Search > am Sea ‘cHaPTER? Means MATERIAL AND ENERGY BALANCES (cont'd) > About the book MATERIAL BALANCES: > Introduction > Material and energy. «Basis and units blanc {otal mass and composition Seen + concentrations luid-flew theory 29S of process situations > Fluid-flow applications ontinuous processes > + blending > + Layout The first stop is to look at the three basic categories: materials in, materials out > Dyving and materials stored. Then the materials in each category have to be considered > Evaporation whether they are to be treated as a whole, a gross mass balance, or whether > Contact-equllbrium various constituents should be treated separately and if so what constituents. Separation processes Tq take a simple example, it might be to take dry solids as opposed to total > Mechanical material; this really means Separating the two groups of constituents, non-water separations and water. More complete dissection can separate out chemical types such as > Size reduction minerals, or chemical elements such as carbon, ‘Size reduction The choice and the detail depend on the reasons for making the balance and on > Mixing the information that is required. A major factor in industry is, of course, the value > Appendices of the materials and so expensive raw materials are more likely to be considered > Index to Figures than cheaper ones, and products than waste materials. > Index to Examples > References > Bibliography > Useful inks > Feedback (emai ink) Basis and Units Having decided which constituents need consideration, the basis for the calculations has to be decided. This might be some mass of raw material entering the process in a batch system, or some mass per hour in a ‘continuous process. It could be: some mass of a particular predominant constituent, for example mass balances in a bakery might be all related to 100 kg of flour entering; or some unchanging constituent, such as in combustion calculations with air where it is helpful to relate everything to the inert nitrogen component; or carbon added in the nutrients in a fermentation system because the essential energy relationships of the growing micro-organisms are related to the combined carbon in the feed; or the essentially inert non-oil constituents of the oilseeds in an oil-extraction process. Sometimes it is unimportant what basis is chosen and in such cases a convenient quantity such as the total raw materials into one batch or passed in per hour to a continuous process are often selected. Having selected the basis, then the units can be chosen such as mass, or concentrations which can be weight or molar if reactions are important, Total mass and composition Material balances can be based on total mass, mass of dry solids, or mass of particular components, for ‘example protein, © EXAMPLE 2.1. Constituent balance of milk Skim milk is prepared by the removal of some of the fat from whole milk. This skim milk is found to contain 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original milk contained 4.5% fat, calculate its composition assuming that fat only was removed to make the skim milk and that there are no losses inprocessing. Basis: 100 kg of skim milk, This contains, therefore, 0.1 kg of fat. Let the fat which was removed from it to make skim milk be x kg Total original fat = (x +0.1) kg Total ariginal mass = (100 + x) kg and as itis known that the original fat content was 4.5% so ot oorx = = 0.085 whence x +0.1 .045(100 + x) x =4.6kg So the composition of the whole milk is then fat = 45% 905 | water = PS = 865% a8 8 protein ib, = 33% wbohydrate~ SI = 4.9% carbohydrate tee = 49% og. andash = OR = 0.8% Concentrations Concentrations can be expressed in many ways: weight/weight fraction (wiw ), weight/volume fraction (wiv), molar concentration (M), mole fraction. The weightweight concentration is the weight of the solute divided by the total weight of the solution and this is the fractional form of the percentage composition by weight. The weight/volume concentration is the weight of solute in the total volume of the solution, The molar concentration is the number of molecular weights of the solute expressed as moles in 1:m° of the solution, The mole fraction is the ratio of the number of moles of the solute to the total number of moles of all species present in the solution. Notice that in process engineering, it is usual to consider kg moles and in this book the term mole means a mass of the material equal to its molecular weight in kilograms. In this book, percentage signifies percentage by weight (wiw) unless otherwise specified. EXAMPLE 2.2. Concentration of salt in water A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a liquid of density 1323 kgm’. Calculate the concentration of salt in this solution as a (a) weight/weight fraction, (b) weightivalume fraction, (c) mole fraction, (d) molar concentration. (a) Weight fraction: 20 = 100+20 = 0167 % weighweight = 16.7% (b) Weightvolume fraction: Adensity of 1323 kg m'® means that 1m? of solution weighs 1323 kg, but 1323 kg of salt solution contains 20 400+ 20. * 1323 kg salt = 220.5 kg salt m* and so 1 m? solution contains 220.5 kg salt. Weightivolume fraction = 220.5 1000 2208.‘and so weightivolume = 22.1% (©) Mole fraction 100 Moles of water +8 = 5.56, = 20 = Moles of salt = as = 0.34, = 0.34 Mole fraction of salt = 5557034 and so mole fraction = 0,058 (@) The molar concentration (M) is 22 = 3.77 moles in 1 m°. Note that the mole fraction can be approximated by the (moles of salt/moles of water) as the number af moles of water are dominant, that is the mole fraction is close to 0.34/5.56 = 0.061. As the solution becomes more dilute, this approximation improves and generally for dilute solutions the mole fraction of solute is a close approximation to the moles of solute/moles of solvent. In solidiliquid mixtures all these methods can be used but in solid mixtures the concentrations are normally ‘expressed as simple weight fractions, With gases, concentrations are primarily measured in weight concentrations per unit volume or as partial Pressures, These can be related through the gas laws. Using the gas law in the form: pV=nRT where p is the pressure, V the volume, n the number of moles, T the absolute temperature, and R the gas constant which is equal to 0.08206 m? atm mole"! K"'. The molar concentration of a gas is then nV = pRT and the weight concentration is then nM/V where M is the molecular weight of the gas. ‘The SI unit of pressure is the N mr? called the Pascal (Pa). As this is of inconvenient size for many purposes, standard atmospheres (atm) are often used as pressure units, the conversion being 1 atm = 1.013 x 10° Pa, or very nearly 1 alm = 100 kPa. ‘EXAMPLE 2.3. Air composition If air consists of 77% by weight of nitrogen and 23% by weight of oxygen calculate the: (2) mean molecular weight of alr, {b) mole fraction of oxygen, (0) concentration of oxygen in mole m’ and kg mr if the total pressure is 1.5 atmospheres and the temperature is 25°C. (2) Taking the basis of 100 kg of air: 7 23 ZZ moles of Nz and 23 les of 0, itcontains ZZ moles of Nand 22 moles of O2 Total number of moles = 2.75 + 0.72 = 3.47 moles. ‘So mean molecular weight = Mean molecular weight of air (b) The mole fraction of oxygen O72 275+072 ~ 347 0.21 ‘Mole fraction of oxygen in air_= 0.21.and this is also the volume fraction (©) In the gas equation, n is the number of moles present, p is the pressure, 1.5 atm and the value of R is 0,08206 m? atm mote"’ K"" and at a temperature of 25°C = 25 + 273 = 298 K, and V= 1m? pV = nRT andso 15x14 nx 0.08206 x 298 n 0.061 mole nx mean molecular weight 0.061 x 28.8 = 1.76 kg and of this 23% is oxygen, weighing 0.23 x 1.76 = 0.4 kg. Concentration of oxygen = 0.4 kg ar 04 3 o 3 0.013 mole m’ When a gas is dissolved in a liquid, the mole fraction of the gas in the liquid can be determined by first calculating the number of moles of gas using the gas laws, treating the volume as the volume of the liquid, and then calculating the number of moles of liquid directly. ‘© EXAMPLE 2.4. Carbonation of a soft drink In the carbonation of a soft drink, the total quantity of carbon dioxide required is the equivalent of 3 volumes of gas to one volume of water at 0°C and atmospheric pressure. Calculate (a) the mass fraction and (b) the mole fraction of the C0, in the drink, ignoring all components other than CO, and water. Basis 1 m3 of water 1000 kg, Volume of carbon dioxide added = 3 m°. From the gas equation pV = nRT 1x3 and so n 1 x 0.08206 x 273. 0.134 mole. Molecular weight of carbon dioxide and so weight of carbon dioxide added 4. 0.134 x44 = 5.9 kg. (a) Mass fraction of carbon dioxide in drink = §.9/(1000 + 5.9) 5.910%, (b) Mole fraction of carbon dioxide in drink = 0.134/(1000/18 + 0.134) = 2.41 x 10 ‘Types of Process Situations Continuous processes In continuous processes, time also enters into consideration and the balances are related to unit time. Thus in considering a continuous centrifuge separating whole milk into skim milk and cream, if the material holdup in the centrifuge is constant both in mass and in composition, then the quantities of the components entering and leaving in the different streams in unit time are constant and a mass balance can be written on this basis. Such an analysis assumes that the process is in a steady state, that is flows and quantities held up in vessels do not change with time. O EXAMPLE 2.5. Materials balance in continuous centrifuging of milk If 35,000 kg of whole milk containing 4% fat is to be separated in a 6 h period into skim milk with 0.45% fat and ‘cream with 45% fat, what are the flow rates of the two output streams from a continuous centrifuge which accomplishes this separation? Basis 1 hour's flow of whole milk Mass in Total mass 000 = 5833 kg6 Fat = 5833 x0.04 = 233kg ‘And so water plus solids-not-fat = 5600 kg, Mass out Let the mass of cream be x kg then its total fat content is 0.45x. The mass of skim milk is (5833 - x) and its total fat content is 0.0045(5833 - x), Material balance on fat: Fat in = Fat out 5833 x 0.04 = 0.0045(5833 - x) + 0.45x. andso x = 485kg. ‘So that the flow of cream is 465 ko hand skim milk (5833 - 465) = 5368 kg hr’ ‘The time unit has to be considered carefully in continuous processes as normally such processes operate continuously for only part of the total factory time. Usually there are three periods, start up, continuous processing (so-called steady state) and close down, and it is important to decide what material balance is being studied. Also the time interval over which any measurements are taken must be long enough to allow for any slight periodic or chance variation. In some instances a reaction takes place and the material balances have to be adjusted accordingly. Chemical changes can take place during a process, for example bacteria may be destroyed during heat Processing, sugars may combine with amino acids, fats may be hydrolysed and these affect details of the material balance. The total mass of the system will remain the same but the constituent parts may change, for ‘example in browning the sugars may reduce but browning compounds will increase. An example of the growth of microbial cells is given. Details of chemical and biological changes form a whole area for study in themselves, ‘coming under the heading of unit processes or reaction technology. © EXAMPLE 2.6. Materials balance of yeast fermentation Baker's yeast is to be grown in a continuous fermentation system using a fermenter volume of 20 m? in which the flow residence time is 16 h. A2% inoculum containing 1.2 % of yeast cells is included in the growth medium. This is then passed to the fermenter, in which the yeast grows with a steady doubling time of 2.9 h, The broth leaving the fermenter then passes to a continuous centrifuge which produces a yeast cream containing 7% of yeast, 87% ofthe total yeast in the broth, Calculate the rate of flow of the yeast cream and ofthe residual broth from the centrifuge, ‘The volume of the fermenter is 20 m? and the residence time in this is 16 h so the low rate through the fermenter must be 20/16 = 1.250 m° h* Assuming the broth to have a density substantially equal to that of water, .e. 1000 kg mr, Mass flow rate of broth = 1250 kg h”* Yeast concentration in the liquid flowing to the fermenter = (concentration in inoculum)(dilution of inoculum) = (1.21100)/(100/2) = 2.4 x 10 kg kg"? Now the yeast mass doubles every 2.9 h, so in 2.9h, 1 kg becomes 1 x 2" kg (1 generation). In 16h there are 16/2.9 66 doubling times 1kg yeast grows to 1 x 255 kg = 48.5 kg. Yeast in broth leaving =48.5x24x 104 kg kg” Yeast leaving fermenter initial concentration x growth x flow rate 4x 104 x 48.5 x 1250 5 kg h? ‘Yeast-free broth flow leaving fermenter = (1250 - 15) = 1235 kg htFrom the centrifuge flows a (yeast rich) stream with 7% yeast, this being 97% of the total yeast: The yeast rich stream is (15 x 0.97) x 100/7 = 208 kgh™ and the broth (yeast lean) stream is (1250 - 208) = 1042 kg h* which contains (15 x 0.03 ) = 0.45 kg ht yeast and the yeast concentration in the residual broth = 0.45/1042 = 0.043% Materials balance over the centrifuge per hour Mass in (kg) Mass out (ka) Yeast-free broth 1235 kg Broth 1042 kg Yeast 18kg (Yeast in broth 0.48 kg) Yeast stream 208 kg (Yeast in stream 14.55 kg) Total 1250 kg Total 1250 kg, ‘A materials balance, such as in Example 2.6 for the manufacture of yeast, could be prepared in much greater detail if this were necessary and if the appropriate information were available, Not only broad constituents, such as the yeast, can be balanced as indicated but all the other constituents must also balance. ‘One constituent is the element carbon: this comes with the yeast inoculum in the medium, which must have a suitable fermentable carbon source, for example it might be sucrose in molasses. The input carbon must then balance the output carbon, which will include the carbon in the outgoing yeast, carbon in the unused medium and. also that which was converted to carbon dioxide and which came off as a gas or remained dissolved in the liquid Similarly all of the other elements such as nitrogen and phosphorus can be balanced out and calculation of the balance can be used to determine what inputs are necessary knowing the final yeast production that is required and the expected yields. While a formal solution can be set out in terms of a number of simultaneous equations, it can often be easier both to visualize and to calculate if the data are tabulated and calculation proceeds step by step gradually filling out the whole detail. Blending Another class of situations which arises includes blending problems in which various ingredients are combined in such proportions as to give a product of some desired composition. Complicated examples, in which an optimum ‘or best achievable composition must be sought, need quite elaborate calculation methods, such as linear programming, but simple examples can be solved by straight-forward mass balances, © EXAMPLE 2.7. Blending of minced meat A processing plant is producing minced meat, which must contain 15% of fat. If this is to be made up from boneless cow beef with 23% of fat and from boneless bull beef with 5% of fat, what are the proportions in which these should be mixed? Let the proportions be A of cow beef to B of bull beet ‘Then by a mass balance on the fat, Mass in Mass out Ax 0.23 +Bx0.05 = (A+B) x0.15. that is A(0.23 - 0.15) = B(0.15 -0.05} ‘A(0.08) | = B(0.10). NB 1018 or A(A+B) 10/18 = 5/9. It is possible to solve such a problem formally using algebraic equations and indeed all material balance problems are amenable to algebraic treatment. They reduce to sets of simultaneous equations and if the number of independent equations equals the number of unknowns the equations can be solved. For example, the blending problem above can be solved in this way.If the weights of the constituents are A and B and proportions of fat are a, b, blended to give C of composition c: then forfat Aa + Bb =Cc and overall A+B =C of which A and B are unknown, and say we require these to make up 100 kg of C then A+B or B 100 100-A and substituting into the first equation ‘Aa + (100- Alb = 100c or Ala-b) = 100(c- b) or ‘A = 100 (cb) (a-b) and taking the numbers from the example 100 (0.15 - 0.05) (0.23 -0.08) 100(0.10) (0.18) = 555k9 andB = 44.4 kg as before, but the algebraic solution has really added nothing beyond a formula which could be useful if a number of blending operations were under consideration, Layout In setting up a material balance for a process a series of equations can be written for the various individual ‘components and for the process as a whole. In some cases where groups of materials maintain constant ratios, then the equations can include such groups rather than their individual constituents. For example in drying vegetables the carbohydrates, minerals, proteins etc., can be grouped together as ‘dry solids’, and then only dry solids and water need be taken through the material balance. EXAMPLE 2.8. Drying yield of potatoes Potatoes are dried from 14% total solids to 93% total solids. What is the product yield from each 1000 kg of raw Potatoes assuming that 8% by weight of the original potatoes is lost in peeling. Basis 1000kg potato entering As 8% of potatoes are lost in peeling, potatoes to drying are 920 kg, solids 129 kg. Mass in (ka) Mass out (9) Raw Potatoes Dried Product Potato solids 140 kg Potato solids 129g Water 860 kg Associated water 10 kg Total product 139k9 Losses: Peelings ~ solids 11g - water 69kg Water evaporated 781 kg ‘Totallosses 861kg Total 1000 kg Total 1000 kg Product yield 43 = 14% Notice that numbers have been rounded to whole numbers as this is appropriate accuracy.Often itis important to be able to follow particular constituents of the raw material through a process. This is just ‘a matter of calculating each constituent, EXAMPLE 2.9. Extraction of oil fom soya beans 1000 kg of soya beans, of composition 18% oil, 35% protein, 27.1% carbohydrate. 9.4%, fibre and ash, 10.5% moisture, are’ (a) crushed and pressed, which reduces oil content in beans to 6%; (b) then extracted with hexane to produce a meal containing 0.5% oil; (c) finally dried to 8% moisture. Assuming that there is no loss of protein and water with the oil, set out a mass balance for the soya-bean constituents. Basis 1000 kg Mass in: Oil = 1000x 18/100 = 180 kg Protein = 1000 x 35/100 = 350 kg Total non-cil constituents = 820 kg Carbohydrate, ash, fibre and water are calculated in a similar manner to fat and protein. Mass out: (a) Expressed oil, In original beans, 820kg of protein, water, etc., are associated with 180 kg of oi In pressed material, 94 parts of protein, water, et., are associated with 6 parts of oil, Total oil in expressed material 820 x 6/94 = 52.3 kg. Oil extracted in press = 180-523, 127.7 kg. (b) Extracted oil. In extracted meal 99.5 parts of protein, water, elc., are associated with 0.5 parts of oil Total oil in extracted meal = 820 x 0.5/99.5 = 4.1 kg, Ollextracted inhexane = 523-41 = 48.2kg. (c) Water. In the dried meal, 8 parts of water are associated with 92 parts of oil, protein, etc. Weights of dry materials in final meal = 350 + 271 +94 + 4.1 = 719.1 kg, Total water in dried meal 719.1 x 8/92 = 62.5 kg Water loss in drying = 105-625 =425kg. MASS BALANCE. BASIS 1000 kg SOYA BEANS ENTERING Mass in (kg) Mass out, (kg) ‘Total oil consisting of 175.9 oll 180 - Expressed oil 127 Protein 360 - Oil in hexane 48,2 Carbohydrate amt ‘Total meal consisting of 781.6 Ash and fbre 94 - Protein 350 Water 105 - Carbohydrate an ~ Ash and fibre 94 - Water 625 - Oil 44 Water lost in drying 428 Total 1000 kg Total 1000 kg > Material & Energy Balances > ENERGY BALANCES 4 Back to the topUnit Operations in Food Processing. Copyright® 1983, RL. Earle. : Published by NZIFST (Inc) ise Fens