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DRYING OF SOLIDS References: Coulson & Richardson Vol 2 (4th Edn) ch. 16 Def: Reduction of liquid content from an initial value to an acceptable final value Solids may be indifferent forms: flakes, powders, granules (crystals), sheets, slabs Drying often preceded by mechanical removal of liquid – filtration, centrifugation, pressing Unassisted or natural drying (slow) vs. Accelerated drying Here we only deal with accelerated drying – in equipment built for the purpose Fundamental mechanism: Thermal vapourization (heat of vapourization) Liquid may be on surface or entirely within, or partly on surface and partly within solid What is ‘dry’? Bone dry

0% water

Dry salt

0.5%

Coal

4%

Food products

5 – 8%

Equilibrium moisture content Moisture content X’ is expressed in terms of kg H2O/ kg Bone dry solid A solid that is in equilibrium with its surrounding air (i.t.o. moisture content) – neither absorbs or loses moisture. The equilibrium moisture content is designated X*. The equilibrium depends on the solid characteristics and the relative humidity of air (which is a function of temperature).

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Question 1. Using the above diagram, what is the lowest moisture content that can be achieved in egg albumin at 25°C if the relative humidity of air is 40%? Typical curve for experimental drying data Mass solid

W0

Critical moisture content

We

Time

Bound moisture is the lowest concentration of moisture in the solid that is in equilibrium with the moisture of saturated air in the dryer. Free moisture (X) is the difference between the total water content (XT) of the solid and the equilibrium water content (X*) X = XT - X *

(1)

All calculations regarding drying use the value X (free moisture content)

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Psychrometric chart – gives the humidity and enthalpy relationships of air at different temperatures at atmospheric pressure.

Adiabatic saturation line Humidity = kg H2O per kg dry air Relative Humidity =

𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑢𝑟 𝑖𝑛 𝑎𝑖𝑟 𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑖𝑛 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑 𝑎𝑖𝑟

𝑥 100

(2)

Question Air at 20% relative humidity and 40°C enters an adiabatic drier and exits the drier at 80% relative humidity. What is its temperature? Why does its temperature change?

Methods of drying 1. Moving heated air at atmospheric pressure over wet solid. 2. Vacuum drying (allows lower temperatures) 3. Freeze drying – water sublimed from frozen material 4. Drying in energy fields – infrared, dielectiric (microwave), acoustic 2 basic methods of drying: Batch or Continuous

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Purpose of drying 1. 2. 3. 4.

Reduce mass Improve handling Improve shelf life (storage for extended periods) Reduce biological/other degradation

Drying equipment – classified according to design and operating feature    

Batch or continuous. Batch usually small, drying cycle long. Continuous usually large, less labour & floor space, product quality more uniform Physical state of feed – liquid, slurry or wet solid Method of conveying solids – belt, rotary, fluidized Heating mechanism – convection, conduction & convection, radiation (microwave)

Selection of dryer – depends on: 1. 2. 3. 4. 5. 6.

Feed condition – solid , liquid, powder, paste Feed concentration – amount of H2O Product spec – dryness required, physical form (grains, sheets, flakes,etc) Heat sensitivity – food products Nature of vapourization – toxicity, flammability Nature of solid – dust explosion hazard, toxicity

Different types of dryers Type Tray dryers Vacuum shelf indirect dryers toxic & volatile solvents Continuous tunnel dryers Rotary dryers Drum dryer Spray dryer

Characteristics Low rates of production < 100 – 250 kg/h Pharmaceuticals, heat sensitive materials Long cycles (4 – 48 h/batch) Food materials Revolving cylindrical shell – granular feed (sugar) Heated metal shell - slurries Short contact time, control of particle size (e.g. milk powder) Granular, crystalline material. Contact ime 30 – 60 sec. Feed conveyed by high velocity gas stream

Fluidized bed dryers Pneumatic dryers Revolving cylindrical shell

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Tray Dryer Tunnel Dryer

Drum Dryer

Spray Dryer

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Mechanism of drying Drying is a very complicated process involving both heat transfer and mass transfer. Essentially, energy has to be supplied (heat transfer) to evaporate water. The water vapour then has to be carried off (otherwise we have saturation conditions) for more evaporation to take place (mass transfer). The driving force for mass transfer is what? Initially drying occurs by evaporation from the surface. The evaporated moisture will be replaced by water moving from within the solid to the surface. Drying occurs at constant rate (BC) as long as there is moisture at the surface. The temperature of the solid during this period is the wet bulb temperature at which the rate of evaporation is balanced by the rate of transfer of heat that makes the evaporation possible. When there is no more moisture at the surface, the falling rate (CD) of drying commences. The mechanisms of the movement of moisture affect the drying during the falling rate periods.

Typical drying rate curve for constant drying conditions

Drying rate R

C

B

[kg H2O/kg dry solid.h] A D

E X*

Xc X’ moisture content [kg H2O/kg dry solid]

A = initial wet condition BC = constant rate period CD = First falling rate period (capillary-controlled) DE = second falling rate period (diffusion-controlled) E = equilibrium moisture content X*

Liquid diffusion theory Diffusion of liquid moisture occurs when there is a concentration difference between the depths of the solid and the surface. This mechanism is found in: 6

-

Non-porous solids where single phase solutions are formed with the moisture (paste, soap, gelatine, glue) Flour, wood, leather, paper, starches and textiles when last portions of moisture are removed.

The rate of moisture diffusion is rate – controlling (the rate of evaporation from the surface is quite fast). Since the moisture diffusivity usually decreases with decreased moisture content, the drying rate appears to be a non-linear function of X. Therefore, very often a diffusion-controlled curve exhibits no first falling-rate period. Capillary movement in porous solids Porous solids (clays, sand, soil, paint pigments, minerals) contain inter-connecting pores and channels of varying pore sizes. As water is evaporated a meniscus of liquid water is formed across each pore in the depth of the solid. This sets up capillary forces by the interfacial tension between solid and water. These capillary forces provide the driving force for moving water through the pores to the surface. Small pores develop larger forces than large pores. Drying rate R

Drying rate R

X, Free moisture content

X, Free moisture content

(a)

(b)

(a) – diffusion rate-controlled

(b) – capillary-controlled

At the beginning of the falling-rate period (point C) the water is being brought to the surface by capillary action, but the surface layer starts to recede below the surface. Air rushes in to fill the voids. As the water is continuously being removed, a point is reached where there is insufficient water left to maintain continuous films across the pores, and the rate of drying suddenly decreases at the start of the second falling-rate period (point D). Then the rate of diffusion of water vapour in the pores and the rate of heat conduction in the solid may be the main factor in drying. Effect of shrinkage The shrinkage of the solid, as moisture is removed, may affect the rate of drying significantly. Materials undergoing shrinkage: colloidal and fibrous materials (e.g. vegetables), other foodstuffs, clay, soap, wood. During drying of these materials, a hard layer may develop on the surface which significantly retards the flow of liquid or vapour moisture. In many foodstuffs a layer of closely packed cells is formed at the surface which presents a barrier to moisture migration. Rigid solids do not shrink appreciably. Shrinkage can be reduced by drying with moist air (i.e. at lower rates of drying). 7

Prediction of the constant rate of drying Assumption: no radiation and conduction. Steady-state mass and heat transfer (vapour at surface is saturated). Temperature at surface = wet bulb temp Tw Q NA Tw; Yw; yw Gas; T1, Y1 ;y Q Moisture to surface

The rate of convective heat transfer from the gas at temp T to the surface of the solid at temp Tw is:

Q = h (T-Tw) A

(3)

H = heat transfer coefficient [W/m2.K];

A = exposed drying area

The mass transfer flux from the surface:

N A = k y (y w - y) =

k' y y Bm

( yw  y)

(4)

A = water; B = air; NA = no mole A (water) transferred per m2 per s; yBm log mean inert (air) mole fraction; yw = mole fraction of water vapour in the gas at the surface; y = mole fraction of water in the flowing air; ky ; k’y mass transfer coefficients (k’y for one-way transfer)

The mole fractions y and yw can be expressed in terms of humidities (H )

Humidity Y 

M A pA M y  A M B Pt  p A M B 1  y

(5)

pA = partial pressure (of water); Pt = total pressure (atm); Y = humidity [kg H2O/kg dry air] Since usually y T1), Tw is also increased (Tw2 > Tw1), but not as much as T. Hence R will increase (R2 > R1). Effect of thickness of solid being dried For heat transfer by convection only, the rate of drying is independent of the thickness of the solid.

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Example 4 Using as basis the data of Example 1 and 2 in the notes, calculate the time required in an industrial tray dryer to reduce the moisture content for the same material from 50% by mass to 5 % by mass (both on a wet basis). The cake will be 25 mm thick. The air in the drier will be 80°C and will have a relative humidity of 20%. Assume that the velocity at the surface of the cake in the industrial drier will be similar to that in the test dryer. Assume that the exposed area of the cake is the same as in Example 1. SOLUTION What remains the same? Velocity, area exposed, humidity (from psychrometric chart) What is different? T = 80°C, Tw = 56°C, ms , density of air (hence mass flow-rate G of air) ms is in proportion to thickness of cake = 48 x (25/10) = 120g

Influence of temperature on density of gas: Density is inversely proportional to absolute temperature :

 2 T1 G T 273  65  338   2  1  1 T2 G1 T2 273  80 353

P1V1 P2V2 V T   1  1  T1 T2 V2 T2

Since air mass flow-rate is different [(using eqn (22)]:

R2  G 2    R1  G1 

m

R  338   2   R1  353 

0.8

 R2  0.255 x 0. 966  0.246 kg H 2 O / kg dry solid

Influence of temperature on drying rate[(using eqn (24)]: with T1 = 65, Tw1 = 55; T2 = 80, Tw2 = 56, ɣw1 = ɣw2

R2 T2  Tw2  w1  . R1 T1  Tw1  w2

R2 80  56 24   R2  0.246 x  0.5904 kgH 2 O / kg dry solid R1 65  55 10

To calculate drying time use equation (20) with X1 = 1.0, X2 = 0.0526, Xc = 0.55 , A = 0.007854 :

tT 

ms  Xc  0.120 0.55     1 . 0  0 . 55  0 . 55 ln  45 min .  X 1  X c   X c ln  ARc  X 2  0.007854 x 0.5904  0.0526 

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Material and Heat balances of Continuous dryers Example: Tunnel dryer

Q (heat lost) 2

1

GAS

TG1, Y1

GS,, TG2, Y2 SOLID TS2, X2

LS,, TS1, X1

X = total moisture content (not free) of solid [kg H2O/kg dry solid] Y= humidity of air (kg H2O/kg dry air] Enthalpy of gas: HG = cH (TG – T0) where T0 = reference temp (0°C) and cH = humid heat

: HG [kJ/kg dr air]

cH = C air + CH2OY = 1.005 + 1.884 Y [kJ/kgK]

(22)

CH2O = spec. Heat capacity of water vapour [kJ/kgK] Thus HG = cH TG

if TG in °C

Enthalpy of wet solid:

HS = (cS + cL X) (TG – T0) = (cS + cL X) (TG )

(23)

if T0 = 0°C TG in °C : HS [kJ/kg dry solid] cS = sp. heat capacity of dry solid; cL = sp. heat capacity of liquid moisture (water) Gs = dry gas flow [kg dry gas /h],

Ls = dry solid flow [kg dry solid /h]

Many of these values can be found form thep sychrometric chart. Heat balance over dryer:

Gs HG2 + LS HS1 = Gs HG1 + LS HS2 + Q

Q = 0 adiabatic process Material balance on the moisture:

(24)

Q > 0 heat losses

G s Y2 + L S X 1 = G s Y1 + L S X 2

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(25)

Temperature profiles in continuous counter-current drying

Preheat Zone

Zone I I Falling rate

Zone I Constant rate

Temperature

ΔT

TG Gas

TS Solid

Distance through dryer

The time of drying in the constant rate period can be calculated from:

t

Gs Y  Yc 1 ln w Ls Ak y M air Yw  Y1

Yc  Y2 

Ls X c  X 2  Gs

(25) where

(26

For the falling rate period the drying time can be calculated from:

t

Gs Xc . Ls Ak y M air

 X Y  Y2  1 ln  c w  Gs X 2 Yw  Yc   Yw  Y2   X 2 Ls

(27)

ky = the mass transfer coefficient for water vapour from wet surface to air

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