡ ) [dimension ≡ ] μ = viscosity (unit in SI ≡ )[dimension ≡ ]
Dimension of Re = ? Unit of Re = ? or (-1) Example 1.12. Use the principle of dimensional consistency to determine the units of the numerical values of 70.5 and 8.27 × 10-7 in the following empirical formula: ρ = 70.5exp(8.27 ×10−7 P) (1.10) where ρ = density (lb 3 m/ft ) P = pressure (lb 2 f/in ) The left hand side (LHS) of Eq. 1.10 has the unit of lb 3 m/ft Thus, the RHS must have the same unit; i.e. lb 3 m/ft Since the number in an exponential (exp) form must be dimensionless, the term exp(8.27 ×10−7 P ) has no unit Accordingly, the numerical value of 70.5 must have the unit of lb 3 m/ft , in order to enable each side of equation to have the same unit
The term 8.27 ×10−7 P must be dimensionless. Hence, the unit of 8.27 ×10−7 can be determined as follows −7 8.27 ×10 P −7 8.27 ×10 −7 8.27 ×10 In summary, the numerical values of • 70.5 is in the unit of • 8.27 ×10−7 is in the unit of
Problem Set 1 Dimensions, Units, and Unit Conversion PROB 1. a) In the Olympics, on average, those who enter the final round of the 100 m (men) track event can run 100 m within 10 seconds. Some said that these men run faster than the car travelling at the speed of 40 km/h. Is this comment correct? b) You are in the purchasing division, which is considering buying a new car. The prices of the cars A and B are almost identical. Other options are also almost the same. However, the fuel consumption rates are presented in different unit systems. The car A consumes the fuel at the rate of 28 miles/gal, while the fuel consumption rate for the car B is 9 km/L. Which car is you going to recommend the company to purchase, and why? c) The current prices of gasoline (ULG-95) in some countries are as follows: 3.361 U.S. dollars/U.S. gallon in California, U.S.A.; 4.93 Danish Krones/Litre in Denmark; 384.0 Japanese Yens/U.S. gallon; and 32.49 Thai Bahts/L. Compare the price of each country in the unit of “Euro/L”. Note that the current foreign exchange rates per U.S. dollar for Danish Krone, Japanese Yen, Thai Baht, and Euro are 5.09, 111.72, 33.81, and 0.706, respectively. d) To lower the fuel cost used by an airplane, it is instructed that the velocity if to be cut down from 525 mi/h to 475 mi/h, which leads to a cut in fuel consumption from 2,200 to 2,000 U.S. gallons/h. How many litre of fuel that can be saved for the distance of 1,500 km. PROB 2. Perform the unit conversions for the following questions a) 554 m4 4 /(day . kg) = ?? cm /(min g) b) 38.1 ft/s = ?? km/h c) 921 kg/m3 3 = ?? lbm/ft d) 42 ft2 2 /h = ?? cm /s PROB 3. A waste-water-treatment basin has the dimension: width × length × depth of 15 m × 50 m × 2 m, and the density of the waste water is 75.6 lbm/ft3. Calculate the total weight, in kg, of the waste water, if the basin is completely filled with it.
PROB 4. When a fluid flows through an area that causes any friction, pressure drop arises, and the pressure drop is proportional to the flow rate or the velocity of the fluid. One the of relationship between the pressure drop and the velocity of the fluid is proposed by Calvert, as follows: −5 ΔP = 5×10 2L , where ΔP = pressure drop, in psi or lb 2 f/in = gas velocity, in ft/s L = liquid flow rate, in US gallon/min a) What is a unit of the constant 5 × 10-5 b) What is the dimension of the constant 5 × 10-5 PROB 5. In a manual from a U.S. company, it is said that the unit of a material’s thermal conductivity (k) is , but in SI system, the unit of k is reported as . Is this possible?
Chapter 2: Introduction to Process Variables and Basic Food Engineering Calculations
Chapter 2 Introduction to Process Variables and Basic Food Engineering Calculations Learning Objectives Upon completing this Chapter, you should be able to: • define what mole is • define gram-mole, kilogram-mole, and pound-mole • convert from moles to mass and vice versa • define density and specific gravity • calculate a density of a substance given its specific gravity and vice versa • convert a composition of a mixture from one concentration unit to another • define mass, molar, and volumetric flow rates; and velocity Before we can proceed to the main content of this course (i.e. learn to carry out material and energy balance calculations), some technical terms from basic Chemistry and Physics courses necessary for material and energy balance calculations should be reviewed, in order to enable you to have a decent understanding of the materials in the remaining chapters. Mole Wilhelm Ostwald (Latvian/German chemist and a 1909 Nobel prize laureate: 1853– 1932) introduced the word “mole” in 1896 as follows one mole as molecular weight of a substance in grams In 1969, International Committee on Weights and Measures defined the “mole” as “the amount of a substance that contains as many elementary entities as there are atoms in 0.012 kg of carbon 12” (note that the word “entities” can be atoms, molecules, ions, or other particles) Element vs Compound Element = a substance that cannot be broken down into smaller substances (e.g., O, H, P, Na, S) Compound = a substance that comprises atoms of more than one elements (e.g., H O, H PO , SO ) 2 3 4 2 Entities for • element = atoms • compound = molecules From a basic Chemistry course: 1 mole ≡ 6.02214 × 1023 3 atoms/molecules ≡ 22.4 dm or L at STP
Atomic Weight or Molecular Weight vs Molar Mass To be exact, • atomic weight (AW) (in some textbooks, atomic mass) is “the compositional average mass of an element, averaged over the distribution of its isotopes in nature” (for example, atomic weight of CARBON = 12.011, which is averaged from the atomic weights of C-12 (AW = 12.0), C-13, and C-14) • molecular weight (MW) is “a summation of the weights of the atoms in a compound” Note that both AW and MW are dimensionless or have no unit • Molar mass (in some textbooks, molar mass and MW is used interchangeably) is the mass in grams equal to atomic/ molecular weight of one mole of a substance (either element or compound) Molar mass (or MW) has a unit of g/mole (g/mol) Example 2.1. • Atomic weight of O = 16.00 • Atomic weight of V = 50.94 (V = Vanadium) Hence, molar mass of • oxygen (O) = 16.00 g/mol , • vanadium (V) = 50.94 g/mol Example 2.2. A superconductivity material is the material that does not have an electrical resistance; an example of the superconductivity material is Ba2Cu16O24Y Calculate the molar mass (or MW) of this superconductor Given AW of Ba = 137.34, Cu = 63.55, O = 16.00, and Y (Yttrium) = 88.91 MW is the summation of AW, and molar mass of each substance is equal to MW, but has a unit of, e.g., g/mol Hence, molar mass (or MW) of Ba2Cu16O24Y can be calculated as follows: [2×137.34] + [16× 63.55] + [24×16.00] + [1× 88.91] = 1,764.39 g/mol Example 2.3. How many moles of MeOH if it weighs 32.04 kg? (do you know what MeOH is?) MeOH is, in fact, “methanol”, CH OH (CH O) 3 4 AW of C = 12.01, H = 1.008, and O = 16.00 Hence, MW of CH OH = CH O = [1×12.01] + [4×1.008] + [1×16.00] = 32.04 3 4 Thus, molar mass of MeOH = 32.04 g/mol
It is given that the mass of MeOH is 32.04 kg, in which 32.04 kg = 32.04 × 103 g 3 32.04 × 10 g = ? mole 3 = 1 × 10 mol = 1 kmol = 1 kg-mol Example 2.4. How many moles of MeOH if it weighs 32.04 lbm? From a conversion-factors table: 32.04 lb 2 m = 32.04 × (4.536 × 10 ) g Thus, for MeOH, 32.04 g ≡ 1 mol 2 = 1× (4.536×10 ) mol = 1× (4.536×102 ) g-mol = 1 lbm-mol So, in conclusion • g mol (or g-mol) • kg kmol (or kg-mol) • lbm lbm-mol Hence, the units of molar mass could be • g/mol or g/g-mol • kg/kmol or kg/kg-mol • lbm/lbm-mol Proposed (New) Method for Calculations Concerning “Mole” and “Mass” Example 2.5. How many moles of C H OH in 100 kg of C H OH ? (MW of C H OH = 46.07) 2 5 2 5 2 5 From the given data, molar mass of
C H OH (or EtOH) is 46.07 kg/kg-mol; in other words, 2 5 1 kg-mol of EtOH ≡ 46.07 kg Thus, = 2.17 kg-mol EtOH 3 = 2.17 × 10 mol = 2,170 mol Example 2.6. How many lb -mol of C H O in 1,000 g of C H O (MW of C H O = 180.16) m 6 12 6 6 12 6 6 12 6 From the given information, molar mass of C H O (monosaccharide; e.g., glucose) is 6 12 6 180.16 lbm/lbm −mol Hence, = 0.0122 lb -mol C H O m 6 12 6 Density Density a “ratio of mass per volume”; in other words, it is mass of a unit volume of a specified substance. Hence, a density can be written in an equation form as follow (2.1) Note that, a specific volume (v ) is defined as volume of a unit mass Accordingly, (2.2) From Eqs. 2.1 & 2.2, dimensions of • • Thus, the units of density and specific volume are, e.g.,
• for density: • for specific volume: Specific Gravity (SG) of LIQUIDS Specific gravity of liquid is defined as follows (2.3) Normally, T o o o 2= Tref = 15.6 C or 60 F or 288.7 K (in some textbooks, Tref = 4 C) Note that SG has no unit; in other words, SG is dimensionless In addition to SG, there are a number of scales that represent specific gravity and/or density, e.g., • Twaddell (Tw ) = 200 (SG288.7/288.7 −1) o o • Baume ( Be) o For liquid heavier than water o For liquid lighter than water • (used mostly for petroleum products) • (commonly used in sugar industry)
Density of GASES Let’s start with an equation of state of ideal gases: (2.4) (Note that, in some textbooks, Rv is used instead of ) We have just learned recently that molar mass (or AW/MW ) = m / n (2.5) where m = mass and n = mole of a selected substance Hence, (2.6) Combining Eq. 2.6 with Eq. 2.4 gives (2.7) (note that ≡ gas constant) Rearranging Eq. 2.7 results in (2.8) Combining Eq. 2.2: with Eq. 2.8 yields (2.9)
Rearranging Eq. 2.9 gives (2.10) This indicates that densities of gases depends directly on T & P So, it is necessary to specify T & P when dealing a density of any gas. Specific Gravity (SG) of GASES “Specific gravity (SG) of gas is the ratio between density of any gas and that of air at the same T & P” Combining Eq. 2.10: with the definition that where = universal gas constant (in some textbooks, may be written as R ) yields ν (2.12) Following the same procedure, the density of air at given T & P can be obtained as follows (2.13) (2.12)/(2.13) results in SG of any gas, as follows (2.14)
Example2.7. Determine molar volume and density of ethane (MW = 30.07) at 101.3 kPa (1 atm) and 25oC (NTP – normal T & P) 3 Given R or = 8.314 m ⋅kPa/(kmol)(K) ν Rearranging an ideal-gas EoS or Eq. 2.4: yields (2.15) Note that ≡ molar volume ( V / n) Molar volume (v ) of ethane can thus be computed, using Eq. 2.15, as follows The density of ethane can be calculated by substituting corresponding numerical values into Eq. 2.10 (or Eq. 2.12), as follows Example 2.8. Calculate the density of air at NTP. Given MW of air = 28.97 The density of air can be calculated using Eq. 2.12, as follows
From the recent Examples, it is worth noting that 1) molar volumes at the same T & P are identical for all gases, as, according to Eq. 2.15 a molar volume (v ) depends only on T and P (this statement is valid only when all gases are assumed to be ideal gases) 2) the density of air (@ NTP – 25 oC & 1 atm) is ~1/1,000 times that of water (note that ρ 3 water = 1,000 kg/m ) Mixtures & Solutions Normally, chemical (food) engineers usually deal with not only pure substance, but we also deal with systems or processes that comprise more than one substance, which are called “mixtures” or “solutions” Hence, it is necessary to learn how to express the concentration of each substance (or species) in a mixture or solution, and how to calculate one form of concentration from the given form of concentration. From a basic Chemistry course, we have learned that a concentration or each species in a mixture and solution (soln) can be expressed as, e.g., • wt% • vol% • %wt/vol • molarity (molar or M) mol of solute/L of soln: or mol/L) • molality (mol of solute/1,000 g of solvent) In this course, in addition to the above concentration expressions, we are going to learn some more concentration expressions commonly used in chemical processes. Such forms of concentration include: • mole fraction and percentage • mass fraction and percentage • ppm & ppb To begin with, we will review how to convert one form of concentration to another.
Example 2.9. Convert a concentration of a 10 wt% H SO solution to the unit of molarity 2 4 (mol H SO /L soln: mol/L) 2 4 From the definition of wt%, we can write the following relationship: In this Example, we want to convert To convert • g H SO to mol (in “g-mol”) H SO 2 4 2 4 • g H SO solution to volume (in L or mL) of soln 2 4 what kind of “data” do we need? • g H SO g-mol H SO H2SO4 2 4 2 4 data needed = MW of H SO 2 4 • g solution L solution data needed = density of solution • MW of H SO = 98.08 g/g-mol 2 4 • ρ of 10 wt% H o SO soln = 1.064 g/mL (@ 25 C) 2 4 Accordingly, Thus, In conclusion, a sulphuric acid solution with the concentration of 10 wt% is equivalent to that of 1.085 g-mol/L .
Example 2.10. Convert the concentration of a 10 wt% H SO solution to the units of vol% 2 4 From a basic Chemistry course, we can formulate the following equation regarding vol% as follows: From our recent calculations , 100 g H SO soln ≡ 93.98 mL H SO soln 2 4 2 4 In order to convert g H SO4 to mL H SO , 2 2 4 what kind of data do we needed? DENSITY of H SO 2 4 From “Perry’s Chemical Engineers’ Handbook” Density of H SO = 1.834 g/mL 2 4 Hence, Thus, vol% of a 10 wt% H SO solution is 2 4 = 5.80 vol% Mass & Mole Fractions/Percentages Mass fraction/percentage and mole fraction/percentage are defined as follows (2.16) (2.17)
(2.18) (2.19) Example 2.11. An industrial drain cleaner contains 6.0 kg of water (H O: MW = 18.02) and 4.0 kg of 2 NaOH (MW = 40.00). Compute: (a) mass fraction and (b) mole percentage of each species in the mixture (a) The total mass of all species in the mixture is 6.0 kg H2O + 4.0 kg NaOH = 10.0 kg mixture Hence, (b) In order to calculate mol% of the mixture, mass of each species must be converted to “mole” first To convert mass to mole, what kind of data do we need? MOLAR MASSES (with the unit of, e.g., kg kg -mol) MW of •H2O = 18.02 kg/kg-mol •NaOH = 40.00 kg/kg-mol Hence, Thus, total moles of all species in the mixture is 0.33 kg-mol H2O + 0.10 kg-mol NaOH = 0.43 kg-mol mixture
Hence, mole percentages of Proposed Method for Calculations Concerning Mass & Mole Fractions/ Percentages From a recent Example, a new way of calculating mass fraction/percentage and mole fraction/percentage, in a table format, is proposed as follows Species Mass Mass Molar Mole Mole fraction mass percentage H O 6.0 0.60 18.02 0.33 76.7 2 NaOH 4.0 0.40 40.00 0.10 23.3 TOTAL 10.0 1.0 0.43 100.0 Example 2.12 A mixture of gases from coal gasification has the following composition • CO 5.3 mol% 2 • CO 27.3% • H2 16.6% • CH4 3.4% • N2 47.4% What is the composition in mass fractionof this mixture? Assuming that the total number of moles of all species in this mixture is 100 g-mol (in other words, we set the basis of calculation as 100 g-mol of the mixture) Hence, the mixture comprises CO = 5.3 g-mol, CO = 27.3 g-mol, H = 16.6 g-mol, CH = 3.4 g-mol, N = 47.4 g-mol 2 2 4 2 Species Moles Molar Mass Mass (g-mol) mass (g) fraction CO2 5.3 44.01 233.25 0.097 CO 27.3 28.01 764.67 0.317 H2 16.6 2.016 33.47 0.014 CH4 3.4 16.04 54.54 0.022 N2 47.4 28.02 1,328.15 0.550 TOTAL 100.0 2,414.08 1.000
Average MW ( ) In order to compute molecular weight (or molar mass) of a mixture (called “average molecular weight”: ), the following equation is employed (2.20) Example 2.13. Calculate average MW of air, which comprises 79 and 21 mol% of N and O , 2 2 respectively (note that this is an approximate composition of air). Given MW: of N = 28.02 and O = 2 2 32.00 To compute an average MW of air ( ) air M , we employ Eq. 2.20 in the table format, as follows Species Mol % Mole fraction( ) Molar mass ( ) N2 79 0.79 28.02 22.136 O2 21 0.21 32.00 6.72 TOTAL 10.0 1.0 28.856 Thus, an average MW of air = 28.86 ( the exact and widely-used value of is 28.97) Example2.14. A mixture of liquefied mixture has the following composition: • n-C H 50 wt% 4 10 • n-C5 H12 30% • n-C6H14 20% Calculate an average MW of the mixture? Basis: 100 g of the mixture Hence, the mixture contains • n-C4 H10 50 g • n-C5 H12 30 g • n-C6H14 20 g Since this is a “mass” basis, or the composition of the mixture is given as “mass percentage”, in order to compute an average MW of this mixture, we have to convert mass fraction to “mole fraction”, before being able to use Eq. 2.20, as follows Species Mass (g) Molar mass ( ) Mole (g-mol) Mole fraction( ) C H 50 58.12 0.860 0.570 33.13 4 10 C H 30 72.15 0.416 0.276 19.91 5 12 C H 20 86.17 0.232 0.154 13.27 6 14 TOTAL 100 1.508 1.0 66.31
Part per million (ppm) and Part per billion (ppb) “The units of ppm and ppb are used to express the concentrations of trace species” For instances, the concentrations of gaseous pollutants (e.g., SO2, NO, CO) inthe air, or the concentrations of heavy metals (e.g., Pb, Hg, Cd) in the waters (e.g., river, cannel, ocean) The concentrations in the forms of “ppm” and “ppb” can be expressed in equation forms as follows (2.21) (2.22) Accordingly, (2.23) & (2.24) (2.25) & (2.26) In case of “gaseous” solutions, ppm (or ppb) may mean (2.27) (2.28) Thus, for clarification, they are denoted as “ppmv” & “ppbv”, respectively Flow rate By saying “rate”, it means “how fast” or it means “per unit time” Thus, • Mass flow rate has a dimension of and the unit of mass flow rate can be, e.g., kg/s (SI) or lbm/s (AE) • Volumetric flow rate is in the dimension of or , and has the unit of, 3 3 e.g., cm /s (cgs), or m /s (SI)
Example A volumetric flow rate of diesel oil in a pipe is found to be 80 US gallons per minute. Find the mass flow rate in kg/s ( ρdiesel = 0.84 kg/L) Volumetric flow rate & Velocity Since volumetric flow rate and area have the dimensions of t and , respectively, we obtain the fact that (2.29) Example 2.15. Velocity of natural gas (NG) in a 300-mm diameter pipe is found to be 5 m/s (using an anemometer). Find 3 (a) volumetric flow rate (in ft /min) & (b) mass flow rate (in kg/s). Given o 3 ρNG ( @25 C & 5 atm = 3.28 kg/m ) Assume that the pipe has a constant diameter (of 300 mm) The cross-sectional area of the pipe can thus be calculated as follows Thus, by rearranging Eq. 2.29 and substituting numerical values into the resulting equation, we obtain Volumetric flow rate = Velocity x Area Performing a unit conversion yields
Mass flow rate of NG can be computed, using the data of volumetric flow rate and density of NG, as follows Example 2.16. The “molar” flow rate of octane (C H ) in a 3-in. diameter pipe is 35 lb -mol/min. Find 8 18 m (a) the mass flow rate, in kg/s and (b) velocity, in m/s. Given: ρ 3 C8H18 = 703 kg/m and MWC8H18 = 114.22 (a) Given MW = 114.22 means that a molar mass of C H = 114.22 lb /lb -mol 8 18 m m Hence, Performing a unit conversion gives Hence, the volumetric flow rate of octane is (b) Performing a unit conversion for a pipe diameter yields Hence, the cross-sectional area of the pipe can be computed as follows Thus, the velocity of octane can be calculated as follows
Problem Set 2 Introduction to Process Variables and Basic Engineering Calculations PROB 1. Determine that NH Cl with the mass of 267.5 kg has 4 a) how many g-mol? b) how many lbm-mol? c) how many molecules? PROB 2. How many kilograms (kg) of silver nitrate (AgNO3) are there in a) 13.0 lb -mol of AgNO ? m 3 b) 55.0 g-mol of AgNO3? PROB 3. Magnesite is the mixture of MgCO 81.0 %, SiO 14.0%, and H O 5.0% by mol. Find the 3 2 2 composition of the magnesite in the form of mass fraction PROB 4. Glass comprises Na O 7.65%, MgO 10.57%, ZnO 7.25%, Al O 1.19%, B O 7.43%, and SiO 2 2 3 2 3 2 65.91% by mass. Calculate the composition of this glass in the form of mole percent PROB 5. The gaseous mixture from a polymer-producing plant contains C H 30.6%, C H 24.5%, O 2 4 6 6 2 1.3%, CH 15.5%, C H 25.0%, and N 3.1% by volume. Determine 4 2 6 2 a) average MW (M) of the gaseous mixture b) density, in the unit of kg/m3, of this mixture (Given that 1 mol of the gaseous mixture has the volume of 0.0224 m3 @ STP) PROB 6. Liquid A flows in the tube with a diameter of 2.5 cm at the flow rate of 30 cm/s. Calculate the flow rate of liquid A in the units of a) L/h b) kg/min (Given the density of liquid A of 1,553 kg/m3)
PROB 7. The mixture of methanol and methyl acetate comprises 15 wt.% methanol. If the flow rate of methyl acetate is found to be 100 kg-mol/h, calculate the flow rate, in g-mol/s, of methanol. (Given MW of methanol and methyl acetate of 32.04 and 74.08, respectively) PROB 8. Flue gas from the combustion of coal is at the temperature and pressure of 190 oC and 100 kPa, respectively, and contains CO 10.0%, O 7.96%, N 82.0% and SO 0.04% by volume. Find 2 2 2 2 a) an average MW (M) of the flue gas b) a concentration of SO2, in the units of ppmv and ppm PROB 9. A gas mixture comprises CH 30 mol%, H 10%, and N , 60%. How many kilograms (kg) of this 4 2 2 mixture when it is in the amount of 3 lbm-mol?
Chapter 3: Introduction to Material Balances
Chapter 3 Introduction to Material Balances Learning Objectives Upon completing this Chapter, you should be able to • understand what process flow sheet (PFS) or process flow diagram (PFD) is • know a standard symbol for each of some important process equipments • draw a simple process flow diagram (or a “block flow diagram” or a flow chart) for the given problem • make a necessary assumption/necessary assumptions pertaining the given problem • set up an appropriate “basis of calculation” (basis) of the given problem • understand what “steady-state” process is and how it affects the establishment/set-up of material-balance equations • understand what “overall” and “species” balances are • establish overall- and species-balance equations for the given problem • solve simple material-balance problems One of the main responsibilities of food engineers is to create/construct/ analyse chemical processes (or, at least, to understand the existing processes) The layout of a chemical process is called “process flow sheet (PFS)” or “process flow diagram (PFD)”. PFS or PFD can be for just a single process unit of for the whole process, either simple or complicated process. PFD for a water-softening by ion-exchange process (from Stoichiometry, 4th ed; Bhatt et.al, 2004)
Normally, a PFS or a PFD comprises • all major process equipments/units • lines entering or leaving the process/unit and/or lines connecting two or more process equipments/units (these lines are called “streams”) • flow rate of each stream • composition of each stream • operating conditions of each stream and/or unit/equipment (e.g., T, P) • energy/heat needed to be added to and/or removed from any particular part of the process or the entire process Some important symbols of process equipment are illustrated as follows (from “Introduction to Chemical Processes” by Murphy, 2007) In order to be able to create or to understand PFS or PFD, the knowledge concerning MATERIAL & ENERGY BALANCES is required. As a food engineer , you need to be able to perform “material and energy balances” for any particular process or for the entire process efficiently/competently. We start our learning by doing MATERIAL BALANCES, using an underlying knowledge of “law of conservation of mass”
Principles of MATERIAL BALANCE In the case that there is NO Chemical Rxn. Total mass entering a process/unit – Total mass leaving a process/unit = Mass accumulation in a process/unit However, the “material balance” problem will be more complicated (but not too difficult – believe me!) when there is/are a Rxn./Rxns. in the process/unit, as follows: Total mass entering a process/unit – Total mass leaving a process/unit + + Mass generating from a Rxn/Rxns = Mass accumulation in a process/unit From the principles above, the following equations can be written: In the case that there is NO Rxn. where = initial mass of a system = final mass of a system In the case that there is/are a Rxn./Rxns. Normally, chemical processes are continuous, the change in mass should, therefore, be written in the “rate” form (i.e. it changes with time) Eqs. 3.1 & 3.2 can, then, be re-written, as follows
Example 3.1. One thousand (1,000) kilograms of a mixture of benzene (B) and toluene (T), containing 40% by mass of B is to be separated into two streams in a distillation column. The top output stream of the column contains 375 kg of B and the bottom output stream contains 515 kg of T. (a) Perform the mass balance for B & T (b) Determine the composition of the top and bottom streams Standard Procedure: 1) Make any necessary assumption(s) For instance, in this case, we make an assumption that the process is “steady state” 2) Draw a flow chart of the process/unit 3) Set a “basis of calculations” In this case, we should set a basis as “1,000 kg of mixture” 4) Determine the numbers of unknowns In this case, there are 3 unknowns: x = ? y = ? % of T of an input stream = ? 5) Establish “material balance” equations In order to be able to solve for unknowns, it is necessary that the # of Eqs. must be equal to the # of unknowns. In this Example, since the input stream consists of only 2 components, we obtain the following equation: wt% of B + wt% of T = 100
Accordingly, % T = 100 – % B 100 = 40 % T = 60% Note that one unknown is eliminated (the # of unknowns are now only 2) From the basis of calculation we have set (in Step 3) and from the percentages of benzene and toluene (the percentage of toluene (%T) has been solved in Step 5), we obtain the information that the input stream comprises: Benzene (B) = (1,000 kg) = 400 kg Toluene (T) = (1,000 kg) = 600 kg In a general form, let mF = mass of the input stream (feed) mB = mass of benzene in the input stream mT = mass of toluene in the input stream yB = mass fraction of benzene in the input stream yT = mass fraction of toluene in the input stream We can, then, write the following equations For the output streams, let m = mass of the top stream top mbottom = mass of the bottom stream Since the process is steady-state (as we made an assumption in Step 1) and has no Rxn., the mass balance equation can be written as follows
From the process flow chart (on Page 4), we obtain the facts that and that Thus, Eq. 3.5 is an example of an “overall mass balance” equation Substituting corresponding numerical values into Eq. 3.5 yields In order to solve for 2 unknowns in Eq.3.6 (i.e. top m & bottom m ), only 1 equation is NOT enough We need to have 2 equations; since we have already got one, we, therefore, need another equation To obtain another equation, we need to do a species balance In this Example, we shall perform a “benzene” balance, as follows Let (mB)top = mass of benzene in the top output stream (mB)bottom = mass of benzene in the bottom output stream we can, then, write the following equation: (3.7) or (3.8)
Eqs. 3.7 & 3.8 are the examples of species balance equations (in this case, it is called “benzene balance” equations) Substituting numerical values into Eq. 3.8 gives It is given that Hence, We can also perform a toluene balance, as follows (3.9) or (3.10) It is given that Thus, we obtain We can summarise our calculations as illustrated in the following Table Species Input Output (kg) (kg) Top Bottom Benzene 400 375 25 Toluene 600 85 515 TOTAL 1,000 460 540 Mole Balance Performing “mole balance” is similar to that of the mass balance, but “overall” mole balances are applicable only for the processes that have no Rxns Thus, in the case that there is NO Rxn.:
Overall balance Species balance where = initial total # of moles of all species in the system = final total # of moles of all species in the system = # of moles of species j in the system at the initial state = # of moles of species j in the system at the final state Eqs. 3.11 & 3.12 can also be written in the “rate” form (i.e. with respect to “time”), as follows Overall balance Species balance In the case that there is a Rxn/are Rxns, mole balance is still applicable, but only for species balances (NOT for an overall balance), as follows or, in the rate form
Example 3.2. An experiment on the growth rate of organisms requires an environment of humid air enriched in oxygen. Three input streams are fed into an evaporation chamber: 3 • liquid water, fed at the volumetric flow rate of 20 cm /min • air (21.0 vol% O & 79 vol% N ) 2 2 • pure O2, with the molar flow rate of one-fifth (1/5) of the stream of air to produce an output stream with the desired composition The output gas is analysed and is found to contain 1.5 mol% of water Calculate all unknowns (Data: Density of water is 1.0 g/cm3; and MW of water = 18.02, of O = 32.00, & of N = 28.02) 2 2 Assumption: Steady-state process Flow chart: Since the process is steady-state, (as we made an assumption), From Eq. 3.13 is then reduced to From the flow chart, we can write an overall mole balance equation, as follows
Let Thus, Eq. 3.17 can be re-written as follows It is given that Combining Eq. 3.19 with Eq. 3.18 gives It is also given that the volumetric flow rate of stream 3 = 20 cm3/min Thus, the mass flow rate of stream 3 can be calculated as follows which can be converted to the molar flow rate as follows Thus, Performing species balances: H O balance: 2 where, = mole fraction of H2O in stream 4 Substituting corresponding numerical values into Eq. 3.21: