4 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.
5 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
10 Drying rate, dW/dt (kg/h) time (hrs)ABCDMoisture content(kg /kg dry solids)Drying rate, dW/dt (kg/h)tc1b1aFig. 1a &b Typical drying rate curve of for food solids.
11 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.
24 Rate of drying curves for constant-drying conditions Conversion of data to rate-of-drying curveData 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).
25 Slope = R = drying rate X = free moisture (kg water/kg dry solid) R = drying rate (kg water/h.m2)Ls = kg of dry solidA = exposed surface area for drying (m2)
26 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 hTime for drying == 1.80 h.
27 Constant drying rate period R = drying rate (kg dry solid/h.m2)Rc = drying rate for constant drying rate period = constant
28 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/m2Rc = 1.51 kg/h.m2
29 Method to predict transfer coefficient for constant-rate period NAMA=RcMA NA A=m=RcAq = mwNA = mass flux = kg mol/s.m2, M = molecular weight, H = humidity ratioy = 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 vaporizationRelate drying rate with heat transfer and mass transfer to determine the coefficient
30 For air temperature of 45-150C and mass velocity G of 2450-29300 kg/h For air temperature of C and mass velocity G of kg/h.m2 or a velocity of m/s and air is flowing parallel to drying surface.For air temperature of C and mass velocity G of kg/h.m2 or a velocity of m/s and air is flowing perpendicular to drying surface.Mass velocity (G) = vTo estimate time of drying during constant-rate period:
40 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.
41 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.
42 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.
43 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)
44 For counter-current system Overall moisture balancema W2 + mp w1 = ma W1 + mp w2Energy balancema Ha2 + mp Hp1 = ma Ha1 + mp Hp2 + qlossq = heat lossesHa = heat content of air (kJ/kg dry air)Hp = heat content of product (kJ/kg dry solids)
45 Ha = CS (Ta – TO )+ WHLCS = humid heat (kJ/kg dry air.K)= WTa = 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
46 ExampleA 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.
47 ExampleA 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.