Drying Foods

Geankoplis

Singh&Heldman

Bound and unbound water in solids

If the equilibrium moisture content of a given material is continued to its intersection with the 100% humidity line, the moisture is called bound water.

If such a material contains more than indicated by intersection with the 100% humidity line, it can still exert only a vapor pressure as high as that of ordinary water at the same temperature. This excess moisture content is called unbound water, and it is held primarily in the voids of solid.

Substances containing bound water are often called hygroscopic materials.

Free and equilibrium moisture of a substance

Free moisture content is the moisture above the equilibrium moisture content.

Free moisture content is the moisture that can be removed by drying under the given percent relative humidity

Drying

Methods of drying

Fig. 1a &b Typical drying rate curve of for food solids.

time (hrs)

A B

C

D

time (hrs)

Moisture content (kg /kg dry solids)

D

Drying rate, dW/dt (kg/h)

A

B C

tctc

1a 1b

AB = Settling down period where the solid surface conditions come into equilibrium with the drying air.

BC = Constant rate period which the surface of the solid remains saturated with liquid because the movement of water vapour to the surface equals the evaporation rate. Thus the drying rate depends on the rate of heat transfer to the drying surface and temperature remains constant.

C = Critical moisture content where the drying rate starts falling and surface temperature rises.

CD = Falling rate period which surface is drying out and the drying rate falls. This is influenced by the movement of moisture within the solid and take time.

Typical drying curve

Drying process

Rate of drying curves for constant-drying conditions

Conversion of data to rate-of-drying curve Data obtained as W total weight of solid at different times t hours in the drying period. Ws is weight of dry

soild.

If X* is equilibrium moisture content (kg moisture/kg dry solid), X is free moisture content (kg water/kg dry

solid).

Slope = R = drying rate

X = free moisture (kg water/kg dry solid)

R = drying rate (kg water/h.m2)

Ls = kg of dry solid

A = exposed surface area for drying (m2)

Time of drying from drying curve

A solid is to be dried from free moisture content X1 = 0.38 kg water/ kg dry soild to X2 = 0.25 kg water/ kg dry soild. Estimate the time required.

t1 = 1.28 h, t2 = 3.08 h Time for drying = 3.08-

1.28

= 1.80 h.

Constant drying rate period

R = drying rate (kg dry solid/h.m2)

Rc = drying rate for constant drying rate period = constant

Time of drying from drying curve

A solid is to be dried from free moisture content X1 = 0.38 kg water/ kg dry soild to X2 = 0.25

kg water/ kg dry soild. Estimate the time required.

Given: Ls/A = 21.5 kg/m2

Rc = 1.51 kg/h.m2

Method to predict transfer coefficient for constant-rate period

NA = mass flux = kg mol/s.m2, M = molecular weight, H = humidity ratio

y = mole fraction of water vapor in gas,

yw = mole fraction of water vapor in gas at surface,

A = water, B = air, ky = mass transfer coefficient, w = latent heat of vaporization

q = mw

MA NA A=m=RcA

NAMA=Rc

Relate drying rate with heat transfer and mass transfer to determine the coefficient

For air temperature of 45-150C and mass velocity G of 2450-29300

kg/h.m2 or a velocity of 0.61-7.6 m/s and air is flowing parallel to drying

surface.

For air temperature of 45-150C and mass velocity G of 3900-19500

kg/h.m2 or a velocity of 0.9-4.6 m/s and air is flowing perpendicular to

drying surface.

To estimate time of drying during constant-rate period:

Mass velocity (G) = v

Calculation methods for falling-rate drying period

Method using graphical integration

Calculation methods for special cases in falling-rate region

Rate is a linear function of X

Calculation methods for special cases in falling-rate region

Rate is a linear function through origin

Heat and Mass Transfer

Drying

Heat transfer occurs within the product structure and is related to the temperature

gradient between product surface and water surface at same location within the

product. The vapors are transported from water

surface within the product to product surface.

The gradient causing moisture-vapor diffusion is vapor pressure at water surface, compared to

vapor pressure of air at product surface. The heat and mass transfer within the product

structure occurs at molecular level, with heat transfer being limited by thermal conductivity of

product structure, while mass transfer is proportional to molecular diffusion of water

vapor in air.

At product surface, simultaneous heat and mass transfer occurs but is controlled by convective processes.

The transport of vapor from product surface to air and transfer of heat from air to product surface is a function of existing vapor pressure and temperature gradients, respectively.

Mass and energy balance for dehydration process

ma = air flow rate (kg dry air / hr)

mp = product flow rate (kg dry solids / hr)

W = absolute humidity (kg water / kg dry air)

w = product moisture content (kg water / kg dry solid)

For counter-current system

Overall moisture balance

ma W2 + mp w1 = ma W1 + mp w2 Energy balance

ma Ha2 + mp Hp1 = ma Ha1 + mp Hp2 + qloss

q = heat losses

Ha = heat content of air (kJ/kg dry air)

Hp = heat content of product (kJ/kg dry solids)

Ha = CS (Ta – TO )+ WHL

CS = humid heat (kJ/kg dry air.K)= 1.005 + 1.88 W

Ta = air temperature (C)

TO = reference temperature (0C)

HL = latent heat of vaporization of water (kJ/kg water)

HP = CPP (TP – TO )+ wCPw (TP – TO)

CPP= specific heat of dry solid (kJ/kg.K)

TP = product temperature (C)

CPW = specific heat of water

Example

A cabinet dryer is being used to dry a food product from 68% moisture content (wet

basis). The drying air enters the system at 54C and 10% RH and leaves at 30C and 70% RH. The product temperature is 25C throughout drying. Compute the quantity of air required for drying on the basis of 1

kg of product solids.

Example

A fluidized-bed dryer is being used to dried. The product enters the dryer with 60% moisture content (wet basis)

at 25C. The air used for drying enters the dryer at 120C after being heated from ambient air with 60%RH at 20C.

Estimate the production rate when air is entering the dryer at 700 kg dry air/hr and product leaving the dryer is

at 10% moisture content (wet basis). Assume product leaves the dryer at wet bulb temperature of air and the specific heat of product solid is 2.0 kJ/kg.C. Air leaves

the dryer 10C above the product temperature.

Drying Model: Thin layer drying Application: Dryer design, Simulation Three forms of model

Diffusion form D = diifusivity constant , D = f(T, RH, etc.)

Log model, modified log model MR = exp(-kt) MR = exp(-ktn) k = f(t, RH, IMC)

Quadratic form t = A ln(MR) + B[ln(MR)]2

A = f(T, RH, IMC), B = f(T, RH, IMC)