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Lesson 11. RATE OF DRYING CONSTANT AND FALLING RATE, EFFECT OF SHRINKAGE
Module 2. Drying
Lesson 11
RATE OF DRYING CONSTANT AND FALLING RATE, EFFECT OF SHRINKING
11.1 Rate and Time of Drying
Drying process can be divided in to three periods
(ii) first falling drying rate period and
(iii) second falling rate period.
Fig 11.1 General rate of drying curve for solid material
Free Moisture is equal to ×_- ×*
(i) Constant drying rate period
Web bulb Temperature(WBT)
The air blown at high velocity (minimum recommended is 300 m/min). It causes evaporation of water from the wick. Evaporation requires latent heat. This heat comes from surface of glass bulb of thermometer. So the temperature of the glass bulb decreases. The the heat comes from the temperature difference between Tw and Ta (large). It is the case of simultaneous heat and mass transfer. This heat is latent heat for phase change of water to water vapor.
q = Mw Nw λw A----------------- (i)
Mw = mol. Mass of water kg/kg mol
Nw = molar flu× of water vapour, kg mol m-2 s-1
λw = Latest heat of vaporization kJ/kg
A = surface area of the bumb m2
Nw = ky (yw-y)
ky = mass transfer coefficient kg mol m-2 s-1
yw = mole fraction of water vapour in the stagnant air layer adjacent to the wet cloth
y = mole fraction of water vapour in the air stream, some distance away from the wet cloth
As long as the rate of surface evaporation continues, the rate of drying is governed by equation.
Where,
w = kg of moisture
Ø = time
h = heat transfer coefficient between air and moisture kcal/ kg hr °C
ta = dry bulb temperature of air, °C
ts = surface temperature, °C
A= area, m2
ΔHv = heat of vapourisation at ts ,Kcal/ kg
ka = mass transfer coefficient (kg/ hr m)
Hs = humidity of saturated air at the surface temperature
Ha = humidity of air
Drying in falling rate period involves two processes:
b) removal of the moisture from the surface
(ii) First falling drying rate period
Point B, the moisture content at the end of the constant rate period, is the ‘critical moisture content’. At this point the surface of the solid is no longer saturated, and the rate of drying decreases with the decrease in moisture content. At point C, the surface moisture film has evaporated fully, and with the further decrease in moisture content, the drying rate is controlled by the rate of moisture movement through the solid.
(iii) Second falling drying rate period
Period C to D represents conditions when the drying rate is largely independent of conditions outside the solid. The moisture transfer may be by any combination of liquid diffusion, capillary movement, and vapour diffusion.
11.2 Estimation of Drying Time
In order to determine the time required to achieve the desired reduction in product moisture content, the rate of moisture removal or drying rate must be predicted. The rate of drying depends on properties of drying air (the dry bulb temperature, RH, and velocity of air and the surface heat transfer coefficient), the properties of food (moisture content, surface to volume ratio and the surface temperature) and rate of moisture loss. The size of the pieces has an important effect on the drying rate in both the constant and falling rate periods. In the constant rate period, smaller pieces have a larger surface area available for evaporation where as in falling rate period smaller pieces have a shorter distance for moisture to travel through the food. Other factors which influence the rate of drying include:
2. The method of preparation of food (cut pieces lose moisture more quickly than losses through skin.
3. The amount of food placed in a dryer in relation to its size (in a given dryer faster drying is achieved with smaller quantities of food).
For constant rate drying period the following general e×pression would apply:
Where,
During falling rate drying, the following analysis would apply.
Where the limits of integration are between critical moisture content wc or end of constant rate drying, tc and some desired final moisture content, w.
On integration:
tf = wc/Rc × ln (wc/w) ------------(3) and
The total drying times becomes
t = (wo - wc) /Rc + Wc/Rc × ln (wc/w) ----------(4)
The above equation indicates that the time for complete drying from some initial moisture content wo to a desirable final moisture content w depends on knowledge of critical moisture content wc, the time for constant rate drying tc, and the rate for constant drying Rc.
Example 11.3
A tunnel dryer is being designed for drying apple halves from initial moisture content of 70 % (wet basis) to final moisture content of 5 % (wet basis). An e×perimental drying curve for the product indicates that the critical moisture content is 25 % (wet basis) and the time for constant drying is 5 min. Based on the information provided, estimate the total drying time for product.
Solution
Initial product moisture content, wo = 0.7 / 0.3 = 2.33 kg H2O / kg solids
Critical moisture content,wc = 0.25 / 0.75 = 0.333 kg H2O / kg solids
Final moisture content,w = 0.05 / 0.95 = 0.0526 kg H2O / kg solids
Time for constant rate drying, tc = 5 min
Required Total drying time
Falling rate drying time tF = wc/Rc × ln(wc/w)
= 1.54 min
Total drying time becomes t = 5 + 1.54 = 6.54 min.
Example 11.4
Estimate the drying rate and time needed to reduce the moisture content of a 100 mm diameter spherically shaped droplet, falling in a spray dryer from 60 to 35 %. The initial density of the droplet is 900 kg/m3. The droplet is in an air stream such that Ta = 200 oC, P = 101.3 k Pa, h = 200 w/m2 oC and Twb = 60 oC. Assume that constant rate drying applies over the total drying process and droplet doesn’t change in size.
Solution
The surface area of the droplet is π D2 = π (0.0001)2 = 3.14 × 10-8 m3
The volume of the droplet is 1/6 π D 3 = π /6 × (0.0001)3 = 5.24 × 10-13 m3
The initial mass of the droplet is r× V = 900 × 5.24 × 10-13
Applying solids balance: mp × 0.4 = mf × 0.65
So final mass of the product mf = mi × 0.4 / 0.65 = 2.9 × 10-10 kg.
Drying rate R = h A (Ta - Ts)
hfg × 1000
where,
hfg = latent heat of evaporation corresponding to given pressure in kJ / kg.
2370.7 × 1000 = 3.71 × 10-10 kg s
The drying time is given by t = (mi - mf) / R = (4.71 – 2.9) 10-10 / 3.71 10-10
Effect of Shrinkage
A factor often greatly affecting the drying rate is the shrinkage of the solid as moisture is removed. Rigid solids do not shrink appreciably, but colloidal and fibrous materials such as milk, vegetables and other foodstuffs do undergo shrinkage. The most serious effect is that there amy be developed a hard layer on the surface which is impervious to the flow of liquid or vapor moisture and slows and drying rate. In many foodstuffs, if drying occurs at too high temperature, a layuer of closely packed, shrunken cells, which are sealed together, forms at the surface. This presents a barrier to moisture migration and is known as case hardening. Another effect of shrinkage is to cause the materials to warp and change its structure. Some times, to decrese these effects of shrinkage, it is desirable to dry with moist air. This decreases the rate of drying so that the effets of shrinkage on warping or hardening at the surface are greatly reduced.
Example to be incorporated. And modify the above paragraph.