• Author / Uploaded
• priyaanka

Citation preview

Drying is the operation by which moisture is removed in relatively smallquantities from solids ot soidlike materials by evaporation. Most solid materialsrequire drying at somestageduringtheirmanufacture. In process industries, drying generallyfollowsoperationssuch asevaporation, filtration or crystallisation. Itis one of the last steps and iscarried outjust abead of packaging and despatch ofthe product. Drying is different from the removal of water from wet sotids by mechanical means such as squeczing, pressing, centrifugation, decantation, filtration,et. ‘Though dehumidification, adsorptionand evaporation processes can be employed for the removal of moisture, our definition of ‘drying’ exclude them from treating as ‘drying operations’. In dehumidification, water vapour present in a gas stream is removed bbycondensation; adsorption employs solid adsorbent such as silica gel or activated alumina to remove ‘water vapour from gas streams. Drying differs from evaporation operation in which water is removed ‘by vaporisation from solutions,in the following respects: |, Inevaporation, water is vaporised in large quantities from a solution whereas in drying, water is removed from solids in relatively smaller quantities. 2. Evaporation occurs at the boiling point of the solution whereas drying can be carried ont at relatively low temperatures and the moisture is removed by vaporisation at temperatures well below the boiling point ofwater. 3, ‘The vapour leaving an evaporator is generally pure water vapour unless itis contaminated by ‘unavoidable leakages the unit. On theother hand,drying operations frequently employ cartier gas such as air or flue gas for removing water vapour. ‘The liquid content ofthe dried product depends on the type ofthe materials, Dried table salt contains 0.5% water, coal about 4% and many food products about 5%. For products such as paper, timber, pharmaceuticals,solid catalysts etc. drying is anintegral part ofthe process, whereasfor products such. as sonp, detergents, dyestuffsfertilizers, coal, salt, sugar, milk,etc. removal of moisture is important for various reasons. Reducing the moisture content reduces the cost of transport of the material. The product must be made suitable for subsequent processing or sale, Matetial should have a certain ‘moisture content for processing, pelletizing or moulding. Its highly desirable that the materials have easy-to-handlepropertieslike the “free flowing’natureofsalt. Mostpharmacuticals and finechemicals require drying to a suitable moisture content to facilitate satisfactory packaging. Removal ofmoisture from the process gas streams such as coal gas, gaseous fuels and benzene is necessary as moisture present in the gas streams may lead to corrosion of process equipment. 667

668 ml Mass Transfer: Theory and Appiications

“Micro-organisms responsible for food spoilage cannot survive, grow and multiplyin the absence of water and drying is therefore used as a preservation technique for food and biological materials. [Enzymatic action that cause certain chemical changes in food and biological materials can be prevented byreducing the moisture content by drying. It is also necessary to remove moisture from food stulTs for preserving their flavour and nutritional values. Removal ofliquid contaminants such as benzene and organic solvents from solids can also be treated as ‘drying’ and the theory and equipmentdiscussed here forthe removal of water from the solids ‘are applicable to such operations as well, with necessary modifications. 9.1. DRYING EQUILIBRIUM

Since the quality of the dried product largely depends on the operating conditions prevailing in the dryer, the choice of the drying conditions is very important in the design and operation of dryers. Contamination of the pharmaceutical products, damage to the crystalline solids, checking or warping of the wood, shrinkage as in the drying of paper, loss of flavour in the drying of fruits, case hardening asin the case ofclay and soap, etc. are some ofthe difficulties thatoccurduring drying and these should be avoided by proper choice of the conditions of drying, Considering the behaviour on drying, the solids may be classified into two broad categories: Category 1: Granular or crystalline solids. Inorganic solids generally belongtothis class. Moisture is contained within the interstices ofparticles or in shallow open surface pores. The solidis relatively ‘unaffected by the presence ofthe liquid and by the dryingprocess. These materials can be dried rapidly to very low moisture contents. The drying conditions can be chosen based on our convenience and economic advantage, asthe final properties and appearance ofthe dried product are not affected by the choice. Examples of this class of materials are crushed rock, titanium dioxide, sand, catalysts, zine sulphate monohydrate and sodium phosphates. Category 2: Amorphous, fibrous or gel like materials. Organic sotids of plant and animal origin which inchide wood, leather, soap, cotton, starch, wool, etc.belong to this class. Moisture is either dissolved by such materials or it exists as trapped in the fibres or very fine pores of the solid. Since moisture is present as an integral part of the solid structure, the moisture removal affects the structure ‘The surface layers dry more rapidlythan the interior of the solid.Ifthe rate ofmoisture removalis very high,the difference in moisture content through the solid, lead to warping orcracking. Case hardening, that is, the formation of an imperviousshell of partly dry solid which prevents moisture removal from the interior, is another problem associated with the drying ofcertain amorphous materials. The drying conditions should be chosen with care as these determine the properties of the dried product. Before we consider the drying behaviour and the drying equilibrium, we should be familiar with the ‘method of expressing the moisture concentration in the solid. Moisture Content

Generally the moisture content of a solid is expressed as weight percent on a wet basis, whichis the kg of water contained in 100 kg of the wet solid. Thus of water in solid 100 ‘Weight percent (wet basis) = weight ‘weight of wet solid "The weight fraction denoted by wis the amount of water present in unit weight of the wet solid. ‘Therefore,the weight percent on wet basis = w x 100. In the design and analysis of drying operations, it is more convenient to express the moisture concentration on dry basis because, only the moisture is exchanged between the solid and the gas phases, and the mass of dry solid and the mass ofdry gas undengo no change. The weight percent of

Drying m 669 moisture on dry bassisthe kilograms ofmoisture associated with 100 kg dry solid. Moisture content ‘on dry basis is denoted by X, which is defined as the weight ofwater in kg associated with one kg dry solid, X is in fact the weight ratio of water in the wet soli, weight ofwater the soli weight of dry soli

Weight percent of water on dry basis is then Xx 100. If X’is the moisture content on dry basis. 1 kg dry solid is associated with X kg water o 1 + X kg wet solid contains X kg moisture. = eight of water in the solid __X weight of wet solid

x

Example9.1: A wet solid weighs 5Okg and contains 10g moisture. Express themoisture concentration fon wet and dry basis. ‘Solution: 10 kg moisture is presentin SO kg wet solid. The moisturecontent on the wet basis,is defined as the kg moisture present per one kg of wet solid. ‘Moisture content on wet basis, weight of water in the solid 10 _g _ke water (Answer) ‘weight of wetsolid SO” ‘kg wet solid Expressed as percent, the moisture content on the wet basis is 20% water, ‘Moisture content on dry basis is denoted by X and is defined as kg moisture present per kg dry solid, Since 10 kg water is associated with 40 kg dry solids, x _ Weight of water in the solid _ 10 _ (Answer) ~~~

weight of dry solid

40

Or, the solids contain 25% moisture on dry basis. ‘The weight percent on the wet basis is calculated as ‘Weight percent (wet basis) = weight of water inthe solid ‘weight of wet solid Equilibrium Moisture

Consider a material being dried in contact with air passed across its surface. Assume that the air is passed in sufficiently large quantities that the humidity and temperature of the air are not altered due to the heat and mass exchange with the wet solid. The moisture in the solid evaporates and diffuses to the air stream and the solid gets dried. This will continne till the moisture in the solid comes in ‘equilibrium with the water vapour present in the air.The drying will come to-a stop thoughthe material still contains significantquantitiesofmoisture.Themoisture inthesolidat thisconditionitheequilibrium ‘moisture content and is denoted by X*.The partial pressure exerted by the moisture in the solid under this condition equals the partial pressure of water vapourin the air. The solid material cannot be dried toa moisture content lower than X* by further exposure to the scune gas for an indefinitely long period. ‘The equilibrium moisture content for a given material isa function ofthe temperature, pressure and the relative humidity of the air in contact with the solid. If the wet material were contacted with air of lower relative humidity at the same temperature asabove,the equilibrium moisture content X" ofthe solid will take a lower value, Figure 9.1 shows the equilibrium moisture content of wood at 298 K plotted asa function ofthe petwent relative humidity (RED ofair. Percent relative humidity indicates the concentration of moisture in the air and is defined as the ratio ofthe partial pressure of water vapour to the vapour pressure at the same temperature expressed as percentage. Prom Fig. 9.1. it can be seen thatthe equilibrium moisture content ofwood incontact with air of 60% RH at 298 K is 0.113 kg waterikg dry solid and that with air of40% RH is0.075 kg water/kg dry solid. Since at equilibrium,the moisture

670 m Mass Transfer: Theory and Applications_ in the solid exerts a partial pressure which is equal to the partial pressure of water vapourin the air, Fig. 9.1 may be treated as the plot ofthe moisturecontent ofthe solid versusthe ratio ofpartial pressure exerted by the moisture in the solid at equilibrium to the vapour pressure of pure water at the given temperanute,

i

8

Relative humidity, %

100

°

000 005 010 015 020 025 030 035 X. hg moistuelhg dry solid Fig. 91: Equilibrium moisture content ofwood at 298 K

Relative hums, 6

est seesa es

Suppose that the wood initially contains 35% moisture on dry basis and the initial condition is represented by pointA in the figure. The moisture presentin the solid at this condition exerts a partial pressure whichis equal to the vapour pressure of water atthe given temperature. I is broughtin contact with the air of60% relative humidity. The partial pressure ofwaterin the aiis only 60% of the vapour pressure of the water at 298 K. Since the partial pressure of water in the air is Jess than that exerted by ‘waterin the wood, moisture is evaporated and transferred to the gas stream. The moisture content of the solid decreasesalong the curveABCD. However, whenthe moisturecontent fills tothat comesponding to point C, the vapour pressure exerted by the moisture in the solid equals the partial pressure of ‘water vapourin the air. The moisture content of the solid at this condition, 0.113 kg moisture/kg dry ‘wood, is X*, the equilibrium moisture content of wood when contacted with air of 60% RH. It is impossible to dry wood to a lower moisture 4o9, content by contacting with air at 298 K and LA (60% RH. To do so, airofa lower relative saturation should be brought in contact with the solid. For example, by using 40% RH air, we can dry the wood to X= 0.075 kg moisture/kg dry solid which is the equilibrium moisture content X* correspondingto the air of relative humidity 40%, represented by point C’ in the figure. Different materials exhibit different equilibrium behaviour as shown in Fig. 9.2. Forany : «given percent relative saturation, the equili000 005 010 015 020 025 030 035 brium moisture content varies greatly with the X-kg moisturerkgdry solid type of materials. Amorphous and fibrous Fig. 9.2: Equilibrium moisture content ofsolids at 298 K

Drying 671 materials of biological origin usually show large equilibrium moisture content (e.g. wool, leather and wood), whereas granular oF crystalline solids and inorganic materials showrelatively low equilibrium. moisture content (eg. sand, china clay, glass wool, kaoline). To dry a material to the bone dry condition, perfectly dry air (RH has to be brought in contact with the wet material. Equilibrium moisture content depends on the temperature and it decreases somewhat with increase in temperature. For example, for raw cotion in contact with air of RH = 50%, X" falls from 7.3 kg water/kg Ury solid at 311 K to 5.3 kg water/kg dry solid at 366.5 K. For a given species of solid, the equilibrium moisture may depend upon the panicle size or specific surface, if moisture is physically adsorbed, Free Moisture

Equilibrium moisture content sets the limit to which a material can be dried under specified set of conditions. Free moisture is the moisture in excess of the equilibrium moisture content X°,that can be removed in any drying operation, as shown in Fig. 9.1. Exposing the material whichisin equilibrium, with the gas of a given humidity and temperature for even an indefinitely Jong period will not cause an additional loss of moisture from the solid. For example,in drying of the wood containing intially 35, ‘weight% moisture on dry basis (X = 0.35) usingair of 60% RH at 298 K, the free moisture = X ~X" =0.35—0.113 =0.237kg/ke dry solid ‘That is,23.7 kg wateris removed from 100 kgofdry solids, which is nearly 68% ofthe water originally presentin the wet solid Bound and Unbound Water

Moisture in the solids may be the bound moisture, the unbound moisture or both. Referring to the drying equilibriumcurve for wood shown in Fig. 9.1, we sce that the moisture in the solid exerts the full vapour pressure of waterforall concentration greaterthan thatindicated by point in the figure, Such moisture,called the unbound moisture, behaves as free water, uninfluenced bythe solid. It is the ‘unbound moisture that is removed firs in any drying operation when wet materials containingsufficiently large amounts of waterare subjected todrying.The removal of unbound water from a wetsolid occurs by the same mechanism as the evaporation of water from a free water surface. Its evidentfrom Fig. 9.1 that only unbound moisture can be removed when the solid is in contact with saturated air, Bound moisture exerts a partial pressure less than the vapour pressure of pure water, For moisture concentrations less than that shown by point B in Fig. 9.1, the partial pressure exerted the moisture in the solid is less than the vapour pressure of pure water. The bound water can remain in equilibrium with the unsaturated air. It can exist in the solid under several conditions. The water held by fine capillariesin the solid or contained within the cell walls of biological materials, water in physical and chemical combination with the natural organic materials, water present as a liquid solution of soluble constituents of the solid, etc. constitute bound moisture.Lowering ofthe vapour pressure may be due to the concave curvature of the surface when water is held byfine capillaries or due to dissolution of soluble constituents when wateris contained withinthe cell or fibre walls. Solids which contain bound ‘water are called hygroscopic substances. The moisture content corresponding to point B in Fig. 9.1 is the highest moisture content ofthe solid that can exist in equilibrium with moist unsaturated air, and for cellular materials such as wood, itis often referred to as thefibre saturation point.

Hysteresis

‘The solid may attain the equilibrium moisturecontenteitherby sorption (condensation) orby desorption evaporation) ofwater. If the original moisturecontentofthe solid which was exposed to an air stream is less than the equilibrium moisture content X", the material will gain moisture from the airtill the

(672 M Mass Transfer: Theory and Applications

‘moisture concentration in the solid reaches the equilibrium value. In this case,the material gains water by somption. In drying of a wet material on the other hand, the material loses moisture by desorption, Generally, the equilibrium moisture content sttained by a solid is independent of the direction of approach to the equilibrium state. However, the equilibrium moistuse content of certain materials depends upon the direction in which the equilibrium is attained. If the material exhibits different equilibrium moisture contents for sorption and desorption, the material is said to exhibit hysteresis. ‘A possible explanation for hysteresis can be provided by assuming thatthe solid consists of necked capillaries such as the ones shown in Fig. 9.3. The material attains equilibrium by desorption in Fig. 9.3 and by adsorption in Fig. 9.3, The similrityin the curvature of the liquid meniscus leads to identical vapour pressure ofthe liquid even when the amount of ‘moisture contained in the pores are different. Desorption ‘Adsorption Figure 9.4 shows the equilibrium behaviour of (Oryng} (Weting) a solid exhibiting hysteresis. Sulphite pulp is (@) ©) aan example of a material exhibiting hysteresis. Fig. 9.3: Retationsh ystresis and pore For a given temperature and relative bumidity ——stmiture far, the equilibrium moisture content reached by desorption (X;) is generally greater than

hat reached by adsorption (Xx). The dry material gains moisture along the adsorption curve and wet material looses moisture along the desorption curve. ‘Toprevent the moisture regain of the dried product when exposed to moist airit would be necessary to store the materialin air-tight containers, immediately after drying. For material showing hysteresis behaviour, if the Aried product is likely to be exposed to air of certain relative humility, drying the material to 4 moisture content below the adsorption equilibrium value, is of no use. Soluble Solids

‘Water of hydration of inorganic crystals is an example of the bound moisture held chemically by the solids. The hydrated form obtained on drying depends on the humidity as well as the temperature of the drying gas. Consider for example, a solution of CuSO, exposed to the air at 298 K. For partial pressure of water vapour in the air exceeding 3.08 kPa, the salt remains completely dissolved in the water(Fig. 9.5). If the partial pressure of water vapour in the air is between 1.04 kPa and 3.08 kPa, CuSO,.5H,O is produced. At 298K, the stable hycirate is the pentahydrate of CnSO,. If the partial pressure is 3.08 kPa, a mixture of pentahydrate and saturated solution can coexist and at a partial pressure of 1.04 kPa, a mixture of penta and trihydrates of any proportion can coexist. For a partial pressure ofwater vapourin the air between 1 10 Pa and750 Pa, itis the monohydrate (CuSO,.H,O)that is formed and for partial pressures below 110 Pa, dry anhydrous CuSO,crystals are formed on exposure to air at 298 K. The equilibrium moisture content falls to zero if bone-dry air is used.

Drying @ 673

o1 3 306 Equitbrium motszure contont, mel watorimol CuSO, Fig. 9.5: Equiibeium moisture content ofCuSO,at 298 K

Figure 9.6 shows theequilibriummoisture contentofsodium nitrateat 298 K.Consideran unsaturated solution ofNaNO,containing, say, 2kg water per kg NaNO,represented by pointA in the figure, Water in the solution exerts a partial pressure, which is less than the vapour pressure of pure water at 298 K. Suppose this solution is exposed to air containing water vapour with a partial pressure of say, 2 kPa, Moisture evaporates and the concentration of water in the solution falls along the curve AB. At point B, the solution becomes saturated. ‘The partial pressure exerted by the water in the solution is about 2.35 kPa. On further exposure of this solution with relatively dry air, evaporation of water

sen

© soo

jzs00 :

vale nendnaboae pee —8 drreease jee

T]7—~}sotuated sctton

& 2000 fo—+

}

se

soe nano, |

-

1 0, a 1 2 3s & S Equitorium moisture content, «a waterkg NaNO, Fig. 9.46: Bquiibeium moistore contentof NaNOat 298 K (notto scale)

continues along the line BC with erystallisation of anhydrous NaNOs, The partial pressure remains constant on this line where the saturated solution and solid crystals coexist. At point C crystallisation is completed and the solidis almost free of water, but for the small traces adsorbed on the surface. On furthercontact with air with partial pressure ofwater2 kPa, the moisture contentfallstothe equilibrium value corresponding to point D. We see that the solids which are soluble in water show insignificant equilibrium moisture content when contacted with air in which the partial pressure of moisture is less than the vapour pressure exerted by a saturated solution of the solid in water.

674 @ Mass Transfer: Theory and Applications

Example 9.2: Fibres with an initial total water content of20% on wet basis are dried in a tunnel dryer operating at 101.3 kPa. Ifthe products leaving the dryer are in equilibrium with air of relative humidity of 50% at 298 K,determine the kg of moisture evaporated per kg of bone- Xc and X; < Xe, the drying involves both constantand fallingrate periods. Since the falling-rate period is linear in moisture content, Eq. (9.22) can be used for estimating the drying time Re ee ye ccs 083080 X, kg moisurokkg dy sold 0= 66 +0, = (X, - Xe) +(Xe XI al ARe Fig. 943: Rate of drying curve for Example 9.6 Bone-dry weight of the solids = 5 = volume of sample x bone-dry density of sol $= (15x12 0.05) x 1200 = 108 kg. Material is dried from both sides. The area exposed to drying is therefore, A=2X(1.5X1.2)=3.6 m? ‘The constantrate of drying Re kg/m? h. Substitute the data in Eq, (9.22). 108 500 0.33)+(0.33-0)In0. Sox (Answer) Example 9.7: I¢ takes 9 hours for a porous solid to reduce the moisture content from 45 to 10% when dried in a batch dryer under constant drying conditions. The critical moisture content was found to be 25% and the equilibrium moisture 3%. All moisture contents are on the dry basis. Assuming that the rate ofdrying during the falling rate period is proportionalto the free moisturecontent,how long should it take to dry a sample of the same solid from 35 to 5% under the same drying conditions? Solution: Case 1 = 045 kg water/kg dry solid; Xz 0.10 kg water/kg dry solid; X- = 0.25 kg water/kg dry solid; x .03 kg, water/kg dry solid and @ = 9 h. Since drying takes place with both constant-ate and falling-rate, atthe falling rate period varies linearly with free moisture content. the drying time calculated using Bq, (9.22) Xo-X" = Xe) (Xe > X" In X,-xX"

Substinutng the given values in Eq.(9.22),

s (45-025) +(0.25-0.03)1n 225008 025-003 =| 8 ael® S294 pn s 904519

a)

c Case 2 ‘The critical moisture content and equilibrium moisture content are the same as in case 1. That is, 125 kg waterikg dry solid: X" 0.03 kg water/kg dry solid. 35 kg water/kg dry solid; X, 005 kg water/kg dry solid: let @ be the time of drying.

Drying m 709

Substituting the values in Eq. (9.22) gives

s -( (0.3559s 0.25-0.03 125 Jen 8 =] 35-0.25)-+(0.25-0.03)In=003)

s o- 0027s5 ® Divide Eq. (B) by Eg. (A): 8 06275 © 9” 0519 Solving Eq, (C), we get @= 12.50b, Time required for drying the second sample is 12.5 h (Answer) Example 98: A traydryer consisting of 12 ways i used to dry raw cotton from 93% moisture to 10% moisture by ross circulation of ar. The trays which are insulated atthe bottom are each 30mm high With an area of 1.8 m? and are completely filled with raw colon which has a bone dry density of 700 kg/m*. The heating medium, which is 30% saturated ar t 344 K and 101.3 KPa, Dowsacross the top surface of the tay at a mass velocity of 12000 p/h m*. Under the piven conditions, the eitical ‘moisture contentwillbe 0.4 kg watertkg hone-dry cotton andthatthe falling rae cryingperiod is near: Equilibrium moisture content is 0035 kg water/kg bone-dry cotton.All moisture contents are on dry basis, Determine: (a) ‘The amount ofraw cotton inkgthat can be dried in one batch, = (b) The drying-time forthe constant-rate perio. (6) The drying time for the falling-rate period. (2) The tol drying ime, Solution: (a) The amount of dry cotton in one tay S volume x bone-dry density = (1.8 x 30x 10") x 700 = 37.8 kg. Since the dryer consists of12 trays, dry solids that can be handled in one batch = 1237.8 = 453.6 hg. Since the raw cotton contains 93% moisture on dry bass, total weightof raw cotton that can be dried in one batch= 453.6422 = 975.45 ke

(©) Drying time in the constant rate period is given by Eq.(9.18)

(Answer)

Sy Kod Bem FRMi

‘The constant rate of drying is estimated using Bq, (9.4)

hey -T) Rox = ih, -Ts For cross-circulation drying, the heat transfer coefficient (Wm? K) may be estimated using Ing, = 0.02046"

where G is the mass velocity of air kg/m? h.

ng, = 0.0208(12000} ForairatT= 344 K with a percent humidity of 30%, the wet-bulb temperature Tyi obtained from psychrometric chat, Ty = 325.7 K. The temperature at the surface 7; s the same as Ty. T;= 325.7 K.The latent heat of vaporisation of water at 325.7 K is read from steam tables, which tiveshy hy = 2376 Kitkg. Now,

710 @ Moss Transfer: Theory and Applications

3741360010" (344-325.7) = 1.04ke/m?h 2376 Initial moisture content X, = 0.93; the critical moisture content X,-= 0.40; area exposed to drying A=1.8 m¥, Bone-dry weightof the material in the tray S = 37.8 kg.Substituting in eq. (9.18),0¢ the rying time in the constant rate drying is obtained. s Xe Te, 378 pg (0.93 040)=10:70h 86 aR (Answer) (©) Timeof drying in the falling rate drying is @ , which is given by eq. (9.19). S(Xe= XN) Xe =X Sie=X),,

RA

a Final moisture contentX; 0.10; Equilibrium moisturecontent X=0.035, Substitutingin Eq,(9.19), 37.80.40 - 0.035), 0.40-0.035 1, 6 Totxis o.o-0035 */271™ (uum) (d) The toualdrying time is obtained es 0 = 8+0, = 10.70 412.71 = 2341h, (Answer) Effect of Conduction and Radiation on Drying Rate ‘The equations presented earlier for the rate of drying under the constant drying conditions assume that the heat transfer by conduction through the metallic wall, and heat radiation from the hot surfaces to the material being dried are negligible compared to the heat transferred by convection from the hot gases. Now consider wet material takenon a tray subjected to cross circulation drying as shownin Fig. 9.44 etal surface, Ty

Fig. 944: Crosscteulaton deying in the presence ofheat conduction, convection and radiation Jn the constant rate-period. the heat transferred to the drying surface (q) is balanced by the heat removed by the evaporation of moisture. q=Rehs

(9.23)

“leat is transferred wo the drying surfaces by convection (gc) radiation (gg)and by conduction (q). Therefore, 9=9c+9R*% +924) “The surface temperature remains constant at Tthroughoutthe constant rate period. The heat is transferred from the hot gases by convection, which is given by Ge = hgTe -Ts) (9.25)

Drying @ 71

Hotsurfacesatthe high temperature 7 emitsradiation and is received by thedrying surface at a rate ge. Now, . Gq =€(5.676 x 10 (Ty - TS)

(9.26)

x= hg(Ty -Ts)

9.27)

Here € is the emissivity of the solid. The heat transferred by radiation may be expressed in terms of ‘a heat transfer coefficient for radiation, fig as

pg= £158T6RIO * UTE -T3)4 (028) GT Forheat transfer by conduction to the drying surface, three resistances are involved—the resistance offered by the gas film atthe bottom surface ofthe try,the resistance offered by the tray material and in which,

the resistance offered by the substance being dried. Thus,

4 =UMTe -Ts) where U; isthe overall coefficient of heat transfer given as 1 Zy ZT" u,-[t+20.2s eh |

(9.29) oe. (9.30)

where Zyy ancl Z; are the thickness of the tray and wet material, respectively, and kiy ancl ky are the ‘thermal conductivity of the material of the tray and the wet solid, respectively. The rate ofdrying in the constant rate period can be written by combining Eqs. (9,23) through (9.30) and Eq. (9.2): Ree 4 he Ui 1Ts~T5)+helTa Tg, y, yy 931) hs iy , To use Eq, (9.31),the surface temperature T; should be known, In the absence of heat conduction and ridiation, 7, is equal to the wel-bulb temperature of the gas, which is equal to the adiabatic saturation temperature for the air-water systems. For the present case, the surface temperature will be sreaterthan the wet-bulbtemperanurebecause ofadditionalheat reachingthe drying surface by conduction and radiation. To estimate the surface temperature Ts, Bq. (9:31) can be putin a more convenient formas: 49.32)

‘The ratio hy; /ky forthe air-water vapour systems is approximately equal tothe humid heat C's. 0 that hig [ky = Cy = 1.005 +1.884¥ (9.33) where Cs isin ki/kg K. Since Ysis the saturation humidity at temperature 7, the surface temperature should be calculatedbytrial using Eq. (9.32) and the saturation humidity curve on the psychro‘metric chart Example 9.9: A layer of wet granular material 25 mm thick is taken in a pan and is being dried. The pan has «1.5 mm thick metal bottomhavingathermal conductivity 50W/m K.The thermal conductivity of the solid in the pan is 1.2 W/m K. Air at a temperatureof 333 K and humidity 0.01 kg water/kg dry air flows parallel to the top exposedsurface and the bottom metal surfaceat a velocity of4.0 mvs.Direct radiation of heat from the steam pipes having a surface temperature of 380 K falls on the exposed top surface, whose emissivity is 0.9. Estimate (a) The surface temperature (b) The drying rate for the constantrate period. Solution: (a) The surface temperature can be evaluated using Eq. (9.32)

712 m Mass Transfer: Theory and Appications -¥ias (ee -1,)+®_-T,) hgtky

hig)

*

hg

in which the humidity of air ¥ = 0.01 kg waterikg dry air.

hig [hy =C,= 1.005 +1884¥ = 1,005 + 1.884 x 0.01 = 1.0238 Lfkg K. ‘The convective heat transfer coefficient h;can be evaluated using Eq, (96)

Here,

hg =0.0204G"*

G where p and w are the density and velocity of air, respectively. u=4 mis Density of air is (1 + ¥) times the reciprocal of its humid volume.

paltt 4

101

sarax(29 00) 333 18 )*t013,

0547kg/m"

G 1.0547 x 4 = 4.2188 kg/m? s or (=15187.83 kg/m? h) Substituting in Eq. (9.6) ng

0.0204(15187.83)"*

162.60 kim? h K)

‘The overall coefficient of heat transfer from the air to the surface of the solid through the bottom. surface is given by Eq. (9.30):

welt]

[2.2 2s

Here, 2y = 1.5 mun,ky = 50 Win K, Zs = 25 mm, ks = 1.2 Wim K u | 1 bsx10 25x10 y = 23.254 W/m? K “(45.168

"50

12

In Eq, (9.32), To = 333 K, and Ty 380 K. Substituting the known values, Eq, (9:32) becomes sO (1+ 23254) =T,)+ 45.168 "8 280-T,) 1.0238 45.168 (0.9768 (Y~ 0.01)k= 1.5148(333 ~ 7, )+0.021380 Tehg “wy “The heat transfercoefficient for radiation is £(5.676%10*)(T¢ -T3) oassro(3 8

(Ty -Ts)

*

son (omss0t-(%-(i)

ms (380-75)

+

(G80=7,)

{7 y

(

@B)

In Eq, (A), As is the heat of vaporisation at the surface temperature 7; and Ys is the saturation humidity at Ts. Thus,2, Ys nd hyto be used in Eg.(9.32) forestimating 7s depend on T; and therefore Eq. (A)is to be solved for T; by trial and error. Assume arbitrarily a surface temperature T;(note that it should be greater than the wet-bulb-temperature ofthe aircirculated) and calculate hg using Eq. B).

Diying m 713 Substitute hy so calculated, 2. read from the steam tables and ¥ read from the psychrometric chart or calculated from the vapourpressure datafortheassumedtemperature 7, in Eq.(9.32) and calculate 7, If the assumed temperature and calculated Ts are identical, the assumed temperature is the correct choice for surfuce temperature. Otherwise, repeatthe calculation with a new assumed temperature.For example, assume T; = 306 K. Then As = 2423.6 ki/kg, ¥, = 0.0324 kg water’kg dry air. he

=8.3416W/m?K

Substinuting in Eq. (A), Ts = 306.9 K. Assumed TK Aw kg ‘Yoke watechke dry air. Wim K 306 223.6 0.032424 34164 307 24712 0.03439 8.375317 306.2 2423 0032817 8.348368

_Caleulated 7K_ 306.8573 304.1842 306.323

Now assume T, = 306.2 K. Then dy = 2423 kg, Ys = 0.0328 kg water/kg dry ai. \! s to3404 -(258) ) A

p=

mluea

=

8.3484 W/m? W/m? K

From Eq. (A), Ts = 306.3 K. This is reasonably close to the assumed value. Hence, the surface

temperature 7s 306.2 K is accepted. ‘The rate of drying in the constant rate period is estimated using Bq. (9.31). (lig +Us)Tg Ts) higTe —Ts) Roo Oe Re

(45.168 +23. 2:

As 5 x (380 = 306, 2) ) 2 3.64kg/m?h }) x 3.6% 3.6 (333 (3 3 — 306.2) + +8. 8.3484 x 3.6

2423

Rate of dryingis 3.64 kg/m? h.

Effect of Process Variables on Re

(Answer)

‘The rate of drying in the constant-rate period is evaluated by Eq. (9.4):

Rox kyle -¥)= "2, -T5) As

‘The heattransfer coefficient for parallel low of aie is given as (Bq. (9.6)) ng, = 0.02046"

and for perpendicular flow (Eq, (9-7)}

hg=117e2”

“The variables affecting the rate ofdrying such as the gas velocity, temperature and humidity ofthe gs and the thicknessofthe drying bed can now be predicted. Gas velocity Equations (9.4),(9.6) and (9.7) reveal that ifthe heat transfer by conduction and radiation through the bed of solidsis negligible, then the rate of dryingin the constant rate period is proportional to G"*for

714 @ Mass Transfer: Theory and Applications

parallel flow and G®"for perpendicular flow. Gas velocitywill not have much influence on the rate of dryingif heat transfer by radiation and conduction are also present. Temperature

‘The change in the drying rate with changes in the gas temperature can be estimated as follows. Let the original conditions of the air are T,,, Ts), ¥;, ¥s, and conditions at the second state are Toa,Tsp Ya, Yon ‘Then using Eq, (9.4), and assuming that the heat of vaporisation is independentof the temperature for ‘moderate changes in temperatures, Roy ky si —

A

x -Ts))

(9.34),

Rea=hy Wen ~Yo)= 92Tg : Te)

49.35)

en 2-2) gp, Tor-T) Rea Regy OT)

(9.36)

Letthe humidity be not changed. Y, = ¥; = ¥. Combining Eqs. (9.34) and (9.35),

radiation effects and the influenceof temperature on the heat of vaporisation2s are negligible,the rate of drying at constant rate period is directly proportional to the temperature difference (qT). By increasing the temperature ofthe gas stream, T,;~ Ts is increased and hence the rate of dryingincreases by increasing the temperature of the gas. Gas humidity

At constanttemperature T,, the humidity of the air used isdecreased, the wet-bulb temperature Tiy is also decreased, and by Eq. (9.4) the rate should increase. For this case, Eq. (9.36) can be written as )

-(937)

It should be noted that since the changes in T,; and involve simultaneous changes in Tand; Re should be estimated using Eq. (9.4). Thickness of the drying bed IF the heat transfer by conduction is negligible relative to that by convection, thickness ofthe solid bed has no influence on the rate of drying. On the other hand, if conduction heat transfer is important, increasing the thickness of the solid bed would result in the decrease ofthe drying rate. Atthe same the increase in the thickness will lead to an increase in the heat conduction through the edge surfaces,if edges are not insulated, and therefore,the rate of drying will increase.If the edge surfaces tre heat insulated and the heat conduction occurs from both faces, the rate of drying Re will be independentofthe bed thickness. However, the time required fordrying from a given initial to thefinal moisture content varies in direct proportion to thethickness, as the mass Sin the equationsforcalculating the time of drying is directly proportional to the thickness.

‘Shrinkage and Case Hardening ‘Some materials,particularly the colloidal and fibrous materials undergo shrinkage when the moisture is removed from them. The extent ofshrinkage is linearly elated to the change in the moisture content. For materials such as vegetables and fooxstuffs, shrinkage alters the extent of surface exposed 10 the drying gas per unit weight ofthe substance. Formaterials such as wood, shrinkage causes them to warp

Dying m 715

or check and change its structure.Shrinkage may lead t0 case hardening in certain materials such as, soap and clay. Case hardeningisthe formation ofa hardened and dry outershel, impervious to the flow of moisture, while the inside of the solid is still wet. In many food stuils, if drying occurs at a high temperature, a layer of closely packed shrunken cells, which are sealed together, is formed at the surface, hich acts asa bartier tothe flow of moisture. The moisture cannot move as liquid or as vapour, to the surface or tothe boundary at which the evaporation is taking place, For matetials that warp or check. or those which undergo case hardening,itis necessary to retard evaporation of the moisture and bring it in tune with the rate of moisture movement from the interior. ‘Thisis achieved by increasing the humidity ofthe air used fordrying. By supplyingrelatively moist air, Yin Eq, 9.4)is increased, the humidity difference (Y, — Yis decreased and the dryingrate is reduced. ‘Thus, wood dryers are supplied in the intial stages ofdrying with humidified air,to reduce the moisture gradient within the solid and hence to minimize the effect of shrinkage. 9.7 THEORETICAL ESTIMATION OF DRYING RATE AND TIME

Empirical approachof estimating the time of dryingpresented in the preceding sectionis applicable to

the particular conditionsfor which the data were obtained. Two mechanismsfor moisture migration

from the imterior of the solids have been proposed which can be used to estimate theoretically, the drying time and the moisture concentration profile in the solid. Liquid diffusion and capillary movement

are the two mechanisms proposed to explain the different types of falling rate curves, as already discussed. The mechanism of the moisture movementaffects the drying duringthe constant rate and the

falling rate periods Liquid Diffusion

|

Drying rate R

‘Thedifferencein the moisture concentration existing between the surface and the interior depths of the solid provides the driving force for diffusion of the moisture to the surface. This is the mechanism of moisture movement in nonporous solids in which the internal moisture is predominantly the bound ‘moisture, Liquiddiffusion prevailsin the last stages of drying ofclay,starch, lour, textiles, paper and ‘wood, and in the drying ofsoluble moisture as in soap, gelatine,glue and paste. In drying many food ‘materials, the movement of water in the falling rate period occurs by diffusion. ‘The moisture profile developed by the transport of moisture,is concave downward throughout shown Fig. 9.45a. The shape of the diffusion controlled rate curve is similar to that shown in 9.45b, Fick's second law for un-steady state diffusion can be written to determine the rate of drying in the falling-rate period.

Free moisture, X Distance from surface o @) Fig. 948: (a) Moise prosile in wet solid during drying when moisture flow is by liguid diffusion (b) Diffusion‘conolled flling-rate period

716 @ Mass Tronsfer: Theory and Applications ax

=(9.38)

where Dyy is the liquid diffusion coefficient and is the distance through the solid. During the diffusion-type drying, the resistance to mass transfer from the surface is usually very small, and the diffusionin the solid controls the rate ofdrying. Then the moisture content atthe surface can be assumed to betheequilibriumvalueX".Itsassumedthatintially auniform moisturecontentX,exists in the wet solid. AC the beginning ofthe falling rate period (0 = 0), the exposed facesare brought to the equilibrium moisture content X*, Equation (9.38) is integrated with the following boundary conditions X= Xp, a0 for-a

Dryingtimerequired in the actual dryi

|, 8(0.50- 0.10)

#7(0.25-0.10) 1289

289h

(Answer)

Example 9.11: The following data on the unaccomplished moisture change X-x versus timeis

Xe-X available for crying certain material of vegetable origin in the falling rate period. The materialis 4 nm thick and drying is from both sides under constant drying conditions. Oh (KXXe

NX}

@ LO

oF 02 03 04 085072062 053

05 06 045 038

07 08 032027

09 0.23

10 0.20

Determine the average diffusivity of moisture up to an unaccomplished moisture change of 0.25

720 @ Mass Transfer: Theory and Applications (a) Using the simptified equation assuming long drying times. (b) Using the general theoretical model. ‘Solution: The given data are plotted on a semi-logarithmic graph with (X—X"V(Xc — X") on the y-axis and time 6, on the a-axis. The result is the upper curve in Fig. 9.48.

1

RQ

_

an

: PKerrr 7

Theoretical

£

3

| a

Seon x 00

>

02

04 =«08 OB 10 aga Fig. 9.48: Analysisofdrying data ofExample 9.11

From this cre, for x =0.25, the time of drying @ is found to be 0.857 h. For long drying fc times the diffusion equation simplifies to Eq. (9.40)

X=" 8 Le Foon! ] mors

Substituting 6 = 0.857 h in the above equation, 025-5 [--eume2ay ]_ 8[,-vgoasaaisnioy ]_ 8[,-s106904] vr

x

r

Solving the above equation for Dag. we get Day 2.23 * 10° mth = 6.19 x 10°"mils (Answer) (b) The lower curve in Fig. 9.48 is the graphical representation of the diffusion Eq. (9.39) plotted using the following theoretical values:

Dyydia® (X-XVO%e-N)__

0 1

From the graph, for

0020S OBE__0.7S

CSS 0642 0563

CSS 0.496 0.387 0.236

' 0.069

= 0.25, the parameter D,,0/a’ 0.4769. « the drying time is found to be @ = 0.857h.

Substituting 0 = 0.857 h in D,,6/a?= 0.4769, we get

Day = 0.4769 x (2 x 10~)? 0.857 = 2.23 x 10% m* h= 6.19 x 10" ms

(Answer)

Drying @ 721 9.8 TRAY-DRYING WITH VARYING AIR CONDITIONS

‘Considerair being passed across a wet material taken in tray as shown in Fig. 9.49. The depth of the airflow space is 6 m. This is the distance between the material on the tray and the bottom surface of thetray above. The length of the tray is Lm. Air enters the space between the trays at temperature Ty ‘and humidity ¥; and leaves at temperature Tz» with humidity ¥;. ‘The air humidity and temperature change along the Jength ofthe «ay. Consider a small section of the tray of length dL. and of unit width. Let Gy be the flow rate of dry air per unit area of eross section, kg/mh. The temperature of the gas falls by dT; due to transfer of heat dg in this section. Writing a heat balance over this section, dq -Gy(b x Cyd 50) where Cisthe humid heat of the air. The rate equation for heat transfer front the gus is dq = hed * KT, =Ts) (951) Fig.949: Heatand material balance in atray dryer ‘where fig i the heat transfer coefficient and T, isthe surface temperature of the solid. Combining Eqs. (9.50) anwl (9.51), we get Gib XC Te = hy(LX INTe ~Ts) (9.52) a h le 0 Or Tg Te hey, Gybty (9.53) Integrating Fq.(9.53), we get

% jw -2- ejdL. BTonTy GsbCs 4

Tyr-Ty wnlar aTs

hh Me

954

(9.55)

Equation (9.

(956) so that

or

9.57)

(9.58) ‘Therate ofdrying in the constant rate period can be calculated from GC AB INT: ~Tea) RALX es

(9.59)

Or (9.60) whereAis the heat of vaporisation ofthe moisture at the surface temperature,Ts.The timeof drying in the constant rate period is given by Eq. (9.18) as

722 @ Mass Transfer: Theory and Applications

‘Substituting Eqs. (9.60)into Eq. (9.18), the time of drying can be written as __S(X, -Xo)Lies

AGCbg, ~To2)

(9.61)

where Sis the kg ofdry solid in the tray and A is the surface exposed to the drying gas. Noting that the

surface exposed to the drying gas is (wL) where w isthe width of the tay, Eq,(9.61) becomes S(X,

Ne dhs

o= GC0bTr Equation (9,62) can be rewritten using the Eq. (9.58) as o=

(9.62)

29.63 Oo)

GoCsbo (Tay

For the falling rate period, we can write an approximate equation for the drying rate using the relation (Eq, (9.19)] ® . SiXe-X),, Xe aX" RA

x-x"

Therefore, for the falling rate period, we get 8

S(Xe =XVy

7 hgh

bw(Ty, -Ts)] 1 exp) - "4

“ue

[

\ GshCs

}

Xo-X ne xox

9.9 THROUGH CIRCULATION DRYING IN PACKED BEDS

9.64 -964)

Suppose that the drying gas is passed through a bed ofwet granular particles as shown in Fig. 9.50. ‘Assume thatthe bed of solids is resting on screens. The gas admitted below the screen flowsthrough the open spaces and voids between the solid particles in the bed. Drying occurs with both constantrae and the falling rate, The evaporation of the unbound moisture occurs in a relatively narrow zone and the gas leaving this zone will be saturated with moisture.The drying in this zone is at & constant rate. ‘This zone moves upward in the direction ofthe gas flow leaving behind.a section ofthe bed in which evaporation of the bound moisture takes place at a lower rate. Therefore, in the initial stages of the drying, the bed consists ofthree zones, zone 1 in which the moisture being removed is bound moisture, zone 2 in which the unbound moisture is rentoved at constant rate and zone 3 in which the solid still contains the original moisture concentration, as shown in Fig. 9.50a. With passage of time, zone 2 ‘moves up increasing the thickness of zone 1 and reducing the thickness of zone3. Rate of Drying of UnboundMoisture Consider a differentia! section in the bed of thickness dz with unit area of cross section as shown in Fig. 9.50b. Let a be the interfacial surface per unit volume ofthe bed so that the total surface available in the differential section is adz. The gas receives moisture and its humidity increases by dY.If the mass flow rate of the dry gas is Gs kg/h m? of the bed cross section, then, Gy¥ = ky (adzX¥ —¥) (9.65)

Drying m 723

[ose

tev

zeaJanes

Toa, War

Zone FANG 2one 2

oP

Sune

ify

Fon

Tra

@)

o)

Fig. 90: Theosgh-circulation dyingin packed beds

Tris assumed that the solids artain a temperature equal to the adiabatic saturation temperature of the gas; the humidity attained by the gas at the interface between the gas and solid is Ys, the saturation humidityat the adiabatic saturation temperature. Rearranging Eq,(9.65) and integrating, we get de

(9.66)

Let Z be the thickness of the bed and ¥, and ¥, are the humidities of the gas entering and leaving the bed.

Yo-¥, kyaz ntsati ney, brat G,

(9.67)

Denoting the number of gas transfer units in the bed by Nj, the above cquation may be writtenas

Now

ten’ — brad

on G,

(9.68)

We can put Eq, (9.68) in a more convenient form as follows:

(9.69)

1

(9.70)

‘The constantrate of drying can be calculated as G0,a-¥,) Substting for (73 ¥;) fom Eq, (9.71), we get »_ Gs RcE Le = ASSL exp) } wg Equation (9.18) can be used to determine the drying time. aah X,-X, -Xe) |S_ muss of dry solid in the bed _ Zs _ Ps Bere,

A” Areaexposedtodrying

aZ

a

49.1) (9.72) .

(9.73) 7

(9.74)

724 @ Mass Transfer: Theory and Applications

where pisthe dry density ofthe bed, kg dry solids per unit volume of the bed. Substituting Eqs, (9.74) and (9.73) into (9.18), weget for the drying time. the following equation. Ps x, y= Psi Be hie) are

(9.75)

G04, -¥)]t-e9|A

Gs ‘We can also develop equations forthe drying time and the rate ofdrying in terms ofthe heat transfer coefficient and temperature difference driving force. The results are a>

PsZAsiX-Xe)

(9.76)

]

aZhs

9.77)

‘The time of drying in the falling rate period can be calculated in a similar manner assuming the fallingrate period to be linear. Then the rate of drying is given by

(9.16) (9.19) 29.78) Similarly, coresponding to Fy, (9.76), we can waite

ea GsCs(Tuy 2 CATs -T5)} a exp] o( - eer]

-

9.19)

For through circulation drying, when the gas is passed through a packed bed ofgranular solids, the following correlations havebeen proposed forthe heattransfercoefficientforadiabatic evaporation of water: hg

0.15169”/DS"for Re > 350

hig = 0.2146"/D)*for Re < 350

(9.8) 199)

where Re DG with fy: in Wim? K, Dp is the particle diameter in m, G is the superficial mass ‘velocity, kg/h m?and pL is the viscosity in kg/m h,

Drying w 725

For packed bed of spherical particles, the surface arca per unit volume ofpacking, « is calculated from

eae) D, where € is the void fraction in the bed. For cylindrical particles a is evaluated from 4(l-eNL +050) DE DL

(9.80) 2981)

where L and D are the length and diameter ofcylindrical particles. The valve of D, tobe used in Eq. (9.8) and (9.9)for cylindrical particles is the diameter of a sphere having the same surtace atea as the cylinder. D,=(DL+0.5D*)'?

(9.82)

‘The equations developed for calculating the rteand time ofdryingin the packed beds are applicable for particles of size 3-20 mm in diameter in shallow beds about 10-65 mm thick. For very fine particles in the size range 10 to 200 mesh (2.03 to 0.074 mun) and bed depth greater than 11 mm,the interfacial area varies with the moisture content and these equations cannot be used Example 9.12: Spherical catalyst pellets are being dried fromthe initial free moisture content of 1.2 kg ‘waterikg solid to a free moisture content of 0.40 kg water/kg solid, which is well abovethecriticalfree ‘moisture content, The pellets are 10 mm in diameterand are arranged in a bed of thickness 70 mm on screen. Thedy solid density is 1400 kg/m’.and the void fraction of the bed is 0.45.Airat 358 K and having humidity of O01 kg watery dry air is passed through the bed with a supertcial velocity 1.0mvs. Calculate the time of drying. Take viscosity of air be 2.1 x 10kg/m s for the given conditions, Solution: the initial and final moisture contents are above the critical moisture content, drying ‘occurs the constant rate-period. The drying time is given by Eq. (9.76)

oe

PsZAs(% Xp)

gaz \ mC. ‘ GCs) y= L400 ky/m); Z = 70 x 10° m; X; X= 1.20 kg water/kg dry solid: X3 X" = 0.40 kg water/kg dry solid, Ty, =358 K, ¥=0.01 kg watertkg dry air: The surface temperature 7; = wet-butb temperature of air 307 K. CU GCs(Tey

Ts) V ov

005 + 1.884 «0.01 = 1.0238 kJ/kg ; As 2421 kI/kg. G puwhere pis the density of air which is calculated as 14 ‘where vyis the humid volume. Therefore,

Yn

ley Lor [1,¥)7T T te 001) 358 garaxf eZ be aasx{ sax| og is) p “(35 18 sions G = pu=0.9811x1= 0.9811 kg/m? s (= 3531.81 kgf? h) G.=6

1 _ossnt = 0.9714 kg/m? s( = 3497 kg/m? hy

1+¥ 1.01

‘The Reynold’s number for flow through the bed is

DG 10107 «0.9811 = 467.19 >350

21x10"

981 1ke/m*

726 @ Mass Transfer: Theory and Applications ‘Theheat transfer coefficienthe; is evaluated using Eq. (9.8)

hg = 0.1516")D8"

=0.151(3531.81)°°/(10 x10)! = 123.68 W/m? K

Th inafcal aca peruvelo of the bed 6(1-e) (1-045) _ 4.4m" D, 001 m ‘Substituting the values, we get 6

1400 x 0.07 x 2421 x(1.2-0.4) =hh 123.683, 623305007) 3497 x 1.0238 x (358. son ~en( 3497 x 1.0238

Time of drying = 1.1 h (Answer) Example 9.13: A wet pigment material is extruded into small eylinders 6 mm in diameter, 50 mm long d are placed on screens to a depth of60 mm. The material is subjected to through- circulation drying ing a stream of air flowing through the bed with mass velocity of 1.2 kg dry aif? s, entering at 393 K with a humidity 0.05 kg water/kg dry air. The surface of the particles is estimated to be 300 m2/m? of the bed and the apparent density 1000 kg dry solid/m®, Take viscosity of the air to be 1.8 x 10° kg/m § forthe given conditions. Estimate the constant rate of drying. Solution: The rate of drying is calculated using Eq, (9.77) )

et CT r)lt-es(- tGCs Rea

ah,

Ty = 393 K, ¥ = 0.05, Ts = Ty = 322 K (from humidity chart).

Cy = 1.005 + 1.884 x0. 0992 ki/ke K ds =2385.3kI/kg. a =300 m2/m*, Z = 60% 10m,

Gy = 1.2 kg/ms (= 4320 kp’? hy

G = Ge + ¥) =1.2 1,05 = 1.26 kg/m? s (= 4536 kg/m? hy

D, =(DL+0.SD*)% =(6 x 50-+05%6")"* s = 17.83 mm

Reynold’s number,

x Re= DG _ 1783x107 126 249.5350, Therefore, “HO 18x10 1151-4536" 1(0.01783)" = 113.095 W/m? K (= 407.141 kim? h K) 0.151G°? DP

The values are substituted into Eq.(9.77)

sap 200 006 4320x 1.0992 093-322) —exp| ( 407.4320 1.0992 R= 300% 0.06 2385. “The constant rate of drying = 6.175 kg/m? h

(Answer)

Drying m 727 9.10 CONTINUOUS DRYING Material and Heat Balances for Continuous Dryers

A countercurrent continuous direct heat dryer is shown in Fig. 9.51. The analysis presented here is applicable to parallel current dryers as well with obvious modifications. Gy and 5 represent the flow rates ofdry gas and moisture-free solid in kg/m?h orkg/h. Xrepresents the moisture content ofthe solid

°

Dryinggas out Wet std tod SXMoy Tn

Drying ges in| Dry sols product SKHaTin

Fig. 951: Material and energy balance in continuous dryers

in kg moisture/kgdry solid and ¥ represents the humidity of the air in kg moisture/kg dry gas. J; and Taare the temperature of the gas and solids, respectively and H,, and H, are,respectively,the enthalpies ‘of the gas and solid. The suffix-I represents the conditions at the end where the wet solids feed is ‘acimitted and the suffix-2 the conditions at the product end of the dryer. ‘Material balancefor the moisture gives SX) + GY, = SX, + GY, (983) which can be rearranged as 5(X, -X,) = Gs (¥,-¥s) (9.84) Heat balance over the dryer gives SH5) +GsHga = S53 + GH +O (985) where Q is the net heat loss from the dryer. kW. The enthalpy of the wet solid in Eg. (9.85) can be written as the sumof the enthalpies of the dry solid and the liquid present asfree moisture. Hy = Cys U5 ~Tq) + XCp(T, ~Ty) + AH (9.86) Cps is the heat capacity ofdry solids and Cry the heat capacity ofthe moisture as liquid. Ty is the reference temperature for estimating the enthalpy and AMT, is the integral heat of wetting or adsorption ‘or solution referred to the pure liquid and the solid at the reference temperature. AH, is usually neglected for want ofreliable data. Enthalpy of the gas can be evaluated using the equation Hy =Cs(Te -Ty)+ Yho 9.87) ,in Bq, (9.87)is the humid heat of the gas. ki/kg dry gas K and is the heat of vaporisation of the moisture at the reference temperature T>, kW/kg. The humid heat for the ait-water vapour system is 005 +1.884¥ (9.88) and the latent heat of vaporisation at Tp 273.15 K is 2502 ki/kg. Heat loss Q = 0 for the adiabatic operation. Q will be negative, ifthe heat supplied to the dryeris in excess of the heat loss from the dryer. Example 9.14: A rotary countercurrent dryer is to be used to dry $0,000 kg/h wet sand from 50% moisture to 2.5% moisture on wet basis, The air is admitted at 400 K with a humidity of 0.007 kg wwaterfkg dry air and it leaves the dryerat 325 K. The wet sand enters at 295 K and leavesat 318 K. Heat lossesfrom the dryer amount to 25 kW kg dry air. Specific heat ofdry sand is 0.9 ki/kg K. Specific heat of the dry airis 1.005 kWtkg K and the specific heat ofthe water vapouris 1.884 kW/kg K. Latent heat Of vaporisation at 273 K is 2502 kI/kg K. Calculate the amountsofdry air passed through the dryerand the humidity of the air leaving the dryer.

728 W@ Moss Transfer: Theory and Applications Solution:

Yt sans cot

o-n6, I

S$ 25000 kgh, X, = 1.0,

6, Y-200r toa ya ce ites ‘$= 25000 kg/h, X,

0.0256,

Fig, 9.52: Comtinaows dyer in Example 9.14 Feedrate of wet same! = 50,000 kg/h Since the feed contains 50% moisture, = 0.5% 50,000 = 25,000 kg/h x, _ e050 = 1.0: X; x _ 0.025 = 0.0256 I=x, 0.50 I=x, 0975 Ty) = 295 K, Typ =318 K Enthalpy of the solids entering and leaving the dryer are calculated using Eg, (9.86) which, in the absence of the heat of solution, is Therefore,

Hy = Cps(Ts ~ Ty) + XCp(T, ~Ty)

He, = (Cs + XsCpg 17g) ~ To) = (0.9 + 1 4.1868 )(295 ~ 273) = 111. 91KI/kg Hoy = (Cys + XgCp XTgq ~My) (0.9 + 0.0256 x 4.1868)(318 ~ 273) = 45.32 KI/kg, ‘Air enters at 400 K with a humidity 0.007 kg water/kg dry air andleaves at 325 K. Tu = 400 K, ¥s 0.007, Te 325 K. Enthalpies ofthe air entering and leaving the dryer are given by Eq. (9.87), Hg =Cs(Tq ~Ty)+¥ho For the air entering the dryer, Cp = 1.0085 + 1.884 x ¥, = 1.005 + 1.884 x 0.007 = 1.0182kI/ke K. Hg) = Csp(Tgy ~To)+ Yoho = 1.0182 x (400 ~ 273) +0007 x 2502 = 146.83KI/kg FFor the air leaving the dryer. Cy = L005 + 1.884 x Y, Hg = C5\(Ty ~Ty) + ¥iPog = (1.005 + 1. 884¥, (325-273) + 2502%, Hy = 52.26 + 2600¥, Material and heat balances for the dryer are given by Eq, (9.84) and (9.85). Substituting the values we get 25000(1 ~ 0.0256) =G,(¥, ~ 0.007) and 25000 x 111.91 + 146.83G, 25000x 45.32 + G(52.26 + 2600Y, } + 25G, ‘These equations get simplified as Gy (¥, - 0.007) = 24360 Gs (2600¥, ~ 69.57) = 1664750 which can be solved simultaneously for Gs and ¥). Dry air flow rate,= 1.2 x 10° kg/h

Exit humidity of air, ¥,

0.02729kg water/kg dry air

(Answer)

(Answer)

Drying m 729

Continuous Drying at Low Temperatures ‘The continuous countercurrent dryer operating at low temperature can be considered to be consisting primarily of two zones as shown in Fig. 9.53a. In zone 1, the surface and unbound moisture are ‘evaporated and the moisture content falls from initial value X; to the eriical moisture content Xe. The temperature of the solid remains almost constant at 7; because the rate of heat transfer to the solidis balanced by the heat required for evaporation. At the inlet of the solidthere is a small preheat zone in whichthe temperature ofthe solid is raised to Ts. In this zone, no significant drying occurs and the heat transferred to the solid merely goes to increase its temperature. In Jow-temperature drying the preheat zone is usually ignored. The mode of drying in zone 1 is identical to that in the constant rate period, ‘except that here the temperature and humidity ofthegas change duringdrying. The temperature profile is as shownin Fig. 9.53, Points B and Crepresent the inlet and exit conditions ofthe solids in zone 1 ‘and points C’and B’ represent the inlet and exit conditions ofthe gas in tis zone, flowing countercurrent to the solids. In zone 2 itis the bound moisture that is evaporated and the solid temperature no longer remains constant;it rises from Ts to Te» along the curve CD as shown in Fig. 9.53b, The moisture ‘contentin the solidfalls from the critical moisture content at point C to the final value X;of the solids product, The temperature of the gas stream decreases along the smooth curve D'C’B‘Aas it moves, wuntercurtent (0 the solids through the length of the dryer. The humidity of the gas streamincreases from ¥; atthe inlet of the gas to ¥; at the exit. Preheat zone GY. TT

Zone1

ee

sxyL_“—1_

Zone 2

GY, poe

Tex,

(a)

Distance twougdryer e) (a) Drrcc-heat low-temperature continuous countercurrent deyer(b) Temperatire profile for a continuous urrent dryer

‘The retention timeof the solids in the dryer 0 can be calculated as 0=0, +8, where 0,is the retention time in zone I and 0, that in zone 2. Equation forretention time in zone 1 Equationfor retention time can be written by rearranging Eq. (9.13).

R +989) where A/S is the surface exposed to the drying gas per unit mass of the solids m"/kg dry solids. Since the rate ofdeying is govemed by the heat and mass transfer from the surface ofthe solid, the rate R in Eq. (9.89) can be replaced using Eq. (9.2)

730 @ Mass Transfer: Theory and Applications (9.90)

Since the moisture balance SAX = Gd¥ is applicable to the dryer, dX in Eq, (9.90) can be replaced » in terms of d¥. Gf5)Aj _ay 8, s lah, 05 -¥)

9.91)

Yc is the humidity of the gas stream leaving zone 2. It can be estimated by a moisture balance for zone or zone 2. For zone 2, we have S(Xe —X3) = GeYo ~ Vo) (9.92) Ss

so that

Yor Vat GKe Xz)

49.93)

¥; in Eq, (9.91) is the humidity of the gas which is in contact with the solid surface. It varies with the humidity of the bulk gas and this variation should be taken into account when determining the integralin Eq. (9.91) graphically. For a given value of gas humidity ¥, Ys is the saturation humidity at the adiabatic saturation temperature ofthe gas provided, the heat transfer by conduction and radiation are negligible. If ¥; is assumed constant, the integral in Eq. (9.91) can be determined analytically. 6, S$)bint “Yo 49.94) S\Aky ¥-¥ Equation (9.94) is applicable for adiabatic drying involving evaporation of water vapour intoair,

Equation for retention time in zone 2 ‘When the drying occurs below the critical moisture content of the solid, the conditions shown in zone 2 prevail and the rate of drying falls. Assuming that the drying in this zone is entirely unsaturated surface drying, the rate varies linearly with moisture content as given in Eq, (9.16). =X") R= R(X Xx" 9.16) where

Rea ky (Ys -¥)

(9.2)

‘Now thetimeofdrying ofthe solid from the initial moisture content X,-to the final moisture content X,, which is the retention time for the solids in zone 2, is given by S 4a

ale (9.95) Substituting Eqs. (9.2) and (9.16)into Eq. (9.95),we get si (Xe =X" )dx 6, Ay -(9.96) «(fp -YXX =X) ‘The integral in Bq. (9.96) can be determined graphically after establishing the relationship between the variables X", X, Ys and ¥. If the equilibrium moisture content X* is negligible, 8.

SXe

ak

Aky £0-0X

‘The moisture balance gives Sax = GAY

(9.97) (9.98)

Drying @ 731 which can be integrated to give S(X=X,}= Gs-%) Equation (9.99) can be rearranged to give G. X=+h)

(9.99)

Substituting Eqs. (9.100) and (9.98)into Eq, (9.97), we get @. (5)Fe! _—— k Ty. Th, ~nx, + GsSry]

(9.100)

0.101)

Assuming Ys to be constant, Eq. (9.101) becomes

Gi(s)ke__1__,Xee=to Ah x, San-yy M0s-1O

(9.102)

Ifthe rate ofdrying iscontrolled by internal diffusion, linear variationofthe rate with free moisture content is notjustifiable. Therefore, Eg. (9.102) should not be used for determining the retention time, if the rate is controlled byinternal diffusion.

Example 9.15: A wet solid with a moisture content 0.5 kg water/kg dry solid is tobe dried to 0.04 kg water/kg dry solid at a rate of 600 kg/h dry solids in a continuous countercurrent dryer. Air flows at a rate 12,000 kydy airh, It eaters the dryer at 378 K with a humidityof0.085kg water/kgdry ai, The solid entersthe dryerat the wet-bulb-temperature of the inlet air. The surface area available for drying {80.06 m'/kg dry solid, Thecritical moisture content ofthe solid is 0.12 kg water/kgdry solid. The mass transfer coefficient ky is [50 kg water/m? h. In the falling rate period, the drying rate was directly proportional to moisture content. Calculate the time ofdrying (a)in the constantrate period (b)inthe falling-rate period,

‘Solution: Refer to Fig. 9.54. The following values are known. 5 kg/kgdry solid, X, = 0.04 kg/kg dry solid, X,

0.12 kg/kg dry solid

1055 ke/kg dry air. A/S = 0.06 mi/kg dry solid

(a) Calculation for zone 1: Constant rate drying Retention time in zone | is given by Eq.(9.94)

S\Alky

YS -¥

‘A molsture balance forte dyer canbe writen S(X,- X,)= G54 -¥5)

600(0.50— 0.04) = 12000(¥, - 0.055)

‘Therefore, ¥,

0.078 kg/kg dry air

Temperature

Sf ) 1 tsn%e

‘The wet-bulb temperature of air at 378 K and

humidity 0.055 kg Ac cry air is Found to be Ty = 322 K from the psychrometric charts and the solids

in zone 1 may be assumed to be at this temperature. ‘The saturation humidity ofair at 322 =0.082 kekky

dry ur (from the psychrometric char). ¥= kg dry air.

0.082 key

Distancethrough dryer

ig. 9.4: riation oftemperature and moisture ‘content in continuous drying (Example 9.15)

732 m Moss Transfer: Theory and Applications

A moisture balance may be written for zone 1 for determining the humidity of air ¥ entering this zone.

SUX, Xe) = Gel, -Yo)

600(0.50— 0.12) =12000(0.078 — ¥,)

which gives Yo = 0.059 kglkg dry air Bquation (9.94) now gives

200 1 \e 0.082 0.059 _, 4g 5 0.06 /150'" 0.082 -0.078

(Answer)

(©) Calculationfor zone 2: Drying withafling rate ‘The retention time in zone 2 is calculated using Fa, (9.102):

(S}%« 1 nXe%s—%) AV ky x, Gy, yy ¥205~Yod 2 Us Vs 6 42000( 1 jose 1 jn in 120.082 -0.055) 600 (0.06 150 go4 + 1222 052-0055) OOMO082 0089) =0.58h

(Answer)

Continuous Drying at High Temperatures. For continuous dryers operated at high temperatures, the design calculations are made basedon heat transfer. In a differential section of the dryer the sensible heat transferred from the gas to thesolid results im a dropin its temperature by dT, We can write OsCsdTq = UalT ~Ts)dz (9.103) where U = Overall heat transfer coefficient between the gas and the solid, WimK. @ = interfacial surface per unit volume of the dryer, m?/m*. Z length ofthe dryer, m

Tq~Ty= temperature difference driving force for heat transfer, between the gas and the solid, K. Cs = humid heat, J/kg K

Gs = Flow rate of dry gas per unit area of cross section of the dryer, kw/tn? h Equation (9.103) canbe rearranged as dT Ua

ih Gc,”

(9.104)

2%,

(9.105)

Assuming that te volumetric overall heat transfer coefficient Ua andthehumid heat Care constant, "eae If the solid temperature remains constant as in the constant rate region zone 1, Eq. (9.105) gives

In

Te Lae GC

(9-106)

If the temperature profiles of the gas and solid streams can be approximated by straight lines, Eq. (9.106) can be written as

Drying m 733

where

ATg UaZ Qn,” GC;

(9.107)

ATs =To2 Tar

(9.108)

and

(9.109)

where Tg, Ty, and Tg, ~ Typare the temperature difference driving forces at the two terminals ofthe different zonesinthe dryer and (AT), is the logarithmic mean temperature difference. For zone 2, Eq, (9.107) is applicable if the falling rate is due to unsaturated surface drying, but not applicable ifthe rate of drying is controlled by internal diffusion. Equation (9.107) can be written in terms of the number and lengths of heat transfer units Moo =

Zz 0G

o(9.110)

where Nog = Number of heat transfer units

(9-110)

Hoc: = Height of heat transfer units = Z-= Length of the deyer. Direct-heat Rotary Dryer

(0.112)

Usingthe continuity equation, the diameter of the dinect-heat rotary dryer can be estimated using the total mass flow rate G atthe exit end of the dryer. aadgOWe G= 9.113)

‘Where wisthe allowable gas velocity and, ithe gas density at the exit. Thus, rearranging Eq.

(9.113), we get

rag

“rung | ‘The residence time of the solids in the dryer ®, is given by o- Bn

eo(9.114) (9.115)

i

In Eg.(9.115), Vy ithe fractional hold up of the solids.It is the fraction ofthe dryer volume that is occupied by the sotid at any instant. £ is the length of the dryer cylinder and Fy is the solids volumetric velocity, m’/s m* cross sectional area. The following relationship maybe usedto estimate the fractional holdup ofsolids.

G

Spe SN" =D. Ses DS ps on9.116) In Fq. (9.116), Sisthe slope ofthe dryer, nvm, Nis the rotational speed, rev/s, and D isthe diameter in m, ps the apparent solids density, kg dry solids/m’, D,is the particle diameter in m, The plus sign (@)corresponds to countercurrent flow and minus sign (-) for cocurrent flowin the dryer. The above ‘equation is valid for allowable gas velocity up to 1.) mis.

734 @ Mass Transfer: Theory and Applications

‘When drying is in the constant rate period, the volumetric heat-transfer coefficient, Ua, can be used to determine the rate of drying. q=UaV(AT),, (9.117) where V is the volume of the dryer cylinder and AT7,, is the logarithmic mean of the temperature differences atthe inlet and exit of the dryer.

aa Ta To In Tox ~Tss

(9.118)

‘The volumetric heat transfer coefficient based on the dryer cylinder volume may be estimated when

the drying gas is air, by the empirical correlation due to MeCommick (1973):

Ua=2376°"'1D

9.119)

where Ua is in W/m’K,G is gas flow rate per unitcross sectional area in kg/m? s and D in m.

Example 9.16: A wet granular substanceisto be dried from 10% to 0.54% moisture in countercurrent adiabatic rotarydryerat a rate of 10,000 kg/h of bone- 03. C5= 1.0085 + 1.884 x 0.03 = 1.061541/ke 9148.19 x 1.0615(810 ~ Te) = 10000(205.73 ~ 139.98) Solving this, we get Toe 742.29 K For the the preheating section, enthalpy balance gives, GCs (Tgp

—Toy) S(Hsg ~ Hs)

"p= 1005 # 1.884% 0.1338 =1.2571 ki/keK 9148.19 x 1.2571 (Tey ~400) = 10000(166.23 -68.00) Solving this, we get Top = 485.42 K “The number of heat transfer units can now be determined: ‘tee (7 Mee Moo = Number of heat transfer units= = For the preheating zone, Ty 485.42 K, Toy = 400 K; Tag = 339 K, Tay = 300 K where

13813

Forthe drying zone (zone 1), Tee = 742.29 K, Ty = 485.42 K; Tye = 339 K, Toy = 339K I,

q

in

339 339

For the heating zone (zone 2), Ty»

(Nog= 0 Tig ~Tey

Toe Tye

= 1.0132

= 742.29 K; Ty = 370 K. Ty = 339 K

_yy 810-370

742.29 -339

= 0.0871

‘Total number of transfer units for all the three sections put together is

Noe = Moa) +(Noa)s + Moa )s

= 0.3813 + 1.0132 + 0.087) = 1.4816

“The maximum allowable gas mass velocity is 7000 kg/m? h and the average gas rate is

oncyietasta

0.03 + 0.1 38) 9897.43 kg/h

‘The minimum area of cross section permissible is A= 209748 1.4139 m*

That is {DP =1.4139 where isthe minimum diameter ofthe dryer.

_(1.4139x4)* 1.34mi= 9897.43 FAORS P= "000 ( = ) “ (The standard sizes may be used for the diameter and the gas and solids mass velocity may be recalculated in the actual design calculations.) ‘The volumettic heat transfer coefficient can be calculated using Eq. (9.119)

Ua =2376°"Db

‘where Ua is in Wim" K, G is the gas flow rate per unit cross sectional area, kg/ms and D in m.

Drying 737 Ua = 237(7000 13600)" /1.34

‘The height of transfer unit is given by Eq. (9.112) Ho:

276.145 W/mK.

ine OSES Height of heat transfer units =F

‘The average humid heatof the gasis

onse4222138 song

276.145%3.6 163 m_ ‘The lengthof the dryer is given by Z = Hogg * Nog 8.163 14816 = m Length of the dryer = 12.1 m, diameter of the dryer 1.34 m

9.11 FREEZE DRYING

(Answer)

In freeze drying, the material to be dried is initially in a frozen state. The moisture gets vaporised when it receives the heat of sublimation from extemal sources. The plane of sublimation, whichis initially

the outer surface of the material,retreats inward with the progress ofdrying, leaving behind a layer of

porous dried material as shown inFg.9.86, Let L be the thickness ofthe solids to be died in the direction of heat transfer and be the thickness ofthe died product at anyinstant. Heat i transferred tothe sold surface by convection and through the solid wo the surface of sublimation,by conduction, Let ht be the heat transfer coefficientat the surface of the material and & be the thermal conductivity of the dried material, Then hea transfer rate at any instant is a" 1-1,

Thsvk

120)

where 7, is the external temperature of the drying medium, 7; is the temperature at the sublimation

Front

v2 —4 Hig. 9.56: Freezedrying model

In a similar manner, the rate of mass transfer from the sublimation frontis given by Pro ~ Pow he Vike + RIDay 49.121) where py and py afe the partial pressure of moisture at the surface of the frozen layer and at the balk sas, respectively. p,. is the vapour pressure of water at the sublimation temperature.kg is the extemal mass transfer coefficient and Dapis an avcrage effective diffusivity in the dry layer. The diffusivity Dap

738 ® Moss Transfer: Theory and Applications

and the thermal conductivity & are properties of the material being dried. The coefficientsf and ke are determined by the gas velocities and the dryer characteristics and are constant. The gas propertiesare determined by the extemal operating conditions T, andPa. At steady state, the heat received at the sublimation front is balanced by the heat removed bythe sublimation ofice, = DHNy = (9.122) Substituting q from Eq. (9.122) into Eq. (9.121),

T.-T,

Tae Nw (9.123) where AM; is the heat of sublimation of ice in ki/kg (or ki/kmol) and Ny is the mass flux of water vapour, kg/m? s (or kmoV/n? 5) Let V; be the volume of material that contains 1 kg of frozen moisture init, Then atany instant,the ‘moisture remaining in the solid per unit area in the direction of heat and mass transferis (9.124)

(9.125)

Substituting Eq. (9.125)into Eq, (9.123).

AH, at dt T.=T, AM, 9126) Ved WhoUR Rearrangingand integrating Eg, (9.126) gives 0, the time required for completely drying the material ® AM, tn

Jer Jomstnya oA ve 1 ET, ( Ley (2h se)

9.127)

9128)

Fornegligible external resistance, h is very large and Eq, (9.128) can be simplified as o- 2 AM 1 “Vv, 7-7, 9.129) Using the mass transfer flux equation (9.121), equation for the drying time can also be derived in

terms ofthe partialpressure difference driving fore Py Pr

What are the benefits of the drying operationsin process industries? Distinguish between the unit operations ofdrying and evaporation. ‘What are the problems encountered with respect to the product quality in drying operations? How do youclassify the solid materials from the view point of the drying mechanisms? Describe ‘with examples. Distinguish between the moisture content expressed on the dry basis and the wet basis 6. Define equilibrium moisture content.

Drying 739

‘What are the factors that determine the equilibrium moisture content of solids? How does the temperature affect the equilibrium moisture content of a solid? Distinguish between the bound and the unbound moisture. ‘What is free moisture? Ilustrate with a numerical example.

|. How are the drying equilibrium curves ofsoluble and insoluble solids different”

Explaintheequilibrium moisturecontentofamaterial in whichthe boundwaterexistsinchemically combined form. ‘What is fibre sanuration point? Explain the phenomena of hysteresis with reference to drying. . What are the advantages and disadvantages of direct heat and indirect heat dryers? Why is batch dryer more suitable for small scale drying’? ‘What are the advantages ofcontinuous dryers? . Whycontinuous countercurrent dryers are not suitable for drying beat seusitive materials?

Describe why the thermal efficiency ofcontinuous countercurrent dryer is generally lower than that of eocurrent dryers. ‘What are the important methods of solids handling employed in commercial dryers? |. Explain with the help ofatleast twoexamples each, the direct heat batch dryers andindirect heat batch dryers, What are the common pre-forming treatments used in preparing the feed to a through circulation type tray dryer? Why is such a treatment necessary? Describe the major advantages and areas ofapplication ofbatch tray dryers. st some examples of indirectbatch dryers. Describe their areas ofapplication. Describe the salient features and uses of agitated pan dryers,rotating vacuumdryers. Describe the importance und advantages offreeze drying. Discuss the construction, working and fields of applications of tunnel dryers and band dryers. Theturbodryersare notaffectedby the temperatureconstraints andproductcontamination problems. Why? What are the type ofmaterials that ean be handled in rotary dryers? Describe the role of the following component parts ofa rotary dryer: spiral flights, longitudinal

fights, knockers, gear and pinions, fans, sctew conveyor and cyclone. Explain why gypsum, iton pyrites, and organic materials such as peat and alfafa are better dried using cocurrent rotary dryers using flue gases. For drying of white pigments at high temperature what type of dryer would you recommend? Chambers with large cross sectional area, small bed height, large froe-board height above the bed are preferred for fluidized bed dryers. Why? ‘Whit are the general characteristics ofthe materialsthat can be dried in fluidized andspouted bed dryers? Describe the construction. working and applications ofthe pneumatic-conveyor dryer. Describe the major components ofa spraydrying system. Whatare the major industrial applications of spray dryers? Feed atomiser can be treated as the most important component of a spray dryer. Justify.

What are the different types ofatomisers used in spray dryers?

1. Discuss the different methods of feeding the drum dryers, their advantages and applications,

Explain the principle behind microwave and diclectric heating. ‘What are the advantages ofmicrowave and dielectric drying over conventional dryers? ‘Whatare the constant drying conditions with reference to the batch drying experiments?

740 @ Mass Transter: theory and Appications

43. 44. 45. 46. 47, 48, 49, ‘50, 51, 52, 53,

Define the rate of drying. How is the rite determined as a function ofthe moisture content by conducting batch drying tests? 85 the different regions ofa typical rate-ofdrying curve. Discuss the mechanism of drying in the constant rate period. Howdo yon justifythe two distinct falling rate regions drying? What unsaturated surface drying? How is the mechanism ofdryingin this period different from thatin the constant rate period”? How would you estimate the constantrate ofdrying from the heat and mass transfer data? Why the heat transfer rate equations are more preferred for calculating Rethan the mass transfer rate equations? Explain the significance of critical motsture content in drying. Explain why materials with smaller thickness have lowercritical moisture content. Describe the effect of gas velocity, humidity, temperature and thickness of the drying bed on the rate of drying under constant rate period, Whatis case hardening? How would you prevent tardening in drying?

9.1 Which one ofthe followingis not trie with regard to amorphous or fibrous materials? |A. Moisture is trappedin fibres or very fine pores. B. Solid structure or properties ere unaffected bythe drying conditions. C. Wood,leather, soap, cotton starch, woo! ete belong to this class ofmaterials. D. The equilibrium moisture content is higher than that of granular materials. 9.2 A maierial contains 20% water on welbasis. What is the moisture content ofthe materialon dry basis? A. 33.3% C. 16.67%

B. 25% D. 80%

9.3 Which one of the following statements is true with regard to equilibrium moisture content? A. The partial pressure of water in the solid is equal to the relative humidity of air, B. The partial pressure of water in the solid is equal to the vapour pressure of water at the temperature of air C. The partial pressure exerted by water in the solid is equal to the partial pressure of water ‘vapour in the air D. The partial pressure of water in the solid is negligible at this condition. 9.4 Which one ofthe following statements is incorrect with regard to equilibrium moisture content oy A.Spongy,cellular materials have high X” values compared to granular inorganic materials B. With increase in temperature, X* of a solid decreases. C. Moisture in excess of X° only can be removed in a drying operation with given supply of gas. D._X" of a solid is independent of the humidity of the air in contact with the solid, 95. A wet solid containing 33% moisture is to be dried in a current ofair of 50% relative humidity ‘The equilibrium moisture content of the solid is 10% and the highest concentration of moisture in the solid that can exist in equilibrium with unsaturated aiis 28%. (Al moisture contents are on dry basis). Which of the following is tree? ‘A. The fice moisture in the solid is 23% B. ‘The bound moisture is 5% C. The unbound moisture is 28% D. The free moisture is 17%

Drying m 741

9.6 In the presence of saturated air ‘A. no moisture can be removed from a solid B.only unbound moisture can be removed. C.only bound moisture can be removed. D. moisture in fine capillaries within the solid can be removed

9.7. Fora material,if the thickness of the material is smaller, the critical moisture content, A. increases B. decreases. C.remains unaffected D.cannot be predicted.

9.8 A250 mm x 250 mm x 10 mm flat wet sheet weighing 1.2 kg initially, was dried from both the sides under constant drying rate petiod. It took 1500 s for the weight of the sheet to reduce to 1105 kg, Another 1 mJ m x 10 mm flat sheet ofthe same material, isto be dried fromone side only. Underthe same constant drying rate conditions,time required for drying in seconds from initial weight of 19.2 kg to 17.6 kg is A. 1000 B. 1500 c. 2000 D. 2500 9.9. A solids beingdried in the lineardrying rate regime from moisturecontent Xjto X,."The drying rate is zero at X =O and the critical moisture contentis the same as the initia moisture, Xp, The drying time for (M = S/AR¢) is A. M(X)~Xp) BL M(X,/Xp) ©. MInXe/Xp) D. MX, In(Xp/Xp) where $ = total inass of dry solids, = (otal surface area for drying, Re = constant maximurn drying rate per unit area, and X = moisture content in mass of water/mass ofdry solids

9.10. A wet paper pulpcontains 75% water.After 100 kg water i removed in the dryer,its found that the pulp is now containing 30% water. The weight of the originalpulp is kg A. 167 B. 2222 C. 296.3

D. 155.6

9.11. Ina laboratory test-nan, the rate ofdrying was found to be 0.5 x 10" kg/m?s when the moisture ‘content reduced from 0.4 to 0.1 on a dry basis, ‘The critical moisture content of the materialis (0.08 on a dry basis. A tray dryer is used to dry 100 kg (ary basis)of the same material under identicalconditions. The surfacearea ofthe material is0.04 m*/kgofdry solid. The timerequired insto reducethe moisture content ofthe solids from 0.3 to 0.2 (dry basis) is A. 2000 B. 4000 . 5000 D. 6000 9.12. A batch ofmaterial is dried under constant drying conditions. When drying is taking place from all the surfuces, the rate of drying under the constant rate period is: ‘A. Directly proportional to the sofid thickness. B, Independent ofthe sotid thickness. . Inversely proportional to the solid thickness. D, Dircetly proportional to the square of solid thickness. 943 Ittakes 6h to dry a wet solid from 50% moisturecontent tothe critical moisture contentof 15%. ‘How much longer willit take to dry the solid to 10% moisture content, under the same drying conditions. The equilibrium moisture content ofthe solid is 5%. Al] moisture contentare on dry basis. A. 15 min B. SI min ©. 71 min. D. 94 min

742 @ Mass Transter theory and Applications 9.14. A sample of wool containing 40% moisture has an equilibrium moisture content of 12.5% in contact with air of RH 50%. The critical moisture content is 20%. All moisture contents are on dry basis. How much water can be removedby drying the sample in acurrent ofairof50% RH? A. 50% B. 15% C. 275% D. 375% 9.15 200 kg ofsolid (on dry basis) is subjected to a drying process for a period of 5000 s. The drying ‘occurs in theconstantrate period withdryingrate as Rc=0.5 x 10” kgm” s."The initial moisture cconten¢ of the solid is 0.2 kg moisture/kg dry solid. The interfacial area available for drying is 4m*/100 kg dry solid. The moisture content atthe end ofthe dryingperiod in kg moisture/kg dry solid is AOS B. 005 C. 0.10 D. 015 9.16 Toprevent shrinkage and case hardening on drying, which ofthe following is recommended? A. Use gas al high temperature B. Use gas at high velocity C. Use gas with high humidity D. Use gus at iow temperature and humidity 9.17 In batch drying of fibrous materials . generally nol observed to occur. ‘A.Falling rate period B. Firstfalling rate period. C. Secondfalling-rate period. D. Constant rate period. 9.18. Ifthe rate of drying is practically independent ofthe velocity ofthe gas, the drying is ‘A.internaldiffusion controlled. BB. in the constant rate period C. capillary flow controlled, D. surface heat transfer resistance controlled 9.19 Which one of the following is likely to be true for granular and crystalline solids? A. Lengthy constantrate period and linear falling rate period B. Short constant rate period andconcave falling rate period CC. Constant rate period extending up to the equilibriummoisture content, D. Firstfalling rate period is concave upward and the second falling rate is linear. 9.20. Which one of the following is likely to be true for amorphous or fibrous materials? ‘A. Lengthy constant-rate period and linear falling rate period B. Short constant rate period and concavefalling rate period C. Constant rate period extending up to the equilibrium moisture content D. First falling rate period is concave upward and the second falling rate is linear. 9.21. The entire drying may occurin the constant rate period for AA. cross circulation dry ing of crystalline solids B. fluidized bed drying of granular and crystalline solids CC. drying of pasty and sticky materials on trays D. continuous drying of amorphous or fibrous materials 9.22 In the constant-rate petiod, the rate-contlling mechanismis ‘A. internal diffusion of moisture B. surface vaporisation . capillary flow D._ vapour diffusion 9.23. Removal of last traces of bound moisture is facilitated better in ‘A. batch dryers B. continuous cocurrent dryers CC. continuous countercurrent dryers. indirect heat dryers 9.24 Whichoneofthe followingis treforcontinuous cocurrentdryers incomparisonwith continuous countercurrent dryers? ‘A. They can handle heat sensitive materials B. They have lower thermal efficiency.

Drying m 743 9.25

9.26 9.27 9.28

CC. They are more suited for removing last traces of unbound moisture D. Itis difficult to control the moisture content of the solid leaving the dryer. Which one of the following is favourable as far as batch tray dryers are concemed? A. Labourcost B, Heat economy C. Uniform moisture distribution in dried product D. Suitability for multi-product plants. Which oneofthe followingis ideal fordryingeasilyoxidizable and heatsensitivepharmaceutical product? ‘A.Batch Tray dryer B. Rotary dryer C. Vacuumshelf dryer D, Freeze dryer For dryingpastes or sluts that tend io agglomerate on drying, the dryer recommended is : B. tray dryer D. vacuum shelf dryer Which one of the following is recommended for a heat sensitive material which is likely to undergo size degradation on drying? A. Agitated pan-dryer B. Double-cone vacuumdryer C. Paddle-agitated cylindrical rotating dryer D. Double-truck tray dryer

9.29 For drying continuous sheets of wet cloth, modified versions of ....... can be used. A. shelf dryers B. tunnel dryers CC. double-cone vacuum dryers D. rotary dryers 9.30 Suggest a continuous dryerthat can be used for all materials that are dried in batchtray dryers. ‘A. Fluidised bed dryers B. Spray dryers C. Rotary dryers D. Turbo dryers 9.31. For large scale continuous dryingof minerals,sand, limestone, clays.ete. usinghot flue gases ‘A. spray dryers are used B. direct rotary dryers are used CC.indirect rotary dryers are used. drum dryers are used 9.32. Rapid drying of gypsum, iron pyrites, and organic materials such as peat and alfalfa using hot flue gases can be done using ‘A. direct heat cocurrentrotary deyers B. indirect heat cocurrent rotary dryers

C. direct heat counter current rotary dryers D.indirect heat counter currentrotary dryers 9.33 Which ofthe following would you recommend for drying materials such as white pigments, precipitated calcium carbonate and cattle feeds? ‘A.Direct heat rotary dryer using flue gas B. Indirect heat rotary dryer C.Direct indirect rotary dryers D. Fluidized bed dryer. 9.34 Free flowing material with almost uniform particle size is to be dried continuously with a very shor residence timeforthe solids inthe dryer. Ideal “yer for this purpose is A. rotary dryer B. tunnel dryer . band dryer D. fluidized bed dryer

7448 Mass Transfer: Theory and Applications 9.38 Name a dryer that combines drying and pulverization in a single unit 9.36 9.37 9.38 9.39

9.40

941

A. rotary dryer B. fluidized bed dryer C. flash dryer D. tunnel dryer Namethe dryer which combines evaporation, crystallisation, filtration, size reduction, classification, and drying. A. Rotary dryer B. Fluidized bed dryer C. Tunnel dryer D. Spray dryer ‘Which one ofthe followingwould you recommend forproducinga high grade productin granular form? A. Spray dryer B. Rotary direct dryer C. Fluidized bed dryer D. Tunnel dryer For continuous drying of thick pasty materials and heavy sludges, we can use a ‘A. spray dryer B. flash dryer C. tray dryer D, drum dryer ‘According to the liquid diffusion theory,the time of drying is, ‘A. directly proportional to the thickness of material B. inversely proportional to the thickness of material C. directly proportional tothe square of the thickness of material D.inversely proportional tothe square of the thickness of material ‘The rate ofdrying during constant rate period ‘A. is unalfected by air humidity B. increases with increase in air humidity C. decreases with increase in air humidity DD. mayincrease or decrease with increase in air humidity Hold up of solidsin rotary driers generally lie between A. 0.1 and 0.2 C. 0.05 and 0.15

B. 0.6 and 0.75 D. 04 and 0S

the diameter of the drier AL 10% C. 60%

B. W% D. 70%

9.82 ‘The flights provided in rotary driers generally extend from the wall t about........ percent of 9.83 Rotary driers are used

‘A. for making milk powder B. for drying pastes and slurries C. for drying free-flowing granular materials D. for drying of fruits 9.44 A filter cakeis dried with air at wet and dry-bulb temperatures of 300 and 323K,respectively. ‘The heat transfer coefficientis 11 W/m? and thelatent heatofvaporisation ofwater is 2500 kI/ kg. Masstransfer does not limit the process. Select the drying rate during constant rate period. [Neglect conduction through the solid and radiation effects, A. 1.32 107 kg/m’s B. 071 x 107 ke/m’s C. 4.53 x 107kg/m’s D. 0.10 x 107 kg/m’s 9.45

batchof 120 kg wet solid has initial moisture content of0.2 kg water/kgdry solid. The exposed

area of drying is 0.05 mv'hkg dry solid. The rate of drying follows the curve given below.

Rate of drying, kg waterremoved? h

Drying 748

00

e023 A kg moisture/xg dry solid

‘The time required (in hours)for drying this batch to a moisture content of 0.1 kg waterikg dry solid is A. 0.033 C. 0.60

B. 043 D. 231

9.46 The equilibrium moisture curve for a solid is shown below:

Mossture content kg waterthg dry solid

‘The total moisture contentofthe solid isX and it is exposed to air of relative humidity #7. In the table below, Group lists the types ofmoisture, andGroup II represents the region in the graph

above.

P._ Q. R. S.

Group I Equilibrium moisture Bound moisture Unbound moisture Free moisture

Group I 1 2 3 4

Which oneofthe following is the correct match? A. P-1.Q-2,R-3, $4 B. P-.Q3, R452 C. Pl, Q4,R2, $3 D. P-l,Q2,R4,S-3 9.47 Consider the drying operation shown in the figure bel. for a solid loading (dry basis)of50 ki? with a constant drying rate of 5 kg/m? h. The falling rate of drying is linear with moisture content.

746 @ Mass Transfer: Theory and Applications 1

g

i

$

|

i

| x

x X, kg moisturaikg dry solid ‘The drying time in hours to reduce an initial moisture contentof25% to a final moisture content of 25 is A. 155 B. 1.75 ©. 3.25 D. 435 Psychrometry reviewed

9.1 Airat 341 K dry-bulb temperature and 303 K wet-bulb temperature is being passed over abed of sliced carrots at the rate of 25 kg dry airs. Ifthe rate ofevaporation of water from the carrots is 0.275 kgs. estimate (a) temperature, (b) humidity and (c) percent humidity ofthe airleaving the dyer (Answer: (a) 318 K (b) 0.021 kg water/kg dry air (c) 32%) 9.2 A continuous countercurrent dryer is used todry a wet material which contains 50% moisture to 20%, both moisture contents being on wet basis. The dried product leaves at the rate of 900 kg/h. ‘The solids enter and leave the dryer at 302 K. Fresh air to the system is at 300 K and has a humidity of0.007 kg water/kg dry air. The moist air leaves the dryer at 312 K and humidity of 0.020and part ofit is recirculated and snixed with the fresh air before entering a heater. ‘The heated mixed air enters the dryer at 340 K with a humidity 0.010 kg water/kgdry air, Calculate: (a) The fresh dry air flow rate, ke’ (b) Thefraction of air leaving the dryer that is recycled (©) ‘The heat added in the heater, kW (Answer: (a) 41,538.5 kg/h (b) 0.23 (©) $71 kW) 9.3. A wet material having an initial moisture content of 25% on wet basis enters the continuous dryer at 295 K at the rate of 2500 kg/h. The product leaves the dryer at 355 K with a moisture content of 5%. Freshair ata temperature of 288 K with 50% humidity is heated in a preheater. ‘The effluent air enters the dryer at 435 K und leaves at 355 K, with 90% humidity. The vapour pressure ofwater at 355 K is S13 KPa. Calculate: (a) The volumetric flow rate of air entering the preheater (b) The volumetric flow rate of air entering the dryer. () The rate at which heat must be supplied to the preheater. (Answer: (a) 761.2 m°M (b)1149.7 m'sh (¢) 1.382 x 10° kim) 94 A granular solid containing 90% water is t0 be dried in a fluidized bed dryer to produce 100 kg/h ‘of dry material. Atmospheric air is passed through a steam-heated exchanger in which itis heated to 350 K and then passed through a bed ofsolid from witich it emenges 90% saturated. Assuming thatthe atmospheric ar is saturatedwith watervapourat 311 K, estimate the quantity

Drying 747

9.3

9.6

9.7

9.8.

9.9

9.10

of air need and the duty of the heat exchanger. (Flint: Assume that the air temperature varies along the adiabatic cooling line) (Answer: 64,769 kg/h wet air, 2.63 x 10°KI/ti) A 1Ostray cabinet dryer carries on cach tray, kg wet solids containing 95% moisture on wet basis. Fresh airat 293 Kcontaining0.01 kg moistare /kgdryairis mixed with recycled air before being heated by passing through a heater. The air leaving the heater is passed through the dryer from which it emerges at 323 K containing0.025 kg water/kg dry air. 90% ofthis ar is recycled. For aircirculation through the dryer at a rate of 1500 kg dry air per hour, determine: (a) The rate at which moisture is removed in the dryer. (b) The heat duty of the exchanger. (©) The time required to dry the material to 45% moisture on wet basis, if the drying is at constant rate (Answer: (a) 2.25 kg/h (b) 4.07 kW (c) 4.04 hy A two-stage adiabatic dryer is used to dry paper. Air at 101.3 kPa, 353 K enters the first-stage dryer where it picks up moisture and leaves saturated. Airis then heated to 348K by passing through a beat exchanger before admitting to the second stage of the dryer. The air leaves the dryer with a percent saturation of $0%. Itis covled to288 K and waters condensed andremoved by passing througha cooler. The air leaving the cooleris heated to 353 K by passing through a second heat exchanger. For every kilogram of dry air circulated (a) Determine kilograms of moisture evaporated in each dryer (b) Kilograms of water condensed and removed in the cooler. (Answer: (a) 0,022 kg, 0.013 ke (b)0.035 ke) An adiabatic dryer is used to dry a wet material. The drying air enters at 381 K and 101.3 kPa with a dew point of 298 K. Measurements show that 2.25 kg wateris evaporated per 100 m' of wot inlet ais. Calculate (a) Humidity of air entering the dryer (b) The exit air humidity and percent humidity (C) The exit air wet-bulb and dry-bulls temperatures, (@) The volume ofexit air per 100 m’ofinlet wet air, (Answer: (a) 0.02 ke water/kg dry air (b) 0.045 ky water/kg dry air, 55% (6) 313.2 K,322.2 K (d)87.92 m’) A laboratory dryer produces 50 kg of dry solid containing 6 weightwater from a wet feed material containing 20.5% water. The dryer operates adiabatically, with the wet solid entering the dryerat the wet-bulb temperature ofthedryer air. A total of 500 m’sh of hot air at 366 K and 101.3 kPa with a dew point of 303 K is supplied to the dryer. Calculate: (a) ‘The temperature of the air leaving the dryer (b) ‘The percent humidity of the air leaving the dryer. (Answer: (a) 318.65 K, (b)70%) Airat 101.3 kPa enters an adiabatic dryer at 372 K with a dew pointof 287 K and leaves at 80 per cent hnumidity. Wet paper enters the dryer with 25% moisture and leaves with SG moisture Determine: (a) ‘The temperature of air leaving the dryer (b)_ Water evaporated in kg per 100 cubic metres of sir entering (©) Kilograms of finished product per 100 cubic metres of air entering (Answer: (a) 311 K,(b) 2.43 kg (6) 9.11 kg) Air supplied toa dryer is at 340 K and 101.3 kPaand has a dew point of278 K. Theair leaves the. dryer at 310 K and 100 kPa with a dew point 298 K. At what rate (m? per hour measured at the

748 @ Moss Transfer: Theory and Applications

dryer inlet conditions)the air should be admitted tothe dryer if itis desired that 100 kg water is removed per hour? The vapour pressure of water at 278 K = 0.87 kPa and at 298 K 3.17 kPa. (Answer: 6794 mh) 9.11 Airata rate of 25 m*per minute is supplied to a dryer in which it removes moisture ata rate of (0.45 kg per minute. The airenters the dryer at 333 K and 101.3 kPa with a percentage humidity of 10 and leaves at 310 K and 100 kPa. The vapour pressure ofwater at 310 K = 6.3 kPaand at 333 K = 19.9 kPa (a), Whatis the percent humidity of air leaving the dryer? (b) At what rate air leaves the dryer in. m* per minute? (Answer: (a) 78.8% (b)24.23 m/min) 9.12 Itis required to evaporate 100 kg/h of water froma wet solid. The drying is carried outin an adiabatic dryer, Airisavailableat 101.3 kPaand 295 K at 70% humidity. Its temperature is raised by passing through a preheater to such a valve that the air leaving the dryer is at 311.7 K and 90% saturated. Calculate: (@) ‘The adiabatic saturation temperature of air leaving the dryer. (b) The temperature of air leaving the preheater. (©) The rate of circulation of moist ait, kg/h. (a) The volume of moist air before preheating. (Answer: (a) 310K, (b) 378 K, (¢) 3614 K/h, (d) 3039 mh) 9.13 Air with a dry-bulb temperature of 363 K and wet-bulb temperature of 308 K is passed over a ‘wet materialin a dryer. The rate of evaporation is found to be 0.015 kg water per kg dry air passed through the dryer. Assuming the process to be adiabatic, calculate using psychrometric chart (a) the absolute humidity, molal humidity ofair entering the dryer, (b)the dry-bulb temperature,the wet-bulb temperature, the absolute humidityand the percent saturation ofair leaving the dryer. (Answer: (a) 0.013 ke/ke dry air,0.0209 kmolV/kmoldry air (b) 329 K, 308 K,0.028, 22.5%)

Drying equilibrium 9.14 One-kg blocks of wet laundry soup containing 20%moisture on dry basis is dried using air at 298 and25% relative humidity inatunnel dryerat 101.3 kPa, Determinethe kg waterevaporated from each block. Following equilibrium data is applicable: Re x

10 0.024

20 0.038

30 0.052__

0 30 00650085

60 70 80 0108013808 (Answer: 0.1292 kx) 9.15 A wet solidcontaining 40% moisture is w be dried in a current ofair of 50% relative humidity. ‘The equilibrium moisture content ofthe solid is 5% and the highest concentration of moisture in the solid that can exist in equilibrium with unsaturated air is 30%. (All moisture contentsare on. ‘wet basis). Determine the free moisture, bound moisture and unbound moisture in the solid. (Answer: 0.6141 kg/kg dry solid, 0.4286 kg/kg dry solid, 0.2381 kg/kg dry solid) 9.46 Fibres with an inital total watercontent of35% on dry basis arc dried in traysin a tune} dryer operating at 101.3 kPa. If the fibres are brought to equilibrium with air of relative humidity of 40% at 298 K.determine the kgofmoistureevaporated per kgof bone-dry fibres. Theequilibrium moisture content varies with relative humidity of air as given below:

Drying w 749 x RH%

0 0

002 3

00% 005 2

008 010 012 O14 O16 O18 SST (Answer: 0.267 kg) 9.17 ‘The equilibrium moisturecontentoflumber varies with humidity ofair asgiven inthe following, table. X 0 002 0.04 006 008 0.1 O12 O14 016 O18 02 022 024 0.26 028 03 RHO 45 15 30 43 S563 7075-8086 88 91-94 +96 100

A dryer is fed with lumber containing 40% moisture on dry basis. The dried product leaves the dryer in equilibrium with atmospheric air at 298 K and 50% relative humidity, Determine: (a) The unbound moisture in the original feed lumber in kg water/kg-bone-dry lumber (b) The bound moisture in the original feed lumber in kg water/kg bone-dry lumber. (c) The free moisture in the original feed lumber, in kg water /kg-bone-dry lumber, (@) ‘The kg moisture evaporated per kg bone-iry kumber. (Answer: (a) 0.1 (b) 0.3 (c) 0.308 (4) 0.308) 9.18 Wet cotton cloth, 100 kg in weight, containing 20% moisture on dry basis is kept in a room which contains 500 m’ of air st 101.3 kPa and 313 K. The wet-balb temperature ofthe air is 293 K, ‘The air in the roomis allowed to come in equilibrium with the cloth keeping the temperature constant at 313 K and without allowing airto enteror leave the room. The equilibriummoisture content ofthe cloth at this temperature is given by: X,ke moisturchkg dry cloth Relative Humidity.%

0.02 0.08006 45S

O.0R8—O.10 01D Oe BSS.

Determine thefinal moisture content of the cloth and the relative humidityof air in the room, Neglect the effect due to the change of pressure of air. Batch drying (Cross-circulation)

(Answer: 7.63% on dry basis,relative humidity = 52% )

9.49 Wet material is dried in a pan0.5 m x 0.5 mand 25 mmdeep. Dryingis from the top surface only. Airis passed across the pan at 6 m/s. Airis at 342 K and hasa humidityof0.01 kg wate/kg dry air. (a) Estimate the rate of drying Ry.

(b) Predict Re if the air velocityis 3 mV/s. (6) Predict Rif the gas temperature is raised to 369 K keeping the humidity constant. (A)Predict the effect of increasing the thickness of the drying bed lo 40 mm onthe time required for drying in the constant rate period between specified initial and final moisture limits (Answer: (a) 3.5324 kg/m? (b) 2.03 kg/m’ h (c) 5.23 kg/m? h (a)timeincreases by 1.6 times) 9.20. A filter cake 600 mm x 600 mm x 50 mm, supported on a screen is dried from both faces using, a stream of air with dry-bulb temperature 348 K and wet-bulb temperature 303 K. Air flows parallel to the surface ata velocity 2.5 nvs. Equivalent diameter ofthe airflow space is 150 mm. ‘Thelatent heat of vaporisation may be taken to be 2431 ki/kg. (@) Assume that the heat transter coefticient is given by 0 nase Dé where ft is in Win?K, G is im kg/m? s and D,is in m. Determine the rate of drying in kg/m? h

750 @ Moss Transter: Theory and Applications

(b) Repeat part (a) using h ~ 0.0208 G°* W/m?K. where Gis the mass velocity ofair in kg/m h.

(Answer: (a) 18 kg/m’ h (b) 2.0 ke/m?h) 9.21 A drying pan, 0.5 m x 0.5 m x 40 mm deep and insulated onthe sides and bottomis filled with ‘wet granular material. Heat transfer is by convection from an air stream at 342 K and having humidity 0.010 kg water/kg dry air, flowing parallelto the top surface at a velocity of 3.0 m/s. ‘The top surface receives radiation from steam heated pipes whose surface temperature Tis 368K, The emissivity of the material being dried is 0.95. Calculate the surface temperature and the rate of drying for the constant-rate period. (Answer: 305.5 K, 2.67 kg/m*h) 9.22. Aninsoluble wet granular material isdried ina pan0.5 m x0.5 mand 25 mm deep. Heat transfer is by convection from an air stream flowing parallel to the surface at a velocity of 5 nvs. The ai is 338 K and 101.3 kPa with a humidityof0.01 kg water/kg dry air. Estimate the rate of drying in kg/n?h given that = 0.02046" where fis in Wm’K and G is in kg/m? h and latent heat of vaporisation is 2433 kVkg. (Answer: 2.84 kg/m*h) 9.23 A sample ofwet material on a tray is exposed to 2 kg/m? h ofdryingair across the top surface, “The areaexposed to drying is4 x10 mi'/kgdry solid and the thickness of the bed is 25 mm, The drying rate during the constant rate period is 2.0 kg water/hm?, The critical moisture contentis 0.25 ke watet/kg dry soli. Calculate the time todry another sample of the same material but ofthickness 50 mm from both top and bottomsurfaces from a moisture content of 0.50 kg water/kg dry solid 10 0.30 kg. watertkg dry solid. (Answer: 2.5 h) 9.24 A wetsolid is to be dried from aninitial moisture content of 25% to a final value of5%. Drying, test shows that the rate ofdrying is constant at IS kg water/m’h in the region X= 0.2 - 0.4 kg. wwater/kg dry solid. The dryingrate falls linearlyin the range X = 0.01 ~0.2 kg water/kg solid.If the equilibrium moisture contentis 0.01 kg water/kg dry solid, calculate the time of drying, The drying surface is | m*/30kg dry weight, All moisture contents are on dry basis. (Anywer: 6.92) 9.25 A batchofgranular solids containing 30% moisture istobe dried ina tray dryerto 15% moisture b y passing a stream ofairacross its surface. Iftheconstant rate ofdrying under these conditions, is 2.5 kg/m? h and the critical moisture content is 15%, calculate the drying time. Assume the drying surface to be 0.03 m’/kg dry solid. All moisture contents are om wetbasis. (Answer: 3.36 h) 9.26 Sheet material SO mm thick with a dry density 800 kg/mis to be dried at constant drying conditions. The drying rate is 3.2 kg/h m?at the initial moisture content of 35% and the rate is 0.7 kghm?atthe final moisture content of 10%. Ifthe drying is from the two large surfacesonly, ‘and the dryingrate in the falling rate period isproportional to the free moisture content, calculate the total drying time. All moisture contents are on the dry basis. (Answer: 3.04 h) 9.27 One hundred kg ofcertain food material is dried from a moisture content of 80% to 50%. The drying surface area is 15 m?. Aircirculated is ata dry-bulb-temperature of 393 K and wet-bulbtemperature of323 K. Under the conditions in the dryer, the heat transfercoefficient to the food surface fromthe air is measured to be 20 W/m? K.Take the latent heat of vaporisation of steam at 323 to be 2383 kW/kg.All moisture contents are on wet basis. Assuming constant-ratedrying, estimate the time of drying. (Answer: 1.89 h) 9.28 Trays used fordrying wet filtercake ofcalcium carbonate has an areaof 1.5 m?andsfilled with 75 kg ofcake. Drying occurs from the top surface by crass circulation of 10% saturated air at a

Drying m_751 pressure of 101.3 kPa and temperature of 351 K. The average velocity ofair passed across the ‘wetsolid is 4 m/s, Ifthe initial water content of the wet cake is 25%, estimate the time in hours to reach the experimentally determined critical moisture content of 10%. All moisture contents, are on dry basis. The preheat period may be neglected. (Answer: 2.75 h) 9.29 In the falling rate period, ifthe rate ofdrying varies as R=aX + bX?

show that the time of drying in the falling rate period is given by T x2, Sin)XR ‘where Ryisthe rate when the moisture content in the solid is X. 9.30 From experiments on batch dryingofsheetmaterial, therate ofdrying in the constant arte period is found tobe 5 kg water’h m?. The critical moisture content is 30% and the equilibriummoisture contentis negligible, Sheets ofsize 1.5 mx 1.2m x $0 mm are tobedried from 60% moisture to ‘5% moisture. The drying is from both sides and under the same drying conditions as in the ‘experiments. Ifthe material has a bone-dry density of1200 ka/m’,calculate the time required for drying. All moisture contents are on dry basis. Assume thatthe falling rate period is linear in the moisture content. (Answer: 5,03 h) 9.31. Ittakes 5 h to dry a wet solid, from 35 to 8% moisture content,usingair at constant conditions. ‘The critical and equilibrium moisture contents are 15% and 5%, respectively. Assume thatthe length of the preheat period is negligible and the falling rate period is linear in free moisture. For the same drying conditions, determine the drying time ifthe inital moisture content is40% and the nal moisture content is 7%. All moisture contents are on the dry basis (Answer: 6h, 25 min) 9.32 A non-porous solid material istobe dried underconstant drying conditions from 0.20 100.05 kg water/kg dry solid. The bone-dry mass ofthesolidis 1000 kg and the area ofsurface drying 5S 1m’.Air at a velocity of0.75 m/s is passed across the surface of the solid and the drying process is surfuce-evaporation controlled. The critical moisture content ofthe material may be taken as (0.125 kg water/kg dry solid and the equilibrium moisture contentis negligible. (a) Ifthe initalrate of dryingis 1.5 kg/mh, how long will it take to dry the material? (b) If the air velocity were increased to 4.0 mvs, what would be the anticipated saving in time’? (Answer: (a) 2.3 h, (b) 1.7 h, 74% saving) 9.33. Air used for drying certain food material has a partial pressure ofwater equal to 2.4 KPa and wetbulb temperature of 303 K. The material is taken in a try with a drying surface of0.05 m*/kg dry solid, The rate in the falling rate period is proportional to the moisture content and the mass transfercoefficient is 5.0 x 10* kw’m? h Pa. Critical moisture content is 0.15 kg moisture/kg dry solid and equilibriummoisture is negligible. Calculate: (a) ‘Thedrying rate in the constant-rate period in kg/m? h. (b) The time required to dry the material from 0.25 ky water/kg dry solid 10 0.05 kg water/ky dry solid. Vapour pressure ofwater at 303 K = 4.232 kPa. (Answer: (a) 0.916 kg/m? h (b) 5.78 h) 9.34 A solid isto be dried from 1 kg water/kg dry solids to 0.01 kg water/kg dry solids in a tray dryer. ‘The trays are 0.7 m square and 0.02 m deep and are completely filled with wet material. The air at 308 K and humidity 0.003 kg water/kg dry air is passed across the trays with a velocity of2.0, m/s. The convective coefficient ofheat transfer is given by

fh =0.0208 G8 W/mK

752 @ Mass Transter: Theory and Applications

where Gis the mass velocityofairinkg/m*h. The critical moisturecontentofthe solid is 0.3 and the equilibrium moisture content is negligible. The density of the solid is 1200 kg/m?. Assume thatthe dying is by convection from the top surfaceofthetraysonly. Determine thedrying time. (Answer: 55 h) 9.38 ‘A wet erystalline solid is tobe dried in a rectangular pan 1 mx 1 mx 25 mm. Air at 338 K and 101.3 kPa with humidity of 0.01 kg water/kg dry air is flowing across at a velocity of 3 m/s, Depth ofair flow space is 100 nm. The latent heat of vaporisation of water may be taken as 2433 kW/kg. (a) Estimate the rate of drying af constant rate given that the convective heat transfer coefficient is on h=59) De

‘where A is in Wim? K and G in kg/m? s and D,is in m. (b) Repeat Part (a)if the convective heat transfer coefficient is given by f= 0.0204 G?* W/m?K

where Gis the mass velocity of air in kg/m? h. (Amswer: (a) 1.155 kg/m? h,(b) 1.89 kg/m? h)

9.36 Ittakes eight hours for a porous solid toreduce the moisture contentfrom 40 to 10%when dried

9.37

9.38

9.39

9.40

in abatch dryer under constant drying conditions. The critical moisture content was found to be 20% and! the equilibrium moisture 3%. All moisture contents are on the dry basis. Assuming that the rate ofdrying during the falling rate period is proportional to the free moisture content, how Jong should it take to dry a sample of the same solid from 35 to 5% under the sane drying ‘conditions? (Answer: 11h, 43 min) It took 6 h when a material was dried from the initialfree moisture content of 30% to the final free moistore content of 8% in a tray-type batch dryer using constant drying conditions. ‘The critical moisture content is 159. If the drying rate in the falling-rate region is linear from the critical point tothe origin, predict the time todry the sample from a free moisture contentof35% to 5%. All moisture contents are on dry basis. (Answer: 8.96 h) ‘A300 mm x 300 mm x 6 mm sample of porous sheet material was dried in a laboratory dryer fromboth flat surfaces. The rate of evaporation of moisture from the solid was found to be 0.27 kg/h until the critical moisture content of 15% is reached. In the falling-rate period the rate of ‘evaporationfell linearly with moisture content until the sample was dry. The weight ofthe dry board was found tobe 1.5 kg. Estimate the time required for drying another sheet ofsimilar material with size 600 mum x 1200 mm x 12 mm under identical conditions from an initial moisture content of 25% to a final moisture content of 2%. All moisture contents are on wet basis. (Answer: 6 h) Airat a dry-bulb temperature of333 K and a wet-bulb temperature of303 K is passed across the top surface ofa granular wet material contained in a pan 0.60 m x 0,60 m to a depth of 25 mm. ‘The air velocity is 3.0 mvs. The sides and bottom of the pan are insulated. The pan contains 12.0kg ofdry solid having free moisture content of0.4 ke water/kg dry solid and the material is to be dried in a constant rate-period to a free moisture content of 0.20 kg water/kg dy solid. Estimate (a) The drying rate and the drying time. (b) ‘The time of drying if the material thickness is 50 mm, (Answer: (a) 1.59 kg/m?4.19 h (by 8.38 h) A wetsolidis dried from 20to9 percent moisture underconstant drying conditionsin 5h. Ifthe critical and the equilibrium moisture contents are 13 and5 percentrespectively, how long will it

Drying m 753

take to dry the solid from 25 to 7 per cent moisture under the same conditions? All moisture ‘coments are on wet basis. (Answer: 9.43 h) 941. Calculate thetotal dryingtime todry abatch ofa wetmaterial from 30% moistureto 5% moisture ‘on wet basis. The drying surfaceavailable m*/40kgdry solids.The falling rate is assumed to ‘be linear. The constant dryingfluxis 1.5 kg/m? h. Thecritical moisturecontent is 0.2 kg moisture/ kg dry solid and the equilibrium moisture content is 0.05 kg moisture per kg sotid. (Answer: 22.3 h) 942A slab made ofCaCO,paste is dried by passing airacross the top and bottomsurfaces. The slab has an area of 0.025 my" on oneside and weighs 1.725 kg. Determine the time of drying if the moisture content is to be reduced from 15% to 2%. The critical moisture content is 8.5%. The first falling rate period is Linear up to a moisture content of4% with R=3X+0.25 nd the second falling rate period is given by R=300X? -2.75X

tll the moisture contentfalls to zero. The rate isin ky/m*hand the moisture contents are on dry basis, (Answer: 10.82 h) 943 Woollen cloth is dried from an initial moisture content of 100% to a final moisture content of 10%,Thedryingrate varieswiththe moisturecontentaccording to the followingrelation: R = Re. for x >0.55 R= 2.0408Re(X ~0.06), for X $0.55 where Ris the constant rate of drying. Calculate the saving in dryingtime,if the material is to be dried to 15% instead of 10%. All moisture contents are ondry basi (Answer: 23.7%) 944 The rate ofdrying R (kg water evaporated/n* h) of a wet materialis given by R=S.40X,for X03 R=1,62, for X>03 where X = kg moisture/kg cry solid. The drying surface is 0.025 m?kg dry solid. How long the ‘material to be dried to reduce the moisture content from 0.6 to 0.25 kg moisture/kg dry solid? ‘The equilibriummoisture is zero. (Answer: 8.76 h) 9AS Flat cakes of an onganic productare to be dried in a current of warmn air maintained at constant humidity and temperature, ‘The original weight of the wetcake is 15 kg and the bone-dry weight is 5 kg. Pilot plant results indicate that the drying under the stated conditions shows a well defined constant drying rate given by

and falling rate given by

Xo2 dt

ax ‘ = 025K (X= =- X")

‘The equilibrium moisture content X° =0.1 kg water/kg dry solid. Determine: (a) ‘The initial moisture content on dry basis. (b) The critical moisture content. (©) The time required to reduce the moisture content to 0.2 kg water/kg dry solid. Assume that the surface exposed to the dry airis Im. (Amswer: (a) 2(b) 0.9 (c) 13.82) 9.46. A tray dryer consisting of 16 traysisused to dry raw cotton from 95% moisture to 10% moisture by cross circulation ofair. The trays which are insulatedon the bottom are each 30 mmhigh with an area of 1.5 m? and are completely filled with raw cotton which has a bone-dry density of

754 @ Moss Transfer: Theory and Applications

947

9.48

9.49

9.50

700 ke/m*. The heating medium, which is 30% sanurated air at 344 K and 101.3 kPa, flows actoss the top surface ofthetray at amassvelocityof 10000 kg/h m?.Under the given conditions, the critical moisture content will be 0.4 kg water/ke bone-