PHARMACEUTICAL ENGINEERING

PHARMACEUTICAL ENGINEERING Unit Operations and Unit Proesses Dr Jasmina Khanam Reader of Pharmaeutial Engineering Division Department of Pharmaeutial Tehnology Jadavpur University Kolkatta-70003 (06--007) CONTENTS Introdution Unit Systems Material Balane Energy Balane Ideal gas law Gaseous Mixtures Dimensional analysis Graph plotting Graphial Integration Graphial differentiation Ternary Plot

Introdution Every industrial hemial proess is based on Unit Operations (physial treatment) and Unit Proess (hemial treatment) to produe eonomially a desired produt from speifi raw materials. The raw materials are treated through physial steps to make it suitable for hemial reation. So, knowledge of unit operations like Mixing and agitation of liquid and heat flow is very muh neessary. The subjet Unit Operations is based on fundamental laws, physiohemial priniples. Unit Operations gives idea about siene related to speifi physial operation; different equipments-its design, material of onstrution and operation; and alulation of various physial parameters (mass flow, heat flow, mass balane, power and fore et.). Examples of Unit Operations are listed in Table. Table : List of some unit operations Heat flow, Fluid flow Drying Evaporation Distillation Crystallization Leahing Extration Filtration Mixing Absorption Adsorption Condensation Vaporization Separation Sedimentation Crushing After preparing raw materials by physial treatment, these undergo hemial onversion in a reator. To perform hemial onversion basi knowledge of stoihiometry, reation kinetis, thermodynamis, hemial equilibrium, energy balane and mass balane is neessary. Many alternatives may be proposed to design a reator for a hemial proess. One design may have low reator ost, but the final materials leaving the unit need higher treatment ost while separating and purifying the desired produt. Therefore, the eonomis of the overall proess also play a vital role to selet a suitable alternative design. Eah hemial proess onsists of series of assembly that are organized systematially to ahieve the goal. The physial and hemial steps in a proess are set with the help of ombined knowledge and experiene of engineers, tehnologists and ost experts to produe a produt. The individual operations have some ommon phenomena and are based on the same sientifi priniples e.g. Heat transfer is the ommon phenomenon in evaporation, drying and rystallization. (Table ) A proess designer designs a hemial proess onsidering () effiieny of proess and equipments () Safety with respet to the proess, raw hemials, finished produts and long term effet on environment (3) finanial viability of the produts as demanded by the purhaser. Following are some examples of physial proesses: (a) Sugar Manufature: Sugar ane rushing sugar extration thikening of syrup evaporation of water sugar rystallization filtration drying sreening paking.

Common Tehnique / Phenomenon Table : Common Tehnique/Phenomenon Appliations Movement of Fluid. Fluid flow from one storage to reation vessel.. Fluid flow involving heat transfer/ exhange. 3. Mixing & agitation in liquid liquid, liquid (solid). 4. Fluidization (solid liquid) 5. Filtration 6. Separation Movement of Solid Mass Transfer. Fluidization (solid liquid). Fluid bed drying 3. Settling 4. Blending of powder 5. Powder flow 6. Conveying Drying, Evaporation, distillation, hemial reation, diffusion, extration, humidifiation, adsorption, hemial kinetis. (b) () Pharmaeutial Manufature: Formulation of hemials, mixing, granulation drying of granules sreening pressing tablet pakaging. Salt Manufature: Brine transportation evaporation rystallization drying sreening onveying pakaging. On the other hand onversion of starh to dextrose with the help of aid atalyst is a typial hemial reation whih involves transportation of raw materials, physial steps of mixing the reatants, heat transfer, reation kinetis, fluid flow, separation of produts, produt purifiation, drying, sreening, onveying and pakaging. In 'Pharmaeutial' urriulum, knowledge of 'unit operations' is very muh relevant with respet to formulated drug produts and basi drugs in Pharmaeutial Industry. Beause of the variety and omplexity of modern proesses, the 'proess' had been lassified for better understanding of the steps in detail. General flow hart exhibiting produt formation has been shown in Fig.. The figure shows the sequene of basi omponents generally used in a typial hemial proess in whih eah blok represents a stage in the overall proess for produing a produt from the raw materials. The design of the proess involves seletion, arrangement of the stages and the seletion of speifiation and design of the equipment required to perform the stages. 3

Reyle of unreated material By produts Raw Materials Storage Feed preparation Reation Produt separation Produt Purifi ation Produt Storage Sales Stage Stage Stage 3 Stage 4 Stage 5 Stage 6 Stage 7 Fig. : Flow sheet of a typial hemial proess Eah stage may be simple or omplex aording to need of the proess. Both theory and pratie should be onsidered to yield designs for equipments that an be fabriated, assembled, operated and maintained. Unit Systems Knowledge of unit systems is very muh neessary to express any physial quantity and to solve/alulate any problem based on physial data. A physial quantity is expressed with its magnitude and unit. For example: 6 feet, the unit (foot) tells us the type of a partiular quantity / magnitude and it is this standard by whih the quantity is measured. Magnitude or numerial value tells us how many units are needed to make that up to required quantity. For example, the statement that the height of a person is 6 feet; it means six ' foot units' are required to over the height of this person. The offiial international system of units is the SI system (system international d' unite's), but older systems, partiularly the entimeter gram seond (gs) and foot pound seond (fps) engineering gravitational systems, are still in use. Beause of growing importane of the SI system in siene and engineering, it beomes neessary to put effort on its universal adoption as the exlusive system for all engineering and siene. The SI system overs the entire field of siene and engineering inluding eletromagneti and illumination. The units are derivable from some basi equations with the help of arbitrarily hosen standards for mass, length, time, temperature, mole and arbitrarily hosen numerial values for the proportionality onstants of the following basi equations. d F = K (m u) () Newton's seond law of motion dt ma.m F = K r b () Newton's law of gravitation Q = K 3 W (3) First law of thermodynamis PV T = K 4 Lim (4) Equation relating absolute temperature, pressure and volume. P 0 m F = Fore, t = time, m = mass, m a, m b = masses of two bodies, u = veloity, r = distane, W = Work, Q = heat, P = pressure, V = Volume, T=Absolute temperature, K, K, K 3, K 4 = proportionality fators 4

These K values an be alulated if all the variables are measurable. Unit of K is derived from the units used for measuring the variables in the equation. In SI system, K = and unit of fore (F) is Newton (N) in equation () IN = kg. m/s Similarly, in FPS system, poundal is one unit fore. poundal = lb ft /s (K = ) lb of mass experienes aeleration of 3.74 ft/s under gravitational fore. Unit fore annot be defined if this value of aeleration due to gravity (g) is put in equation (), onsidering K =. So, it is ustomary to use pound fore instead of poundal and to use K as. So equation () beomes, F = mg/g (5) g varies with latitude. But g is onsidered a onstant, alled 'Newton's law proportionality fator', sine the ratio g/g an be taken as unity for all pratial purposes. So, unit fore (lb fore, gm fore) an be defined using unit mass. Numerial value of 'g ' approximates 3.74 that is average value of g. lbmass ft Unit of g is ; it is derived from equation () lbfore s 'g ' is very important in the onversion of unit of fore. Conversion of x poundal to pound fore unit is done as follows: ft x poundal = x lb m ; lb mass and lb fore are written as lb m and lb f respetively. s or, x poundal g x ft lbf s = lb m = 3. s lbm ft Some examples of units: BTU g.al Convert to o o (h)(ft)( F) (s)(m)( C) Conversion fators BTU = 5 gm.al; ft = 30.48 m; h = 3600 s; o 5 F = o C 9 BTU (h)(ft)( o F) = 5 gm.al. o 5 s.m. C 3600 30.48 9 = 4.3385 x 0 3 gm.al o s.m. C g.al s.m. o C = 4.904 ~ 4 BTU o h.ft. F x 3. lb f g 5

Physial quantities are expressed in terms of primary and seondary units. Length, mass, time, heat and temperature are primary or fundamental units. Seondary units are expressed in terms of primary ones. The units of fore, aeleration are seondary type. Table 3: List of some physial quantities and their onversion units Quantity To onvert from To Multiply by Energy BTU kj.05506 Speifi enthalpy BTU/lb kj/kg.36 Speifi heat apaity BTU/lb o F kj/kg o C 4.868 Heat transfer oeff. BTU/ft h o F W/m o C 5.678 Visosity entipoise kg/m.s 0-3 Surfae tension dyne/m N/m 0-3 Pressure lb f /in N/m 6.894 0 3 Density lb/ft 3 kg/m 3 6.090 Volume m 3 gal (US) 64.7 US gallon = 0.84 imperial gallons (UK) Material Balane Mass balane or material balane is the expression of the onservation of mass that involves aounting of materials in a proess. This age old mass balane onept is useful for assessing a proess and sometimes improving management praties inluding waste redution, on site traking of toxi hemials and transportation into and out of a faility. Moreover, material balane over a proess helps to hek performane against design, instrument alibration and to identify the soure of material loss. It also helps to extend often the limited data available from the plant instrumentation. Therefore, a good understanding of material balane is neessary. Einstein showed that mass and energy are equivalent. Energy an be onverted into mass and mass into energy. The loss of mass assoiated with the prodution of energy is signifiant in nulear reations. In proesses of nulear fission and ondensation the law of onservation of mass beomes invalid, but in ordinary industrial proesses mass balane is highly pratied and it is aepted that energy and matter are to be separately onserved. The onservation of mass an be written as: Rate of flow of reatant into volume element = Rate of flow of reatant out of volume element + Rate of reatant removal by reation within volume element + Rate of aumulation of reatant within volume element. Material balane is arried over a ertain element of volume. In steady state proess i.e. ontinuous proess, the aumulation will be zero. If there is no hemial reation the steady state balane redues to: 6

Rate of flow of reatant into volume element = Rate of flow of reatant out of volume element. A mass balane an be written for individual elements, ompounds and for total materials. Diret measurements of quantity of omponents of both entering and leaving streams in a proess during a given time interval needs no alulation. Calulation of material balane is indispensable when diret measurement of one or more omponents is not possible. The measurements require knowledge of onservation of mass and of the standard units for expressing mass data, omposition, onentration, flow rate of flowing streams in a proess. If there is any hange of physial or hemial properties of flowing streams, that must be aountable. Appropriate judgement must be exerised in seleting the methods for sampling, analysis and in hoosing the frequeny and duration of data aquisition of mass flows within a proess. Gross deviation from mass balane indiates that errors have been made in sampling and quantifying one or more mass balane omponents or else it needs some relevant information. So, preise evaluation is needed. In a mass balane operation, all hemial inputs to a manufaturing proess and outputs from it and aumulations within are first identified and the masses are measured. The mass of inputs should be equal or losely approximate the mass of outputs plus aumulations. The following step by step proedure helps to get effiient solution of material balane problems:. Draw a blok diagram of the proess to show signifiant steps by flow sheet diagram.. List all the available data. Indiate known quantities of the parameters on the blok diagram. 3. List all the information. 4. Deide the system boundaries. 5. Write out all the hemial reations involved for the main produts and byproduts. 6. Note any other onstraints if any, like azeotropes, phase equilibria, tie substanes. 7. Chek the number of onservation equations that an be written and ompared with the number of unknowns. Deide the basis of alulation. The order of steps may vary aording to need of the problem. With simple problems, having only one or two reyle loops; the alulation an often be simplified by areful seletion of the basis of alulation and the system boundaries. Following is an example of blok diagram showing the main steps in the balaned proess for the prodution of vinyl hloride (VC) from ethylene (C H 4 ). Eah blok represents a reator and several other proessing units (Fig.). Blok A, Chlorination C H 4 + Cl C H 4 Cl (di hloroethane, DCE) Yield: based on ethylene 98% Blok B, oxy hydrohlorination C H 4 + HCl + / O C H 4 Cl + H O Yield : based on ethylene 95%; on HCl 90% 7

Blok C, Pyrolysis C H 4 Cl C H 3 Cl (vinyl hloride) + HCl Yields: based on DCE 99%, on HCl 99.5% X Reyle DCE Cl A Chlorinatio C VC Pyrolysis C H 4 Y B Oxyhydro hlorination O Z Reyle HCl Fig. : Flow sheet of a hemial onversion. X,Y,Z represent flow of ethylene to Blok A, flow of ethylene to blok B and flow of HCl in reyle loop. In this problem flow of C H 4 and DCE are unknown that an be alulated on the basis of the prodution rate of vinyl hloride with the help of algebrai method. Basis of Calulation: The initial step in takling a mathematial problems starts with seletion of a basis of alulation. This makes the solution of problem easier and simple. If the ompositions of substanes involved in a hemial reation are given in weight perent, 00 pound or 00 gm of one of the substanes entering or leaving system of a ontinuous proess may be hosen as the basis of alulations and finally, the alulated values may be onverted to other basis as per the desired statement of problem. A hoie of basis is made before alulation depending upon desired information on parameters like time, mass, mole, volume (for gas), and bath (for bath proess). Example: Carbon dioxide is added at a rate of 0 lb/hr to an air stream and the air sample at the outlet shows 0.45% (v/v) CO.Calulate the airflow rate. Normal ontent of CO in air is 0.03% (v/v). Solution: Inlet air ontains 0.03% (v/v) CO Basis: one hour Number of mole of CO added to the air stream = mass of CO / moleular weight of 0 CO = = 0.73 lb mole/hr 44 Let M lb mol/h be the airflow rate, CO being the tie omponent. Balane an be made on CO, using the onept of ideal gas law, volume %=mole% CO in = 0.0003M + 0.73 CO out = 0.0045M 8

By balaning CO, 0.0045M = 0.0003M + 0.73 M = 54.9 lb mol/hr = 54.9 /9=569.45 lb/hr Moleular weight of air is 9. Exess Reagent In most hemial reations arried out in industry, the omponents are seldom fed to the reator in exat stoihiometri proportions. A reagent may be supplied in exess of the amounts theoretially required for ombination with the others in order to maximize the use of an expensive reagent or to ensure omplete reation of a reagent. As a result, the produts ontain some of the unreated reatants. The amount of desired ompound is determined by the amount of limiting reatant. The perentage exess of any reatant is defined as the perentage ratio of the exess to the amount theoretially required by the stoihiometri equation for ombination with the limiting reatant. So the 'Exess' refers to limiting reagent. The perentage exess reagent is expressed by the following: Quantity supplied - stoihiometri quantity Perentage exess = 00 stoihiometri quantity Complete reation of the limiting material may not be ahieved even in the presene of exess reagent. It happens due to insuffiient time or opportunity for ompletion to the theoretially possible equilibrium. The degree of ompletion of a reation is ordinarily expressed as the perentage of the limiting reating material, whih is onverted or deomposed into other produts. In ase of 00% ompletion, outlet stream ontains produts and exess reatant. Example: 0% exess air is supplied to a furnae burning natural gas (Methane 95% (v/v), Ethane 5% (v/v)). Calulate the moles of air required per mole of fuel. Reations: CH 4 + O CO + H O C H 6 + 3.5 O CO + 3 H O Solution: Basis :We may use any number of mole of gas as basis, let us take 000 mole of fuel gas, as the given omposition is in volume perentage, stoihiometri moles of O required = 950 x + 50 x 3 = 075 0 with 0% exess, moles of O required = 075 x = 490 00 00 number of mole of air (% O ) = 490 x = 857 Number of mole of air per mole fuel = 857 / 000 =.86 Conversion and yield Conversion refers to the reatants and yield refers to the produts formed. Conversion is a measure of the fration of the reagent that onverts due to reation. Conversion is expressed by the following: Amount of reagent onsumed Conversion = amount supplied 9

(amount of reagent in feed stream) - (amount of reagent in produt stream) = (amount in feed stream) This definition indiates the total onversion of the partiular reagent to all produts. The onversion of a partiular reagent is often less than 00 perent in order to minimize by produt formation. If the onversion of a valuable reagent in a reation proess is appreiably less than 00 perent, the unreated material is reyled. Yield is a measure of the performane of a reator or plant. Yield should be defined on learly stated basis. mole of produt produed stoihiometri fator Yield = moles of reagent onverted Stoihiometri fator =Stoihiometri mole of reagent required per mol of produt produed. 'Plant yield' is a measure of the overall performane of the plant and inludes all hemial and physial losses. mole of produt produed stoihiometri fator Plant yield = mole of reagent fed to the proess When more than one reagent is used or produt produed it is essential that the produt and the reagent to whih the yield value refers are learly stated. Example: Reation: C H 4 + H O C H 5 OH C H 5 OH (C H 5 ) O + H O Ethanol is produed by the hydrolysis of ethylene and diethyl ether is the by-produt formed. C H 4, 55% C H 4 5.6% Inert, 5% Reator C H 5 OH 5.49% H O, 40% (C H 5 ) O 0.6% Fig. 3: Problem on yield of ethanol. H O 36.8%, Inert 5.8% Calulate perentage of onversion of ethylene and the yield of ethanol and ether based on ethylene. Solution: Basis: 00 moles of feed (Easier alulation on the basis feed omponents) Inert remains unhanged. So amounts of produt omponents an be alulated on the basis of inert material s amount. Yield of produt (ethanol) and byprodut are alulated on the basis of ethylene, not on water as water is relatively heaper than ethylene. Water is fed in exess. Ethylene is the reagent. 0

Feed stream omponents Ethylene 55 mole Inert 5 mole Produt stream omponents 5.6 Ethylene x5 = 49. 49 mole 5.8 Ethanol 5.49 5 = 5. mole 5.8 Water Total 40 mole 00 mole Basis of alulation 0.6 Ether x(5) = 0. 5 mole 5.8 36.8 Water x5=34.85 mole 5.8 5.8 Inert 5 = 5 mole 5.8 Calulation is done with respet to quantity of inert. Amount ethylene onverted = Amount of reagent in feed stream - Amount of reagent in produt stream =55 49.49 = 5.5 mole mole fed - mole out Conversion = 00 mole fed 5.5 = 00 = 0% 55 Yield of ethanol based on ethylene = (Mole of produt formed/ Mole of reagent 5. onverted) Stoihiometri fator 00 = x 00 = 94.4% 5.5 Stoihiometri fator is. 0.5 Yield of ether based on ethylene = x 00 = 5.45% 5.5 Stoihiometri fator is, as moles of ethylene produe mol of ether. Problems on Material Balane. The omposition of gas derived by the gasifiation of oal is, volume perentage: arbon dioxide 4, arbon monoxide 6, hydrogen 50, methane 5, ethane 3, and benzene, balane nitrogen. If the gas is burnt in a furnae with 0 per ent exess air, alulate: (a) The amount of air required per 00 kmol of gas (b) The amount of flue gas produed per 00 kmol of gas, () The omposition of the flue gases, on a dry basis Assume omplete ombustion.. Ammonia is removed from a stream of air by absorption in water in a paked olumn. The air entering the olumn is at 760 mmhg pressure and 0 o C. The air ontains 5.0 per ent v/v ammonia. Only ammonia is absorbed in the olumn. If the flow rate of the ammonia air mixture to the olumn is 00 m 3 /s

and the stream leaving the olumn ontains 0.05 per ent v/v ammonia, alulate: (a) The flow rate of gas leaving the olumn. (b) The mass of ammonia absorbed. () The flow rate of water to the olumn, if the exit water ontains % w/w ammonia. 3. Allyl alohol an be produed by the hydrolysis of allyl hloride. Together with the main produt, allyl alohol, di allyl ether is produed as a by produt. The onversion of allyl hloride is typially 97 per ent and the yield to alohol 90 per ent, both on a molar basis. Assuming that there are no other signifiant side reations, alulate masses of alohol and ether produed, per 000 kg of allyl hloride fed to the reator. Energy Balane Energy balane is a mathematial or numerial expression of 'Conservation of energy' (also alled first law of thermodynamis). Energy balane and mass balane are enountered often in the problems related to proess design and operation. Priniple of onservation of energy states that energy is indestrutible, but an be transformed to other forms of energy and the total amount of energy entering any system must be exatly equal to that of leaving plus any aumulation within this system. Energy out = Energy in + generation onsumption aumulation For steady state proess the aumulation of both mass and energy will be zero. In mass balane, the total mass flow into a proess unit is generally equal to the flow out at the steady state exept in biomass formation within a bioreator where generation is onerned. In energy balane, the total enthalpy of the outlet streams will not be equal to that of the inlet streams if energy is generated (exothermi) or onsumed (endothermi). Energy balane is arried out to know the energy needed in heating / ooling and supplying power in a proess. 'Energy balane' exhibits a pattern of usage and suggests areas for onservation and savings. Energy an exist in many forms making 'energy balane' more omplex than mass balane, suh as potential energy, kineti energy, flow energy, heating energy, mehanial energy, eletrial energy et. Total energy is onserved as per the law of onservation of energy. External Form It is attributed due to relative position / height of matter above datum point (z). It is gz quantified by, g Table 4: Potential Energy Internal Form It is attributed due to inherent omposition, struture and state of matter, fore of attration / repulsion in moleular or atomi level.

External Form It is attributed due to motion of matter and u quantified by per unit mass; u is the g average veloity of the stream. Table 5: Kineti Energy Internal Form It is attributed due to presene, relative position, and movement of moleules, atoms, and subatomi units. The translational, rotational and vibrational motion of the atoms, eletrons, ontributes it. The kineti portion of total 'internal energy' is determined by the temperature of the substane and by its moleular struture. The remainder of the internal energy remains as potential energy due to attrative and repulsive fores between moleules, atoms. All translational energy disappears at temperature absolute zero, but energy is reserved as potential energy. 'Absolute Zero', temperature as referene is used to alulate kineti part of internal energy (U = f(t). The total internal energy of a substane annot be determined. Flow energy (PV) often appears in flow proess. It is external form of energy and determined by the produt of pressure and volume. Another form of external energy is surfae energy, whih is often enountered in making of emulsion. Heat energy is the energy in transition under the influene of temperature differene ( t). When heat flows from a soure (hot) to a reeiver (old) body, it is stored as internal energy and so, internal energy of reeiver is enhaned. Another form of energy in transition is work. Work done is manifestation of the transition of one form of energy into another. When a fore (F) ats on a body through a distane, work is expressed by, w = l 0 F dx x, l = distane, Work is onsidered negative when it is done on a system by its surrounding. It is positive when work is done by the system on its surrounding.when work is done due to hange in pressure or volume, w = Pdv; P = pressure, pasal v = volume per unit mass, m 3 /kg Following figure illustrates energy balane in a steady state proess Q W Inlet stream Leaving stream Energy input () Energy output () U internal energy U P V flow energy z z P V u Kineti energy u g datum plane g z g/g potential energy Fig. 4: Energy balane. z g/g 3

P is pressure, V is volume, u is the average veloity, z is the distane from datum plane, g is the aeleration due to gravity and g is gravitational onstant. For unit mass of material, following energy balane equation an be written, overing both flow proess (steady state) and nonflow or bath proess (unsteady state). U + P V + u /g + z g/g + Q + E = U + P V + u /g + z g/g + W In the above expression Q is the heat transferred to the system and W is the work done by the system. E is the speial terminology used for bath proess only, where u zg terms like U, PV,, will not appear in the equation, sine all these terms are g g aounted in E. E represents the hange in energy ontent of the system as a result of any hange in inventory (temperature, omposition) of the system. / / / + g g g z g u z g u + = g / / E= U + ( U + ) ( ) + ( ) g + / / / / / / u u U U z z g g g Prime sign is used for bath proess. / In hemial proesses (ontinuous or flow), the kineti and potential energy terms are usually small ompared to heat and work terms, so these an be omitted from the generalised equation. Here, E = 0. Therefore, we an write Q W = (U + P V ) (U + P V ) or Q W = H H Here, H (= U + PV) is the enthalpy that is funtion of temperature and pressures. Enthalpy an be alulated by speifi and latent heat. The above equation helps to estimate heating and ooling requirements of various unit operations involved in hemial proesses. In nonflow proess, we may write, Q W = E. Considering hanges in potential energy and kineti energy due to stirring negligible, the above equation an be simplified to / / Q W = (U U ) We know that, W = P V At onstant volume, W = 0 / / Q = U = U = Change in internal energy U 4

At onstant pressure, if work of expansion is performed against pressure (P) / / W = PV PV, putting this equation in the earlier equation obtained in non flow proess, we get / / Q W = (U U ) / / / / / / Or, Q = W + (U U) = PV PV + ( U U ) / / / / Q = ( U PV ) ( U + ) / / + = H H = H PV Therefore, heat added to the system at onstant volume hanges internal energy and heat added to the system at onstant pressure hanges enthalpy of the system. Units used in energy balane (a) erg (dyne m), Joule (0 7 erg), Newton meter (Joule), foot pound or pound fore (3.74 poundal), foot poundal. (b) gm alorie, kilogram alorie, British Thermal Unit (BTU), Centigrade heat unit (CHU) or pound alorie. Thermohemial gram alorie = 4.84 absolute Joule I. T. gm alorie = 4.8605 International Joule = 4.8674 Absolute Joule (I. T. g Cal) /gm = 9/5 BTU/lb BTU = 5 g al = 5/5 CHU. Example: One hundred pound of pure C is burnt to CO using dry air (79% N, % O ) theoretially neessary to supply O for ombustion. The air and C enters the burner at 8 o C and the produt gases leave at 000 o C. The burner is operated at a onstant pressure of 760 mm of Hg. Calulate the amount of heat unaounted for. When lb mole of C ombined with lb mole of O at atmospheri pressure and 8 o C, 6950 BTU is evolved. Mean heat apaity of CO and N in the temperature range (8 000 o ) at onstant pressure are 3.45 BTU / lb mole o C and 4.38 BTU / lb mole o C respetively. Solution: Amount of C = 00/ = 8.34 lb mole Amount of O neessary = 8.34 lb mole 00 Amount of Air neessary = 8.34 = 39.7 lb mole Amount of CO in produt = 8.34 lb mole Amount of N in outlet stream= Amount of N in inlet stream= 0.79 x 39.7 = 3.36 lb mole N is tie substane. 5

Heat Balane of Input Sensible heat of air, arbon at 8 o C is taken as zero energy level. Potential heat energy ontent of C in onversion of CO = 8.34 x 6950 = 4545 BTU Heat input = 4545BTU Heat Balane of Output Total sensible heat in gases = 8.34 x 3.45 (000 8) + 3.36 x 4.38 (000 8) = 84 BTU Total heat energy = 84+ unaounted for losses 4545= 84+ unaounted for loss. Heat unaounted for losses = 303 BTU. Ideal gas law Ideal gas law is a useful relationship among pressure, volume, mass and temperature that has been utilized in many pratial problems: in solvent reovery system, ammonia reovery from air ammonia mixture, humidifiation, drying and ombustion et. Although real gas or vapour does not follow aurately ideal gas law, but ideal gas law is suffiiently useful at ordinary temperature and pressure in many engineering alulations. Ideal gas law is expressed by PV = nrt, Where, P = pressure, V = volume, T = Absolute temperature, n = number of moles of gas. R = Gas onstant (equal for all gases) = 8.34.47 J / kg mole / o K =.9873 al/gm mole / o K = 8.056 m 3 atm / gm mole / o K Ideal gas law is based on some assumptions of kineti theory: Gas moleules behave like elasti spheres and these are free to move and ollide with eah other and surrounding wall randomly. No hange in kineti energy during ollision. Energy due to intermoleular fore of attration is negligible in omparison to kineti energy. Number of moleules per unit volume in spae is onstant. Total translational kineti energy possessed by one mole of gas is determined by temperature. Based on the above assumptions the following equation may be derived from the priniples of mehanis: PV = n x N mu = nrt 3 PV = n x 3 x kineti energy possessed by a mole = nrt n = number of mole N = Avogadro's number u = Average veloity of gas moleule. 6

When a real gas expands, it has to overome fore of attration at the expense of its own internal energy. Let us suppose, pressure of a gas (P ) is hanged to P so that volume expands from V to V. Now energy due to P V is not exatly same as P V, sine gas spends its own internal energy during expansion in order to overome intermoleular fore of attration. At standard ondition, volume of ideal gas is.4 litres per gm mole or 359 ubi feet per pound mole or.4 ubi meter per kg mole. But real gas has volume less than.4 litre per kg mole at N.T.P. Real gas deviates from ideal gas law at high pressure due to ompression, so ompressibility fator (Z= PV/RT) must be introdued. For ideal gas behavior this fator is unity. At standard ondition, following speifiations are followed. Table 6 Volume / Mole Temperature Pressure 400 m 3 73.5 o K atm or 760 mm of Hg 359 u ft 49.69 o R 9.9 inh of Hg or 4.7 p.s.i. Gas law holding the relationship among mass, pressure, temperature and volume of gaseous substane, often is desired to alulate any one unknown quantity of the above mentioned variables when others are speified with the help of well known Boyles' law and Charles law expression: P V PV = nr = T T P T V = V x P T, indiate initial and final onditions. If one ondition refers to N.T.P., number of moles (n) an be determined on the basis of volume oupied at standard ondition. Let V u ft. hanges to V u ft. at standard ondition due to hanges of pressure and temperature. V Now, n = ; n an also be alulated also if P, V and T are known 359 P V = nrt, unit of R should be onsistent with P, V and T. When volume of gas expands from V to V due to hanges in pressure or P T temperature, the ratio > and > P T When volume of gas dereases from V to V due to hanges in pressure or temperature, the ratio P /P is < and T /T is <. 7

Gaseous Mixtures Problem enountered in many pratial situations like 'vapor reovery by ondensation', srubbing of ammonia from air ammonia mixture', fuel gas prodution by ombustion' et. gases produed in hemial reation' often deal with mixture of gaseous substanes. In a mixture of different gas moleules, the total pressure is equal to the sum of the pressures (partial) exerted by the moleules of eah omponent gas. Dalton's law of partial pressures states that the total pressure exerted by a mixture of ideal gases may be onsidered to be the sum of the pressures that would be exerted by eah of the ideal gases if it alone were present and oupied the total volume. Amagat's law of partial volumes, whih states that, in a mixture of ideal gases, eah gas an be onsidered to oupy the fration of the total volume equal to its own mole fration and to be at the total pressure of the mixture. Ideal gas law is appliable at low pressure, high temperature orresponding to large molal volumes. Let us onsider three gas omponents (a, b, ) are onfined in a volume V at temp T. Aording to Dalton's law of partial pressure, the following equations an be written: n RT p a = a n (); brt p b = (); V V Total pressure = P p = n RT (3) V RT nrt P = p a + p b + p = (n a + n b + n ) P = (4) V V n a, n b, n are the number of moles of a, b, respetively. Dividing () by (4) p n we get, a a = = N a = mole fration of a ; n is the sum of number of moles of a,b P n and. p b n b p n Similarly, we get == and = P n P n Pressure % = mole % Aording to Amagat's law, it an be written that PV a = n a RT.(5) PV b = n b RT (6) PV = n RT (7) v a, v b, v are pure omponent volumes at pressure P, temperature T. P(V a + V b + V ) = (n a + n b + n )RT or, PV = nrt (8) Va n a Dividing (5) by (8), we get = = N a ; V is the sum of the volume of a,b and. V n V n Similarly, we get b b V = and n = V n V n Volume % = mole %. 8

Combining Dalton's law of partial pressure and Amagat's law of partial volume, the following expression an be written: pressure % = mole % = volume % This expression is widely used in solving problems of gaseous system. Problem on gas laws. It is desired to onstrut a drier for removing 00 lb of water from wet granules per hour. Air is supplied to the drying hamber at 5 o F at atmospheri pressure (humidity of air 8.35 x 0 3 lb mole / lb mole. If the air leaves the drier at temperature of 95 o F at atmospheri pressure (humidity of air = 30 x 0 3 lb mole / lb mole), alulate the volume of air at the initial onditions that must be supplied per hour. Solution: Basis: hr. In tablet manufaturing by wet granulation method, drier is indispensable equipment and humidity of air is a serious parameter to be onsidered both for drying operation and designing of drier. Let us know about the definition of humidity number of mole of moisture Molal humidity is defined as number of mole of moisture free gas 00 Number of moles of moisture to be removed from wet material = lb mole 8 From humidity data 3 Number of mole of moisture removed (30 8.35) 0 = lb/mole. Number of mole of moisture free air = 0.065 lb mole/lb mole So we an write, 0.065 mole of moisture is removed from lb mole of moisture free air 00 00 lb mole of moisture is removed from / 0.065 lb mole of moisture 8 8 free air. Therefore, 56.6076 lb mole of moisture free air enters the drier. Volume of inlet air (V ) an be alulated by using ideal gas law. 760 V = 760 56.6076 359 (459.6+5) (459.6+3) or V = 44.83u ft (349 u ft is the volume of mole at standard onditions). Answer: 44 u ft (approx.) of air is supplied per hour to remove 00 lb of moisture from wet material. Following is another example of appliation of gas law.. A ombustion gas at temperature of 0.6 o C and pressure of atm is humidified. The gas leaves the humidifier at 94.93 o C and a pressure of atm. Table 7: Composition (mole %) of gas Component Inlet gas Outlet gas Nitrogen 79. 48.3 Oxygen 7. 4.4 Carbon dioxide 3.6 8.3 Water 0 39 Total 00 00 9

Calulate (a) weight of water evaporated per 00 u ft of inlet gas (b) volume of gas leaving the humidifier per 00 u ft of inlet gas. Solution Basis: gm mole of inlet / entering gas Table 8 Component Mole fration number of mole Volume of mole N 0.79 gm mole (n a ) V a O 0.07 gm mole (n b ) V b CO 0.36 gm mole (n ) V Total gm mole N V Now, volume of inlet gas (V) at initial onditions is alulated with the help of 'Amagat's law' equation ( n a + n b + n ) 8.06 (73 + 0.6) V = RT = P = 39684... =.40435 u ft ~.40 u ft. Let x be the number of mole of water added to entering air during humidifiation. x 00 = 39 or x = 0.6393 gm mole x + 0.6393 8 = = 0.05369 lb 453.6 453.6gm = lb, Moleular weight of water is 8 for,.40 u ft of entering air 0.05369 lb of water is needed 0.05369 00 00 u ft of entering air of water is needed..40 ~.8 lb Number of mole of outlet / leaving gas per gm mole of inlet gas is (+0.6393) Volume of leaving gas.6393 8.06 (73 + 94.93) = = 49494.96.. ~.7478 u ft For.40 u ft of inlet gas, volume of outlet gas =.7478 u ft..7478 For 00 u ft of inlet gas, volume of outlet gas = 00.40 = 4.75 u ft. Answer: Weight of water evaporated during humidifiation of gas and volume of gas leaving the humidifier per 00 u ft of entering gas are respetively.8 lb. (approx.) and 5 u.ft (approx.) Problem Pure methane gas (CH 4 ) is ompletely burnt with air. Exess air is not onsidered here. The mixture of gases leaving the ombustion hamber is passed through a ondenser where 30 lb of water is ondensed. Calulate the following: a.(i) Composition of mixture of gases leaving the ondenser. 0

(ii) Mole fration of nitrogen in the gas mixture leaving the ondenser. (iii) The average moleular weight of the mixture of gases leaving the ondenser. (iv) The partial pressure of moisture that exists with the leaving gases at total pressure of 8 psia. b. How muh amount of water is ondensed if mole fration of nitrogen in the gas mixture leaving the ondenser is 0.8? Dimensional analysis Dimensional analysis is a oneptual tool often applied in Engineering and Siene to attak a ompliated problem for whih no formal mathematial equation ould be derived. It is often the basis of mathematial models of real situation. This method is intermediate between formal mathematial development and ompletely empirial method. Dimensionless groups are generated after analysis of variables involved in a phenomenon by dimensional analysis method. Fluid flow when analyzed, Reynolds number is generated. Problems of this type are speifially found in fluid flow, heat flow, powder flow and diffusional operations. In empirial method, effet of eah independent variable on dependent variable is studied by systematially varying that variable while keeping rest of the variables onstant. The draw baks of this method are that the proedure is lengthy and laborious and diffiult to organize the results to a useful orrelation for further alulation and the orrelation is usually not dimensionally homogeneous. Engineers and physiists often utilize Dimensional analysis method to develop relationships among variables of physial quantities that are involved in a partiular phenomenon. There are two ways in dimensional analysis method.. Raleighs method of indies (Algebrai summation method). Bukingham method (dimensional groups an be shown as zero power of a group of variables). Dimensional analysis is an algebrai treatment of the symbols for units onsidered independently of magnitude. This method rapidly simplifies the task of fitting experimental data to derive a dimensionally homogeneous equation. Dimensional analysis of a partiular problem is made omplete by knowing entire physis of the situation that reveals all possible variables involved in that problem and related basi laws. In this method, many variables are assembled in a group that is dimensionless. Dimensionally homogeneous equation onsists of more than one dimensionless group. However, it does not give numerial equation. These groups may arry exponents of any magnitude, not neessarily whole numbers but a pure number of dimensionless. Numerial values of exponents, proportionality onstant are determined by putting experimental data in that orrelation. The final orrelation helps to find value of unknown variable if other values of variables / physial quantities are known. Any system of units (fps, gs, SI) may be used to alulate numerial value of dimensionless groups. For example, Reynolds group or number, N Re = Duρ/µ ; it is used in fluid flow, Where, D = diameter of pipe u = Veloity of fluid

e = Density of fluid µ = Visosity of fluid. Value of N Re would be same irrespetive of system of units. Now, N Re is found dimensionless putting gs and fps system of units ft lb ft N = Duρ Re µ = m m/s gm/m3 3 s ft = or N Re = gm lb m.s ft.s = This method is advantageous in heking the onsisteny of the units in equations, in onverting units and in sale up of data obtained in sample experiments to predit the performane of full sale equipment, in reduing number of independent variables. This is based on onept of dimension and dimensional formulas. Following steps are followed in Raleigh's method of indies. Step. All the variables / parameters (dependent and independent) are entered as a funtion in a orrelation. Step. (a) Independent variables are assigned exponents, (b) all the variables are expressed with their respetive standard dimensions (Mass M, Length L, Time θ). Step 3. The sums of the exponents relating to any given dimension (Mass, for example) must be same on both sides of the equation. So, algebrai equations are obtained by balaning exponents. Step 4. The above equations are solved to obtain simplified value or expression of one exponent in terms of others. Step 5. Exponents obtained in Step 4 are put in Step. Step 6. Variables are assembled in-groups in suh a way, that these are found as dimensionless groups. Finally these groups are entered in an equation, whih is dimensionally homogeneous. Following is an example of dimensional analysis method: Dimensional analysis of variables involved in mass transfer of fluid flow operation. From the mehanism of mass transfer it an be expeted that the mass transfer oeffiient,k would depend on the diffusivity, D v and on other ontrolling variables like veloity, u; visosity, µ; density, ρ of fluid flowing through a pipe and linear dimension, D. K = f(d v, D, u, µ, ρ) or, K = A D v a D b u µ d ρ e K Standard dimension L/θ D v Standard dimension L /θ D Standard dimension, L u Standard dimension, L/θ µ Standard dimension, M/Lθ ()

ρ Standard dimension, M/L 3 Equation () may be written as a d L L b L M M = A.L 3 θ θ θ Lθ L A is proportionality onstant and a, b,, d and e are exponents. Now, balaning exponents of M, L, θ we get the following equations 0 = d + e () = a + b + d 3e (3) = a + +d (4) From equation () d = e. From equation (3), we get or, or, or, = a + b + a + d + 3d = a + b + (a + + d) + d = a + b + + d a + b = d = e From equation (4), we get d = (a + ) or, e = a + Sine a + b = d b = a d = a ( a ) or, b = Now the exponents are arranged serially. a,,, a, a + Putting the above exponents in equation (), we get a a a+ K = AD D (u) (µ ) ) or, K v (ρ Duρ = A µ µ Dvρ a e µ Dρ Both L.H.S. and R.H.S. are divided by u or, K u Duρ = A µ µ Dvρ a µ Duρ Duρ µ Both L.H.S. and R.H.S. are multiplied by µ ρd v 3

K u Duρ µ µ ρd v Duρ = A µ µ D vρ a µ Duρ Duρ µ µ ρd v or, KD D v Duρ = A µ b+ µ ρd v a K D Therefore, we get 3 groups whih are known as Sherwood number, N sh ; D v Duρ µ Reynolds number, N Re ; Shmidt number, NS, µ ρd v These groups are found dimensionless by putting respetive units. N sh K D = D v m/s.m m /s N Re= N s = Duρ = µ µ ρd v = m * m/s.gm/m gm/m3/s gm/m/s gm/m 3.m Total number of variables is 6. Total number of standard dimension is 3 (M, L, θ). Total number of dimensionless groups is 6 3 = 3. /s The numerial values of A, are determined with the help of experimental data. Thus, dimensionless orrelation is developed by dimensional analysis method. Here, M, L, θ are the standard dimensions used for primary physial quantities. Seondary physial quantities are expressed by dimensional formulas; for example, for Fore, F ML/θ used. In the mixing of liquid, some fators like speed of impellor (n), diameter of impellor (Da), Newton's law proportionality fator (g ), aeleration due to gravity (g) visosity (µ) and density of fluid (ρ) and shape fators ontrol power (P) delivered by Pg the impeller. By applying dimensional analysis, power number N p, 3 ρn Da nda ρ expressed as funtion of Reynolds number, N Re µ Froude number N Fr ( n D a /g ) while ignoring shape fators. Following are some other examples of dimensionless groups. Fannings frition fator, f ; Nusselt number, N Nu, Prandtl number, N Pr. 5 is and 4

Reynolds number ( Duρ / µ ) is a measure of type of fluid flow ( laminar, turbulent) through a tubular pathway. This expression is different in mixing of liquid ( N Re =nd a ρ / µ), here veloity term is proportional to nd a where n is the r.p.m. of the impellor and D a is the diameter of impellor. Froude number is a measure of the ratio of the inertial stress to the gravitational fore per unit area ating on the fluid. Nusselt number (N nu =hd/k =D/x) is the ratio of the tube diameter to the equivalent thikness (x) of the laminar layer. This number has appliation in heat transfer phenomenon. h is the heat transfer oeffiient (BTU/ft -h- F) and k is the thermal ondutivity (BTU/ft-h- F). Prandtle number is a measure of momentum diffusivity relative to that of the thermal diffusivity. Its numerial value depends on temperature, pressure of fluid. Power number is proportional to the ratio of the drag fore ating on a unit area of the impellor and the inertial stress. Fannings frition fator is defined as the ratio of the wall shear stress to the produt of the density and veloity head. Different types of Graph plotting Graph plotting is done on graphial sheets in many ways, along its axes so that plotted data at various loation of sheet are displayed. This may help in many ways Exhibit relative position / loation of various data that represent one variable with respet to another variable. Develop orrelation between variables. Extrapolation of urve to find unknown data. There are different types of plotting system. So, various types of graph papers with axes of different types of graduation are available, suh as. Cartesian graph papers in whih both the axes are graduated evenly or linearly.. Semi logarithmi graph paper and logarithmi graph paper or log log graph paper. 3. Triangular diagrams or ternary plotting. In semi logarithmi graph paper one axis is graduated with logarithmi value and another axis is graduated linearly. In log log paper both the axes are graduated with logarithmi value. So, logarithmi sale is not uniform. Presentation of data on a logarithmi sale an be helpful when the data overs a large range of values. Before the advent of omputer graphis, logarithmi graph paper was a basi sientifi tool. Logarithmi graph plotting redues the tedious task of onverting a good number of physial quantities to its log value as required by the orrelation. It was diffiult to loate a point speifially when log values of data were very small and negative and data range is omparatively wide. There are numerous exponential equations suh as first order rate equation, mirobial growth equation, ell death kinetis and nutrient degradation kinetis equation et. 5

One famous equation known as Arrhenius equation, K = Ae Ea/RT an be used for explaining method of semi logarithmi graph plotting. The above equation an be written as: Ea log 0 K =. + ln A ; This is straight-line equation..303 RT Now, suppose we want to determine unknown quantity Ea (energy of ativation). Therefore, data obtained for the reation rate onstant, K and the reiproal of temperature (/T) are plotted on y axis (log sale) and on x axis (linear sale) respetively. Ea Q = Slope of the straight line.303r y log log y y log y = = x x x x where (x, y ) and (x, y ) are the o ordinates of two points on the straight line and R is gas onstant. Average value of Ea is determined with the help of at least three pairs of o ordinates of points that exist on different parts of straight line. 'A' is determined by putting 'Ea' value and any o ordinate of point that lie on the straight line, in Arrhenius equation. Another equation used for Non Newtonian flow, F N = µg an be written as, log G = N log F log µ Where, F = shear stress, G = Rate of shear strain, µ = Co effiient of visosity of fluid, N is exponent. In the above equation, both the variables (F, G) are expressed in log form. Then µ an be determined by plotting data of F on x-axis and data of G on y-axis of log log graph paper. Here, both the axes are graduated aording to logarithmi sale. When the plot appears as straight line slope is determined by the following expression: log y log y log(y / y) N = slope = = log x log x log(x / x ) Putting the average value of N and o ordinate value of a point on the equation of straight line, 'µ' is determined. Logarithmi sale an be developed along the axes aording to values of log 0 x (x = l to 0). A network of lines is then ruled onto the paper based on the non linear sales of antilogarithm. That is, antilogarithmi values (anti log 0 x = x, for example, it is written as,,3.) are shown on the axes. The same style of graduation is repeated after one suh yle (x= to 0) is ompleted. Eah yle has 9 nonlinear divisions. Eah division is further graduated into smaller divisions. End of one yle is 0 times the starting value of this yle. Axis is graduated aording to data range. Following is an example of axis graduation: 0., 0., 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, (),, 3, 4, 5, 6, 7, 8, 9, (0), 0, 30, 40, 50, 60, 70, 80, 90, (00); 6

Here, value (), (0), (00) indiate end of yle or starting of next yle. Numerous data an be plotted easily and diretly on graph paper without onverting these to orresponding logarithmi values. The only disadvantage of the logarithmi plot is that the sales normally used annot be read too losely, but in most ases points an be plotted with auray omparable to the auray of ordinary engineering data. An advantage of log log plotting is that deviations of point from a urve of a given distane represent deviation of a onstant perent of the total value of the variable at that point, irrespetive of the part of the plot in whih it exists. Following is an example of log log plotting of two variables (x, y). y = kx n this is a orrelation between x and y and k and n are onstants. Table 9 y 3.4 4.5 5.5 5.9 7 7.55 0 x 30.3 55 75 93.5 37 75 80 The data are plotted on the log log sheet and the straight line urve is obtained. Fig. 5: Plotting on log log graph paper Average slope (n) is found as 0.473. The data annot be extrapolated to determine the interept, where x = goes far beyond the sale of plot. By putting n = 0.473 and any o ordinate (70, 7.8) in the equation, K is found as 0.693. Graphial Integration Graphial Integration is sorted when no mathematial orrelation helps to determine one of the variables when a variable hanges with respet to another. The addition or integration of small hanges (dy) in dependent variable with respet to small hange (dx) in independent variable gives the value of total hange in y within some limits. From the first priniple of integral alulus, we know the value of a definite integral is the area bound by the urve of f(x) vs. x, the ordinates at x = x a and x = x b and x axis. y x b / dy = f (x)dx () f / (x) is the derivative of original funtion, y = f(x) 0 xa 7

If the derivative f / (x) is very simple type i.e. y = x 3 or by analytial method of integration. dy dx 4 x =, it an be integrated 4 There are equations that annot be integrated so easily. For example: Rayleigh's equation in differential distillation, Wo x o dw W 0 dx = ln = W W ; W 0 and W are the initial and final weight in moles (y x) W x respetively and x and y are the liquid and vapor omposition of the omponent respetively. In the above equation, right hand side an be integrated graphially with the help equilibrium data (x, y) and x 0, x. Though, x and y are related with eah other but there is no mathematial orrelation between x and y so that it an be integrated by analytial method of integration, therefore the area under the orrelating urve (/ (y-x) vs. x) within limits x 0, x are measured. Right hand side / (y-x) (dx) gives area onept. So, numerial value of area may be taken same as integral value. The area under the urve within the limits may be determined by dividing the area into a number of retangular strips and adding area of all retangles. Eah of these may not be perfet retangle due to nonlinearly of urve. Eah retangular, strip is made a perfet 'retangle' in suh a way the area of a triangle exluded due to new retangle formation is approximately equal to the area of new triangle inluded in the retangle above the urve. Area of eah retangle is alulated by multiplying its height and width and then all the areas are added to get the desired value of integral. Following example will help undergraduate students to understand the method of graphial integration. A given variable z is a funtion of the two variable x and y suh that z = ydx. Integrate the funtion over the range from x = 0. to x = 0.4 by graphial method. Table 0 x 0 0.05 0. 0. 0.5 0.3 0.35 0.4 0.45 0.5 y 0 0.98 0.384 0.67 0.75 0.768 0.74 0.576 0.34 0 x x Solution: Following steps to be followed:. Plot the urve y vs. x.. Draw vertial lines at x = 0. and x = 0.4 interseting the above urve. 3. Split the area under the urve, vertial lines and axis into number of retangles with width, dx = 0.05. 4. Draw a vertial dotted line through the mid point of eah width that intersets the urve at a point. Through this point of intersetion, draw a short horizontal line along eah width. Now, a perfet retangle informed. Height of eah retangle is tabulated in a olumn. 8

Table Retangle Width of retangle Height of Area of retangle Total area = retangle umulative area a 0.05 0.75 0.03575 0.03575 b 0.05 0.760 0.03800 0.07375 0.05 0.745 0.0375 0.0 d 0.05 0.650 0.035 0.435 Therefore, Z = 0.48 0.8 0.7 0.6 0.5 y 0.4 a b d 0.3 0. 0. 0.0 0.0 0. 0. 0.5 0.3 0.35 0.4 0.5 y plot for Graphial Integration Fig. 6: Plot for Graphial Integration Graphial differentiation When two variables (x, y) are simply related with eah other and then differentiation dy an be done easily by analytial method. If it is not so, then it an be done with dx the help of mathematial orrelation. If it is not so, then it an be done by graphial method with the help of experimental data. First, a histogram like plot is made y relating finite derivative against x, where width of eah retangle is x and x y height is y/ x. Finite derivative is alulated with the help of experimental x data (x, y). A partiular y value is onstant for partiular range x. x is not dy infinitesimally small value. Infinite derivative value at any value of x an be dx obtained by drawing a urve that onnets all the mid points of range ( x) orresponding to eah y/ x value. Now derivative dy/dx at any value of x an be obtained with the help this urve. Ternary Plot The ternary plot or triangular diagram is a speialization of bary entri plot for three variables. It graphially depits the ratios of three proportions. It is often used in petrology, mineralogy and other physial sienes to show the relative omposition of 9

soils and roks, but it an be more generally applied to any system of three omponents. Pharmaeutial formulation often deals with multiomponent liquid system, whih may not appear homogeneous as desired if omposition is not properly formulated. This an be ahieved by altering omposition of hemials. This needs a trial. This kind of plotting is useful in extration problems also. Ternary diagram and plotting of three omponent system help to determine amount of solvent needed to extrat a omponent from another solvent Let us suppose two omponent A and B are soluble with eah other and we want to add to a ertain quantity of C. After addition of C, the liquid may be found as heterogeneous or homogeneous system depending upon the respetive amounts of A, B and C. It needs time onsuming trial to prepare desired system. It is better to make a graphial hart. This an be done by a method known as Ternary plotting or plotting in triangular diagram. Different ompositions are prepared first and its homogeneity or heterogeneity is observed and then a urve is plotted within this triangle. The urve is a boundary between homogeneous omposition and heterogeneous ompositions. One the plot is ready, omposition is seleted aording to the need of the formulation. Table A 0 6 8 36 40 3 0 0 B 86 80 66 4 30 8 0 0 C 4 4 6 30 50 80 90 In dealing with three omponent systems, ompositions may be represented by points within a triangle, eah point representing one unique omposition. While an equilateral triangle is frequently used but a triangle of any shape may be employed. In a ternary plot, eah variable (a, b, ) is represented as (as fration) or 00 (as %) when it exists as single omponent. Therefore, if omposition (%) of two variables (a, b) are known in the system (a, b, ), then an be easily determined. Eah base or side of the triangle represents a proportion of 0% of a partiular omponent, with the point (apex) of the triangle opposite to that base represents a proportion of 00% of that omponent in the system (a/b/). As a proportion inreases in any one of the samples, representing point moves from base towards its apex up to inreased proportion. Now, a straight line passing through this point intersets another line passing through the point orresponding to another variable's proportion. The point of intersetion represents omposition of a, b and in that partiular sample. In this way, point is loated within a triangle. Following figure illustrates triangular or ternary plot of 3 omponent system (a, b, ). A graphial plot is made with the help of following ompositions. 30

B A 0.00.00 0.5 b 0.75 0.50 a P 0.50 a' A 0.75 0.5.00 B 0.00 0.00 0.5 ' b' 0.50 0.75.00 C C Ternary plot of a three omponent system Fig. 7: Ternary plot of a three omponent system In the above figure, eah apex represents 00% of a omponent. The apex (A) of above triangle represents 00% of omponent a. Points on the line opposite to an apex ontain none of the material orresponding to the apex. So, the points on the line B C have no omponent of a. Eah edge represents binary omposition of two omponents; for example, line A B represents ompositions of mixture of A and B. The loser a point is towards apex A, the more of the omponent a in the mixture. The mole % omponent a at C is zero. The lines a /, b /, / are parallel to the edges of the triangle and interset at the point P. P represents the omposition (a 40%, b 30%, 30%) of a 3 omponent system that is presented in the above table by enirled data. The mole fration of a at P is given by the following expression: / The perpendiular distane from line B C to line a - a = a / /a = 0.4/=0.4 The perpendiular distane from line B C to A Aaa / and ABC are similar triangles. The zone under the urve (Fig.7) represents some physial harateristis of the system. This type of urve is exhibited in liquid liquid extration.when omposition falls within the urved area, two solvent phases ( extrat and raffinate) separate out. So, the omposition is seleted in suh a way that liquid phases an be separated out after extration is over. Referenes:. Walter L. Badger, Julius T.Banhero, Introdution to Chemial Engineering. MGraw Hill Series in Chemial Engineering.. Olat A. Hougen, Kenneth M. Watson, Roland A. Ragatz. Chemial proess priniples, Part, Asian Students edition, Asia Publishing House. 3. J M Coulson and J F Rihardson,Coulson & Rihardson s Chemial Engineering, Volume 6, th ed; Pergamon Press, Oxford. 4. J.H. Perry, Chemial Engineer s Handbook, 4 th ed, MGraw Hill Book Company, 963). 3