Gp-Cpc-01 Units – Composition – Basic Ideas
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GP-CPC-01 UNITS – BASIC IDEAS – COMPOSITION 11-06-2020 Prof.G.Prabhakar Chem Engg, SVU GP-CPC-01 UNITS – CONVERSION (1) ➢ A two term system is followed. A base unit is chosen and the number of base units that represent the quantity is added ahead of the base unit. Number Base unit Eg : 2 kg, 4 meters , 60 seconds ➢ Manipulations Possible : • If the nature & base unit are the same, direct addition / subtraction is permitted 2 m + 4 m = 6m ; 5 kg – 2.5 kg = 2.5 kg • If the nature is the same but the base unit is different , say, 1 m + 10 c m both m and the cm are length units but do not represent identical quantity, Equivalence considered 2 options are available. 1 m is equivalent to 100 cm So, 100 cm + 10 cm = 110 cm 0.01 m is equivalent to 1 cm 1 m + 10 (0.01) m = 1. 1 m • If the nature of the quantity is different, addition / subtraction is NOT possible. Factors used to check equivalence are known as Conversion Factors. GP-CPC-01 UNITS – CONVERSION (2) • For multiplication / division, there are no such restrictions. They give rise to a set called derived units Even if there is divergence in the nature, multiplication / division can be carried out. Eg : Velocity ( length divided by time ) Mass flow rate (Mass divided by time) Mass Flux ( Mass divided by area (Length 2) – time). Force (Mass * Acceleration = Mass * Length / time 2) In derived units, each unit is to be individually converted to suit the requirement Density = 500 kg / m3 . What is its equivalence in g / cm3 1 kg = 1000 g ; 1 m = 100 cm 500 kg * 1000 g * m3 = 500 * 10-3 g / cm3 m3 1 kg (100)3 cm3 GP-CPC-01 Vander waals equation ( P + a/V2) ( V-b) = RT Units of b : Since it is being subtracted from V, b will have the same units as V Units of a : Since a/V2 is being added to P, this group shall have the units of P. Further a is divided by V2 and V is not available in the units of the group, units of a will be P – V2 Equations that have equal units on either side are labelled dimensionally consistent Mass Transfer Flux = Coefficient * Driving Force ; Coefficient will carry the units of flux divided by driving force Reaction Rate = Rate Constant * Concentration n Units of Rate constant = Rate / Concentration n Certain equations developed from experimental data are called empirical equations. Constants present in such empirical equations will have dimensions and they must be strictly adhered to. Proper representation of quantities / conditions , number followed by base unit, is a good habit. It helps to check the correctness of the calculational approach. Sometimes , they even suggest the direction of calculation GP-CPC-01 Concept of ‘Mole’ ➢ Certain events in chemical processing happen in direct proportion to the number of molecules of the system. Eg : Molecular diffusion , Chemical Reaction… It is convenient and purposeful to use a dimension in such cases. MOLE ➢ Avagadro’s number defines a mole of a substance , that contains 6.022 *1023 units ( molecules) of that substance. Further, it is related to mass through molecular weight by a simple relation 1 Mole = Mass / Molecular weight ➢ Mole, thus, can be considered as a secondary unit of mass The units of mole are dependent on the units of mass Mass in grams - g mol or mol Mass in pounds( lb) - lb mol Mass in kilograms - kilogram mole or kilo mole ➢ Conversion of one to another is possible with the conversion factors used for mass 1 kg = 2.205 lb = 1000 grams 1 k mol = 2. 205 lb mol = 1000 g mol GP-CPC-01 COMPOSITION ➢ All steps in industrial processing involve mixtures of substances ➢ All the members ( components) of a mixture are in the same phase - A homogeneous mixture or solution. If they remain in different phases - A heterogeneous mixture. ➢ Composition indicates the relative amounts of components present in a mixture, homogeneous or heterogeneous ➢ Composition is a critical parameter in all the computations related to industrial processing Composition keeps varying in most of the processing steps – unit operations & unit processes GP-CPC-01 COMPOSITION ➢ Composition is specified in a number of ways. A clear understanding of the ideas and their inter relationship is of paramount importance. For any specification, a basis is necessary. Hence, different ways Consider a mixture consisting of components – A, B & C ➢ Mass fraction – Mass percent Mass is the basis Mass fraction of A = Mass of A / Total mass = Mass of A / Mass of (A + B + C ) Sum of Mass fractions = 1 Mass Percent of A = 100 * Mass of A / Total mass = 100 * Mass of A / Mass of (A + B + C ) Sum of Mass Percent s = 100 Independent of temp & Pres. Usually used to express the compositions of solid & liquid mixtures . Default Composition ( 2 / 3) GP-CPC-01 ➢ Mole fraction – Mole percent Mole is the basis Mole fraction of A = Moles of A / Total moles = Moles of A / Moles of (A + B + C ) Sum of Mole fractions = 1 Mole Percent of A = 100 * Moles of A / Total moles = 100 * Moles of A / Moles of (A + B + C ) Sum of Mole Percent s = 100 Independent of temp & Pres. Suitable for any mixture, more useful in case of gas mixtures ➢ Volume fraction – Volume percent Volume is the basis Volume Percent of A = 100 * Volume of A / Total Volume = 100 * Volume of A / Volume of (A + B + C ) Sum of Volume Percents = 100 Volume is temp. and pres. Dependent, but it is present in both the numerator and the denominator. It will be affected in the same way for both Ratio is considered to be independent of temperature & pressure . Default for gas mixtures. Volumetric Composition coincides with molar composition for IDEAL GASES GP-CPC-01 Composition ( 3 / 3) ➢ Mass Ratio Mass of A / Mass of (B+C) , Mass of B / Mass of (C + A ) ➢ Mole Ratio Moles of A / Moles of (B + C ) Moles of B / Moles of (A + C) ➢ Hybrid Moles of A / Mass of ( B+ C ) Mass of B / moles of ( C +A) These are normally preferred in case of one component affected ( Eg : Gas Absorption, Humidification) These can sometimes assume a value greater than 1 ➢ Density – Molar Density These give the mass or mole contained per unit volume Their reciprocals – specific volume and molar volume, give volume of unit mass or mole These are defined to suit specific requirements of the case ➢ Concentration units used for Liquid Mixtures Molarity , M = Number of g mol contained in a Liter of solution Normality , N = Number of gram equivalents present in a Litre of solution Molality , m = Number of gram moles of solute present per 1000 grams solvent It is to be reiterated that a good grasp of composition specification will be of great help in all process calculations, mass transfer, reaction, everywhere… GP-CPC-01 • Acceleration of a body is given as 1 m/ s2. Express the value in ft / min2 1 m = 3.8 ft , 1 min = 60 sec Acceleration = 1 m * 3.8 ft * 60 * 60 s2 = 13680 ft s2 m min2 min2 • Thermal conductivity of a steel is given as 14.7 BTU / hr – ft – OF. Express the value in cal / sec-cm-OC 1 BTU = 252 cal 1 hr = 3600 sec 1 ft = 30.48 cm 1.8 OF = 1 OC Thermal conductivity = 14.7 BTU * 252 cal * 1 hr * 1 ft * 1.8 OF = 1.85 Cal hr – ft – OF 1 BTU 3600 sec 30.48 cm 1 OC sec-cm-OC • Convert 1 .987 cal / g – OC to BTU / lb - OF 1 BTU = 252 cal 1 lb = 453.6 g 1 OC = (5 / 9 ) 1 OF 1 .987 cal * 1 BTU * 453.6 g * 1 OC = 1.987 BTU g – OC 252 cal 1 lb 1.8 OF lb - OF GP-CPC-01 ➢ Air has a composition of 79 % Nitrogen and 21 % Oxygen, by volume. Express the composition on the basis of mass Volumetric composition is taken as molar composition for ideal gases Required : Molecular weight of nitrogen (28) , molecular weight of oxygen (32) Basis : 1 mol of Air Mass = Moles * Molecular weight Component Moles Mol.Wt Mass Mass % Nitrogen 0.79 28 22.12 100 *22.12/28.84 = 76.7 Oxygen 0.21 32 6.72 100 * 6.72/28.84 = 23.3 28.84 • 28.84 obtained in this calculation is the mass of 1 mol of air. For mixtures, such values are called Average Molecular Weight. They are useful to convert moles to mass or vice versa of mixtures. Average Molecular Weight = Σ xi * wi ( mole fraction of a component multiplied by its molecular weight, summed over all the components) • Atmospheric nitrogen is a mixture of nitrogen, argon, carbon dioxide…. Its molecular weight is usually taken as 28.2 GP-CPC-01 ➢ An aqueous solution has 25 g of salt per g water. What is the mole fraction of the salt? Use Molecular weight of salt = 40 Molecular weight of water = 18 Data provided is mass of salt / mass of water . Mole per total moles is required. Moles = Mass / Mol.wt Component Mass Mol.Wt Moles Mole fraction Salt 25 g 40 25 / 40 = 0.625 0.625/0.68 = 0.919 Water 1 g 18 1 / 18 = 0.055 0.055/0.68 = 0.081 0.680 26 g ( 25 g salt + 1 g water ) are equivalent to 0.68 g moles. Average Mol.Wt = 26 / 0.68 = 38.23 ➢ An aqueous solution has 25 g of salt per 100 g of solution.