85
Penyebab Perpindahan: Perbedaan potensial Perpindahan Massa Contoh potensial: Kecepatan : perpindahan momentum Suhu : perpindahan panas Konsentrasi : perpindahan massa Mekanisme Perpindahan: Difusi : gerakan -random- molekuler (atomik) medium ataupun gerakan partikel turbulen Konveksi : aliran medium

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Page 1: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

Penyebab Perpindahan: Perbedaan potensial

Perpindahan Massa

Contoh potensial: Kecepatan : perpindahan momentumSuhu : perpindahan panasKonsentrasi : perpindahan massa

Mekanisme Perpindahan:Difusi : gerakan -random- molekuler (atomik) medium

ataupun gerakan partikel turbulen

Konveksi : aliran medium

Page 2: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

DifusiPersamaan-persamaan Konstitutif Difusi:

Momentum: Hukum Newton (fluida Newtonian) y

vyv xxyx

Energi: Hukum Fourier xT

xT

ck

c pp

x

Massa: Hukum Fick xDj

Am A

ABAB

Page 3: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

Difusi Gas (Sistem Biner) DAB vs DBA

Dalam keadaan tunak (steady): setiap molekul A diganti oleh setiap molekul B dan sebaliknya, sehingga:

BA NN B

B

A

A

MAm

MAm //

xMD

xMD B

B

BAA

A

AB

xp

TRM

x g

xp

TRM

MD

xp

TRM

MD B

g

B

B

BAA

g

A

A

AB

xp

Dx

pD A

BAA

AB

DAB = DBA

Page 4: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

Difusifitas massa: DD berbanding lurus dengan vrms dan jejak bebas rata-rata :

rmsvD 2/1

MTvrms pA

TnA

1

avBA pAT

MT

MTD

2/1

BABA

AB MMVVp

TD 117,435 23/13/1

2/3

Rumus Semiempirik dari Gilliland:

Page 5: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

ContohKoefisien difusi CO2 didalam udara pada tekanan 1 atmosfir dan suhu 25 oC adalah:

Dari tabel volume molekul: VCO2= 34 [lt/kgmole] dan Vud= 29,9 [lt/kgmole]

Massa molekul: MCO2=44 dan Mud=28,9

9,28

1441

9,29341013252987,435

23/13/1

2/3

,2

UdCOD

]/[cm 132,0 2,2 sD UdCO

]/[cm 164,0 2,2 sD UdCO Dari tabel:

Page 6: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

BABA

AB MMVVp

TD 1110.3,4 23/13/1

2/39

Dari tabel volume molekul: VCO2= 34.10-3 [m3/kgmole] dan Vud= 29,9.10-3 [m3/kgmole]

Massa molekul: MCO2=44 dan Mud=28,9

9,28

1441

)10.9,29()10.34(129810.3,4

23/133/13

2/39

,2

UdCOD

]/[m .10 32,1 2-5,2 sD UdCO

Page 7: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

Fuller et al. (polar & non polar):

BABA

AB MMVVp

TD 1110.0,123/13/1

75,19

Hirschfelder et al. (non polar):

BADABAB MMp

TD 1110.858,12

5,127

2/BAAB 2/1

*

kkkBAAB

Jika tidak ada dalam tabel:

3/18,11 VAB bTk

21,1

Tb = normal boiling point

Page 8: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

Pengaruh Tekanan dan Temperatur (Hirschfelder)

2

1

2/3

1

2

2

11,2,2,2,

T

TTpABTpAB T

TppDD

Langkah-langkah (Hirschfelder) :Tabel: Hitung:Tabel:

Hitung: DAB

.....;.....;.....; .....; kkBA

BA

.....;.....;.....; AB

ABAB

kTk

.....;D

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Beberapa definisi:Hubungan konsentrasi dan fluks:

iii ppmm ; ;

Fraksi massa: 1 ; iii

i mm

Konst molar [kgmole/m3]:i

iiii MV

MmC

/

Fraksi mole: 1 ; ; ixCCCCx i

ii

Fluks molar Ni dan Fluks massa ni:

iii uCN iii un Fluks Ni dan ni relatif terhadap kecepatan rujukan:

oiiio uuCJ oiiio uuj

io uu i

Page 10: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

Kecepatan rata-rata (rujukan):

i

iii m

VMV

Fluks Ni dan ni relatif terhadap kecepatan rujukan:

miiim uuCJ miiim uuj

Kecepatan rata-rata massa:

Kecepatan rata-rata molar:

Kecepatan rata-rata volume:

iim uu

iM uxu i

iiv uCVu i

1/i VVCV ii

MiiiM uuCJ

viiiv uuCJ

MiiiM uuj

viiiv uuj

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Perpindahan Massa KonveksiAnalisa Dimensional

0,,,,, lhDuf mABfe

mdAB

cba lhDu 6 buah pangkat dgn 3 persamaan (M,L,T)

akan menghasilkan 3 buah bilangan tak berdimensi, diantaranya:

ulπ

uhDSc mAB Re321 ; ;

/

Bilangan tak berdimensi yang lain bisa diturunkan dari kombinasi nya, diantaranya 23/1:

AB

m

DlhSh

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Analogi:Perpindahan panas dalam pipa:

8Pr f

cuh

pm

Perpindahan massa dalam pipa:

8fSc

uh

m

m

3/23/23/2

Pr/ Lec

DcScchh p

ABppm

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Single droplet evaporation and drying[A S Mujumdar and Lixin Huang, 2005]

Problem statement Water droplet evaporation experiment is carried out. Variation of the mass of droplet vs. drying time is shown in Figure 1. The experimental conditions are as follows:Temperature of air surrounding water droplet, oC = 48.4Temperature of droplet, oC = 27.2Wet bulb temperature of ambient air, oC = 21.5Dry bulb temperature of ambient air, oC = 26.0Velocity of air, v (m/s) = 1.0Initial diameter of water droplet, D (m) = 0.002052Initial mass of droplet, m (g) = 0.004522The observed drying time, t (s) =354(s)

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From above data:

1. Determine the evaporation rate using energy balance and mass transfer/diffusion methods. Also, compare the numerical results with the attached experimental results.

2. Discuss the effects of air humidity, air velocity, air operating pressure and temperature on drying performance

Figure 1 Mass of water droplet vs. drying time at air temperature 48.4oC and air velocity 1.0m/s

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Solutions: (Method-1: based on the energy balance)

From tables,Pr = 0.7027µa = 19.46 x 10-6 kg/msρa = 1.098kg/m3

Kd = 27.7895 x 10-3 W/m.KFor estimation of the convective heat transfer coefficient, we have the following correlation:

33.05.0

Pr6.00.2

a

a

d

c DvK

Dh

Substituting the corresponding values, Eq 1 yieldshc= 105.28W/m2.KDroplet surface area: A= πD2= 1.323 x 10-5 m2

Temperature difference between drying air and droplet: ∆T= 48.4 – 27.2 = 21.2 oC

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Latent heat of water: λ= 2417.44 kJ/kgThe evaporation rate of this droplet is calculated from

ThcANc

Nc= [(105.28) * (1.323 x10-5) * (21.2)] / (2417.44)= 1.22 x 10-5 g/s

From the graph in Figure 1, the evaporation rate is obtained as 1,28 10-6 g/sThus, the error by computation is = [(1.28 x10-5- 1.22 x 10-5)/ 1.28 x10-5] x 100%= 4.7%, which is negligible. In fact the error in estimation of h using empirical correlations can exceed 10%

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2/13/12/13/1 )()(6.00.2Re6.00.2a

a

ga

ap

g

c DvD

ScD

DkSh

Solution: (Method-2: Based on mass transfer/diffusion equation)

For mass transfer from a spherical droplet subjected to a relative velocity of a drying medium, we have the Sherwood number correlation as follows:

From property tables,Pr = 0.7027µa = 19.46 x 10-6 kg/msρa = 1.098kg/m3

The diffusion coefficient for vapor in air at 50oC is 2,91 x 10-5 m2/s

Page 18: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

)/(0776.0

)1046.19

098.10.1002052.0()1091.2098.1

1046.19(6.00.2[002052.0

1091.2

])()(6.00.2[

2/16

3/15

65

2/13/1

sm

DvDD

Dk

a

a

ga

agc

Based on the initial dry-bulb and wet-bulb temperatures of air, the humidity is found from the psychrometric chart as 0.013 kg H2O/kg dryair.Then, the vapor concentration at the droplet surface (Cs) (kmol/m3);

vapor concentration in the bulk gas (Cg). They are defined as

)/(10443.13008314

3600)( 33 mmolRT

TpC

p

psats

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)/(1078.7

4.3218314101325

18/013.029/0.118/013.0

34 mmolC

RTp

XC

g

g

opig

Based on the mass transfer equation, the mass transfer rate from the water droplet surface to the bulk air is calculated as

)/(1023.1

)1078.710443.1(101810323.10776.0

)(

5

4335

sgN

N

CCMAkN

c

c

gsppcc

)(6.3701022.1

100.1002052.066

5

633

stN

Dt

c

w

%9.3

%1001028.1

1023.11028.15

55

The drying time is then given byThe calculation error is then:

Page 20: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

)/(1053.9

4.3218314101325

18/016.029/0.118/016.0

34 mmolC

RTp

XC

g

g

opig

)/(1006.9

)1053.910443.1(101810323.10776.0

)(

6

4335

sgN

N

CCMAkN

c

c

gsppcc

Discussion about the effects of the operating parameters

(a) Effect of air humidityAssume the absolute humidity is increased to 0.016 kg H2O/kg dry air.

Then,

Now the evaporation rate becomes

Page 21: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

and the drying time becomes:

)(4991006.9

100.1002052.066

6

633

sN

Dt

c

w

%3.26%1001023.1

1006.91023.15

65

The reduced percentage of the evaporation rate is due to the increase of humidity in air is computed by

The increase of drying time is (499-370) = 129(s). Reduction in mass transfer driving potential is responsible for this reduction.

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)/(1089.3

4.321831450663

18/013.029/0.118/013.0

34 mmolC

RTp

XC

g

g

opig

)/(1095.1

)1089.310443.1(101810323.10776.0

)(

6

4335

sgN

N

CCMAkN

c

c

gsppcc

(b) Effect of the operating pressureAssume the operating pressure is only half of the normal ambient atmospheric pressure, i.e., 50663Pa. Therefore,

Then the evaporation rate becomes

Page 23: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

)(2321095.1

100.1002052.066

5

633

sN

Dt

c

w

%5.58%1001023.1

1095.11023.15

55

The drying time becomes:

The increased percentage of the evaporation rate due to the decrease of the operating pressure in air is computed by

The decrease of drying time is (370-232) = 138(s)

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)/(2019.0

)1046.19

098.10.5002052.0()1091.2098.1

1046.19(6.00.2[002052.0

1091.2

])()(6.00.2[

2/16

3/15

65

2/13/1

smk

k

DvDD

Dk

c

c

a

a

ga

agc

)/(1020.3

)1078.710443.1(101810323.12019.0

)(

5

4335

sgN

N

CCMAkN

c

c

gsppcc

(c) Effect of the air velocityAssume the air velocity is increased to 5.0m/s. Then,

The evaporation rate now becomes

Page 25: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

)(1411020.3

100.1002052.066

5

633

sN

Dt

c

w

%160%1001023.1

102.31023.15

55

The drying time therefore is:

The increased percentage of the evaporation rate due to increase of air velocity is computed from:

The decrease of drying time is (370-141) s i.e. 229(s)

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(d) Effect of the air temperature

Assume the air temperature is increased to 71.85oC. From the corresponding tables, the following parameter values for air can be obtained

Pr = 0.698µa = 20.52 x 10-6 kg/msρa = 1.023kg/m3

Dg =3.24x10-5m/s

Then the mass transfer coefficient is calculated by

)/(113.0

)1052.20

023.10.1002052.0()1024.3023.1

1052.20(6.00.2[002052.0

1024.3

])()(6.00.2[

2/16

3/15

65

2/13/1

smk

k

DvDD

Dk

c

c

a

a

ga

agc

Page 27: 3,4,5,6_Fenomena Perpindahan dalam Pengering1.pptx

)/(1079.1

)1078.710443.1(101810323.1113.0

)(

5

4335

sgN

N

CCMAkN

c

c

gsppcc

)(2521079.1

100.1002052.066

5

633

sN

Dt

c

w

%5.45%1001023.1

1079.11023.15

55

Then the evaporation rate becomes

The drying time becomes:

The increased percentage of the evaporation rate due to the increase of air temperature is computed by

The decrease of drying time is (370-252) =118(s)

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From the above computation, we can conclude that:

(1) When the air humidity is increased, the evaporation rate is decreased.

(2) When the air operating pressure is decreased, the evaporation rate is increased and vice versa.

(3) When the relative velocity between air and droplet is increased, the evaporation rate is increased, as well.

(4) When the air temperature is increased, the evaporation rate is increased.

(5) Among the four affecting factors, we can see that the air temperature and relative velocity between air and droplet play a significant role on the drying performance.

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1. The relative velocity between air and droplet is not easy to control

2. The air temperature is always very important in spray dryer

a) Easy to control b) Affects the drying performance significantly

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Geoffrey Lee, 3rd Report to Niro, June 2006

Droplet Drying Kinetics of Water atTda = 25 oC, 40 oC & 60 oC

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Geoffrey Lee, 3rd Report to Niro, June 2006

Droplet Evaporation Kinetics: d2 Law

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The d2 Law: A Gas-Phase Model

Sphere

constant temperature

no convection

saturation vapor pressure at droplet surface

steady state vapour diffusion in gas phase

Geoffrey Lee, 3rd Report to Niro, June 2006

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Geoffrey Lee, 3rd Report to Niro, June 2006

The d2 Law:

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Geoffrey Lee, 3rd Report to Niro, June 2006

Droplet Drying Kinetics of Water:Comparison with d2 Law

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Drying EquationsFluidized beds and spray driersUse correlation for flow past an individual spherical particle

Example-1Wet granules (density = 1.5 g/cm3) are spread onto a screen at an amount of 10 kg/m2 of screen. The bed porosity is 45% and the wet granules have an average diameter of 300 m. Air at 60ºC(dry bulb) and a wet bulb temp of 25ºC is passed through the bed at a velocity of 0.15 m/s. How long will it take to dry the granules from 20 to 10 % moisture? The critical moisture content of the granules is 9 %.

33.05.0

Pr6.00.2

a

a

d

c DvK

DhNu

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Untuk berkembang penuh:

Nud=0.023Red0,8Prn

dengan analogi:

Shd=0.023Red0,8Scn

Untuk daerah masuk:

Nud=0.036Red0,8Pr1/3(d/L)0,055

dengan analogi:

Shd=0.036Red0,8Sc1/3(d/L)0,055).

Example-2: Falling Film

Example-3: Flat Plate

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1. Effective moisture diffusivity2. Effective thermal conductivity3. Air boundary heat and mass transfer

coefficients4. Drying constant5. Equilibrium material moisture content

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Moisture Diffusivity

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Permeation Method A steady-state method applied to a film of material. A material of known thickness. Surface concentrations: constant, well defined. Based on Fick’s diffusion equation.

Concentration–Distance CurvesThe concentration–distance curves method is based on the measurement of the distribution of the diffusant concentration as a function of time. Light interference methods, as well as radiation adsorption or simply gravimetric methods, can be used for concentration measurements. Various sample geometries can be used, for example semi infinite solid, two joint cylinders with the same or different material, and so on. The analysis is based on the solution of Fick’s equation.

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Other MethodsModern methods for the measurement of moisture profiles lead to diffusivity measurement methods. Such methods discussed in the literature are radiotracer methods, nuclear magnetic resonance (NMR), electron spin resonance (ESR), and the like.

Drying MethodsSimplified MethodsFick’s equation is solved analytically for certain sample geometries under the following assumptions: Surface mass transfer coefficient is high enough so that the material moisture content at the surface is in equilibrium with the air drying conditions. Air drying conditions are constant. Moisture diffusivity is constant, independent of material moisture content and temperature. The analytical solution for slab, spherical, or cylindrical samples is used in the analysis. Several alternatives exist concerning the methodology of estimation of diffusivity using the above equations. They are discussed in the COST 90bis project of European Economic Community (EEC) [16]. These alternatives differ essentially on the variable on which a regression analysis is applied.

Regular Regime MethodThe regular regime method is based on the experimental measurement of the regular regime curve, which is the drying curve when it becomes independent of the initial concentration profile. Using this method, the concentration-dependent diffusivity can be calculated from one experiment.

Numerical Solution—Regression Analysis MethodThe regression analysis method can be considered as a generalization of the other two types of methods. It can estimate simultaneously some additional transport properties; it is analyzed in detail in Section 4.7.

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Steady-State MethodsIn steady-state methods, the temperature distribution of the sample is measured at steady state, with the sample placed between a heat source and a heat sink. Different geometries can be used, those for longitudinal heat flow and radial heat flow.

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Steady-State MethodsIn steady-state methods, the temperature distribution of the sample is measured at steady state, with the sample placed between a heat source and a heat sink. Different geometries can be used, those for longitudinal heat flow and radial heat flow.

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Thermal Conductivity

Methods of Experimental Measurement

Thermal Conductivity

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Longitudinal Heat Flow (Guarded Hot Plate)The longitudinal heat flow (guarded hot plate) method is regarded as the most accurate and most widely used apparatus for the measurement of thermal conductivity of poor conductors of heat. This method is most suitable for dry homogeneous specimens in slab forms. The details of the technique are given by the American Society for Testing and Materials (ASTM) Standard C-177 [82].

Radial Heat FlowWhereas the longitudinal heat flow methods are most suitable for slab specimens, the radial heat flow techniques are used for loose, unconsolidated powder or granular materials. The methods can be classified as follows: Cylinder with or without end guards Sphere with central heating source Concentric cylinder comparative method

Unsteady State MethodsTransient-state or unsteady-state methods make use of either a line source of heat or plane sources of heat. In both cases, the usual procedure is to apply a steady heat flux to the specimen, which must be initially in thermal equilibrium, and to measure the temperature rise at some point in the specimen, resulting from this applied flux [83]. The Fitch method is one of the most common transient methods for measuring the thermal conductivity of poor conductors. This method was developed in 1935 and was described in the National Bureau of Standards Research Report No. 561. Experimental apparatus is commercially available.

Pro be Method The probe method is one of the most common transient methods using a line heat source. This method is simple and quick. The probe is a needle of good thermal conductivity that is provided with a heater wire over its length and some means of measuring the temperature at the center of its length. Having the probe embedded in the sample, the temperature response of the probe is measured in a step change of heat source and the thermal conductivity is estimated using the transient solution of Fourier’s law. Detailed descriptions as well as the necessary modifications for the application of the above-mentioned methods in food systems are given in Refs. [83,89,90].

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Interphase Heat and Mass Transfer Coefficients

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Factors Affecting the Heat and Mass Transfer Coefficients

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Theoretical Estimation

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Drying Constant

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Methods of Experimental MeasurementThe measurement of the drying constant is obtained from drying experiments. In a drying apparatus, the air temperature, humidity, and velocity are controlled and kept constant, whereas the material moisture content is monitored versus time. The drying constant is estimated by fitting the thin-layer equation to experimental data

Factors Affecting the Drying Constant

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Theoretical Estimation

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Equilibrium Moisture ContentSorption isotherms can be determined according to two basic principles, gravimetric and hygrometric.

Gravimetric Methods The air temperature and the water activity: kept constant until the

moisture content of the sample attains the constant value. The air may be circulated (dynamic methods) or stagnant (static). The material weight may be registered continuously (continuous

methods) or discontinuously (discontinuous methods).

Hygrometric Methods The material moisture content: kept constant until the surrounding

air attains the constant equilibrium value. The air–water activity is measured via hygrometer or manometer. The working group in the COST 90bis Project: a reference material

(microcrystalline cellulose, MCC) and a reference method.• A detailed procedure for the resulting standardized method.• The factors influencing the results of the method were discussed

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Factors Affecting the Equilibrium Moisture Content

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Experimental Drying Apparatus

Typical experimental drying apparatus: (1) sample; (2) air recirculating duct; (3) heater; (4) humidifier; (5) fan; (6) valve; (7) straighteners; FCR, airflow control and recording; HCR, air humidity control and recording; TCR, air temperature control and recording; WR, sample weight recording; TR, sample temperature recording; PC, personal computer, for on-line measurement and control.

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The Drying Model

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Drying Model

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Regression Analysis

Results

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Moisture Diffusivity

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Predicted values of moisture diffusivity of fruits at 250C.

Predicted values of moisture diffusivity of corn at 250C.

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Predicted values of moisture diffusivity of vegetables at 250C.

Predicted values of moisture diffusivity of cereal at 250C.

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Thermal Conductivity

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Predicted values of thermal conductivity of fruits at 250C.

Predicted values of thermal conductivity of vegetables at 250C.

Predicted values of thermal conductivity of miscellaneousat 250C.

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