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Material & Energy Balances CHM 322 UNITS AND DIMENSIONS • Chemical processes involve variables that need to be expressed in some way. • A measured variable has a numerical value and a unit. Examples: 2 meters, 4.29 kilograms • Measurable units are specific values of dimensions that have been defined by convention: grams for mass, seconds for time, centimeters or feet for length • Numerical values of same units can be added or subtracted: 3 cm -1 cm = 2 cm • A dimension is a property that can be measured : length, time, mass, temperature • OR it can be calculated by multiplying or dividing other dimensions : velocity (length/time ), volume (length3), density (mass /length3) • Numerical values of different units may always be combined by multiplication or division 3 N X 4 m = 12 N . m 5.0 km / 2.0 h = 2.5 km /h 7.0 km /h X 4 h = 28 km 3 m X 4 m = 12 m2 CONVERSION OF UNITS • To convert a variable expressed in terms of one unit to its equivalent in terms of another unit, multiply the given variable by the conversion factor (new unit/old unit): 36 mg x (1g/1000 mg) = 0.036 g 36 mg 1 g 1000 mg Example 2.2-1 Conversion of Units Convert an acceleration of 1 cm/s2 to its equivalent in km /y2 1 cm 36002 s2 242 h2 3652 day2 1m 1 km s2 12 h2 12 day2 12 yr2 102 cm 103 m = (3600 X 24 X 365)2 km/ (102 X 103) yr2 =9.95 X 109 km /yr2 SYSTEMS OF UNITS A system of units has the following components : 1. Base units : mass, length, time, temperature 2. Multiple units: multiples or fractions of base units , ex: minutes, hours, milliseconds 3. Derived units obtained by: A. By multiplying and dividing base or multiple units. These are called compound units. (cm2, ft /min, kg.m/s2,...) B. As defined equivalents of compound units (1 erg ≡ (1g. cm / s2), 1 lbf ≡ 32.174 lbm.. ft/s2) • • • • There are three systems of units: S I, base units: m, kg, s CGS, base units: cm, g, s American engineering system, base units: ft, lbm, s Example 2.3-1 Conversion Between Systems of Units Convert 23 lbm.ft/min2 to its equivalent in kg.cm/s2 Solution • Set the dimensional equation as before: • Fill in the units of conversion factors to cancel old units • Fill in the numerical values of the conversion factors • Do the arithmetic 23 lbm. ft 0.453593 kg 100 cm 12 min2 min2 3.281 ft 1 lbm 602 s2 = (23 ) (0.453593)(100)/(3.281)(3600) = 0.088 kg.cm/s2 FORCE AND WEIGHT • Newton’s second law of motion: force is proportional to the product of mass and acceleration •Therefore , natural force units are compound • To avoid using compound units in calculations involving forces, derived force units are defined in each system of units • In SI system, 1 N ≡ 1 kg. m/s2 • In CGS system, 1 dyne ≡ 1g. cm/s2 • In AES system, 1 lbf ≡ 32.174 lbm. ft/s2 • The conversion factor from natural to derived force unit is denoted gc (gc is called Newton’s conversion factor) • The value of gc depends on the system of units • In SI system, gc = 1 kg . (m /s2) /1 N • In CGS system, gc = 1 g . (cm /s2) / 1 dyne • In AE system, gc = 32.174 lbm. (ft /s2) / lbf • The weight of an object is the force exerted on the object by the gravitational attraction • W = mg, m is the mass of the object, g is the acceleration of gravity usually measured at sea level Example 2.4-1 Weight and Mass Water has a density of 62.4 lbm/ft3. How much does 2.000 ft3 of water weigh at sea level and 45o latitude and in Denver, Colarado, where the altitude is 5374 ft and the gravitational acceleration is 32.139 ft/s2? Solution • Mass of water (M) = (62.4 lbm /ft3 ) (2 ft3) = 124.8 lbm • Weight of water ( W) = 124.8 (lbm) g (ft /s2) X (1 lbf /32.174 lbm . ft /s2) • At sea level g = 32.174 ft / s2, so W= 124.8 lbf DIMENSIONAL HOMOGENITY AND DIMENSIONLESS EQUATIONS • Every valid equation must be dimensionally homogenous: all additive terms on both sides of the equation must have the same dimensions • u (m/s) = uo (m /s) + g (m /s2) t(s) This equation is dimensionally homogeneous, since the dimensions of u, uo , and gt is (Length /Time) • The above equation is dimensionally homogenous and consistent in units • Terms of inconsistent equations may be made consistent by applying the necessary conversion factor(s) • It is not necessarily true that each homogenous equation must be valid: M = 2M Example 2.6-1 Dimensional Homogeneity • Consider the valid equation D(ft) = 3t(s) + 4 1. What are the dimensions of the constants 3 and 4 ? 2. If the equation is consistent in units, what are the units of 3 and 4? 3. Derive an equation for distance in meters in terms of time in minutes Solution 1. Because of dimensional homogeneity, the constant 3 must have the dimension of length/time. Also the constant 4 must have the dimension of length 2. For consistency, the units of the constants must be 3 ft/s and 4 ft • Define new variables D’(m) and t’(min) in terms of the old variables • Find the relations between the old and the new variables • Substitute the relations in the orginal equation and simplify D(ft) = D’ (m) 3.2808 ft = 3.28 D’ 1m t(s) = t’ (min) 60 s 1 min = 60 t’ • Substituting in the original equation 3.28 D’ = (3)(60 t’) + 4 • Simplify by dividing by 3.28 D’(m) = 55t’(min) + 1.22 Dimensionless Quantities • Pure numbers are dimensionless • Multiplicative combination of variables with no net dimensions D(cm) u(cm/s) ρ(g /cm3)/ μ[g/(cm . s)] • Exponents, functions as log, exp, sin and their arguments are dimensionless Example 2.6-2 Dimensional Homogeneity and Dimensionless Groups A quantity k depends on the temperature T in the following manner: k(mol /cm3.s) = 1.2 X 105 exp (-20,000/(1.987 T)) The units of the quantity 20,000 are cal/ mol, and T is in K( kelvin), what are the units of 1.2 X 105 and 1.987 ? Solution • Equation must be consistent in units • Exp part is dimensionless • 1.2 X 105 should have the same units as k (mol / cm3.s) 20,000 cal mol 1 mol. K T(K) 1.987 cal • Therefore 1.987 has the units cal /(mol.K) Validating Results • A question arises as how to check numerical solutions of problems • There are several ways. 1. Back-substitution: substituting answers in to equations, to see if they are valid 2. Order-of-magnitude estimation: obtaining a rough or an approximate answer to the problem close enough to the exact answer 3.Testing by reason (solution makes sense): for example, the temperature of a reactor is higher than the interior temperature of the sun Order-of –magnitude Estimation Procedure: • Substitute simple integers for all numerical quantities • Do the calculations by hand, rounding off all intermediate answers • The final estimated answer should be of the same magnitude as the exact answer Example 2.5-1 •Calculation of a stream volumetric flow rate is given by: 254 13 V (0.879) (62.4) ( 0 . 866 )( 62 . 4 ) 1 ( 31.3145)(60) Estimate Volumetric flow rate without using a calculator. (exact solution is 0.00230) 250 10 1 V 1 (1) (50) ( 1 )( 60 ) (4 10 )(60) 5 2 0.2 10 0.002 2 25 10 Processes and Process Variables MASS AND VOLUME • Density of a substance is the mass per unit volume of this substance (kg/m3, g/cm3, lbm / ft3). It depends slightly on temperature for liquids and solids. • Specific volume is inverse of density (volume occupied by a unit mass of the substance) • Density is the conversion factor to relate mass and volume Examples For a volume of 20.0 cm3 of CCl4, if the density of CCl4 is 1.595 g/ cm3, the mass is: 20 cm3 1.595 g cm3 = 31.9 g The volume of 6.20 lbm of CCl4 is: 6.20 lbm 454 g 1 cm3 = 1760 cm3 1 lbm 1.595 g • The specific gravity of a substance is the ratio of the density (ρ) of the substance to the density (ρref) at a specific condition: SG = ρ / ρref • The reference most commonly used for liquids and solids is water at 4oC ρH2O (4 oC) = 1 g/ cm3 = 1000 kg / m3 = 62.43 lbm /ft3 Example 3.1-1 Mass , Volume, Density Calculate the density of mercury in lbm/ ft3 from a tabulated specific gravity and calculate the volume in ft3 occupied by 215 kg of mercury Solution • Specific gravity of mercury at 20 oC is 13.546 (Table B.1) • ρHg = 13.546 X 62.43 lbm/ ft3 = 845.7 lbm/ ft3 215 kg 1 lbm 1 ft3 0.454 kg 845.7 lbm Example 3.1-2 Effect of Temperature on liquid Density In Example 3.1-1, 215 kg of mercury was found to occupy 0.56 ft3 at 20 oC. (1) What volume would the mercury occupy at 100 oC? (2) Suppose the mercury is contained in a cylinder having a diameter of 0.25 in. What change in height would be observed as the mercury is heated from 20 oC to 100 oC? Solution 1. V(100 oC) = Vo [1+ 0.18182 X 10-3 (100) + 0.0078 X 10-6 (100)2] • V(20 oC) = 0.560 ft3 = Vo [1+ 0.18182 X 10-3 (20) +0.0078 X 10-6 (20)2] • Solving for Vo (volume of mercury at 0 oC) from second equation and substituting in first equation: V(100oC) = 0.568 ft3 2. The volume of the mercury equals πD2 H/ 4 , where D is the cylinder diameter and H is its height, Since D is constant = 0.25 in D = 0.25 /12 H(100 oC) - H (20oC) = (V(100oC) – V(20 oC) ) /( πD2/4) , = 23.5 ft FLOW RATE • Flow rate is the rate at which a material is transported through a process line • Flow rate is expressed as mass flow rate or volumetric flow rate • Mass flow rate and volumetric flow rate are related through the density in the same way as the mass and volume m m ρ V V • Density of a fluid can be used to convert a known volumetric flow rate of a process stream to the mass flow rate of that stream CHEMICAL COMPOSITION There are different ways to express mixture compositions 1.Mass fraction 2.Mole fraction 3.Concentration Moles and Molecular Weight • A gram- mole or mol of a species is the amount of that species whose mass in grams is numerically equal to its molecular weight • Mol applies to both compounds and atoms • Other units for moles are k mol, lb-mol, ton - mol • If molecular weight of a substance is then there is M kg/ kmol, M lb/ lb-mol, M g/mol • Molecular weight is used as a conversion factor that relates mass and number of moles of a substance • 34 kg of NH3 is equivalent to: 34 kg NH3 1 kmol NH3 = 2 kmol 17 kg NH3 • 4 lb-moles of NH3 is equivalent to 40 lb-moles NH3 17 lbm NH3 = 68 lbm 1 lb- mole NH3 • Same factors used to convert mass units to one another are used to convert molar units to one another • Factor to convert from lbm to g is 454 g /lbm Also to convert from lb-moles to mols , the factor is the same, i.e. 454 mol / lb-mol • One mol of any substance contains 6 X 1023 molecules Example 3.3-1 Conversion between Mass and Moles How many of each of the following are contained in 100 g of CO2 (M = 44)? Report your answer in: (1) mol CO2, (2) lb-moles CO2 (3) mol C (4) mol O (5) mol O2 (6) g O (7) g O2 (8) molecules of CO2 Solution (1) 100 g CO2 1 mol CO2 44.01 g CO2 = 2.273 mol CO2 (2) 2.273 mol CO2 1 lb-mol CO2 = 5.011 X 10-3 lb-mole 453.6 mol CO2 Each 1 mol CO2 contains 1 mol C, 1 mol O2 , and 2 mol O (3) 2.273 mol CO2 1 mol C 1 mol CO2 = 2.273 mol C (4) 2.273 mol CO2 2 mol O 1 mol CO2 = 4.546 mol O (5) 2.273 mol CO2 1 mol O2 1 mol CO2 = 2.273 mol O2 (6) 4.546 mol O 16.0 g O 1 mol O = 72.7 g O (7) 2.273 mol O2 32.0 g O2 1 mol O2 = 72.7 g O2 (8) 2.273 mol CO2 6.02 X 1023 Molecules = 1.37 X 1024 molecules 1 mol CO2 • The molecular weight of a species can be used to relate the mass flow rate of a stream to its molar flow rate For example, if carbon dioxide flows through a pipeline at a rate of 100 kg/h, the molar flow rate of CO2 is: 100 kg CO2 h 1 kmol CO2 44.0 kg CO2 = 2.27 kmol /h If the output from a chemical reactor contains CO2 flowing at a rate of 850 lbmoles/ min, then mass flow rate is: 85 lb-moles CO2 min 44.0 lbm CO2 = 37,400 lbm/ min lb-mole CO2 Mass and Mole fractions and Average Molecular Weights • Process streams are usually mixtures of liquids or gases or solutions of one or more solutes in a liquid solvent: • Composition of a mixture in terms of a species A is given as: mass of A Mass fraction x A total mass Mole fraction y A moles of A total moles Mass percent 100x A Mole percent 100y A Example 3.3-2 Conversions Using Mass and Mole Fractions A solution contains 15% by mass A (xA = 0.15) and 20 mole% B (yB = 0.20) 1. Calculate the mass of A in 175 kg of the solution 2. Calculate the mass flow rate of A in a stream of solution flowing at a rate of 53 lbm /h 3. Calculate the molar flow rate of B in a stream flowing at a rate of 1000 mol/min 4. Calculate the total solution flow rate that corresponds to a molar flow rate of 28 kmol B/s 5. Calculate the mass of the solution that contains 300 lbm of A Solution (1) 175 kg solution 0.15 kg A = 26 kg A kg solution (2) 53 lbm solution h 0.15 lbm A 1 lbm solution = 8 lbm A /h (3) 1000 mol solution min 0.2 mol B = 200 mol B/ min 1mol solution (4) 28 kmol B s 1 kmol solution = 140 kmol Solution /s 0.2 kmol B (5) 300 lbm A 1 lbm solution = 2000 lbm solution 0.15 lbm A Conversion from mass fractions to mole fractions or vice versa 1. Assume a mass of the mixture as a basis of calculation 2. Calculate the mass of each component using the known mass fraction 3. Convert the masses into moles 4. Calculate mole fractions • A similar procedure is followed to convert mole fractions to mass fractions. In this case taking total number of moles as a basis. Example 3.3-3 Conversion from a Composition by Mass to a Molar Composition A mixture of gases has the following composition by mass : O2 16 % CO 4 % CO2 17 % N2 63 % What is the molar composition? Solution Basis: 100 g of the mixture Set up calculations in table format comp. i xi mi Mi ni yi O2 CO 0.16 0.04 16 4 32 28 0.5 0.143 0.150 0.044 CO2 0.17 17 44 0.386 0.120 N2 0.63 63 28 2.250 0.690 Total 1.00 100 3.279 1.00 Calculation of Average Molecular Weight • The average molecular weight of a mixture is the ratio of the mass of a sample of the mixture to the total number of moles of all species in the sample • If the mole fractions of components are given, then : M y1M1 y 2 M 2 .... y M i all components i • If the composition of the mixture is given in terms of mass fractions, then the average molecular weight is given by the equation: 1 x1 x2 xi .... M M1 M 2 all components M i Example 3.3-4 Calculation of an Average Molecular Weight Calculate the average molecular weight of air: 1. If the approximate molar composition is 79% N2 , 21% O2 2. If the approximate mass composition is 76.7% N2 , 23.3%O2 Solution (1) • Using yN2 = 0.79, yO2 = 0.21 and the equation: M y1M1 y 2 M 2 .... y M i all components i 0.79 kmol N2 28 kg N2 kmol kmol N2 M + 0.21 kmol O2 32 kg O2 kmol kmol O2 = 29 kg / kmol = 29 lbm /lb-mol = 29 g / mol (2) From the Equation: 1 x1 x2 xi .... M M1 M 2 all components M i = 0.767 g N2 + 0.2333 g O2/ g 28 g N2 / mol 32 g O2 / mol M = 29 g / mol Concentration • Mass concentration of a component in a mixture or solution is the mass of this component per unit volume of the mixture (g/cm3, lbm / ft3, kg/ m3) • Molar concentration of a component in a mixture or solution is the number of moles of the component per unit volume of the mixture (kmol / m3, lb-moles / ft3,….) • Molarity of a solution is the value of the molar concentration of the solute expressed in gram-moles /liter solution Concentration of a component is used as a conversion factor to relate: 1. Mass or moles of a component to a mixture volume 2. Mass or molar flow rate of a component to volumetric flow rate of a stream Examples What is the number of moles in 5L of a 0.02 molar solution of NaOH? 5L 0.02 mol NaOH = 0.1 mol L 2L min 0.02 mol NaOH = 0.04 mol L min Example 3.3-5 Conversion between mass, molar and volumetric flow rates of solution • A 0.5- molar aqueous solution of sulfuric acid ( SG =1.03 )flows into a process unit at a rate of 1.25 m3/ min Calculate: (1) Mass concentration of H2SO4 in kg/m3 (2) Mass flow rate of H2SO4 in kg/s (3)Mass fraction of H2SO4 Solution 1. cH2SO4 (kg H2SO4/ m3) = 0.5 mol H2SO4 98 g 1kg 103 L L mol 103g 1m3 = 49 kg H2 SO4 m3 2. m H2 SO4 ( kg H2SO4 ) = s 1.25 m3 49 kg H2SO4 1 min = 1 kg min m3 60 s s 3. The mass fraction of H2SO4 equals the ratio of the mass flow rate of H2SO4 to the total mass flow rate, which can be calculated from the total volumetric flow rate and the solution density solution = 1.03 (1000 kg) = 1030 kg m3 m3 • m solution (kg) s = 1.25 m3 solution 1030 kg 1 min min m3 solution 60s = 21.46 kg s • xH2SO4 = mH2SO4 = 1 kg H2SO4 /s msolution 21.46 kg solution /s = 0.0 48 kg H2SO4 kg solution Parts per Million and Parts per Billion • These units are used to express trace concentrations of species in gas or liquid mixtures • The definitions usually mean grams of some component per million grams or moles of some component per million moles • Conversions to mole or mass fractions can be done through the following equations: ppmi = yi X 106 ppbi = yi X 109 Temperature • Temperature of a substance either liquid, gas, or solid is defined as a measure of the internal energy possessed by the substance • A temperature scale is obtained by assigning two known temperatures (freezing point of water = 0 oC , boiling point of water = 100 oC) Temperature Scales • Celsius scale: 0, 100 • Fahrenheit scale: 32, 212 • Kelvin scale: 0-273.15 oC • Rankine scale: 0-459.67 oF Conversion of Temperature Units Use the following relationships: T(K) = T(oC) + 273.15 T(oR) = T(oF) + 459.67 T(oR) = 1.8 T(K) T(oF) = 1.8 T(oC) + 32 • A degree is both a temperature and a temperature interval • There are conversion factors relating temperature intervals on different scales: 1.8 oF , 1.8 oR , 1oF , 1o C 1 oC 1 K 1oR 1K Example 3.5-2 Temperature Conversion Consider the interval from 20 oF to 80 o F 1. Calculate the equivalent temperature in oC and the interval between them 2. Calculate directly the interval in o C between the temperatures. Solution 1.T ( o C) = T ( o F) -32 1.8 T1 (20 o F) = (20 - 32)oC = -6.7 oC 1.8 T2 (80 o F) = (80 - 32) o C = 26.6oC 1.8 and T2 - T1 = 26.6 – (-6.7 ) oC = 33.3oC 2. ∆T( oC) = (80 - 20) o F 1o C 1.8 oF = 33.3 o C Example 3.5-3 Temperature Conversion and Dimensional Homogeneity The heat capacity of ammonia is given by the equation : Cp ( Btu / lbm o F) = 0.487 + 2.29 X 10-4 T (o F) Determine the expression for Cp in (J/g.oC) in terms of T ( oC) Solution Two steps, 1. Substitute for T (o F) and simplify the resulting equation Cp (Btu ) lbm oF = 0.487 + 2.29 X 10-4 [ 1.8 T (o C + 32] = 0.494 +4.12 X 10-4 T (o C) • Convert cP to the desired units and simplify the resulting equation Cp J 1.0 o C 9.486 X10-4 Btu 454 g g. oC 1.8 o F 1 J 1 lbm Cp (J/ g. oC) = 2.06 + 1.72 X 10-3 T(o C) Question (2) A liquid mixture of water and phenol. The mixture contains 125 ppb phenol by mass. 1. What is the mass fraction of phenol ? 2. How many milligrams of phenol are contained in one kilogram of the liquid? ppbi = yi * 10 9 Question (3) • Which is a higher temperature 1oC or 1oF? Question (4) A solution with volume V(L) contains n (mol) of a solute A with a molecular weight of MA (g A/mol) What is the molar concentration of A in terms of V, n, and MA Fundamentals of Material Balances • Law of conservation of mass is the basis of mass or material balances • Constraints imposed by material balances must be taken into account when designing a new process or analyzing an existing process • Inputs and outputs of an individual process unit or the entire process should satisfy material balance equations General Procedure for Performing Material Balances 1. First, given a process description, classify it to either batch, continuous, or semibatch process. Also classify the process as either transient or steady state 2. Draw and fully label a process flow block diagram 3. Choose a basis of calculation 4. For a multiple- unit process, identify units for which balances may be written • Do a degree of freedom analysis for the overall system and for each individual units • Specify the necessary number of process variables • Solve the equations and obtain unknowns • Systems involve recycle, purge, bypass streams and chemical reactions Process Classification • Batch process: no continuous flow of inputs and inputs. For example: adding reactants rapidly to a tank, product is removed after some time. • Continuous process: continuous flow of both inputs and outputs. For example: Pumping a mixture of liquids into a distillation column and withdraw bottom and top products steadily. • Semibatch process: either the input or output flows continuously but not both. For example: blending two liquids in a tank with no withdrawal of liquid. Steady State and Transient Processes • Steady process: where all the process variables do not change with time • Transient process: if any process variable changes with time • Continuous processes are run at steady state • Batch processes are necessarily transient General Balance Equation • Mass balance may be written in the following general way: Input (streams entering through system boundaries) + generation ( mass produced within system) - output (streams leaving through system boundaries) – consumption (consumed within system) = accumulation (buildup within system) Rules to Simplify General Material Balance Equation • If total mass is balanced, then generation is 0 and consumption is 0 • If nonreactive species is balanced, then generation is 0 and consumption is 0 • At steady state, accumulation is 0, regardless of what is being balanced Balances on Continuous Steady-State Processes • Type of balance is differential • This type of balance is done at an instant in time. • Each term in the balance equation is a rate ( rate of input, rate of output, rate of generation, rate of consumption) • The material balance equation simplifies to: input + generation = output + consumption Example 4.2-2 Material Balances on a Continuous Distillation Process Given: • 1000 kg benzene + toluene • Steady state distillation process • Two products • Xbenzene mixture = 0.5 • mbenzene top = 450 kg /h • mtoluene bottom = 475 kg /h Required • Write balances on benzene and toluene • Calculate unknown component flow rates in the output streams Solution • Process at steady state there is no accumulation • No chemical reaction involved there is no generation or consumption • Form of mass balance equation is input = output • The flow chart of the process is first drawn 450 kg B/ h m1 kg T /h 500 kg B/ h 500 kg T /h m2 kg B/ h 475 kg T /h Balance Equations Benzene Balance: mbenzene feed = mbenzene top + mbenzene bottom 1000 X 0.5 = 450 kg /h + m2 m2 = 50 kg B /h Toluene Balance: mtoluene feed = mtoluene top + mtoluene bottom 1000 X 0.5 = m1+ 475 kg T/h m1 = 25 kg T/h Need to check solution: • Substitute back the answers in the total mass balance equation: 1000 kg/h = 450 + m1 + m2 + 475 Use m1 = 25 kg /h , m2 = 50 kg /h 1000 kg/h = 1000 kg/h Integral Balances on Batch Processes • An example of a batch process is the production of NH3 from N2 and H2 in a batch reactor. • At to = 0 no mol NH3 in reactor • At tf nf mol NH3 in reactor • Besides, no ammonia enters or leaves the reactor • Material balance equation simplifies to: generation = accumulation • Also the accumulation of ammonia itself is the difference nf – no (final amount of ammonia - initial amount) • Therefore, In general for a batch process where the balanced quantity is total mass or some species mass, the material balance equation is: Accumulation = final output -initial input = generation - consumption initial input + generation = final output + consumption Example 4.2-3 Balances on a Batch Mixing Process Given: • • • • • Two flasks batch process 2 mixtures methanol –water Mix1 40 wt % methanol Mix2 70 wt % methanol 200 g Mix1 + 150 g Mix2 product Solution • Process is batch with no reaction involved, then no generation and consumption terms in material balance equation 200 g 0.4 CH3 OH 150 g 0.7 CH3 OH mg x CH3 OH • Total Mass Balance 200 +150 = m = 350 g Methanol Balance: 200 g 0.4 g CH3OH + 150 g 0.7 g CH3OH g g = m (g ) x (g CH3 OH) g • Substituting m =350 g in the last equation x = 0.529 g CH3OH / g • Use water balance to check the solution Input = Output (200)(0.600) +(150)(0.300) =165 g H2O = (350)(1-0.529) = 165 g H2 O Integral Balances on Semibatch Processes Example 4.2-4 n (kmol /min) 0.1 kmol C6H14 / kmol 0.9 kmol air / kmol 0.1 kmol air / min Given: 1. Semibatch process, air is bubbled continuously through hexane ( air rate = 0.1 kmol/min) 2. Output mixture of hexane vapor & air (0.1 kmol hexane/kmol mixture) Required • Time required to vaporize 10 m3 of hexane Solution: 1. Air Balance: accumulation = 0 (air doesn’t dissolve in hexane), generation = 0,consumption = 0 (air doesn’t react) ِ Air Balance Input = Output 0.1 kmol air = 0.9 kmol air n kmol min kmol min n = 0.111 kmol /min • Integral Balance on hexane: generation = 0, consumption = 0, Input = 0 accumulation = - output OR depletion = - accumulation = output Depletion of hexane = -10 m3 0.659 kg 103 L 1 k mol L m3 86.2 kg = -76.45 kmol • The balance on hexane is therefore: -76.45 kmol C6 H14 = -0.1 n tf n = 0.111 k mol /min tf = 6880 min Example 4.3-1 Flowchart of an Air Humidification and Oxygenation Process Given: • • • • Evaporation chamber Three feed streams Out put gas stream (1.5 mol% H2O) Required 1. Draw and label a flowchart 2. Calculate all unknown stream variables n3(mol gas /min) 0.015 mol H2 O /mol 0.2 n1(mol O2 /min) y mol O2 / mol (0.985 –y) mol N2 / mol n1(mol air /min) 0.21 mol O2 /mol 0.79 mol N2 / mol 20 cm3H2 O/min n2 (mol H2O/min) • n2 = 20 cm3 H2O 1.00 g H2O 1 mol min cm3 18 g = 1.11 mol H2O / min H2O Balance n2 (mol H2O) =n3 mol 0.015 mol H2O min min mol • Substitute the value of n2 = 1.11 mol/ min n3 = 74.1 mol/ min • Total mol balance 0.2 n1 + n1 + n2 = n3 • Substitute n2 and n3 and obtain n1 = 60.8 mol min • N2 Balance n1 mol 0.79 mol N2 min mol = n3 mol (0.985-y) mol N2 min mol 0.79 n1 = n3 (0.985 – y) • Substitute the values of n1and n3 y = 0.337 mol O2 / mol Question (5) A stream has a flow rate of 250 kg/h. The stream contains x kg C6H6 / kg stream and the rest is C7 H8 . Calculate the mass flow rate of the C7H8 in terms of x by using a single dimensional equation. Flow Chart Scaling and Basis of Calculation For any balanced process: • Masses or number of moles of input and output streams could be multiplied by the same factor and the process remains balanced • Stream compositions (mass fractions or mole fractions) are left unchanged • Stream masses could be changed to mass flow rates. Also moles changed to molar rates. The process would still be balanced • Mass units could be changed to other mass units. The process would still be balanced. • Mole units could be changed to other mole units. The process would still be balanced. • The procedure of changing flow rates and leaving compositions unchanged is called scaling Basis of Calculation • It is an amount or flow rate of one stream or a component in a process • If a stream or component amount or flow rate is given, then take this as a basis of calculation • If not, then assume a flow rate of a stream of known composition Example 4.3-2 Scale-up of a Separation Process Given: • 60-40 (mole%) A&B 50 mol 100 mol 0.6 mol A / mol 0.95 mol A / mol 0.05 mol B / mol 0.4 mol B / mol 12.5 mol A 37.5 mol B Required: Achieve Same separation with a continuous feed of 1250 lb-moles /h. Scale the flow chart. Solution: • Originally batch process • Scale factor is 1250 lb-moles/h 100 mol = 12.5 lb-moles/h mol • The masses of all streams of a batch process are converted to flow rates by multiplying by the scale factor • Feed: 100 mol 12.5lb-moles = 1250 lb-moles mol h • Top product stream: 50 X 12.5 = 625 lb-moles /h • Bottom product stream: 12.5 X 12.5 = 156 lb-moles A/h 37.5 X 12.5 = 469 lb-moles B/h 625 lb-moles/h 0.95 lb-mol A /lb-mol 1250 lb -moles / h 0.05 lb-mol B /lb- mol 0.6 lb-mol A /lb- mol 0.4 lb-mol B /lb- mol 156 lb-moles A/h 469 lb-moles B/h Rules for Balancing a Nonreactive Process 1. The maximum number of independent equations that can be derived by writing balances equals the number of chemical species in the input and output streams 2. Write balances first that involve the fewest unknown variables Example 4.3-3 Balances on a Mixing Unit Given: • Initially: NaOH aqueous solution • 20% NaOH by mass • Dilute by pure water • Product: 8% NaOH solution 100 kg 0.2 kg NaOH / kg 0.8 kg H2 O/ kg m2 kg 0.08 kg NaOH /kg 0.92 kg H2O / kg m1 kg H2O V1 liters H2O Required: V1 /100? M2/100? Solution 1. The process could be considered continuous or Batch 2. Examine the unknown variables and available balance equations i. Unknowns: m1, m2 and V1 ii. Equations: can write 2 balance equations only because there are 2 species (NaOH, H2O) iii. Need a third equation to solve for three unknowns (It is the equation relating volume and mass of water) • Plan your solution NaOH Balance (0.2 kg NaOH)(100kg) = (0.08 kg NaOH /kg) m2 m2 = 250 kg NaOH Total mass balance 100 kg +m1 = m2 = 250 kg m1 = 150 kg H2O Volume of diluent water V1 = 150 kg 1.00 liter = 150 liters kg Required ratios: 1. V1/100 kg = 1.50 liters H2O/ kg feed solution 2. m2 /100 kg = 2.50 kg product solution/ kg feed solution Example 4.3-4 Given: • Stream of humid air • Condenser • 95% of H2O v condenses • Flow rate of condensate = 225 L /h mol O2 /h mol N2 /h mol dry air 21 mol% O2 79 mol % N2 mol H2O(v) mol H2O(v) /h 225 liters H2O mol H2O (l) /h 95% of water in feed Required Solution 1. Process: • Steady state • Continuous • Non-reactive • Single unit 2. Basis of solution: 225 L/h Condensate 3. Degree-of-freedom analysis: • Unknown variables: 6 (counted on chart) n3 is given as the basis of calculation ˙ ˙ ˙ • Number of independent equations: 3 (= number of species in the problem) Need to specify three other variables: ˙ • n2 is calculated as a process condition (liquid water condensate is 0.95 of n2) No other variable could be specified • The problem is underspecified and another piece of information must be given • Assume that the entering air contains 10 mole% of water vapor in air • Then the number of degrees of freedom is zero (five unknowns and five equations) and the problem is solvable. • Start writing out 3 equations in the order of solving the equation (s) with one unknown variable first, then simultaneous equations • First calculate the molar flow rate of condensate (mol /h) from the volumetric flow rate (225 L/h) and the density of water (1 kg/L) • Then proceed with the species balances • n˙3 (mol)/ h = 225 L H2O 1.00 kg 1 mol H2O h L 18 X10-3 kg = 12500 mol / h ˙ = 0.95 n˙ , n˙ = 13158 mol / h • n 3 2 2 • n˙˙2 = 0.1 n˙ 1+ n˙ 2˙ • 0.9 n˙2 =0.1 n˙ 1 • n˙1 = 118422 mol / h ˙ 4 = 0.21 X n˙1 , ˙n4 = 24869 mol / h • n ˙ = 0.79 X n˙ , n˙ = 93553 mol / h • n 5 1 5 ˙ 6 = 0.05 X n ˙ 2 , n˙6 = 658 mol / h • n • yO2 = 0.209 • yN2 = 0.786 • yH2O = 5.50 X 10-3 Example 4.3-5 Given: • Distillation column • Feed: 45 wt % benzene (B), rest toluene (T) • Overhead: 95 mol% Benzene • Bottom: 8% benzene in feed • Feed volumetric flow rate : 2000L/h • Feed specific gravity : 0.872 Overhead ˙m2 (kg/h) 0.95 mol B/mol ˙ 0.05 mol T/mol xB2 kg B / kg Feed 1-xB2 kg T/kg 2000 L/h ˙m1 (kg/h) Bottom 0.45 kg B/kg ˙ B3 (kg B/h) m ˙ 0.55 kg T/kg 8 % benzene in ˙ feed m ˙ T3 (kg T/h) Required ˙ 1. m2 ˙ ˙ ˙ 2. m3 = mB3 +mT3 xB = mB3 / m3 ˙ ˙ 3. Solution 1. Process: • • • • Steady state Continuous Non-reactive Single unit 2. Basis of solution: 2000 L/h Feed • Feed composition is given in terms of mass fraction, and it is also required to determine the mass composition of the bottom product • First step in the solution is to convert mole fractions in the top product to mass fractions Conversion: Basis: 100 kmol overhead Number of moles of toluene = 5 kmol Number of moles of benzene = 95 kmol Mass of benzene = 95 X 78.11= 7420 kg B Mass of Toluene = 5 X 92.13 = 461 kg T Total mass = 7420 kg B + 461 kg T = 7881 kg xB2 = 7420 kg B / 7881 kg mixture = 0.942 kg B / kg xT2 = 1-0.942 = 0.058 kg T/ kg 3. Degree-of-freedom analysis: • Unknown variables: 4 (counted on chart) • • Number of independent equations: 2 (= number of species in the problem) Need to specify two other variables: density relation Process specification: (Benzene split -bottom product contains 8% of benzene in feed) 4. Balances: Mass flow rate of feed calculated from Volumetric flow rate and Feed S.G. ˙ 2000 L 0.872 kg = 1744 kg/h = m 1 h L • Benzene split: ˙ B3 =0.08 (0.45 ˙m1) = 62.784 kg /h m • Benzene Balance ˙ =m ˙ x +m ˙ 0.45 m 1 2 B2 B3 Substituting: m˙ 1 = 1744 kg/h ˙ ˙ B3 = 62.8 kg/h m xB2 = 0.942 m˙ 2 = 766 kg/h • Toluene Balance ˙ +m ˙ 0.55 m˙ 1 = (1-xB2)m 2 T3 ˙ m T3 = 915 kg/h ˙ = 62.8 +915 = 978 kg/h • m˙ 3 = m˙B3 + mT3 ˙ ˙ • xB3 = m B3 /m3 = 62.8 / 978 = 0.064 kg B/ kg • xT3 = 1- xB3 = 0.936 kg T/kg General Procedure for SingleUnit Material Balance Calculation (P. 101 &102) 1. Choose a basis of calculation a. An amount or flow rate of stream or several streams given in the problem b. If no information, take as an arbitrary basis, the amount or the flow rate of a stream with known composition 2. Draw a completely labeled flowchart a. Fill in all known variable values including the basis of calculation b. Label unknown stream variables on the chart 2. Write the requirements of the problem in terms of labeled variables. 3. Problems with mixed mass and mole units must be solved in only one set of units. 4. Do the degree-of-freedom analysis. 5. Solve the equations in an efficient order. 6. Scale the balanced process if necessary. Balances on Multiple-Unit Processes • Chemical processes usually involve more than one process unit (chemical reactors, mixing units, separation processes…. etc.) • Material balance might be done on different subsystems (In a single-unit processes, there is only one system) • Systems may involve subsystems : The entire process Interconnected combination of some units A single unit A mixing point for two or more process streams A splitting point into branches • Inputs and outputs to a system or a subsystem are the process streams that intersect the system boundary Procedure for Material Balance Calculations on Multiple-Unit Processes • Do a degree-of-freedom analysis on the overall system and on each subsystem • Isolate and write balances on subsystems of the process as needed to solve for unknowns Example 4.4-1 Two-Unit Process Given • Steady-state two-unit process • Two components: A &B Required • Mass flow rates of streams 1, 2 and 3? • x’s or y’s of the same streams? 1. Process: Continuous, steady-state 2. Basis: Given flow rates collectively 3. • • • Subsystems: Overall process Individual process units Mixing point 4. Degree-of-Freedom Analysis Do the analysis for the overall system and for each subsystem • Include only in each analysis the variables in the streams intersecting a system boundary • Overall system: . 2 unknowns (m3, x3) -2 balances (2 species) = 0 degrees of freedom . Possible to solve for m3, x3. These two variables become known . • Mixing point: . . 4 unknowns (m1, x1 , m2 , x2 ) - 2 balances (2 species) = 2 degrees of freedom • Unit 1: . 2 unknowns (m1,x1) -2 balances (2 species) = 0 degrees of freedom . Possible to solve for m1,x1. These two variables become known • Back again to mixing point: . It is possible now to determine m2 , x2 Calculations Overall Mass Balance: . 100 + 30 = 40 + 30 +m3 . m3 = 60 kg / h Overall Balance on A: (0.500) (100) + (0.3)(30) = (0.9) (40.0) +(0.6)(30) +x3(60) x3 = 0.0833 kg A/kg • Total mass Balance on Unit 1: . 100 = 40 + m1 . m1 = 60 kg/h • Species A Balance on Unit 1: (0.5) (100) = (0.9)(40) + x1 (60) x1 = 0.233 kg A/kg • Total mass balance on Mixing Point: . . m1 + 30 = m2 = 90 kg/h • Species A Balance on Mixing point: . . x1 m1 +(0.3)(30.0) = x2 m2 . . using the values of m1 , x1 , m2 x2 = 0.255 kg A /kg Example 4.4-2 Extraction-Distillation Process Given: • • • • Mixture : Acetone (A) & Water (W) Composition: 50 wt% (A), 50 wt% (W) 2 stage extraction +Distillation Details given as flow chart Required: • • • • • • • Mass of stage 1 extract Mass of stage 1 raffinate Mass of stage 2 extract Mass of combined extract Mass of overhead product from distillation Mass of bottoms product from distillation Compositions of all above streams Solution 1. Process: batch and non-reactive 2. Basis: 100 kg feed mixture 3. Degree-of-freedom analysis: • Overall process 4 unknown variables (m5 , mA6 , m m6 , m w6) - 3 equations = 1 degree of freedom • Extractor 1 5 unknown variables (mA2 , mM2 , m W2, m1, xm1) - 3 equations = 2 degrees of freedom • Extractor 2 4 unknown variables (mA2 , mM2 , m W2, m3) - 3 equations = 1 degree of freedom • Mixing point 6 unknown variables (m1 , xm1 , m3 , m A4 , mM4 , mW4) - 3 equations = 3 degrees of freedom • Combined extractors 3 unknown variables (m1 , xm1 , m3 ) - 3 equations = 0 degrees of freedom • Distillation column 7 unknown variables (mA4 , mM4 , mW4 , m5 , mA6 , mM6 , mW6 ) - 3 equations = 4 degrees of freedom Start balances on combined extraction units •Total mass balance: (100 + 100 +75) kg = 43.1 + m1 + m3 • Mass balance on A : 100 (0.500) kg A = (43.1) (0.053) + m1(0.275) +m3(0.09) Solve simultaneously m1 = 145 kg , m3 = 86.8 kg • Mass balance on M (100 +75) kg M = = (43.1) (0.016) + m1 xM1 + m3 (0.88) Substitute the values of m1 and m3 xm1 = 0.675 kg MIBK / kg • Balances Around Extract Mixing Point Mass balance on A: m1 (0.275) + m3 (0.09) = mA4 Substitute m1, m3 mA4 = 47.7 kg acetone • Balance on M m1 xM1 + m3 (0.88) = mM4 Substitute the values of m1 , m3 , xM1 mM4 = 174 kg MIBK • Balance on W m1 (0.725 - xM1) + m3 (0.03) = mW4 Substitute the values of m1 , m3 , xM1 mW4 = 9.9 kg water • Balances Around First Extractor Balance on A : 100 (0.50) = mA2 + m1 ( 0.275) Substitute m1 mA2 = 10.1 kg acetone Balance on M : 100 = mM2 + m1 xM1 Substitute m1 , xM1 mM2 = 2.3 kg MIBK • Balance on W 100 (0.500) = mW2 + m1 (0.725 – xM1) Substitute m1 , xM1 mw2 = 42.6 kg water • The remaining unknowns m5 , mA6 , mM6 , mW6 may be obtained through overall balance on the whole process or the distillation column. In either case, we have only three independent equations and 4 unknowns and hence 1 degree of freedom. There is no more information and the calculations are stopped at that point. Recycle Streams • One situation involving recycle streams is chemical reactions in reactors • A chemical reaction rarely proceeds to completion. • It is therefore necessary to separate any unreacted material and recycle it to the inlet of the reactor • This way, the cost of fresh reactant is less and the product being more pure is sold at a higher price. • Recycling doesn’t disturb the overall process material balance Other reasons for using recycle in chemical Processes • Recovery of catalyst and recycling it to reactor • Increasing rate of filtration by diluting concentrated slurry with a recycled part of the filtrate • Controlling a temperature of a reactor where a highly exothermic reaction takes place (see textbook for details P.110) Example 4.5-1 Given • • • • • Dehumidifcation process Mole fraction of fresh humid air Mole fraction of blended stream Mole fraction of dehumidified air Basis: 100 moles of dehumidified air Required • Moles of fresh feed (n1)? • Moles of water condensed (n3)? • Moles of recycled dehumidified air (n5)? Solution 1. Process: Batch, no reaction 2. Basis: 100 moles of dehumidified air 3. Degree-of-freedom analysis • Overall System 2 unknowns (n1 , n3) - 2 independent equations = 0 degrees of freedom • Mixing point 3 unknowns (n1, n5 , n2)- 2 independent equations = 1 degree of freedom • Process 3 unknowns(n2 ,n3 , n4 ) -2 independent equations = 1 degree of freedom •Splitting point 2 unknowns (n4, n5 ) - 2 independent equations = 0 degree of freedom Be Careful now! The splitting point does not follow the rule as shown above. Why? The reason is that only one independent balance equation can be written for any splitting point in any mass balance problem, because the streams entering and leaving this point have the same composition. Therefore, the correct degree of freedom analysis is: 2 unknowns (n4, n5 ) -1 independent equation=1 • Overall balance on dry air n1 (0.96) = 100 (0.983) n1 = 102.4 mol fresh feed • Overall mole balance n1 = n3 + 100 Substitute n1 n3 = 102.4-100 = 2.4 mol H2O • Balances on the mixing point (since the problem asks for n5): Dry air n1 (0.960) + n5 (0.983) = n2 (0.977) Total balance n1 + n5 = n2 Solve simultaneously after substituting n1 n 5 = 290 mol recycled , n2 = 392.5 mol Example 4.5-2 Given: • • • • • 4500 kg K2 Cr2 O4 solution xK2Cr2O4 = 0.33 (feed to evaporator) xK2Cr2O4 = 0.494(feed to crystallizer) Recycle stream (filtrate) : 36.4% K2 Cr2 O4 Filter cake: 95% by mass crystals + 36.4% K2Cr2 O4 solution Required Case1: . •Rate of evaporation (m2)? . . . •Rate of production of crystals (m4)? . •Feed rate to evaporator (m1)? . •Feed rate to crystallizer (m3)? . •Recycle ratio (m6 / 4500)? Case 2: If the filtrate was not recycled, calculate the production rate of crystals. What are the benefits and costs of recycling? Solution 1. . Process: Continuous, non-reactive 2. Basis: 4500 kg/h 3. Degree-of- freedom Analysis • Overall System . . . 3 unknowns (m4 , m5 , m2) - 2 independent equations - 1 more equation . . = 0.95 (m. + m m 4 4 5) =0 • Mixing point . . Evaporator. . 3 unknowns (m1 , x1 , m6 ) - 2 independent equations = 1 degree of freedom • . 4 unknowns (m1 , x1 ,, m2 , m3) – 2 independent equations = = 2 degrees of freedom • Crystallizer and Filter . . .. . . 4 unknowns (m3, m4, m5, m6) – 2 independent equations -1 more equation . . . m4 = 0.95 (m4 + m5)=1 remaining degrees of freedom = 1 Solution Overall System • Process specification . . . m4 = 0.95 (m4 + m5) . . m5 = 0.05263 m4 • Overall chromate balance . . 4500 (0.333) = m4 + 0.364 m5 Solving simultaneously both equations: . m4 = 1470 kg crystals/h . m5 = 77.5 kg entrained solution/h Overall mass balance . . . 4500 = m2 + m4 + m5 . . Substitute m4 , m5 . m2 = 2950 kg H2O vapor Crystallizer Total Mass Balance . . m3 = m 4 + m5 + m 6 . . . . Substitute m , m . 4 5 . . m3 = 1550 + m6 Water Balance Around Crystallizer . . . 0.506 m3 = 0.636 m5 + 0.636 m6 . Substitute m5 . . m3 = 97.4 + 1.257 m6 Solving both equations . m3 = 7200 kg/h crystallizer feed . m6 = 5650 kg/h . Recycle ratio = m6 /4500 = 5650/4500 = 1.26 kg recycle/ kg fresh feed • Case 2: No recycle stream. Degree-of-freedom Analysis • Overall system has 1 degree-of-freedom • Evaporator has zero degrees-of-freedom • Crystallizer has 1 degree-of-freedom Total mass balance and either water or . chromate balance will yield the values of m1 . . and m2. Once m2 is known, balances on the . . . . . crystallizer are solved for m3 , m4 , m5 . • m3 = 622 kg crystals/h. . • With recycle m3 = 1470 kg crystals/h • Mass flow rate of discarded filtrate . (m5) = 2380 kg/h • Discarded filtrate contains = 2380 X 0.364 = 866 kg/h potassium chromate wasted • Recycling enables to recover most of the salt and selling it. This would outweigh the cost of a pump pumping power and recycle pipe after some time. Bypass A fraction of a feed is diverted around a process unit and combined with the product. By varying the fraction of feed bypassed, the product properties are varied. Bypass and recycle streams balance calculations are exactly similar. Mass Balances On Reactive Systems Basic Concepts • Stoichiometric equation: is a balanced chemical reaction equation • Stoichiometric coefficients: are the numbers appearing by molecular species in a balanced chemical reaction equation • Stoichiometric ratio: is the ratio of the stoichiometric coefficients of two molecular species in a balanced chemical reaction equation Example 2 SO2 + O2→2SO3 • Stoichiometric coefficient of SO2 and O2 are 2 and 1 respectively • Stoichiometric ratio (SO2 : O2) = 2:1 • Limiting Reactant: Is the reactant present in less than it stoichiometric ratio in the feed relative to every other reactant •It would run out first if a reaction proceeds to completion • Excess Reactant : Other reactants that are not limiting • If all reactants are present in stoichiometric proportion, then there is no limiting reactant • Fractional Conversion: Applies to a limiting reactant It is the ratio : f = (moles reacted)/ moles fed • I f 100 moles of a reactant are fed and % conversion is 80% then 80 moles react and 20 moles remain • Fractional Excess of reactant: for reactant A : [(nA)feed - (nA) stoich] (nA) stoich Examples: 1. For the reaction: 2 SO2 + O2 →2SO3 If 100 mol of O2 and less than 200 moles of SO2 are present in the feed, then SO2 is the limiting reactant 2. For the reaction C2 H2 + 2H2→ C2 H6 If the feed consists of 20 kmol/h C2 H2 and 50 kmol/h H2 then: A. B. C. D. Stoichiometric ratio (H2: C2 H2 ) = 2:1 Feed ratio (H2: C2 H2 ) = 2.5:1 H2 is in excess, C2 H2 is limiting % excess H2 = [(50 -40)/ 40] X 100 = 25 3. For the same reaction in the past example: If feed to a batch reactor consists of 20 kmol C2 H2 , 50 kmol H2 , 50 kmol C2 H6 and 30 kmol H2 reacted. Find the number of moles of each species at the end of the reaction? • Number of reacted moles of C2 H2 = 15 kmol (from stoichiometric ratio) • Number of remaining moles of H2 = 20 kmol ( 50 -30) • Number of remaining moles of C2 H2 = 5 kmol (20 -15) • Number of moles of C2 H6 formed = 15 ( from reaction -by stoichiometry) + 50 ( in the original feed) = 65 kmol Example 4.6-1 Reaction Stoichiometry Given For the reaction: C3 H6 + NH3+ 1.5 O2 →C3 H3N +3H2O • Feed (100 mol): 10 mole% C3 H6 , 78 mole% Air, 12 mole% NH3 • Fractional conversion of limiting reactant : 30% Required • Determine which reactant is limiting • % excess other reactants? • Molar amounts of all product gas components? Solution • Basis : 100 mol feed (given) • (nNH3/ nC3H6)feed = 12/10 = 1.2 • (nNH3/ nC3H6)stoich = 1/1 = 1 • 1.2 > 1 NH3 is in excess • (nO2/ nC3H6)feed =16.4 / 10 = 1.64 • (nO2/ nC3H6)stoich = 1.5 /1 = 1.5 • 1.64 > 1.5 O2 is in excess Propylene is limiting • (nNH3)stoich = 10 mol (according to stoichiometry) • (nO2)stoich = 1.5 X 10 = 15 mol • % excess NH3 = (12-10)/10 X 100 = = 20 % • % excess O2 = [(78 X 0.21)- 15]/ 15 X 100 = 9.3% • • • • • • • (nNH3)react = 3 mol (nO2)react = 1.5 X3 = 4.5 mol (nNH3)remain = 12-3 = 9 mol (nO2)remain = 16.4 - 4.5 = 11.9 mol (nN2) = 78 X0.79 = 61.6 mol (nH2O)formed = 3 X 3 =9 mol (nC3H3N)formed = 3 mol Multiple Reaction Systems • Reactions are not usually single, but accompanied by side reactions giving undesired products • Side reactions represent economic loss for example C2 H4 + 0.5 O2 →C2 H4 O C2 H4 + 3 O2 →2 CO2 + 2 H2 O • The degree to which a side reaction is important can be described by yield and selectivity • Yield = moles of desired product formed moles formed with no side reactions and if limiting reactant was completely converted • Selectivity = moles of desired product formed moles of undesired product formed • High values of yield and selectivity mean that side reactions are suppressed Example 4.6-3 Yield and Selectivity in Dehydrogenation Reactor Given • C2 H6 → C2 H4 +H2 • C2 H6 + H2→ 2 CH4 Feed: 85 mole%, ethane, 15 % inerts Conversion of ethane: 0.501 Fractional yield ethylene : 0.471 Required • Molar composition of product gas? • Selectivity of ethylene to methane ? Solution • Fractional conversion of ethane = 0.501 n1 (remaining ethane) = 85 - 0.501X 85 = 42.4 mol • Ethylene yield = 0.471 Stoichiometric no. of moles of ethylene = 85 mol n2 ( moles formed of ethylene) = 0.471 X 85 = 40.0 mol • Ethane consumed by the second reaction = 0.501 X 85 - 40 = 2.6 mol • Hydrogen consumed by the second reaction = 2.6 mol • Hydrogen formed by the first reaction = 40 mol • n3 (Hydrogen leaving in the product gas) = 40-2.6 = 37.4 mol • n4 (Moles of methane formed) = 2 X 2.6 = 5.2 mol • n5 (Moles of inerts formed )= 15 mol • Composition of product gas Total moles of product gas = 140 mol 30.3 mol% C2 H6 28.6 mol% C2 H4 26.7 mol% H2 3.7 mol% CH4 10.7 mol% I Independent Species and Independent Reactions • If two molecular species occur in the same ratio to each other everywhere in a process and this ratio is needed in calculations, balances on those species will not be independent equations • If two atomic species occur in the same ratio everywhere in a process, balances on those species will not be independent • Chemical reactions are independent if the stoichiometric equation of any one of them cannot be obtained by adding and subtracting multiples of the stoichiometric equations of the others Example: the following reactions are not all independent [1] A→2B [2] B→C [3] A→2C [3] = [1] + 2 X [2] Possible Balances For A Reactive System 1. Molecular species balances • form: Input + generation = output + consumption • Degree- of - freedom analysis: No. of unknown labeled variables + No. independent chemical reactions - No. independent molecular species balances - No. other equations relating unknown variables = No. degrees of freedom 2. Atomic species balances • form: Input = output • Degree-of-freedom analysis: No. unknown labeled variables - No. independent atomic species balances - No. molecular balances on independent nonreactive species - No. other equations relating unknown variables =No. degrees of freedom Approaches to Material Balances on reactive systems • Atomic species balances are the most used especially if more than one reaction is involved • Molecular species balances should be used for simple systems Example (Text P.125+126+129) • Consider the dehydrogenation of ethane as shown in the figure below Required: Obtain the values of the molar flow rates of ethane and ethylene in the product gas. Solution: Process- continuous steady state Basis -given 100 kmol C2H6/min Degree of freedom analysis (molecular species): No. unknown variables [2] + No. independent chemical reactions [1] -No. independent molecular species balances [3] = 0 Use Stoichiometric equations to find the generation or consumption H2 Balance: generation = output GenH2 = 40 kmol /min C2 H6 Balance: Input = output + consumption 100 = n1+ 40 n1 = 60 kmol /min C2 H4 Balance: generation = output n2 = 40 kmol/min Degree of freedom analysis (Atomic Species Balances) No. unknown variables [2] - No. independent atomic species balances [2] - No. molecular balances on independent nonreactive species [0] -No. other equations relating unknown variables [0] = 0 No consumption or generation terms appear in atomic species balances Input = Output C Balance: 100 kmol C2 H6 /min 2 kmol C/kmol C2 H6 = n1 kmol C2 H6 /min 2 kmol C/kmol C2 H6 + n2 kmol C2 H4 /min 2 kmol C/kmol C2 H4 n1 + n2 = 100 (1) H Balance 100 kmol C2 H6 /min 6 kmol H /kmol C2 H6= 40 kmol H2/min 2 kmol H /kmol H2+ n1 C2 H6 kmol /min 6 kmol H /kmol C2 H6 + n2 kmol C2 H4 /min 4kmol H/ kmol C2 H6 600 kmol H/min = 80 kmol H/min + 6 n1 +4 n2 520 = 6 n1 +4 n2 Solving eqns 1&2 simultaneously: n1= 60 kmol C2 H6 /min n2 = 40 kmol C2 H4 /min (2) Example 4.7-1 Incomplete combustion of methane CH4 +3/2 O2 → CO +2H2O CH4 + 2O2 →CO2 +2H2O Feed : 7.8 CH4 mol%, 19.4%O2 , 72.8% N2 Conversion methane: 90% Gas leaving reactor:8 mol CO2 / mol CO Required • Degree-of-freedom Analysis • Calculate Molar Composition of the product gas • Use: atomic species balances Atomic Species Balances: 5 unknown variables - 3 independent atomic species balances -1 nonreactive molecular species balance - 1 specified methane conversion = 0 degrees of freedom nCH4 = 0.1 X7.80 = 0.78 mol CH4 N2 Balance, input = output nN2 = 72.8 mol N2 Now determine nCO, nH2O , nO2 Atomic Species Balances General form : input = output • Start with Carbon and hydrogen balances then followed by oxygen balance C Balance 7.8 mol CH4 X (1mol C/1mol CH4) = = 0.78 mol CH4 X (1 mol C/ 1 mol CH4) + nCO (mol CO)X (1 mol C/1 mol CO) + 8 nCO (mol CO2) X (1 mol C/1 mol CO2) nCO = 0.78 mol CO nCO2 = 8 nCO = 8 X 0.78 mol CO2 = = 6.24 mol CO2 H Balance 7.8 mol CH4 (4 / 1) = 0.78 (4/1) + nH2O (2/1) nH2O = 14 mol H2O O Balance 19.4 mol O2 (2/1) = nO2 (2/1) +0.78 (1/1) + 6.24 (2/1) + 14 (1/1) nO2 = 5.75 mol O2 •The molar composition of the gas: 0.78 % CH4, 0.78 % CO, 14% H2O, 5.7% O2 , and 72.5 % N2 Product Separation and Recycle 1. Overall Conversion: (reactant input to process - reactant output from process) / reactant input to process 2. Single -pass Conversion: (reactant input reactor -reactant output from reactor)/ reactant input to reactor Example Consider the reaction A →B • The overall conversion of A is: [(75 mol A/ min)in - (0 mol/min)out /(75mol A/min)in ] X 100 = 100% • The single-pass conversion is : [(100 mol A/ min)in - (25 mol/min)out /(100mol A/min)in ] X 100 = 75% • Reason for recycle is evident, the overall conversion reached 100% because of perfect separation. If separation was not not perfect, overall conversion will still be higher than singlepass conversion Example 4.7-2 Dehydrogenation of Propane Required • Composition of product mole fractions of all components • Recycle ratio (moles recycled/ moles fresh feed) [n9 + n10 ]/100 mol • Single-pass conversion [(n1 - n3)/n1] X 100 Solution 1. Process: Reactive, single reaction 2. Basis : 100 mol fresh feed 3. Degree-of- freedom Analysis Overall system 3 unknown variables(n6, n7, n8) - 2 atomic balances ( C and H) -1 additional relation (95 % overall propane conversion) = 0 degrees of freedom Note : Analysis done based on atomic species balance Separator: (physical system, do analysis in the usual way) 5 unknown variables (n3, n4 , n5 , n9 , n10) (n6 through n8 are known from balances on the overall system) - 3 balances - 2 additional relations (n6 = 0.00555n3 , n10 = 0.05 n7) = 0 degrees of freedom Mixing point: ( physical system, do analysis in the usual way - be careful!) • No need to analyze the reactor because no more unknowns exist on the chart First obtain n6 (mol C3 H8) from the Overall conversion: n6 = 0.05 (100 mol) = 5 mol C3 H8 Second write two balances on carbon and hydrogen • Overall C balance (involving only one unknown) 100 mol C3 H8 (3 mol C/ mol C3 H8) = n6 mol C3 H8 (3 mol C/ mol C3 H8) + n7 mol C3 H6 (3 mol C/ mol C3 H6) Substitute n6 = 5 mol n7 = 95 mol C3 H6 • Overall H balance (100) (8) = n6 (8) + n7 (6) +n8 (2) • Substitute n6 = 5 mol, n7 = 95 mol n8 = 95 mol H2 Product composition : 2.6 mol% C3 H8 , 48.7 mol % C3 H6 , 48.7 mol% H2 Separator n6 = 0.00555 n3 , n6 = 5 mol, n3 = 900 mol C3 H8 n10 = 0.0500 n7 , n7 = 95 mol n10 = 4.75 mol C2 H6 Propane balance about separation unit n3 = n 6 + n 9 , Substitute n3 = 900 mol, n6 = 5 mol n9 = 895 mol Propane balance about mixing point 100 mol + n9 = n1 Substitute n9 = 895 mol n1 = 995 mol C3 H8 Now calculate recycle ratio and single-pass conversion • Recycle ratio =(n9 +n10)/ 100 =9.00 • Single-pass conversion = = (n1 - n3)/n1 X 100% = 9.6 % Means of Achieving High Overall Conversions in Reactors 1. Design the reactor to yield a low-single pass conversion, followed by separation of unreacted material and recycling it. • This scheme leads to small volume of the reactor and a decrease in cost. • The savings in reactor cost may be overcome by the costs of the separation process unit and the pump, pipes and fittings in the recycle line. 2. Design the reactor to yield a high singlepass conversion • This scheme leads to much larger reactor volume than the previous one • No separation unit and no recycling The option that should be taken is decided by detailed economic analysis Purging • If some material entering in the feed or produced in a reaction, remains fully in a recycle stream, then it keeps accumulating and steady state is not achieved. • To prevent this accumulation, part of the recycle stream must be removed from the process as a purge stream Example 4.7-3 Recycle and Purge in the synthesis of Methanol Given CO2 +3H2→CH3OH +H2O • Fresh feed to process : H2 , CO2 , 0.4 mole % inerts (I) • Condenser : water and methanol removed as liquids • Unreacted materials + inerts: recycled to reactor • Purge stream taken from recycle • Feed to reactor: 28 mole% CO2, 70 mole% H2 , 2 mole% inerts • Single-pass conversion of hydrogen: 60 % Required Molar flow rate and molar composition of: • Fresh feed • Total feed to reactor • Recycle stream • Purge stream Solution 1. Basis: Neglect given basis 155 kmol CH3OH/hTake 100 mol feed to reactor (combined feed) since composition is known-Scale the answer at the end (155/n3) 2. Degree-of-freedom Analysis i. Overall system: 7 unknowns (no , xoc , n3, n4, np , x5C , x5H) + 1 reaction - 5 independent balances (CO2 , H2 , I, CH3OH,H2 O) = 3 degrees of freedom ii. Recycle -fresh feed mixing point 5 unknowns (no , xoc , nT , x5c , x5H) - 3 independent balances (CO2 , H2 , I) = 2 degrees of freedom iii. Reactor 4 unknowns (n1, n2 , n3 , n4 ) +1 reaction - 4 independent balances (CO2 , H2 , CH3 OH, H2O) - 1 single pass conversion = 0 degrees of freedom iv. Condenser (considering n1, n2, n3, n4 are known from balances on reactor) 3 unknowns (n5 , x5C , x5H) - 3 independent balances = 0 degrees of freedom v. Purge - recycle splitting point (considering n5 , x5C , x5H are known from balances on condenser) 2 unknowns (nr , np) - 1 independent balance = 1 degree of freedom 1. Reactor Analysis: Using single-pass conversion of H2: 60% n2 = 0.4 (70 mol H2 fed) = 28 mol H2 H2 balance: Input + generation = output + consumption generation = 0 consumption = input - output = 70 - 28 = 42 mol H2 consumed CO2 Balance: output = input - consumption = 28 - 42/3 (from stoichiometry) = 14 mol CO2 CH3 OH Balance: output = generation n3 = 42 /3 = 14 mol CH3 OH H2 O Balance: output = generation n4 = 42/3 = 14 mol H2O 2. Condenser Analysis input = output Total Mole Balance: n1 +n2 +n3 + n4 +2 mol = n3 +n4 +n5 Substitute n2 = 28 mol, n1 = n3= n4 = 14 mol n5 = 44 mol CO2 Balance: n1 = n5 x5C Substitute n1 = 14 mol, n5 = 44 mol x5C = 0.3182 mol H2 Balance: n2 = n5 x5C Substitute n2 = 28 mol, n5 = 44 mol x5H = 0.6364 mol CO2/mol xI = 1-x5C -x5H = 0.04545 mol I/mol 3.Mixing point Balances: Total Mole Balance: n0 + nr = 100 mol I balance: no(0.00400) +nr (0.04545) =2.0 Solving the two equations simultaneously, then no = 61.4 mol, nr = 38.6 mol CO2 Balance: nO xOC+ nr x5C = 28 Substitute no = 61.4 mol, nr = 38.6 mol, x5C = 0.3182 mol CO2 /mol xOC = 0.256 mol CO2 /mol xOH = (1-xOC-xOI) =0.740 mol H2 /mol 4. Recycle -Purge Splitting Point Total mole balance: n5 = n r + n P Substitute n5 = 44 mol, nr = 38.6 mol np = 5.4 mol 5. Flowchart Scaling Scale factor = 155 kmol CH3OH/h 14 mol CH3 OH = 11.1 kmol /h mol Fresh feed = 61.4 X 11.1 = 681 kmol/h Feed to reactor = 100 X 11.1 = 1110 kmol/h Recycle = 38.6 X 11.1 = 428 kmol/h Purge = 5.4 X 11.1 = 59.9 kmol/h Combustion (P.142) • Combustion is rapid reaction of fuel with oxygen. • Heat energy resulting from combustion is used to generate electricity. • Products of combustion are CO2 only (if combustion is complete) or CO & CO2 ( for partial combustion), H2O, SO2 (if any sulfur exists in the fuel) • Composition of the product gas is either on a wet basis (mole % of components including H2O). Composition could also be done on dry basis (mole % of components without H2O) • Composition on wet basis could be converted into dry basis or the reverse by assuming 100 mol for the given composition and calculating the desired basis • Read Example 4.8-1 P. 143-144 Theoretical and Excess Air • In combustion reactions, oxygen or air is usually the excess reactant because they are less expensive than fuel • Theoretical oxygen: the moles or molar flow rate of O2 needed for complete combustion of all fuel fed to the reactor assuming all carbon is oxidised into CO2 and Hydrogen is oxidised into H2O • Theoretical air: the quantity of air in moles that the conatains the theoretical oxygen • Percent Excess Air: (moles air)fed - (moles air)theroretical X 100 (moles air)theoretical • If fuel flow rate and stoichiometric equations for complete combustion of fuel are known, then theoretical oxygen and theoretical air feed rates can be calculated. If actual feed rate is also known then the percent excess air can be calculated • If theoretical oxygen or air feed rate , together with the % excess air is given, then the actual feed rate may be calculated. If 50% excess air is supplied, for example, then (moles air)fed = 1.5 (moles air)theoretical Example 4.8-3 Combustion of Ethane Given: • Fuel: Ethane, conversion 90% • Air : 50% excess • CO: 25% reacted ethane • CO2: 75% reacted ethane C2H6 + (7/2) O2→2CO2 + 3H2O C2 H6 + (5/2) O2 →2CO +3H2O Solution Basis : 100 mol C2 H6 Degree-of-Freedom Analysis • 7 unknowns (no, n1, n2, n3 , n4 , n5 , n6 , n7) • 3 atomic balances (C,H, O) • 1 N2 balance • 1 excess air specification • 1 ethane conversion specification • 1 CO/CO2 ratio specification • (nO2)theroretical = 100 mol C2H6 X 3.5 (mol O2 / mol C2 H6) = 350 mol O2 0.21 no = 1.50(350 mol O2) no = 2500 mol air fed • 90 % Ethane conversion n1 ( unreacted ethane) = = 0.1 (100 mol C2 H6) = 10 mol C2H6 reacted ethane = 90 mol • n4 (mol CO formed) = 0.25 X 90 mol C2 H6 X (2 mol CO/1mol C2 H6) = 45 mol CO • Nitrogen balance input = output 2500 X 0.79 = n3 = 1975 mol • Atomic carbon balance: input = output 100 mol C2H6 X (2 mol C/ 1 mol C2 H6 ) = n1 mol C2H6 X (2 mol C/ 1 mol C2 H6) + n4 mol CO X (1mol C/ 1mol CO) + n5 mol CO2 X (1mol C/ 1mol CO2) Substitute n1 = 10 mol, n4 = 45 mol n5 = 135 mol CO2 • Atomic Hydrogen balance: input = output 100 mol C2H6 X (6 mol H/ 1 mol C2 H6 ) = 10 mol C2H6 X (6 mol H/ 1 mol C2 H6) + n6 mol H2O X (2mol H/ 1mol H2O) n6 = 270 mol H2O • Atomic Oxygen balance: 1.5 X350 mol O2 X (2 mol O/ 1 mol O2) = n2 mol O2 X (2 mol O/ 1 mol O2) + 45 mol CO X (1mol O/ 1mol CO) + 135 mol CO2 X (2mol O/ 1mol CO2) + 270 mol H2O X (1mol O/ 1mol H2O) n2 = 232 mol O2 • The analysis of stack gas is: n1 = 10 mol C2 H6 n2 = 232 mol O2 n3 = 1974 mol N2 n4 = 45 mol CO n5 = 135 mol CO2 2396 mol dry gas + n6 = 270 mol H2O 2666 mol total The stack gas composition on a dry basis is: y1= 10 mol C2 H6 = 0.00417 mol C2 H6 / mol 2396 mol dry gas y2 = 232 mol O2 = 0.0970 mol O2 / mol 2396 mol dry gas y3 = 1974 mol N2 = 0.824 mol N2 / mol 2396 mol dry gas y4 = 45 mol CO = 0.019 mol CO /mol 2396 mol dry gas y5= 135 mol CO2 = 0.0563 mol CO2 / mol 2396 mol dry gas • Mole ratio of water to dry stack gas 270 mol H2O = 0.113 mol H2O 2396 mol dry stack gas mol dry gas Energy Balances • For designing a process, an engineer has to account for the energy flowing into and out of process unit and therefore determine the energy requirement of a process • Examples: cooling of CSTR with highly exothermic reaction, pumping, burning of fuel to produce energy, Forms of Energy of a System • Kinetic Energy • Potential Energy • Internal Energy: due to motion of molecules, atoms, electrons, interaction between molecules, interaction between atoms, interaction between electrons……. • Energy transferred from or to a system: A. Work done by a system on surroundings (example: expansion of a gas to move a piston), or work done by surroundings on a system ( example: mixing of a liquid in a tank) or work due to electric energy or due to magnetic field B. Heat transferred from a system to surroundings or from surroundings to a system (examples: condensers and evaporators) Differences Between Energy and Mass Balances • Energy balance equation could be written as: Energy input = Energy output + Energy accumulation • Note: that there is no generation or consumption terms like total mass balance since energy can neither be destroyed nor created • Note: Unlike material balance, there is no component energy balances, only one energy balance equation involving all streams and all components Energy Balances On A Closed System • A system is closed when mass doesn’t cross the system boundaries during the period of time covered by the energy balance (example: a closed tank, a batch reactor) • The integral energy balance equation on a closed system between two instants of time is: ∆U + ∆Ek + ∆Ep = Q-W • ∆ signifies change between final and initial state • U = Internal Energy • Ek = kinetic Energy • Ep = Potential Energy • W = work of different forms as mentioned before • Q = heat transferred from the system to surroundings or from the surroundings to the system Energy Balances On Open Systems at Steady State • An open system has mass crossing its boundaries. (examples: continuous reactors, distillation units, extractors…..) • The steady- state energy balance equation takes the form: Rate of Energy Input = Rate of Energy Output . . . . . ∆ H+ ∆Ek + ∆Ep = Q-Ws • Ws = shaft work due to the presence of a pump, compressor or turbine • ∆H = change in enthalpy between the outlet and inlet of a system Note that the work for an open system has two components : • Shaft work (Ws) appearing in the final equation • Flow work (Wfl) used to push streams in and out from a system and this is implicitly contained in the enthalpy . •∆H = . ^ . ^ ( ∑mj Hj )output streams - ( ∑mj Hj )input streams . •∆Ek = . 2 2 . ( ∑mj uj /2 )output streams -( ∑mj uj /2)input streams . . •∆Ep = . . ( ∑mj g zj )output streams -( ∑mj g zj )input streams So finally the energy balance equation could be written as : . . . . . ∆H + ∆ Ek + ∆Ep = Q-Ws Two simplifications: 1) If the process has a single input stream and a single . output stream having the same mass flow rate (m), then for example, . . ^ ^ ∆H = m (Hout-Hin) 2) For the same value of any of the energy forms for all input and output streams, the corresponding term to this form drops from the energy balance equation . . ^ . ∆H = H (∑ mj output streams - ∑mj input streams) . At steady state, ∆H = 0 Example 7.4-2 Given: Steam turbine, mass flow rate = 500 kg/hr Inlet conditions • Steam pressure and temperature : 44 atm, and 450 oC • Velocity = 60 m/s Outlet conditions Exit point 5 m below inlet Steam pressure: atmospheric Velocity = 360 m/s Shaft work = 70 kW, heat loss = 104 kcal/h steam steam 1 atm 500 kg/hr saturated 44 atm, 360m/s 450 o C 60m/s Q = -104 kcal/ h Ws = 70 kW Required: Change in specific enthalpy of steam between inlet and outlet Solution (P.324&325) Use the Energy balance equation on steam turbine • Convert mass flow rate to kg/s: 500 kg/h * (1/3600 s/h) = 0.139 kg/s • Convert heat loss to kW -104 kcal/h*(1J/0.2390*103)*(1h/3600s)*(1W/103kW) . in kW • Calculate rate of change of kinetic energy 0.139 kg/s*(3602-602)/2 J/kg *(1kW/103W) • Calculate rate of change of potential energy in kW 0.139 kg/s*[9.81*(-5)] J/kg*(1kW/103 W) • Substitute shaft work (positive value, as is) in the energy balance equation • Substitute heat loss ( as negative of the given value) in the energy balance equation . . . . . ∆H = Q-Ws- ∆Ek- ∆Ep = -90.3 kW . .. ^ ^ • Finally, calculate H2-H1 by dividing ∆H by m . . ^ ^ • H2-H1 =∆H/m = -90.3 kW/0.139 kg/s = -650 kJ/kg Tables of Thermodynamic Data State • The state of any material is characterized by temperature, pressure, phase Values of enthalpy and internal energy • The absolute value of internal energy or enthalpy is impossible to determine. • The change in enthalpy or internal energy corresponding to a specified change of state could be measured from the energy balance equation. ^ ^ • The values of U or H listed in thermodynamic tables are changes measured relative to a reference state • A value of zero may be assigned to the ^ ^ reference state, then any ∆H = H-0=H ^ • Tables are constructed for different materials to read the enthalpy and the internal energy values. (for example, steam tables) • Reference states of a table may be known, or unknown, but this is not important since we are always interested in calculating some the changes in enthalpy or internal energy between tabulated states. • Care must be taken to avoid using different property tables for the same material (because enthalpy or internal energy might be based on different reference states) Example 7.5-1 • Use of Tabulated Enthalpy Data for saturated methyl chloride (P. 326 & 327) Steam Tables • As water (subcooled liquid) is heated at constant pressure, it starts to evaporate and after a while, it boils. • Boiling occurs at different temps. and press. These temperatures and pressures are termed saturation conditions. • At the saturation conditions, water may be a saturated liquid, saturated vapor, or a mixture of saturated vapor and a saturated liquid. • At temperatures above the saturation temperature but at the same saturation pressure, water exists as a superheated vapor. • The properties of water (saturation temp, saturation press, superheating temp specific enthalpy, specific internal energy specific volume of liquid water and water vapor) are tabulated in saturated or superheated steam tables Example 7.5-2 Use of steam tables (P.328 and 329) Example 7.5-3 Given: • Steam mass rate = 2000 kg/h •Inlet Steam pressure : 10 bar a and superheat 190 oC (steam temperature is above the saturation temperature at the given pressure by 190 oC) • Q= 0 (turbine operation adiabatic) • Outlet Saturated steam pressure: 1 bar • Neglect P.E. & k .E. changes Required . • Ws in kW Solution • The equation simplifies to: . energy. balance . ^ -^ Ws = -∆ H = m (H out Hin) • First convert mass flow rate to kg/s2000/3600 kg/s • Second obtain the inlet steam enthalpy from superheated steam table B.7: Saturation temp. at 10 bar = 180 oC (from saturated steam tables) Superheated temp. at 10 bar = 180 +190 = 370 oC ^ Hin (10 bar, 370 oC)= 3201 kJ/kg (by interpolation) • Details of interpolation at P = 10 bar ^ kJ/ kg t oC H 350 3159 370 x? 400 3264 370-350 50 x- 3159 3264-3159 x = (3264-3159)*(370-350)/50 + 3159 = 3201 kJ/kg • Third obtain the outlet steam enthalpy from saturated steam tables: ^ Hout (1 bar, saturated)= 2675 kJ/kg . . W= -∆H = 2000 kg/3600 *(2675-3201)= = 292 kW Energy Balance Procedure A. Draw and label a flow chart for the energy balance problem B. Include the labels in the chart : 1. Pressure of each stream 2. Temp of each stream 3. Phase of a stream (liquid, gas or solid) 4. Indicate if stream is pure or multicomponent 5. Specific enthalpy 6. Mass flow rates (material balances could be used to obtain missing m.f.r. of streams) 7. Apply the energy balance equation and find the unknown Example 7.6-1 Energy Balance on one- component Process Given: Inlet • Two water streams entering a boiler: feed stream1 120 kg/min, Temp.= 30oC feed stream2 175 kg/min, Temp. = 65oC Boiler pressure 17 bar (absolute) Outlet • Steam flowing in a 6-cm ID pipe • Neglect kinetic energies of inlet streams • Neglect potential energy change (a usual assumption) 120 kg/min H2O (l) 30oC H^= 125.7 kJ/kg 175 kg/min H2O (l) 65oC H^= 271.9 kJ/kg Boiler 295 kg/min H2O (v) 17 bar (saturated) ^H= 2793 kJ/kg 6-cm ID pipe Required . • Q (kJ/min)? Solution • Obtain the mass flow rate of steam from a simple material balance • Use steam tables to determine the specific enthalpies of liquid water ( read the tabulated values neglecting the effect of pressure), and the specific enthalpy of the produced saturated steam • Apply the energy balance equation . . . . . Q- Ws = ∆ H + ∆Ek + ∆Ep . . . Q = ∆ H + ∆Ek . ∆H = 295 kg/min*(2793 kJ/kg) - 120 kg/min*(125.7 kJ/kg) - 175 kg/min *(271.9 kJ/kg) = 7.61*105 kJ/min . ∆Ek Must first obtain the velocity of steam in the exit line . u(m/s)= V( volumetric flow rate of steam, m3/s) / A (Area of pipe, m2) Volumetric flow rate is obtained from the given mass flow rate and density of steam. Note that the density of steam is the inverse of steam specific volume that could be found from steam tables Area of pipe = πR2 = 3.1416 *32cm2* (1m2/104cm2) = 2.83*10-3m2 . V = 295 kg/min *(1min/ 60s)*(0.1166 m3/ kg) u (m/s)= 295 kg/min *(1min/ 60s)*(0.1166 m3/ kg)/2.83*10-3 m2 = 202 m/s . . ∆Ek = m u2/2 = 295 kg/min*(2022/2) J/kg (1kJ/103J) = 6.02 *103 kJ/min . . . Q =∆ H + ∆Ek = 7.61*105+ 6.02*103kJ/min = 7.67 *105 kJ/min It is typical of energy balance problems on processes with chemical reactions, phase changes or large temperature changes to neglect kinetic energy changes as they represent a small fraction of the total energy requirement for the process. Example 7.6-2 Two-component Process Given: Heating Inlet • Liquid:60 wt% ethane, 40% n-butane • 150 K, P = 5 bar Outlet • Same liquid • 200 K, P = 5 bar • Neglect potential and kinetic energy changes • Use tabulated enthalpy data for ethane & n-butane from Perry’s Chemical Engineer’s Handbook on P. 223 & P. 234 Required • Heat input per kg of the mixture? 1 kg /s (150 K, 5 bar) 1 kg /s (200 K, 5 bar) 0.6 kg /s ethane (l) 0.6 kg /s ethane (l) 0.4 kg /s butane (l) 0.4 kg /s butane (l) Solution • No material balance is necessary since there is only one stream . . . . Q = ∆ H = Hout - Hin =[ 0.6 kg /s C2 H6* 434.5 kJ /kg + 0.4 kg /s C4 H10 * 130. 2 kJ /kg] -[0.6 * 314.3 + 0.4*30] = 112 kJ/s ^ Q = 112 kJ/s / 1kg /s = 112 kJ /kg Example 7.6-3 Simultaneous Material and Energy Balances Given: Adiabatic Mixing Inlet Saturated Steam 1 atm, 1150 kg /h Superheated Steam 400oC, 1atm Outlet Superheated 300oC, 1atm 1150 kg /hr H2O (v) 1 atm, 100o C ^ H = 2676 kJ /kg . . m2 kg /hr H2O (v) m1 kg /hr H2O (v) 1 atm, 300o C 1 atm, 400o C ^ = 3074 kJ /kg H ^ = 3278 kJ /kg H Required • Mass flow rate of steam at 300oC? • Volumetric flow rate of steam at 400oC? Solution • • • • Heat transfer, Shaft work = 0 Negligible change of K.E. & P.E. Two unknowns Must use both material and energy balance Mass balance equation . . 1150 kg/h + m1 = m2 Energy balance equation . ∆H=0 . . Hout = Hin (1) . 1150 kg/h * 2676 kJ /kg + m1*3278 kJ/kg . = m2* 3074 kJ /kg (2) Solving eqs. 1 & 2 simultaneously . m1 = 2240 kg/h . m2 = 3390 kg/h Obtain the specific volume of superheated steam at 400oC and 1 atm from table B.7 ^ V = 3.11 m3/kg • The volumetric flow rate of this steam is 2240 kg/h *3.11 m3/kg = 6980 m3/h • If specific volume data were not available, the ideal gas equation may be used to calculate the density of steam. Enthalpy Changes Based on Inlet or Outlet Conditions Until now, we considered reading enthalpies from tables of thermodynamic data (see past examples on steam turbines, heating ethane and butane) and the reference state is that used to generate the table What if no available thermodynamic data table exists for a particular species in the process? • Choose one of the inlet or outlet states as ^ the reference state for the species so that at least one H may be set equal to zero. • Calculate the enthalpy of each component in each inlet or outlet stream relative to the chosen reference state by using hypothetical paths having constant pressure, constant temperature or constant pressure and temperature. This will be explained in the next few slides. Calculation of a Species Enthalpy • To perform the calculation of the enthalpy of a species at a particular state (inlet or outlet), choose any convenient path from the reference state to the required state of ^ or ∆H ^ for that the species and determine H i i path. • Paths chosen are either constant pressure or constant temperature or both. Constant Temperature Path • For a solid or liquid species undergoing pressure changes at constant temperature and volume: ^ = ^V ∆P ∆H • For an ideal gas species undergoing pressure changes at constant temperature and volume: ^ =0 ∆H • For a real gas ( at temperatures well below 0oC or well above 1 atm), either use thermodynamic tables or thermodynamic correlations. Constant Pressure Path: For ideal gases ˆH T2 c (T) dT T p 1 For real gases at constant pressure For solids or liquids Constant temperature and pressure (phase Changes) ˆ H ˆ H v (latent heat of vaporizati on) For example: change of phase from liquid to vapor or from vapor to liquid usually at the normal boiling point General Procedure For Energy Balance Calculations on Closed and Open Systems (Text, P.361) 1. Draw and label fully a chart for the energy balance problem as mentioned earlier 2. Write the appropriate form of the energy balance equation (closed or open system) and delete any of the terms that are either zero or negligible 3. Check if the values of the specific internal energy or specific enthalpy at the inlet and the outlet for all components are given or could be obtained from information in the problem or could be read from thermodynamic tables • If not, choose one of the inlet or outlet conditions for each species as its reference state 4. For a closed system, construct a table showing the reference states chosen at the top and with columns for initial and final masses or number of moles of each components and with other columns for the components’ final and initial specific internal energies • For an open system, a similar table is constructed but with columns showing molar or mass flow rates of components and their inlet and outlet specific enthalpies 5. Calculate the values of the specific internal energies or enthalpies corresponding to each component by using hypothetical paths from the reference state to a particular initial or final state if the system is closed or from the reference state to a particular inlet or outlet condition if the system is open 6. Insert the calculated values of internal energies or enthalpies in their appropriate places in the table 7. Calculate the changes in internal energies or the changes in enthalpy based on the values in the table 8. Calculate other terms in the energy . . balance equation that have not been neglected (i.e. Δ K.E, Δ P.E, Ws, Q) 9. Solve the energy balance equation for the required unknown term Example 8.1-1 Energy Balance on a Condenser Given Condenser Inlet: Mixture (Ac (v), N2) 100 mol/s, 65oC, 1 atm, known mol fraction Outlet: Mixture (Ac (v), N2) 36.45 mol/s, 20oC, 5 atm, known mol fraction AC (l) 63.55 mol /s, 20oC, 5 atm 36.45 mol/s 100 mol/s 0.092 mol Ac(v) /mol 0.669 mol Ac(v) /mol 0.908 mol N2 /mol 0.331 mol N2 /mol 20oC, 5 atm 65oC, 1 atm Condenser 63.55 mol AC(l) /s 20oC, 5 atm Required: Cooling rate? Solution 1. No need for material balance calculations 2. Application of energy Balance: H Q 3. To calculate the enthalpy change, reference states must be chosen for both Ac and N2: • Table B.8 Lists specific enthalpies of N2 relative to N2(g, 25oC, 1 atm) [this the reference state chosen for nitrogen] • No tabulated enthalpy data for acetone, so one of the process stream conditions and let ^ o it be Ac (l, 20 C, 5 atm) so that HAc at this state is equal to zero Construct an inlet -outlet enthalpy table having references written at the top of the table as follows: References: Ac (l, 20oC, 5 atm), N2(g, 25oC,1 atm) . Substance nin ^ H . ^ out H kJ/mol H3 mol/s kJ/mol nout mol/s Ac (v) 66.9 H1 3.35 Ac (l) _ _ 63.55 0 N2 33.1 H2 33.1 H4 in 5. Calculate all unknown specific enthalpies • As mentioned before, specific enthalpies are obtained relative to the reference states. • For N2, these are directly read from table B.8 • For Acetone, the specific enthalpies are calculated as enthalpy changes between the process state and the reference state. For example : H^1 = ∆H^ ( Ac(l), 20oC, 5 atm) (Ac (v), 65 oC, 1 atm) The following process path from the reference state makes the calculation of the inlet specific enthalpy of Acetone possible: Ac(l, 20 o C, 5 atm) ˆ H 1a Ac(l, 20 o C, 1atm ) ˆ H 1b Ac(l, 56 o C, 1atm ) ˆ H 1c Ac(v, 56 o C, 1atm ) ˆ H 1d Ac(v, 65 o C, 1atm ) Ĥ 1 ˆ H path ˆ H 1a ˆ H 1b ˆ H 1c ˆ H 1d V̂Ac ( l ) (1atm 5atm ) 56 o C 65 o C 20 o C 56 o C Cp( Ac (l ) dT ( Ĥ v ) Ac Cp( Ac ( v ) dT C p ( Ac(l )) 0.123 18.6 10 5 T C p ( Ac( v )) 0.07196 20.10 10 5 T 12.78 10 8 T 2 34.76 10 - 12 T 3 Substituti ng the specific volume and doing the integratio ns Ĥ 1 (0.0279 4.68 30.2 0.753) 35.7kJ / mol Proceeding in a similar manner, values of all other unknown specific enthalpies of Ac are calculated Substance .n in ^in H .n out ^out H kJ/mol mol/s kJ/mol mol/s Ac (v) 66.9 35.7 3.35 32.0 Ac (l) _ _ 63.55 0 N2 33.1 1.16 33.1 -0.10 6. Calculate H ( 3.35 mol / s)( 32.0 kJ/mol) (63.55)(0) ( 33.1)(-0.1 ) - (66.9)(35. 7) - (33.1)(1.1 6) H - 2320 kJ/s 7. Solve the energy balance for Q H -2320 kJ/s Q • To calculate enthalpy changes at constant pressure, formulas for heat capacities of different components are needed. • In the absence of tabulated formulas or data, use is made of kopp’s rule to estimate the at capacity of a solid or liquid compound at or near 200C. Cp Ca (OH)2 = (Cp)Ca +2(Cp)O +2(Cp)H = 26 + 2X17 +2X9.6 = 79 J/mol oC Enthalpy changes for mixtures • For mixtures of gases or liquids (streams), calculate the total enthalpy change as the sum of the enthalpy changes for the pure components. Neglect enthalpy changes due to mixing, this is true for gases at 1atm and for liquids of similar structure. • Also for highly dilute solutions of solids or gases in liquids, neglect enthalpy change of a solute • Enthalpy changes for heating or cooling a mixture can be simplified by using a heat capacity of the mixture given by the equation: (Cp)mix (T) = ∑all components yi Cpi(T) yi could either be the mol fraction or the mass fraction of a component depending on the units of Cp T2 Ĥ (C p ) mix (T)dT T1 Example 8.3-4 Heat capacity of a mixture Given Heating Inlet 60%(by volume) C2H6 , 40% C3H8 150 mol /h, 0oC Outlet same mixture at 400oC Required: ? Q Use heat capacity of a mixture? Solution •Cp data at 1 atm from table B.2 (Cp)mix = 0.6 (0.04937 + 13.92X10-5 T - 5.816 X 10-8T2 + 7.28 X10-12T3) + 0.4 (0.06803 + 22.59 X 10-5 T -13.11 X 10-8T2 + 31.71 X 10-12T3) = 0.05683 +17.39 X10-5 T - 8.734 X 10-8 T2 + 17.05 X 10-12 T3 T2 Ĥ (C p ) mix (T)dT 34.89 kJ / mol T1 H n H 150 mol / h * 34.89 kJ /mol Q 5230 kJ/h Usually for gas mixtures, It is reasonable to assume the mixture behaves as an ideal gas if A. The inlet or outlet pressure or both are not specified B. The mixture of gases may be nearly ideal C. There might be moderately small changes in pressure (of few atmospheres) Therefore, for all of these cases, the specific enthalpy change is zero at constant temperature Example 8.3-5 Energy Balance on a Gas Preheater Given: • Heating Inlet • Mixture 10% CH4, 90% air (by volume) 20oC Outlet • Same mixture 300oC • Gas flow rate 2 X 103 L (STP) /min 2000 L (STP)/min, 2000 L (STP)/min, 0.1 mol CH4 /mol 0.1 mol CH4 /mol 0.9 mol mol air /mol 0.9 mol mol air /mol 20oC, n mol / min Heater 300oC, n mol / min Required ? Q Rate of heating kW? Solution • Assume ideal gas behavior (neglecting effect of pressure because it is not specified for both streams) • Obtain molar flow rate: 2000L (STP)/min *(1 mol/22.4 L(STP)) = 89.3 mol/min • No need to consider mass balances since there is only one stream • No Shaft work , No changes in kinetic or potential energy H n Ĥ n Ĥ Q i i i i out in • Construct enthalpy table References: CH4 (g, 20oC,1atm), air (g, 25oC, 1 atm) Substance CH4 Air .n ^ H .n mol/ min kJ/mol mol/min in 8.93 80.4 in 0 ^ H2 out 8.93 80.4 ^ H out kJ/mol ^ H1 ^ H3 • The reference condition of air is chosen such that both values of specific enthalpy at the inlet and the outlet are read directly from Table B.8 • The reference condition of methane was chosen at the inlet assuming it exists as a pure component •Enthalpies of air are read directly from table B.8, ^ H2 = -0.15 kJ/mol ^ H3 = 8.17 kJ/mol • Specific enthalpy change of methane in the outlet gas mixture at 3000C relative to pure methane at the reference temp. of 20oC CH4 (g, 20oC, 1 atm) CH4 (g, 300oC, P) • Effect of pressure on enthalpy is neglected • Heat of mixing is neglected specific enthalpy change of methane is for heating pure methane from 20oC to 300oC H 300o C (Cp )CH 4 dT , Substitute for Cp from table B.2 20 o C 300o C 5 8 2 12 3 (0.03431 5.469 10 T 0.3661 10 T 11 10 T ) dT 20 o C 12.09 kJ / mol H (8.93 mol / min)( 12.09kJ / mol ) Q [( 80.4)( 8.17) (8.93)( 0) (80.4)( 0.15) 776 (kJ / min)( 1 min/ 60s )(1kw / 1kJ / s) 12.9 kW Example 8.4-2 Vaporization and Heating Given Heating and vaporization at constant pressure Inlet 100 g-moles / h Liquid n-C6 H14 25oC and 7bar Outlet 100 g-moles / h vapor n-C6 H14 300oC and 7bar Required ? Q Rate of heating ? Solution • No need to consider mass balances since there is only one stream • No Shaft work , No changes in kinetic or potential energy H Q • Construct enthalpy table References: C6 H14 (l, 25oC,7atm) . Substance nin C6 H 14 ^ H in .n out mol/ min kJ/mol mol/min 100 0 100 ^ H out kJ/mol H1 The following process path from the reference state makes the calculation of the outlet specific enthalpy of hexane possible: hexane(l, 25o C, 7 atm) ˆ H hexane (l, 25o C, 1atm ) 1a ˆ H 1b hexane (l, 69o C, 1atm ) ˆ H 1c hexane (v, 69o C, 1atm ) ˆ H 1d hexane (v, 300o C, 1atm ) ˆ H 1e hexane (v, 300o C, 7atm ) Ĥ 1 ˆ H path Q n ˆ H ˆ H 1b ˆ H 1c ˆ H 1d 85.5 kJ / mol path 2.38 kW neglecting pressure effects ˆ H 1a , ˆ H 1e 69 o C 25 o C C p(hexane ( l )) dT 300o C C p(hexane ( v )) dT ˆ H v 69 o C C p (hexane (l )) 0.2163kJ / mol .o C C p (hexane ( v )) 0.13744 40.85 10 5 T 23.92 10 8 T 2 57.66 10 - 12 T 3 ENERGY BALANCES ON REACTIVE SYTEMS Basic Concepts 1. Reactions are either exothermic or endothermic 2. The heat of reaction, is the enthalpy change for a process in which stoichiometric quantities of reactants at temperature T and pressure P react completely in a single reaction to form products at the same temperature and pressure For the reaction: υ A A + υ BB υ CC Where υ is the stoichiometric coefficient of a reactant or a product Ĥ (T , P ) H r o o Ĥ r (To , Po ) specific enthalpy of reaction or heat of reaction n reac tan t out - n reac tan t in n product out - n product in extent of reaction (mol /s) reactant product Properties of heat of reaction • Heat of reaction is negative for exothermic reactions and positive for endothermic reactions • Heat of reaction is independent of pressure • Value of heat of reaction depends on the way the stoichiometric equation is written and also on the states of aggregation of both reactants and products • Standard heat of reaction is the heat of reaction when both reactants and products are at 1 atm and 25oC Heat of Formation Standard heat of formation is the enthalpy change associated with the formation of one mol of a compound at 25oC and 1 atm from its constituent elements as they normally occur in nature • Standard heats of formation for many compounds are listed in table B.1 in your text book. More information could be found in Perry’s Chemical Engineers’ Handbook • Standard heat of formation for an elemental species is zero Heat of Combustion The standard heat of combustion is the enthalpy change associated with the combustion of that substance in presence of oxygen to give CO2 and H2O, both reactants and products exist at 25oC and 1 atm • Table B.1 lists standard heats of combustion for a number of compounds. More information could be found in Perry’s Chemical Engineers’ Handbook Measurement of Heat of Reaction • Heat of reaction may be measured in calorimeter (a closed reactor immersed in a fluid contained in a well insulated vessel) • The rise or fall of fluid temperature can be measured to determine how much is the heat transfer to of from the reactor • By knowledge of the heat capacities of reactants and products and the amount of heat transfer, the standard enthalpy of the reaction may be calculated • The calorimeter technique has limitations, for example, if the reaction doesn’t proceed at a high rate, or if a mixture of products form rather than a pure compund. Calculation of Heat of Reaction Hess’s law: If the stoichiometric equation for some reaction1 can be obtained by algebraic operations (like addition, subtraction and multiplication by constants) on the stoichiometric equations for other reactions 2,3 …….etc., then heat of reaction1 can be obtained by performing same algebraic operations on reactions 2,3,……….. Hess’s law can be applied to obtain heat of reaction by knowledge of heats of formation of reactants and products: Ĥ o r i Ĥ o f i i o Ĥ i fi products o Ĥ i fi rectants When applying the above equation, the Standard heats of formation of all elemental species should be set equal to zero Hess’s law can also be applied to obtain standard heats of reactions that involve only combustible substances and combustible products : Ĥ o r i ( Ĥ o c ) i i i (Ĥ o c ) i rectants i (Ĥ o c )i products If any of the reactants or products are themselves combustion products, then their heats of combustion should be set equal to zero in the above equation. Energy Balances On Reactors (Text, P. 450 & 451) • Follow the steps of the general energy balance procedure explained for a physical system • The only differences are: A) The choice of the reference states for reactants and products are taken at 25 oC and 1 atm for both of them and their states as indicated by the chemical reaction equation • Need to calculate the extent of reaction (Text, P.451) • Need to calculate the enthalpy change for the reactor in a different way than a physical system (Text, P.451)