INTRODUCTION

Sugar industry is one of the most important agro-based industries in India and is highly responsible for creating significant impact on rural economy in particular and countrys economy in general. Sugar industry ranks second amongst major agro-based industries in India. As per the Government of Indias recent liberalised policy announced on 12th December, 1986 for licensing of additional capacity for sugar industries during 7th five-year plan, there will be only one sugar mill in a circular area of 40 sq km. Also the new sugar mill is allowed with an installation capacity of 2500 TCD (Tonne Sugar Cane crushed per day) as against the earlier capacity norms of 1250 TCD. Similarly, the existing sugar mills with sugar cane capacity of about 3500 TCD can crush sugar cane to the tune of 5000 TCD with a condition imposed that additional requirement of sugar cane be acquired through increased productivity and not by expansion of area for growing sugar cane. Cane sugar is the name given to sucrose, a disaccharide produced from the sugarcane plant and from the sugar beet. The refined sugars from the two sources are practically indistinguishable and command the same price in competitive markets. However, since they come from different plants, the trace constituents are different and can be used to distinguish the two sugars. One effect of the difference is the odor in the package head space, from which experienced sugar workers can identify the source. In the production scheme for cane sugar, the cane cannot be stored for more than a few hours after it is cut because microbiological action immediately begins to degrade the sucrose. This means that the sugar mills must be located in the cane fields. The raw sugar produced in the mills is item of international commerce. Able to be stored for years, it is handled as raw material shipped at the lowest rates directly in the holds of ships or in dump trucks or railroad cars and pushed around by bulldozers. Because it is not intended to be eaten directly, it is not handled as food. The raw sugar is shipped to the sugar refineries, which are located in population centers. There it is refined to a food product, packaged, and shipped a short distance to the market. In a few places, there is a refinery near or even within a raw-sugar mill. However, the sugar still goes through raw stage. The principle by-product of cane sugar production is molasses. About 10 15% of the sugar in the cane ends up in molasses. Molasses is produced both in the raw-sugar manufacture and also in refining. The blackstrap or final molasses is about 35 40% sucrose and slightly more than 50% total sugars. In the United States, blackstrap is used almost entirely for cattle feed. In some areas, it is fermented and distilled to rum or industrial alcohol. The molasses used for human consumption is of a much higher grade, and contains much more sucrose.

Sugarcane characteristics:

Sugarcane contains not only sucrose but also numerous other dissolved substances, as well as cellulose or woody fibre. The percentage of sugar in the cane varies from 8 to 16% and depends to a great extent on the variety of the cane, its maturity, condition of the soil, climate and agricultural practices followed.

The constituents of ripe cane vary widely in different countries and regions but fall generally within the following limits:

Constituent Percentage range Water 69.0 75.0 Sucrose 8.0 16.0 Reducing sugars 0.5 2.0 Organic matter other than sugar 0.5 1.0 Inorganic compounds 0.2 0.6 Nitrogenous bodies 0.5 1.0 Ash 0.3 0.8 Fibre 10.0 16.0

Organic matters other than sugar include proteins, organic acids, pentosan, colouring matter and wax. Organic acids present in cane are glycolic acid, malic acid, succinic acid and small quantity of tannic acid, butyric acid and aconitic acid. These vary from 0.5 to 1.0% of the cane by weight. The organic compounds are made up of phosphates, chlorides, sulphates, nitrates and silicates of sodium, potassium, calcium, magnesium and iron chiefly. These are present from 0.2 to 0.6%. The nitrogenous bodies are albuminoid, amides, amino acids, ammonia, xanthine bases, etc. These are present to the extent of 0.5 to 1.0%. Fibre is the insoluble substance in the cane. Dry fibre contains about 18.0% lignin, 15% water-soluble substances, 45% cellulose and the rest hemicellulose. The juice expressed from the cane is an opaque liquid covered with froth due to air bubbles entangled in it. The colour of the juice varies from light grey to dark green. Colouring matter is so complex that very little is known about them and there is a great need for research in this direction. Colouring matters consist of chlorophyll, anthocyanin, saccharatin and tannins. Canes which have been injured or which are over-ripe contain ordinarily invert sugar as well. When severe frost damages sugarcane, all buds are killed and the stalk split. Then the juice produced has low purity, less sucrose, high titrable acidity, and abnormal amounts of gum, which make processing difficult and at times impossible. Frost is generally not a very common phenomenon in Indian crops. Insects and pests cause a greater damage. Cane juice has an acidic reaction. It has a pH of about 5.0. The cane juice is viscous owing to the presence of colloids. The colloids are particles existing in a permanent state of fine dispersion and they impart turbidity to the juice. These colloids do not settle ordinarily unless conditions are altered. The application of heat or addition of chemicals brings about flocculation or coagulation. They may be coagulated by the action of electric current and adsorption by sucrose attractions using porous or flocculent material. Some colloids are flocculated easily while others do so with great difficulty. Each colloid has a characteristic pH at which flocculation occurs most easily. It is known as the isoelectric point of the colloid. The cane juice is turbid owing to the presence of such colloidal substances as waxes, proteins, pentosans, gums, starch and silica.

PROPERTIES AND USES

Properties of cane sugars:

Sugar consists mainly of sucrose and to a certain extent of glucose and fructose. Thus properties of sugars are listed below: (A) Physical properties

Property Taste Sweet Crystal Mono-clinic Solubility Very soluble in cold water and dilute

alcohol. Solubility increases with increase in temperature. It is insoluble in chloroform, ether and glycerine.

Specific gravity at 20oC 1.05917 Optical activity Dextro-rotatory

(B) Chemical properties

Property Action of heat Perfectly dry sugar can be heated to 160oC

without decomposition. It then melts forming a non-crystallizing substance. In the presence of moisture it decomposes at 100oC, becoming a caramel and liberating water. On further heating changes to CO2 and formic acid.

Action of heat on dilute solutions By prolonged heating at the boiling point the dissolved sucrose slowly combines with water and breaks up into glucose and fructose.

Uses:

All sugars from whatever source are used almost entirely for food. In the United States, only 1% of sugar consumed is used for nonfood purposes. Technology exists to convert sucrose into many other substances by fermentation, esterification, hydrogenation, alkaline degradation, and many others, but the price is prohibitive. In 1981, the value of sugar as a food was 2-4 times its value as a chemical feedstock. As a food, sugar is all energy, and even brown and raw sugar contains virtually no protein, minerals or vitamins. The per capita consumption of sugar is an indicator of degree of economic advancement of a country. Apparently, human nature is such that one of the first uses of income above the subsistence level is to satisfy the sweet tooth. The consumption of

sugar in all forms, beet, cane, and corn, in Western Europe and North America is approximately the same, 60 kg per person each year, and holding steady. However, eastern countries consume much less sugar, and the poorer third-world countries, much less. In some isolated and remote areas, the figure is less than one kg per person-year. The world average is about 20 kg per person each year and increasing. In the United States, there is a shift toward more use of corn sugar and less of cane and beet sugar because of price, but this applies only in areas where starch is in great surplus.

The Importance of Sugar in the Diet of Man:

As everybody should know, man cannot live without a certain minimum level of sugar in his blood, the diabetic providing a good example of what happens when the blood sugar is not converted into energy, as should be the case in a healthy body. Before the discovery that insulin is essential for normal body function, a person with a defective pancreas (where the hormone, insulin, is formed) would have a short, or very short, life span, depending on the severity of his condition. Such a person was diabetic; his blood sugar was constantly very high, because only a small proportion of it was converted into energy, and his body reaction was therefore the same as that of a healthy person with a low level of blood sugar. He was always hungry, and also thirsty because of the high osmotic pressure in the blood stream, and since the sugar did not burn as it normally does, it was treated by the body as foreign matter and was expelled by the kidneys. As the malfunction became worse, he would slowly starve to death, commonly at a young age. Let it be remembered, therefore, that sugar is absolutely necessary, and if not eaten as such, it must be made from other carbohydrates which exists in food; but since sugar is the lowest in calorific value of all carbohydrates in an ordinary diet, as well as the cheapest, it should be used liberally, but wisely.

VARIOUS COMMERCIAL PROCESSES

(A) CHOICE OF PROCESS

In sugar manufacture, there are no specific processes, which are different from each other. Instead sugar manufacture has a generalized process consisting of various unit operations such as milling, evaporation, drying, crystallization etc. Hence a selection can be made inly with respect to particular choice of equipment and unit operation. Thus a variety of combinations can be made by differing the number of mills, number of effects in evaporator, the kind of dryer etc.

(B) PROCESS DESCRIPTION

At the sugar factory, the cane is piled as reserve supply in the cane yard so that the factory, which runs, 24 hr/day will always have cane to grind. The delivery of the cane to the factory depends upon the time of day, weather, and some other factors. Very closely controlled operations never have more than a few hours worth of cane in the cane yard, but more generally, the cane yard is fairly full toward evening and nearly empty the next morning. The cane is moved from the cane yard or directly from the transport to one of the cane table. Feed chains on the tables move the cane across the tables to the main cane carrier, which runs at constant speed carrying the cane into the factory. The operator manipulates the speed of the various tables to keep the main carrier evenly filled. In order to remove as much dirt and trash as possible, the cane is washed on the main carrier with as much water as is available. This includes decirculated wash water and all of the condenser water. Of the order of 1 2 % of the sugar in the cane is washed out and lost in the washing, but it is considered advantageous to wash. In areas where there are rocks in the cane, it is floated through the so- called mud bath to help separate the rocks. The sugar recovered is normally 10-wt % of the cane, with some variation from region to region. Sugar cane has the distinction of producing the heaviest yield of all crops, both in weight of biomass and in weight of useful product per unit area of land.

Extraction of juice:

The juice is extracted from the cane either by milling, in which the cane is pressed between the heavy rolls, or by diffusion, in which the sugar is leached out with water. In either case, the cane is prepared by breaking into pieces measuring a few centimeters. In the usual system, the magnets first remove the tramp iron, and the cane then passes through two sets of rotating knives. The first set, called cane knives turns at about 700 rpm, cuts the cane into pieces of 1 2 dm length, splits it up a bit, and also act as a leveler to distribute the cane more evenly on the carrier. The second set, called shredder knives turn faster and combine a cutting and a hammer action by having a closer clearance with the housing. These quite thoroughly cutter and shred the cane into a fluffy mat of pieces a few centimeters in the largest dimensions. In preparing cane for diffusion, it is desirable to break every plant cell. Therefore the cane for diffusion is put through an even finer shredder called a buster or fiberizer. No juice is extracted in the shredders. In

milling, the cane then goes to the crusher rolls, which are similar to the mills, but have only two rolls, which have large teeth and are widely spaced. These complete the breaking up of the cane to pieces of the order of 1 3 cm. The large amount of juice is removed here.

Milling:

The prepared cane passes through a series of mills called a tandem or milling train. These mills are composed of massive horizontal cylinders or rolls in groups of three, one on the top and two on the bottom in the triangle formation. The rolls are 50 100 cm diameter and 1 3 m long and have grooves that are 2 5 cm wide and deep around them. There may be anywhere from 3 7 of these 3 roll mills in tandem, hence the name. These mills, together with their associated drive and gearing, are among the most massive machinery used by industry. The bottom two rolls are fixed, and the top is free to move up and down. The top roll is hydraulically loaded with a force equivalent about 500 t. The rolls turn at 2 5 rpm, and the velocity of the cane through them is 10-25cm/s. After passing through the mill, the fibrous residue, from the cane, called bagassae, is carried to the next mill by bagassae carriers and is directed from the first squeeze in a mill to the second by turn plate. In order to, achieve a good extraction, a system of imbibition is used, bagassae going to the final mill is sprayed with water to extract whatever sucrose remains; the resultant juice from the last mill is then sprayed on the bagasse mat going to the next to last mill, and so on. The combination of all these juices is collected from the first mill and is mixed with the juice from the crusher. The result is called the mixed juice and is the material that goes forward to make the sugar. The mills are powered with individual steam turbines. The exhaust steam from the turbines is used to evaporate water from the cane juice. The capacity of the sugarcane mills is 30 300 t of cane per hour.

Diffusion:

Diffusion is used universally with sugar beets but is little used with sugarcane. The process in cane is mostly lixiviation (washing) with only a little true diffusion from unbroken plant cells. Since the lixiviation is much faster, great effort is expanded in preparing the cane by breaking it so thoroughly that nearly all of the plant cells are ruptured. In many instances, diffusers were added to an already existing mill, and, therefore, the diffuser unit was placed after the crusher rolls. In the diffusers, the shredded cane travels countercurrent to hot (75oC) water. In the ring diffuser, the cane moves around in an annular ring. In tower diffusers, the cane moves vertically, and in rotating drum diffusers, it travels in a spiral. Whatever the apparatus, the juice obtained is much like juice from mills. Milling achieves 95% extraction of the sucrose in the cane, diffusion 97% extraction. Diffusion juice contains somewhat less suspended solids (dirt and fibre), and is of higher purity (sucrose as percent of solids). The diffusion plant costs much less and takes much less energy to run. The bagasse from diffusion contains much more water.

Bagasse:

The bagasse from the last mill is about 50-wt% water and will burn directly. Diffusion bagasse is dripping wet and must be dried in a mill or some sort of bagasse press. Most bagasse is burned in the boilers that run the factories.

Clarification:

The juice from either milling or diffusion is about 12 18% solids, 10 15 pol (polarization) (percent sucrose), and 70 85% purity. These figures depend upon geographical location, age of cane, variety, climate, cultivation, condition of juice extraction system, and other factors. As dissolved material, it contains in addition to sucrose some invert sugar, salts, silicates, amino acids, proteins, enzymes, and organic acids; the pH is 5.5 6.5. It carries suspension cane fibre, field soil, silica, bacteria, yeasts, molds, spores, insect parts, chlorophyll, starch, gums, waxes, and fats. It looks brown and muddy with a trace of green from the chlorophyll. In the juice from the mill, the sucrose is inverting (hydrolyzing to glucose and fructose) under the influence of native invertase enzyme or an acid pH. The first step of processing is to stop the inversion by raising the pH to 7.5 and heating to nearly 100oC to inactivate the enzyme and stop microbiological action. At the same time, a large fraction of the suspended material is removed by settling. The cheapest source of hydroxide is lime, and this has the added advantage that calcium makes many insoluble salts. Clarification by heat and lime, a process called defecation, was practiced in Egypt many centuries ago and remains in many ways the most effective means of purifying the juice. Phosphate is added to juices deficient in phosphate to increase the amount of calcium phosphate precipitate, which makes a floc that helps clarification. When the mud settles poorly, polyelectrolyte flocculants such as polyacrylamides are sometimes used. The heat and high Ph serve to coagulate proteins, which are largely removed in clarification. The equipment used for clarification is of the Dorr clarifier type. It consists of a vertical cylindrical vessel composed of a number of trays with conical bottoms stacked one over the other. The limed raw juice enters the center of each tray and flows toward the circumference. A sweep arm in each tray turns quite slowly and sweeps the settled mud toward a central mud outlet. The clear juice from the top circumference overflows into a header. Diffusion juice contains less suspended solids than mill juice. In many diffusion operations, some or all of the clarification is carried out in the diffuser by adding lime. The mud from clarification is filtered on Oliver rotary vacuum filters to recovery the juice. The mud mostly consists of field soil and very fined divided fibre. It also contains nearly all the protein (0.5 wt% of the juice solids) and cane wax. The mud is returned to the fields. Although the clarification removes most of the mud, the resulting juice is not necessarily clear. The equipment is often run at beyond its capacity and control slips a little so that the clarity of the clarified juice is not optimum. Suspended solids that slip

past the clarifiers will be in the sugar. Clarified juice is dark brown. The colour is darker than raw juice because the initial heating causes significant darkening.

Evaporation:

Cane juice has sucrose concentration of normally 15%. The solubility of sucrose in water is about 72%. The concentration of sucrose must reach the solubility point before crystals can start growing. This involves the removal by evaporation of 93% of the water in the cane juice. Since water has the largest of all latent heats of vapourization, this involves a very large amount of energy. In the energy crunch of the late 1970s, the DOE found that the sugar industry was one of the largest users of energy. The sugar industry already knew this very well and had been using multiple-effect evaporators for saving energy for more than a century. The working of multiple-effect evaporator can be seen in fig. In each succeeding effect, the vapours from the previous effect are condensed to supply heat. This works only because each succeeding effect is operating at a lower pressure and hence boils at lower temperature. The result is that 1 kg of steam is used to evaporate 4 kg of water. The steam used is exhaust steam from the turbines in the mill or turbines driving electrical generators. The steam has therefore already been used once and here in the second use it is made to give fourfold duty. The usual evaporator equipment is a vertical body juice-in-tube unit. Several variations are in use, but the result is the same. The only auxiliary equipment is the vacuum pump. Today, steam-jet-ejectors are general, although mechanical pumps were formerly used. Since the cane juice contains significant amounts of inorganic ions, including calcium and sulfate, the heating surfaces are quick to scale and require frequent cleaning. In difficult cases, the heating surfaces must be cleaned every few days. This requires shutting down the whole mill or at least one heat-exchanger unit while the cleaning is done. Inhibited hydrochloric acid or mechanical cleaners are usually employed. Magnesium oxide is sometimes used instead of lime as a source of hydroxide. Magnesium costs more, but it makes less boiler scale on the heaters. It is also easier to remove because it is more soluble; however, for the same reason, more gets into the sugar. Whether it is used or not depends upon the influence, standing, and persuasiveness of the chief engineer who must keep the plant running and the chief chemist who must make good sugar. The evaporation is carried on to a final brix of 65 68. The juice, after evaporation, is called syrup and is very dark brown, almost black, and a little turbid.

Crystallization:

The crystallization of the sucrose from the concentrated syrup is traditionally a batch process. The solubility of sucrose changes rather little with temperature. It is about 68 brix at room temperature and 74 brix at 60oC. For this reason, only a small amount of sugar can be crystallized out of solution by cooling. Evaporating the water must instead crystallize the sugar. Sucrose solutions up to a super saturation of 1.3 are quite stable. Above this super saturation, spontaneous nucleation occurs, and new crystals form. The

sugar boiler therefore evaporates water until the supersaturation is 1.25 and then seeds the pan. The seeding consists of introducing just the right number of small sugar crystals (powdered sugar) so that, when all have grown to the desired size, the pan will be full. After seeding, the evaporation and feeding of syrup are balanced so that the supersaturation is as high as possible in order to achieve the fastest possible rate of crystal growth, without exceeding 1.3. The boiling point of a saturated sugar solution at 101.3 kPa (1 atm) is 112oC. Sugar is heat-sensitive and, at this temperature, thermal degradation is too great. The boiling is therefore done under the highest practical vacuum at a boiling point of 65oC. The sugar boiler therefore must manipulate the vacuum along with the steam and feed. A proof stick on the vacuum pan allows the contents of the pan to be sampled while under vacuum. When the pan is full, the steam and feed are stopped, the vacuum is broken, and the batch, or strike, is dropped into a receiver below. A strike is 50 metric tons of sugar and it is boiled in 90 min. at the end of this time, the mixture of crystals and syrup, called massecuite, must still be fluid enough to be stirred and discharged from the pan. In practice, about half of the sugar in the pan is in crystal form and half remains in the syrup. In this case, the pan yield is said to be 50%. Some very good sugar boilers are able to achieve as much as 60% yields on first strike.

Vacuum pans:

Vacuum pans have a small heating element in comparison to the very large liquor and vapour space above it. The heating element was formerly steam coils but is now usually a chest of vertical tubes called calandria. The sugar is inside the tubes. There is a large center opening (downcomer) for circulation. The vacuum pan has a very large discharge opening: typically 1 m dia. At the end of a strike, the massecuite contains more crystals than syrup and is therefore very viscous. This large opening is required to empty the pan in a reasonable time. At the top or dome of the pan, there are viscous entrainment separators. The pan may also be equipped with a mechanical stirrer. This is usually an impeller in or below the central downcomer, driven by a shaft coming down all the way from the top. The strike is started with liquor just above the top of the calandria. The strike level cannot be very near the top because of vapour space must be allowed for separation of entrainment. In operation, the boiling is very vigorous with much splashing of liquid. The vacuum is maintained mostly by condensing the vapours in a barometric condenser. In some cases, a surface condenser is used. This serves as a source of distilled water and recovers heat. More often, however, a jet condenser is used in which the cold condensing water is sprayed into the hot vapour and both condensate and condenser water are mixed. A supplementary vacuum pump is required to remove noncondensable gases.

Centrifuging:

The massecuites from the vacuum pans enter a holding tank called a mixer that has a very solely turning paddle to prevent the crystals from settling. The mixer is a feed

for the centrifuges. In a batch-type centrifuge, the mother liquor is separated from the crystals in batches of about 1 t at a time.

Boiling systems:

In raw-sugar manufacture, the first strike of sugar is called the A strike, and the mother liquor obtained from this strike from the centrifuges is called A molasses. The pan yield in sugar boiling is about 50%. Because crystallization is an efficient purification process, the product sugar is much purer than the cane juice and the molasses much less pure. As an approximation, crystallization reduces the impurities by factor of 10 or more in the product sugar. Therefore, almost all of the impurities remain in the molasses. Enough molasses accumulates from boiling two first strikes to boil a second strike. The B sugar from the second strike is only half as pure as that from the first strike, but the B molasses is twice as impure. This can go on to a third strike. At this point, 7/8 of the sugar from the cane juice is in the form of crystals and 1/8 in the C molasses. In practice, three strikes is about all that can be gotten from cane juice. The trick is to maneuver to obtain good sugar, but at the same time have the C or final molasses as impure as possible. The purity of the feed to the final strike is adjusted to obtain the lowest possible purity of final molasses. Some of the C sugar is redissolved and started over, some is used as footing for A and B strikes. The C sugar is of very small crystal size so it is taken into the A or B pans as seed and grown to an acceptable size. This practice is actually a step backward because it hides impure C sugar in the center of better A and B sugars. The product raw sugar is a mixture of A and B sugars. There are many variations in the boiling scheme, such as two and four billings, blending molasses, and returning molasses to the same strike from which it came. All of these tricks are used, depending on cane purity and capabilities of the equipment available.

Crystallizers:

When the steam is turned off at the end of a sugar boiling, evaporation ceases immediately and the mixture of crystals and supersaturated syrup in the pan starts toward equilibrium, which is the po9int of saturation. In relatively pure sugar solutions, this equilibrium is reached in few minutes well before the syrup crystallization is slower and reaching equilibrium can take a significant amount of time. In the final strike, the time an amount to days, so final strikes are not sent directly to the centrifuges, but instead to crystallize, holding tank is in which the crystals grow as much as possible and the super saturation in the molasses is reduced to 1.0. Since the intention in handling the final molasses is to remove as much sugar as possible, advantage is taken of the small temperature coefficient of solubility and the massecuite is also cooled. The crystallizers are large tanks, some open-top, with a slow-moving stirrer that is sometimes also a cooling coil. At the end of the holding time, the massecuite is warmed slightly as it enters the centrifuge to lower the viscosity and achieve better separation. The limiting factor in exhaustion of masses is the viscosity. A little more water can always be boiled out, but the molasses must remain fluid enough to run out of the pan, into the centrifuge and to flow between the sugar crystals on the centrifuge screens.

Refining:

Sugar refineries are located in large cities. They are near seacoast with harbors and facilities for receiving raw sugar by ship. They thus can receive sugar from anywhere in the world, although each refiner has favorites that suit the refinery, market, or have been the traditional supplier. Refineries are open all year, although the busy season is in the summer. Refineries are always large. Their capacity is expressed in terms of daily melt. Melt is the sugar term for dissolving, and means the amount of sugar melted or processed each day. The smallest refineries have a daily melt of 450 t, and large ones have as much as ten times that amount. The yield of refined sugar is nominally 93% of the raw-sugar input. Raw sugar is light to da4rk brown in color and sticky. The size of the sucrose crystals is ca 1 mm. Refiners would like to have raw sugar that is high in sucrose and of uniform quality; however, they must be prepared to refine anything. Raw sugars are about 98 pol, although they are always described in terms of the equivalent raw value expressed as 96 pol, a base value from the 1920s when raw sugars were of this pol. Terminology changes slightly in the refinery. In raw sugar, syrup is a concentrated solution going to the pans. After the boiling, the solution separated from the crystals is molasses. In the refinery, it is liquors that are fed to the pans, and syrups that are separated from the crystals. Local jargon adds to the confusion with such terms as greens and jets, meaning syrups, and barrel syrup, meaning final molasses.

Affination:

The first step in refining is to remove the molasses film from the outside of the raw-sugar crystals. This is done by a washing process known as Affination. Syrup that is not quite saturated with sucrose is mingled with the incoming raw sugar in a large trough containing a mixer paddle and scroll. This mixture is then centrifuged and washed in the centrifuge rather more than less. A uniform crystal size is important in raw sugars because a mixture of different sizes or broken crystals does not wash well in the affination centrifuge. The syrup formed is called affination syrup not wash well in the affination centrifuge. The syrup formed is called affination syrup and is used for mingling. The sugar is called washed sugar and is ten shades lighter in color than the raw sugar. It is estimated that 90% of refining is done in this first step. About 10% of the sugar becomes apart of the affination syrup, which thus keeps increasing in volume and is sent to the recovery house. The recovery house is route through a set of equipment in the same building. IT uses the same processes that are used in the main refinery, but in manner more like a raw-sugar operation. As the name suggests, sugar is recovered in the recovery house, but the main object is to transfer impurities into molasses that contains the least possible amount of the sucrose. The recovered sugar is called remelt and is sent back to process.

Melting:

The washed sugar is melted in hot water, and usually the pH is adjusted with lime. Water that contains a little sugar from anywhere in the refinery is called sweetwater, and if it does not contain much impurity, is used in the melter. The washed sugar liquor coming from the melter is adjusted to the operating concentration, usually about 65 brix. The trend is to operate refineries at higher brix up to 68, because if water is not added, it does not have to be boiled away later. The washed sugar liquor is dark brown and quite turbid, and appears much darker than the sugar from which it came. The melter liquor is strained through a plain screen to catch debris in the raw sugar.

Clarification:

The object of clarification is the complete removal of all particulate matter. The particles in the sugar come from all sources, eg, field soil and fiber which escaped clarification in the raw-sugar factory; all microbiological life, including yeasts, molds, bacteria, and their spores; colloids and very high molecular weight polysaccharides; and foreign contaminants such as insect and rodent dropping. The very diversity of the nature of the particulate matter and wide range of particle sizes makes clarification a difficult and critical step in the refining of sugar. One of three processes is used: filtration, carbonization, or phosphatation.

Phosphatation:

In the phosphate clarification scheme, lime and phosphoric acid are added simultaneously with good mixing. The phosphoric acid is added in proportion to the milt at about 0.01-0.02%. The lime is added to bring the pH to 7.8. The calcium phosphate precipitate forms a floc of no particular crystal structure. It is even better at scavenging impurities by entrapment than the carbonate precipitate. It also has the useful property of attaching or entrapping air bubbles. Thus, at the same time that the floc is being formed, some air is injected, mixed, or pumped into the system. Raising the temperature a few degrees also helps tiny air bubbles materialize throughout the liquor. The precipitate then floats to the surface as a scum of 80% organic matte and is scraped off without any filtration. The mixing is very thorough just as the reagents are added, gentle in a floc-development section, and then minimal in the flotation zone. The phosphate clarifiers are also called frothing clarifiers and have many sizes and shapes with scrapers going forward, backward, and around. Some are heated and some are deep. Sugar is recovered from the scum by clarifying again. The phosphate system uses only about 1/10 as much reagents as the carbonate system, and so produces only 1/10 the scum volume. No matter what the method of clarification, the clarified liquor is brilliantly clear without any sign of turbidity. It is, however, dark, rather like a cup of weak coffee.

Decolorization:

The key process in sugar refining is decolorization. Color is the principal control in every sugar refinery. It is the main property that distinguishes refined sugar from raw sugar. The word color is used loosely. It usually means visual appearance, but in technical sugar work it means colorant, the material causing the color. It can be classified in three groups: plant pigments; melanoidins resulting from the reaction of amino acids with reducing sugars; and caramels resulting from the destruction of sucrose. Many, but not all, compounds in each of these classes have been identified. In sugar work, color refers collectively to the optical sum of all the colorants. Bone char and granular carbon behave similarly in decolorization of sugar. For the contact with sugar liquor, both are contained in helds called cisterns ca 3 m-dia and 7 m tall and holding 30-40 t carbon. The liquor flows downward with a contact time of 2-4 h. The first liquors are water white with a very gradual yellowing. The cistern stays on stream until the color of the liquor becomes too great to be handled by the remainder of the refining process. The decolorization is always greater than 90%. The bone-char cycle is about 4 d; for granular carbon, it is 4 wk. The first liquors from bone-char treatment are lighter than the first liquors from carbon. However, from the adsorbent point of view, the two systems are different. Heating degreased cattle bones to about 7000C in the absence of air makes bone char. IT is about 6-10% carbonaceous residue and 90% calcium phosphate from the bone with an open pore structure supplied by the bone. The surface area available to nitrogen is 100 m2/g

. The particle size is about 1 mm. besides being a carbon absorbent; it has ion-

exchange properties that permit removal of considerable ash from the sugar. These same ion-exchange properties result in a buffering effect that keeps the pH of the sugar liquor from falling. After the decolorization cycle, the sugar is washed out of the bed and then the bed is washed with the cool water to remove as much as possible of the adsorbed inorganic salts. The sweet water is of low purity and cannot be recycled. The organic coloring matter is adsorbed so tightly that no amount of washing will remove it. The water is therefore drained from the char and the char moved from the cistern to a kiln where the organic matter is burned off at 5000C, with a little oxygen in the kiln atmosphere to burn away freshly deposited carbon and keep the pores open. On bone char, it has been observed ionic constituents in the liquor affect ionic color removal process. High calcium liquors decolorize well, but high sulfate liquors decolorize as much as ten times more poorly for an ion-concentration change of only 0.01 N.

Crystallization:

The color of the washed, clarified, and decolorized liquor going into the crystallization process ranges from water white to slightly yellow. Many refiners polish-filter the sugar liquor at this stage to make sure that it is sparkling clear with no turbidity. Others rely on good operation upstream and do not polish-filter. In many cases, the brix has become too low, either on purpose or by error; these liquors first go to the evaporators to bring the brix to >= 68.The vacuum pans are the same as were described

under Raw Sugar Manufacture, and their operation is the same. They are operated even more carefully to produce crystals of the desired size. Great care is taken to avoid conglomerates and fines. Boiling rate and throughput are important. A new strike of some 50 metric tons must be dropped every 90 min to keep up with the production schedule. The boiling schemes used in the refinery are more extensive and more extensive and variable than those used in the raw house. This is because the starting material is of much higher purity. Ordinarily, three, four, or five strikes of refined sugar are obtained.

The syrup from the fourth strike may handle in different ways. It may be used in the recovery house, but is more likely used in making specialty syrups or brown sugars. It may also be sent back to decolorization or clarification, and recycled. The refined sugar centrifuges are always batch type because they leave the crystals intact. The centrifuging is easy and the cycles are short. The drying of the sugar from the centrifuges is done by rotary dryer using hot air. This dryer is universally misnamed the granulator because by drying in motion, it keeps the sugar crystals from sticking together, or keeps them granular. The hot sugar from the granulator is cooled in an exactly similar rotary drum using cold air.

Conditioning:

The sugar from the coolers would appear to be finished, but after a few days storage it becomes wet with water trapped inside the grain because of the very high rate crystallization and drying. After a few days, this moisture migrates outside the crystal and the sugar is wet again. A process known as conditioning removes the moisture, in which the sugar is stored for four days with a current of air passing through it to carry away the moisture. In one system, a single silo is used with sugar being continuously added to the top and removed from the bottom, and a current of dry air blowing upward. In another system, the sugar is stored in a number of small bins. It is continuously transferred from bin to bin with dry air blowing around the conveyors that move the sugar.

Packing, storing and shipping:

Sugar is sometimes stored in bulk and then packaged as needed. Others package the sugar and then warehouse the packages. The present trend is away from consumer-sized packages and toward bulk shipments.

MATERIAL BALANCE

5000 tpd of sugarcane are to be processed. Assuming that the factory operates for 20 hours per day. Thus we shall require processing 250 tons of sugarcane per hour.

Note: 1. All units where otherwise mentioned are in tons per hour. 2. The word analyses wherever use implies particular materials following per hour in tons.

Basis: 250 tons per hour

Choosing standard Indian cane quality as below: Water 70.0 % Sucrose 14.0 % Reducing sugars 0.5 % Fibre 13.0 % Ash and other impurities 2.5 %

Thus 250 tons cane feed per hour will contain: Water 175.00 Sucrose 35.00 Reducing sugars 1.25 Fibre 32.50 Ash and other impurities 6.25 250.00 Raw sugar manufacturing:

(1) Milling plant

Water used in milling operation is 25 30% of sugar cane. Assume it as 25%. Therefore imbition water used = 0.25 250 = 62.5 Assuming milling efficiency 95% i.e. 95% of sucrose goes into the juice. Thus sucrose content in juice = 0.95 35 = 33.25 Unextracted sucrose = 0.05 35 = 1.75 The final bagasse from last mill contains the unextracted sucrose, woody fibre and 40 55% (assume 50%) water. Thus water content in bagasse = 32.5 + 1.75 = 34.25 Amount of bagasse = 32.5 fibre + 1.75 sucrose + 34.25 water = 68.5 Overall output of mill or juice entering the clarifier will have the following composition:

Water = 175 + 62.5 34.25 = 203.25 Sucrose = 33.25 Reducing sugars = 1.25 Impurities = 6.25 Thus total juice = 244

(2) Clarifier

Reagent used: lime (0.5 kg lime/ton of sugar cane) Thus lime to be added = 0.5 250 = 125 kg = 0.125 ton Assume 96% efficiency of clarifier to remove impurities. Therefore impurities to be removed = 0.96 6.25 = 6.00 Total sludge from clarifier = 0.125 + 6.0 = 6.125 Amount of clarified juice entering in the evaporator = 244 6.0 = 238 This clarified juice is fed to the first effect of the quadruple effect evaporator and analyses as follows:

Water 203.25 Sucrose 33.25 Impurities 0.25 Reducing sugars 1.25 Thus total juice 238.00 % of solids in this juice = {(33.25 + 0.25 + 1.25) 100}/238 = 14.60

(3) Evaporator

Typical evaporator load = 75 80% of clarified juice Assume 76% of clarified juice as evaporator load. Thus evaporator load = 0.76 238 = 180.88 Input to the evaporator = (203.25 water + 33.25 sucrose + 0.25 impurities + 1.25 reducing Sugars) = 238 Water removed in the evaporator = 180.88 Water remaining in the juice = 203.35 180.88 = 22.37

Hence output of evaporator analyses as follows:

Water 22.37 Sucrose 33.25 Impurities 0.25 Reducing sugars 1.25 Thus total solution 57.12 % of solids in this concentrated solution = {(33.25 + 0.25 + 1.25) 100}/57.12 = 60.84

(4) Crystallizer 1

Crystallization of concentrated juice is done in a vacuum pan crystallizer. Crystallization is done at vacuum not exceeding 25 in (635mm). Hence assume 580 mm Hg vacuum in crystallizer. Therefore absolute pressure = 760 - 560 = 280 mmHg = (280 1.013)/760 bar

= 0.2666 bar Thus boiling point (from steam table) at this pressure = 61.420C Boiling point rise = 40C Solubility of sucrose in water is given by Y = 68.18 + 0.1348 t + 0.000531 t Where Y is % sucrose at saturation t is temperature in 0C At 65.42oC, Y becomes = 73.03% Therefore sucrose per kg of water = Y/(100 Y) = 73.03/(100 - 73.03) = 2.71 kg Sucrose/ kg of water But since impurities are present. Therefore purity can be calculated as Purity = wt. of sucrose/ total wt. of solids = 3.25/(33.25 + 0.25 + 1.25) = 0.957 i.e. 95.7%

Hence the solubility is reduced by factor called solubility coefficient.

For purity of 95.7%, Solubility coefficient = 0.97 Thus effective solubility = 2.71 0.97 = 2.6287 kg of sucrose/kg of water Input to crystallizer = (22.37 water + 33.25 sucrose + 1.25 reducing sugars + 0.25 impurities) Assume 92% of sucrose recovery as crystal with respect to initial sucrose content in feed. Weight of sucrose crystal formed = 0.92 33.25 = 30.59 Weight of reducing sugar crystals = 0.33 1.25 = 0.4125 Moisture associated with crystals = 1% of crystal weight = 0.01 30.59 + 0.01 0.4125 = 0.3059 + 0.004125 = 0.3100 Sucrose in molasses = 33.25 - 30.59 = 2.66 Reducing sugar left in molasses = 1.25 0.4125 = 0.8375 Total sugar in molasses = 2.66 + 0.8375 = 3.4975 Water required to dissolve this = 3.4975/2.6287 = 1.331 Hence water to be evaporated = initial water associated water water in molasses = 22.37 0.31 1.331 = 20.729 Out put of crystallizer: (a) Solids: 31.4375 (30.59 sucrose + 0.3059 water) crystals + (0.4125 reducing sugars + 0.004125 water) crystals + 0.125 impurities (b) Molasses: 4.9535

(1.331 water + 2.66 sucrose + 0.8375 reducing sugar + 0.125 impurities) (a) + (b) = Input to centrifuge = 31.4375 + 4.9535 = 36.3910

(5) Centrifuge

Assume 10% molasses adheres to the crystal. Output of centrifuge = (30.59 sucrose + 0.3059 H2O) + (0.4125 reducing sugars + 0.004125 H2O) + 0.125 impurities + 0.4954 molasses = 31.9329 = input to affination step

CANE SUGAR REFINING:

(6) Affination

The first step in refining process is called affination or washing and consists of removing the adhering film molasses from the surface at the raw sugar crystal. The separation process involves mingling the raw with a heavy syrup (about 75% solids) then purging the mixture in centrifugals and washing with hot water after the syrup has been spun off. Best magma temperature is 430C. Lyle warned sugar loss in affination may be very large especially if the syrup is stored in tanks where it is heated. Assuming 5% of sucrose is lost in affination along with molasses.

Output of affination step (i.e. washed sugar) analyses as follows:

Sucrose = 30.59 0.05 30.59 = 29.0605 Water = 0.31 Reducing sugars = 0.4135 Impurities = 0.125 Total = 29.909

(7) Melting or dissolving the washed sugar

The washed sugar is dissolved in about one-half its weight of water in a tank provided with mixing arms and called a melter, exhaust steam being applied from a perforated coil to aid solution. Therefore water added for melting = 29.0605/2 = 14.5302 Therefore melt sugar analyses as follows: Sucrose = 29.0605 Water = 0.31 + 14.5303

= 14.8403 Reducing sugars = 0.4135 Impurities = 0.125 Total = 44.4393

(8) Defecation or clarification

(a) Prescreening of liquors - screening of melt liquors to remove strings, jute, twine and other coarse material has always been customarily but only during the past two decades has it become standard practice to subject liquors to fine screening. But here assume that there is no such types of material in melt liquor.

(b) Lime in the batch system, first a part of the lime was added, neat the required P2O5 then the remainder of the lime, to give a pH of 7.0 to 7.3 on the clarified liquor. The amount of lime used ranges from 400 to 500 lb CaO per 106 lb melt, the average yearly figure for one refinery using Williamson clarification being 460 lb of CaO with 175 lb of P2O5 i.e. 635 ton per 106 ton melt.

Therefore lime to be added = (44.4393 635)/106 = 0.0282 tons/hr The reducing sugar does not dissolve in water and it goes along with sludge. Assuming all impurities are removed and efficiency of the clarifier as 100%. Thus the solution from clarifier contains sucrose and water & the sludge contains phosphate precipitate and reducing sugars. The solution from this clarifier analyses as follows: Water 14.8403 Sucrose 29.0605 Total 43.9008

(9) Decolourization

Assuming the colouring matter content in the solution from clarification as 2% of sucrose.

Therefore colouring matter = 0.02 29.0605 = 0.5812 Assume all colour is removed in the bone char bed decolourizer. Thus colour removed = 0.5812 Solution from decolourizer analyses as follows: Sucrose = 29.0605 0.5812 = 28.4793 Water = 14.8403 Total = 43.3196

(10) Crystallizer 2

Assuming same pressure as that of crystallizer 1. Therefore solubility of sucrose in water as calculated before = 2.71 kg sucrose/kg water

Here no impurities are present, hence there is no need to consider solubility coefficient. Assume 92% sucrose recovery as crystals with respect to initial sucrose content in feed. Therefore weight of crystals formed = 0.92 28.4793 = 26.2010 Moisture associated = 0.01 26.2010 = 0.26201 Sucrose in molasses = 28.4793 26.2010 = 2.2783 Therefore water in molasses = 2.2783 2.71 = 0.8407 Hence water to be evaporated = initial water associated water water in molasses = 14.8403 0.26201 0.8407 = 13.7376 Assume 10% of molasses adheres to crystal in centrifuge. Therefore molasses adhered = 0.10 (2.2783 + 0.8407) = 0.10 3.1190 = 0.3119 Wet crystals from centrifuge analyses as follows: Sucrose = 26.2010 + {(0.3119 2.2783)/3.119} = 26.4288 Water = 0.26201 + {(0.3119 0.8407)/3.119} = 0.3461 Total = 26.7749

(11) Drying

The deterioration of sugar is retarded and the loss in test is reduced if the moisture content of the sugar is reduced. With a dryer, the moisture content may be reduced to between 0.2 and 0.5%. Drying by contact with hot air involves heating the air, to increase the capacity for absorbing water and bringing it into intimate contact with the sugar from which it evaporates the moisture.

Assume that the final moisture content is 0.2%.

Weight of dry crystal is 26.4288 as calculated before. Therefore final sugar produced = 26.4288 (1 0.002) = 26.4818 Thus overall yield of refined sugar based on cane crushed = (26.4818 250) 100 = 10.59% Thus moisture to be removed = m = 26.7749 26.4818 = 0.2931 tons/hr = 293.1 kg/hr There are two possible methods of circulation of air and sugar. i.e. parallel flow and countercurrent flow. For safety, the calculation is based on most unfavourable condition, i.e. it is assumed that the ambient air is saturated. On the other hand, the air leaving a dryer is generally not saturated; it is assumed that in case of countercurrent flow, it has absorbed only two-

thirds of the quantity of water that it could have absorbed if it had left in a saturated condition. We have then, in the case of countercurrent condition, A = (100 m) {(2 3) (H1 H0)} = (1500 m) (H1 H0) Where A weight of air to be passed through dryer (kg/hr) m moisture to be removed (kg/hr) H0 weight of water vapour contained in saturated air at a temperature t0 of entry air to the heater (ambient temperature) in kg/1000 kg Taking ambient temperature as 30oC. From fig. 36.3 (Hugot) H0 = 26 kg water/1000 kg saturated air H1 weight of water vapour contained in saturated air at the temperature t1 of exit From the dryer in kg/1000 kg The temperature of air leaving the dryer is in between the 45 52oC.

Assuming it as 50oC.

From fig. 36.3 (Hugot) H1 = 85 kg water/1000 kg saturated air Therefore A = (1500 293.1) (85 26) = 7451.69 kg/hr Thus weight of air to be passed through the dryer = A = 7451.69 kg/hr Hence volume of air required = V = A (a0 + e0) From fig. 36.3 and 36.4 (Hugot) a0 = density of air at t0 = 1.12 kg/m3 e0 = weight of vapour contained in saturated air at t0 = 0.03 kg/m3 Therefore V = 7451.69 (1.12 + 0.03) = 6479.73 m3/hr

HEAT BALANCE (1) Crusher

The normal power required to crush sugar cane is given by PN = 0.15 F n D --------------------------------------(1) Where PN normal power (h.p) F load of crusher ( tons) n economical rotational speed (rpm) D mean diameter of the rollers (m) Here we will take 2 roller crusher. Many types of crushers have been designed and tried but only two have achieved general importance. These were

(a) The Krajewski and (b) The Fulton

The Fulton type is named after the Firm which has predominantly contributed to its design and wide use. It is the only type used at the present day. For this type usual roller dimensions are 660 1220 mm Here D = 660 mm and L = 1220 mm Load of crusher = F = 250 tons To calculate speed:

The peripheral speed generally adopted for the crusher is 30 40% greater than that of the mills. The economic peripheral speed of mill is given by VE = (30 D) (D + 0.73) m/min D is calculated in milling tandem calculation = 1016 mm Therefore VE = (30 1.016) (1.016 + 0.73) = 17.46 m/min Hence peripheral speed for crusher is given by VE crusher = 1.35 VE mill = 1.35 17.46 = 23.57 m/min Now the relation between peripheral speed and rotational speed is given by nE = VE ( D) = 23.57 ( 0.66) = 11.37 rpm Therefore from equation (1) PN = 0.15 250 11.37 0.66 = 281.31 h.p

(2) Milling tandem

Assume 5 mill tandems and each mill consists of three rollers. Hence number of rollers = N = 15

To find L (length) and D (diameter) of rollers:

The capacity of a train mills is the quantity of cane, which that train, is capable of treating in unit time. It is generally expressed in tons of cane per hour and it is given by A = [0.9 C n (1 0.06nD) L D2 N] f ------------------------------------(1) Where A = capacity of tandem = 250 tons/hr f = fiber content per unit of cane = 0.13 n = speed of rotation of rollers in rpm Generally used speed is 6 rpm. Hence n = 6 rpm N = number of rollers in tandem = 15 C = coefficient of preparatory plant For 2 knife sets of wide pitch, C = 1.15

Thus above equation (1) becomes 250 = [0.9 1.15 6 (1 0.06 6 D) L D2 15] 0.13 (1 0.36 D) L D2 = 1.3513 ---------------------------------------------------------(2) Standard combinations of L and D as given by Hugot are taken as follows:

D L (mm mm) LHS of equation (2) 813 1675 0.7831 920 2000 1.1321 970 2100 1.2859 1016 2134 1.3971 970 2134 1.3067

Thus choose dimensions as D L 1016 2134 (mm mm)

Power requirements of mills:

The determination of the power consumed by a mill is rather complex because a number of factors enter into it. To begin with this, the power may be split into 3 different principle terms. (a) Power absorbed by compression of bagasse P1 = [0.4 (6 r 5) F n D +A] [U ^ U 1)}] Where +A = 0.0164 For r = 1.2, (6 r 5) [U ^ U 1)}] = 1.388 F = 250 tons/hr n = 6 rpm and D = 1.016 m Therefore P1 = 0.4 1.388 250 6 1.016 = 108.36 h.p

(b) Power absorbed by friction P2 = (0.075 F n D) + (2 L n D)

= (0.075 250 6 1.016) + (2 2.134 6 1.016) = 140.32 h.p

(c) Power absorbed for intermediate carrier drive P3 = 2 L n D = 2 2.134 6 1.016 = 26.02 h.p

Thus total power consumed by mill is given by P = P1 + P2 + P3 = 108.36 + 140.32 + 26.02 = 274.70 h.p But usually the efficiency of mill is 80%. Hence actual power consumed by mill = 274.7 0.8 = 343.37 h.p Thus power required for a five mill tandem = 5 343.37 = 1716.86 h.p

(3) Heater (clarifier)

Assuming the raw juice from milling tandem is available at 30oC. Before adding lime to juice, the juice has to be heated. Thus the final temperatures to which juices are heated in clarification varies from extremes of 90 to 115oC, although by far the commonest practice is to heat slightly above the boiling point. Superheating was advocated by some in earlier studies, but it is generally believed today that superheating is not advantageous and temperatures just above the boiling point say 103oC are the maximum for good practice. Assume it is heated to 96oC and steam is available at 1 atm for heating. /DWHQW KHDW RI YDSRXULVDWLRQ DW WKLV SUHVVXUH LV N-NJ

Heat required to raise the temperature of raw juice from 30 to 96oC is given by Q = m Cp T = ms -----------------------------------------------(1) Cp = heat capacity of raw juice = 4.19 2.35 X kJ/kgoK Where X is fraction of dissolved solids.

Feed to the clarifier is 244 tons/hr and % of solids content is 16.70 (from material balance). Therefore Cp = 4.19 2.35 0.167 = 3.7976 kJ/kgoK And m = (244 1000)/3600 = 67.78 kg/s T = 96 30 = 66oC Thus from equation (1), Q = 67.78 3.7976 66 = 16988 kW Steam requirement is = ms 4 NJV

(4) Evaporator

Saturated steam required, which is calculated in process design part is S = 12.3625 kg/s Saturation temperature of steam = 113.89o& DW ZKLFK N-NJ Therefore heat supplied by this steam = S = 12.3625 2218.2 = 27422.5 kW Steam economy = amount of water evaporated (i.e. evaporator load)/steam used = 50.245/12.3625 = 4.064

(5) Crystallizer 1

As we known from material balance calculation, crystallizer operates at 580 mmHg vacuum. At this pressure, boiling point of water is 61.42oC. But boiling point rise (BPR) is 4oC. Temperature in crystallizer = 65.42oC Heat balance in crystallizer can be written as FhF + mss = LhL + VH ---------------------------------------------(1) F = feed rate to crystallizer = (22.37 + 33.25 + 1.25 + 0.25) (1000 3600) = 15.87 kg/s Cpf = 4.19 2.35 X Here X = 0.6084 Therefore Cpf = 2.7603 kJ/kgoK Feed temperature = Tf = 55.45oC Taking base temperature as 0oC. But hF = Cpf (Tf 0) = 2.7603 55.45 = 153.06 kJ/kg ms = steam rate required in kg/s Assume steam is available at same pressure as that of clarifier. +HQFH s = 2256.9 kJ/kg L = output of crystallizer = (31.4375 + 4.9535) (1000 3600) = (36.391 1000) 3600 = 10.11 kg/s Cpl = 4.19 2.35 X Where X = fraction of solids in crystallizer output = 31.4375/36.391 = 0.955

Thus Cpl = 4.19 2.35 0.955 = 1.95 kJ/kgoK Temperature of output of crystallizer = Tl = 65.42oC But hL = Cpl (Tl 0) = 1.95 65.42 = 127.31 kJ/kg V = amount of water evaporated in the crystallizer

= (20.729 1000) 3600 = 5.76 kg/s

H = DW oC + (BPR Cpsteam) DW

oC = 2618 275.8 = 2342.2 kJ/kg Cp steam at 65.42oC = 1.928 kJ/kgoK Therefore H = 2342.2 + (4 1.928) = 2349.91 kJ/kgoK Now equation (1) becomes, (15.87 153.06) + (ms 2256.9) = (10.11 127.31) + (5.76 2349.91) Hence ms = 12394/2256.9 = 5.49 kg/s Steam to be supplied to the crystallizer is = ms = 5.49 kg/s

(6) Centrifugal separator

Feed to the separator = 31.4375 + 4.9535 = 36.391 tons/hr For this capacity and for speed of 1500 rpm, the power required for an operation as suggested by Hugot = 45 kW

(7) Affination

For every 45 kg of raw sugar, 9 to 13 kg of green syrup is required. Green syrup is a low purity sugar solution has an affinity for sugar and impurities and can dissolve and retain more sugar & impurities in solution than a pure sugar solution, it will also dissolve non-sugars more readily, and when recirculated through the affination station will gradually decreases in purity.

Therefore for 31.932 tons/hr, green syrup has to be added = (10 31.932) 45 = 7.096 tons/hr

43&

Raw sugar 31.932 T/hr 97.48 % DS, Temp = 65.4&

Green Syrup 75% DS, Temp = 72&

Washed Sugar 29.909 T/hr 98.96% DS Temp = 43&

Wash water (m) Kg/s Temp = 30&

Affination Greens

Cp of water at 30oC = 4.184 kJ/kgoK Cp for raw sugar = 4.19 2.35 0.9748 = 1.9105 kJ/kgoK Cp for green syrup = 4.19 2.35 0.75 = 2.4275 kJ/kgoK Assume base temperature 43oC. Now the heat balance can be written as [31.932 (1000 3600) 1.9105 (65.42 43)] + [7.096 (1000 3600) 2.4275 (72 43)] = m 4.184 (43 30) 379.93 + 138.76 = m 54.392 Hence m = 9.54 kg/s Therefore wash water to be added = m = 9.54 kg/s = 34.33 tons/hr Amount of affination green = (31.932 29.909) + 7.096 + 34.33 = 43.449 tons/hr Analysis of this affination green Water 38.127 Solids 5.322 % of dissolved solids = (5.322 43.449) 100 = 12.25 Amount of water to be removed in process to get a recycle of 75% dissolved solids = 43.449 (5.322 0.75) = 36.353 tons/hr

(8) Melting

Steam coils in the melter regulate the temperature of the melt and maintain it between 82 88oC. Assuming it as 86oC. Feed to the melter is 29.909 tons/hr and its temperature is 43oC. Water entering the melter is 14.5302 tons/hr (as calculated in material balance) and temperature of water is 30oC.

Cp of water at 30oC = 4.184 kJ/kgoK Cp of crystals = 4.19 2.35 0.9896 = 1.8644 kJ/kgoK The heat to be supplied to raise the temperature of mixture (melt) to 86oC is given by Q = [29.909 (1000/3600) 1.8644 (86 43)] + [14.5302 (1000/3600) 4.184 (86 30)] Therefore Q = 666.05 + 945.69 = 1611.74 kW

Assume steam is available at 1 atm for heating. 7KXV N-NJ

Hence flow rate of steam required = ms = Q 2256.9 = 0.7141 kg/s

(9) Crystallizer 2

Assume operation temperature and pressure same as that of crystallizer1. At this pressure, boiling point of water is 61.42oC. Sucrose solution from decolourizer = (28.4793 sucrose + 14.8403 water) = 43.3196 tons/hr = 12.03 kg/s

Weight % of sucrose = (28.4793 43.3196) 100 = 65.74 Cp of feed to crystallizer = Cpf = 4.19 2.35 0.6574 = 2.65 kJ/kg Therefore solids per 100 parts of water = (65.74 100) (100 65.74) = 192 Thus boiling point rise (BPR) = 4.5oC Assume the sucrose solution from decolourizer is at ambient temperature i.e. 30oC. Thus Tf = 30oC

Heat balance in crystallizer can be written as FhF + mss = LhL + VH --------------------------------(1) F = feed rate to crystallizer = 12.03 kg/s Cpf = 2.65 kJ/kgoK

Feed temperature = Tf = 30oC Taking base temperature as 0oC. But hF = Cpf (Tf 0) = 2.65 30 = 79.5 kJ/kg ms = steam rate required in kg/s Assume steam is available at same pressure as that of clarifier. Hence s = 2256.9 kJ/kg L = output of crystallizer = (26.4288 + 0.3461) (1000 3600) = (26.7749 1000) 3600 = 7.44 kg/s Cpl = 4.19 2.35 X Where X = fraction of solids in crystallizer output = 26.4288/26.7749 = 0.9871 Thus Cpl = 4.19 2.35 0.9871 = 1.87 kJ/kgoK Temperature of output of crystallizer = Tl = 61.42 + 4.5 = 65.92oC But hL = Cpl (Tl 0) = 1.87 65.92 = 123.27 kJ/kg V = amount of water evaporated in the crystallizer = (13.7376 1000) 3600 = 3.816 kg/s H = DW oC + (BPR Cpsteam) DW

oC = 2618 275.8 = 2342.2 kJ/kg Cp steam at 65.92oC = 1.928 kJ/kgoK Therefore H = 2342.2 + (4.5 1.928) = 2350.88 kJ/kgoK Now equation (1) becomes, (12.03 79.5) + (ms 2256.9) = (7.44 123.27) + (3.816 2350.88) Hence ms = 3.958 kg/s Steam to be supplied to the crystallizer is = ms = 3.958 kg/s

(10) Drying

To calculate the heating surface of air heater: The air heater generally consists of tubes, which are supplied with steam at about 5 kg/cm2. The condensate leaves at bottom. The heating surface of this small heat exchanger will be given by A = Q [U {T (t1 + t0)/2}] Where A heating surface of air heater (m2) Q quantity of heat to be transmitted in kcal/hr U heat transfer coefficient in kcal/(m2hroC) T temperature of saturated steam employed in oC. t0 temperature of cold air entering = 30oC t1 temperature of hot air leaving = 50oC

Assume steam is available at 5 kg/cm2 i.e. 490.5 kPa Therefore T = 152oC (from steam table) As suggested by Hugot, U = 10 kcal/(m2hroC) The quantity of heat Q to be transferred is calculated from the fact that it consists of three principle terms.

(I) The heat necessary to heat the weight of air is given by q1 = M C (t1 t0) M = amount of air entering = 7451.69 kg/hr C = specific heat of air = 0.24 kcal/kgoK Therefore q1 = 7451.69 0.24 (50 30) = 35768.11 kcal/hr

(II) The heat necessary to evaporate the water contained in the sugar is given by q2 = m [607 + 0.3 t1 t0] m = amount of water evaporated = 293.1 kg/hr Thus q2 = 293.1 [607 + 0.3 50 30] = 173515.2 kcal/hr The quantity q2 of heat correspond to that given up by the hot air between T| reached at the outlet from the air heater and t1, which has not been included in q1.

(III) The heat necessary to heat the vapour contained in the weight M of air, assumed saturated is given by q3 = M H0 C| (t1 t0) C | = specific heat of vapour = 0.475 kcal/kgoK H0 = it is expressed as kg of water per kg of air = 0.026 Therefore q3 = 7451.69 0.026 0.475 (50 30) = 1840 kcal/hr The heat lost to ambient air is taken into account by writing the expression for Q as follows: Q = 1.25 (q1 + q2 + q3) = 1.25 (35768.11 + 173515.2 + 1840) = 263.90 103 kcal/hr

Therefore heating surface of heat exchanger will be A = 263900 [10 {152 (30 + 50)/2}] = 235.63 m2 The steam consumption will be ms = Q ms = steam consumption for air heater (kg/hr) ODWHQW KHDW RI VWHDP XVHG NFDONJ

Therefore ms = 263900 500 = 527.8 kg/hr = 0.147 kg/s

1. QUADRUPLE EFFECT EVAPORATOR

From material balance, Feed to the first effect = F = 238 tons/hr = (238 1000) 3600 = 66.11 kg/s Fraction of solids in this feed = Xf = 0.146 Liquid output from the last effect = L4 = 57.12 tons/hr = (57.12 1000) 3600 = 15.865 kg/s Fraction of solids in this output = X4 = 0.6084 Therefore evaporator load = 66.11-15.865 = 50.245 kg/s Assume equal amount of water is vapourized in each effect. Therefore V1 = V2 = V3 = V4 = 50.245 4 = 12.5613 kg/s

Material balance in each effect is as follows:

I Effect

Liquid outlet from first effect = L1 = F V1 = 66.11 12.5613 = 53.5487 kg/s Now the solid balance is, F Xf = L1 X1 Therefore X1 = (66.11 0.146) 53.5487 = 0.1803 In this liquid outlet, the solids per 100 parts of water = (18.03 100) (100 18.03) = 22 Therefore boiling point rise1 = BPR1 = 0.5oC

II Effect

Liquid outlet from second effect = L2 = L1 V2 = 53.5487 12.5613 = 40.9874 kg/s Now the solid balance is, L1 X1 = L2 X2 Therefore X2 = (53.5487 0.1803) 40.9874 = 0.2355 In this liquid outlet, the solids per 100 parts of water = (23.55 100) (100 23.55) = 30.80 Therefore boiling point rise1 = BPR2 = 0.7oC

III Effect Liquid outlet from third effect = L3 = L2 V3 = 40.9874 12.5613 = 28.4261 kg/s Now the solid balance is, L2 X2 = L3 X3 Therefore X3 = (40.9874 0.2355) 28.4261 = 0.3396

In this liquid outlet, the solids per 100 parts of water = (33.96 100) (100 33.96) = 51.42 Therefore boiling point rise1 = BPR3 = 1.2oC

IV Effect

Liquid outlet from fourth effect = L4 = 15.865 kg/s Now the solid balance is, L3 X3 = L4 X4 Therefore X4 = 0.6084 In this liquid outlet, the solids per 100 parts of water = (60.84 100) (100 60.84) = 156.36 Therefore boiling point rise1 = BPR4 = 3.6oC

To calculate t

Steam is available to first effect at 9 psig. i.e. Ps = 9 + 14.695 = 23.695 psia But 1 psia = 6894.8 N/m2 Thus Ps = 23.695 6894.8 = 163.372 kN/m2 = 1.634 bar Therefore T1s = 387oK = 113.89oC

The pressure in the vapour space of the 4th effect is at a vacuum of 26 in Hg. i.e. P4 = 30 26 = 4 in Hg absolute = 1.96 psia = 13.56 kPa = 0.1356 bar Thus T5s = 325oK = 51.85oC Therefore t = T1s T5s = 113.89 51.85 = 62.04oC Effective t = t (BPR1 + BPR2 + BPR3 + BPR4) = 62.04 (0.5 + 0.7 + 1.2 + 3.6) = 56.04oC

To calculate t in each effect

Neglecting the sensible heat necessary to heat the feed to the boiling point, approximately all the latent heat of condensing steam appears as latent heat in the vapour. Hence q1 = q2 = q3 = q4 U1A1t1 = U2A2t2 = U3A3t3 = U4A4t4 Usually, the areas in all effects are equal. Therefore U1t1 = U2t2 = U3t3 = U4t4 According to Hugot, the overall heat transfer coefficients in each effect are given as follows:

Effect 1 2 3 4 U [Btu/hrft2oF] 400-500 275-375 200-275 125-150

Assuming overall heat transfer coefficient in each effect as follows:

Effect 1 2 3 4 U [Btu/hrft2oF] 450 325 250 140

U (w/m2oK) 2555 1845 1420 795

Therefore (t2t1) = U1U2 = 25551845 = 1.385 t2 = 1.385t1 t3t2 = U2U3 = 18451420 = 1.299 t3 = 1.299t2 = 1.2991.385t1 t4t3 = U3U4 = 1420795 = 1.7862 t4 = 1.7862t3 = 1.78621.2991.385t1 But t1+t2+t3+t4 = 56.04 t1 (1 + 1.385 + 1.299 1.385 + 1.299 1.385 1.7862) = 56.04 Therefore t1 = 7.5753 = 7.58 t2 = 10.4918 = 10.49 t3 = 13.6289 = 13.63 t4 = 24.3439 = 24.34 56.04

Boiling point of solution in each effect

To calculate the actual boiling point of the solution in each effect

I Effect

Boiling point of solution = t1 = T1s t1 Where T1s = saturated temperature of steam to first effect = 113.89oC Therefore t1 = 113.89 7.58 = 106.31oC

II Effect

T2s = saturated temperature of steam to second effect = t1 BPR1 = 106.31 0.5 = 105.81oC Boiling point of solution = t2 = T2s t2 = 105.81 10.49 = 95.32oC

III Effect T3s = saturated temperature of steam to third effect = t2 BPR2 = 95.32 0.7 = 94.62oC Boiling point of solution = t3 = T3s t3 = 94.62 13.63

= 80.99oC

IV Effect T4s = saturated temperature of steam to fourth effect = t3 BPR3 = 80.99 1.2 = 79.79oC Boiling point of solution = t4 = T4s t4 = 79.79 24.34 = 55.45oC T5s = saturated temperature of steam to condenser = t4 BPR4 = 55.45 3.6 = 51.85oC

Effect 1 Effect 2 Effect 3 Effect 4 Condenser T1s = 113.89oC T2s = 105.81oC T3s = 94.62oC T4s = 79.79oC T5s = 51.85oC t1 = 106.31oC t2 = 95.32oC t3 = 80.99oC t4 = 55.45oC

Heat Balance

Heat capacity of sugar solution The heat capacity of sugar solution in each effect is calculated from equation Cp = 4.19 - 2.35 X (kJ/kgoK)

Weight fraction of dissolved solids Heat capacity (kJ/kgoK) Xf = 0.146 Cpf = 3.8469

X1 = 0.1803 Cp1 = 3.7663 X2 = 0.2355 Cp2 = 3.6366 X3 = 0.3396 Cp3 = 3.3919 X4 = 0.6084 Cp4 = 2.7603

Enthalpy data

Temperature (oC) Enthalpy of saturated water vapour (Hs kJ/kg)

Enthalpy of saturated liquid water (hs kJ/kg)

T1s = 113.89 H1s = 2698.0 h1s = 479.8 T2s = 105.81 H2s = 2638.8 h2s = 438.8 T3s = 94.62 H3s = 2667.5 h3s = 396.7 T4s = 79.79 H4s = 2643.0 h4s = 329.7 T5s = 51.85 H5s = 2595.0 h5s = 217.0

Heat capacity of steam in each effect (from steam table)

Temperature in each effect (oC) Specific heat of steam (kJ/kgoK) t1 = 106.31 2.0480 t2 = 95.32 2.0090 t3 = 80.99 1.9630 t4 = 55.45 1.9075

I Effect

The heat balance for first effect is as follows: S1s + Fhf = V1H1 + L1h1 1s = H1s h1s = 2698.0 479.8 = 2218.2 kJ/kg F = 66.11 kg/s Taking reference temperature as 0oC and feed to the first effect is at 96oC. Thus Tf = 96oC hf = Cpf (Tf 0) = 3.8469 96 = 368.504 kJ/kg H1 = H2s + Cp BPR1 Where Cp is heat capacity of steam. H1 = 2638.8 + 2.048 0.5 = 2684.82 kJ/kg h1 = Cp1 t1 = 3.7663 106.31 = 400.4 kJ/kg L1 = 66.11 V1 Therefore S 2218.2 + 66.11 368.504 = V1 2684.82 + (66.11 V1) 400.4 2218.2 S 2284.42 V1 = 2042.53 S 1.0299 V1 = 0.9208 ------------------------------------------------------------------(1)

II Effect

The heat balance for second effect is as follows: V12s+ L1h1 = V2H2 + L2h2 2s = H1 h2s = 2684.82 438.8 = 2246.02 kJ/kg H2 = H3s + Cp BPR2 Where Cp is heat capacity of steam. H2 = 2667.5 + 2.009 0.7 = 2668.91 kJ/kg h2 = Cp2 t2 = 3.6366 95.32 = 346.641 kJ/kg L2 = 66.11 V1 V2 Therefore V1 2246.02 + (66.11 V1) 400.4 = V2 2668.91 + (66.11 V1 V2) 346.641 V1 1.0591 V2 = 1.6212 -------------------------------------------------------------------(2)

III Effect

The heat balance for third effect is as follows: V23s+ L2h2 = V3H3 + L3h3 3s = H2 h3s = 2668.91 396.7 = 2272.21 kJ/kg H3 = H4s + Cp BPR3 Where Cp is heat capacity of steam. H3 = 2643.0 + 1.963 1.2 = 2645.36 kJ/kg h3 = Cp3 t3 = 3.3919 80.99 = 274.71 kJ/kg L3 = 66.11 V1 V2 V3 Therefore V2 2272.21 + (66.11 V1 92) 346.641 = V3 2645.36 + (66.11 91 92 93) 274.71

V3 = 2.0059 91 + 0.9281 V2 ------------------------------------------------(3)

IV Effect

The heat balance for third effect is as follows: V34s+ L3h3 = V4H4 + L4h4 4s = H3 K4s = 2645.36 N-NJ H4 = H5s + Cp BPR4 Where Cp is heat capacity of steam. H4 = 2595.0 + 1.9075 3.6 = 2601.867 kJ/kg h4 = Cp4 t4 = 2.7603 55.45 = 153.06 kJ/kg L4 = 66.11 91 92 93 94 Therefore V3 2315.66 + (66.11 91 92 93) 274.71 = 2601.867 V4 + (66.11 91 92 V3 94) 153.06 V4 = 3.2842 91 92 + 0.8960 V3 --------------------------------(4)

But V1 + V2 + V3 + V4 = 50.245 -----------------------------------------------------------(5)

By solving equations (1) to (5), we will get

V1 = 11.3015 kg/s V2 = 12.2015 kg/s V3 = 12.9880 kg/s V4 = 13.7540 kg/s S = 12.5602 kg/s

To calculate areas in each effect

q1 = S 1s = 12.5602 2218.2 = 27861.04 kW q2 = V1 2s = 11.3015 2246.02 = 25383.4 kW q3 = V2 3s = 12.2015 2272.21 = 27724.4 kW q4 = V3 4s = 12.9880 2315.66 = 30075.79 kW

A1 = q1 (U1 t1) = (27861.04 103) (2555 7.58) = 1438.59 m2 A2 = q2 (U2 t2) = (25383.4 103) (1845 10.49) = 1311.53 m2 A3 = q3 (U3 t3) = (27724.4 103) (1420 13.63) = 1432.44 m2 A4 = q4 (U4 t4) = (30075.79 103) (795 24.34) = 1554.28 m2

The mean area = Am = (A1 + A2 + A3 + A4) 4 = 1434 m2

TRAIL 2

Now for this trail, taking the calculated values of vapour evaporated in each effect in last trail as follows: V1 = 11.3015 kg/s

V2 = 12.2015 kg/s V3 = 12.9880 kg/s V4 = 13.7540 kg/s Material balance in each effect is as follows:

I Effect

Liquid outlet from first effect = L1 = F V1 = 66.11 11.3015 = 54.8085 kg/s Now the solid balance is, F Xf = L1 X1 Therefore X1 = (66.11 0.146) 54.8085 = 0.1761 In this liquid outlet, the solids per 100 parts of water = (17.61 100) (100 17.61) = 21.37 Therefore boiling point rise = BPR1 = 0.5oC

II Effect

Liquid outlet from second effect = L2 = L1 V2 = 54.8085 12.2015 = 42.6070 kg/s Now the solid balance is, L1 X1 = L2 X2 Therefore X2 = (54.8085 0.1761) 42.6070 = 0.2265 In this liquid outlet, the solids per 100 parts of water = (22.65 100) (100 22.65) = 29.28 Therefore boiling point rise = BPR2 = 0.7oC III Effect

Liquid outlet from third effect = L3 = L2 V3 = 42.6070 12.9880 = 29.6190 kg/s Now the solid balance is, L2 X2 = L3 X3 Therefore X3 = (42.6070 0.2265) 29.6190 = 0.3259 In this liquid outlet, the solids per 100 parts of water = (32.59 100) (100 32.59) = 48.35 Therefore boiling point rise = BPR3 = 1.15oC

IV Effect Liquid outlet from fourth effect = L4 = 15.865 kg/s Now the solid balance is, L3 X3 = L4 X4 Therefore X4 = 0.6084 In this liquid outlet, the solids per 100 parts of water = (60.84 100) (100 60.84) = 156.36 Therefore boiling point rise = BPR4 = 3.6oC

To calculate t

Steam is available to first effect at 9 psig. i.e. Ps = 9 + 14.695 = 23.695 psia But 1 psia = 6894.8 N/m2 Thus Ps = 23.695 6894.8 = 163.372 kN/m2 = 1.634 bar Therefore T1s = 387oK = 113.89oC

The pressure in the vapour space of the 4th effect is at a vacuum of 26 in Hg. i.e. P4 = 30 26 = 4 in Hg absolute = 1.96 psia = 13.56 kPa = 0.1356 bar Thus T5s = 325oK = 51.85oC Therefore t = T1s T5s = 113.89 51.85 = 62.04oC Effective t = t (BPR1 + BPR2 + BPR3 + BPR4) = 62.04 (0.5 + 0.7 + 1.15 + 3.6) = 56.09oC To calculate W LQ HDFK HIIHFW

W1| =

t1 A1) Am = (7.58 1438.59) 1434 = 7.60oC W2

| = (

t2 A2) Am = (10.49 1311.53) 1434 = 9.59oC W3

| = (

t3 A3) Am = (13.63 1432.44) 1434 = 13.61oC W4

| = (

t4 A4) Am = (24.34 1554.28) 1434 = 26.38oC But

t =

t1| + t2| + t3| + t4| = 56.09oC Let us take

t1| = 7.52oC

t2| = 9.00oC

t3| = 13.48oC

t4| = 26.09oC 56.09oC

Boiling point of solution in each effect

To calculate the actual boiling point of the solution in each effect

I Effect

Boiling point of solution = t1 = T1s W1| Where T1s = saturated temperature of steam to first effect = 113.89oC Therefore t1 = 113.89 oC

II Effect

T2s = saturated temperature of steam to second effect = t1 %351 = 106.37

= 105.87oC Boiling point of solution = t2 = T2s W2| = 105.87

= 96.87oC III Effect

T3s = saturated temperature of steam to third effect = t2 %352 = 96.87

= 96.17oC Boiling point of solution = t3 = T3s W3| = 96.17

= 82.69oC

IV Effect T4s = saturated temperature of steam to fourth effect = t3 %353 = 82.69

= 81.54oC Boiling point of solution = t4 = T4s W4| = 81.54

= 55.45oC T5s = saturated temperature of steam to condenser = t4 %354 = 55.45

= 51.85oC Effect 1 Effect 2 Effect 3 Effect 4 Condenser

T1s = 113.89oC T2s = 105.87oC T3s = 96.17oC T4s = 81.54oC T5s = 51.85oC t1 = 106.37oC t2 = 96.87oC t3 = 82.69oC t4 = 55.45oC

Heat Balance

Heat capacity of sugar solution The heat capacity of sugar solution in each effect is calculated from equation

Cp = 4.19 ; (kJ/kgoK)

Weight fraction of dissolved solids Heat capacity (kJ/kgoK) Xf = 0.146 Cpf = 3.8469

X1 = 0.1761 Cp1 = 3.7662 X2 = 0.2265 Cp2 = 3.6577 X3 = 0.3259 Cp3 = 3.4241 X4 = 0.6084 Cp4 = 2.7603

Enthalpy data

Temperature (oC) Enthalpy of saturated water vapour (Hs kJ/kg)

Enthalpy of saturated liquid water (hs kJ/kg)

T1s = 113.89 H1s = 2698.0 h1s = 479.8 T2s = 105.87 H2s = 2685.5 h2s = 443.6 T3s = 96.17 H3s = 2669.3 h3s = 401.8 T4s = 81.54 H4s = 2647.0 h4s = 342.7 T5s = 51.85 H5s = 2595.0 h5s = 217.0

Heat capacity of steam in each effect (from steam table)

Temperature in each effect (oC) Specific heat of steam (kJ/kgoK) t1 = 106.37 2.0550 t2 = 96.87 2.0170 t3 = 82.69 1.9710 t4 = 55.45 1.9100

I Effect

The heat balance for first effect is as follows: S1s + Fhf = V1H1 + L1h1 1s = H1s K1s = 2698.0 N-NJ F = 66.11 kg/s Taking reference temperature as 0oC and feed to the first effect is at 96oC. Thus Tf = 96oC hf = Cpf (Tf N-NJ H1 = H2s + Cp BPR1 Where Cp is heat capacity of steam. H1 = 2685.5 + 2.055 0.5 = 2686.53 kJ/kg h1 = Cp1 t