dWqh
A (Tdry - Twet)
--- =
- ----- = -------------------
dtll
1. Peas (average diameter = 6 mm, bulk density(a = 610 kg/m3) have to be dried in a conveyor drier from 78% to 16% moisture (wet basis). Air at 85°C (rh = 10%) is blown, at 0.9 Kg/m2 s, perpendicularly through the bed of peas. The bed is 10 cm deep and has a voidage(b of 40%. The drier belt is 0.75 m wide and 4 m long.
Additional dataequilibrium moisture content of peas = 9%critical moisture content = 300% (dry basis)
average diameter of peas = 6 mm
bulk density of peas = 610 kg/m3
mass transfer coefficient = 0.015 Kg /m2 s
Assuming that drying take places from the entire surface of the peas and there is no shrinkage, calculate the drying time and energy consumption in both the constant and falling -rate periods.
Notes:
(a Bulk density is defined as the mass of solids divided with the bulk volume" where the bulk volume is the volume including the air spaces between the solids (peas in this case).
(b Voidage: this is the average distance between particles (peas in this case)
In order to find the drying time in the constant-rate period we will use the following equation (previously derived):
dWqh
A (Tdry - Twet)
--- =
- ----- = -------------------
dtll
As first let' s calculate the heat transfer coefficient, since the air is blown perpendicularly we use:
h = 24.2 (G0.37)
where G is mass flow rate of air per unit area (Kg/m2 s) = 0.9.
Thus
h = (24.2) (O.9)0.37 = 23.2 W/m2 °C
The wet bulb temperature can be derived from the humidity chart. For air at 85° C and 10% r.h., we have that
Twet = 42 °C
Now let's calculate the surface of heat transfer, this is accomplished by calculating the surface area of 1 pea (assuming that each pea is a sphere) and then calculating the number of peas in the bed. The product of the area of 1 pea with the number of peas in the bed will give the total area of heat transfer.
Surface area of a single pea:
average area of 1 pea = 4pr2 = 4p (0.003)2 = 113 (10-6) m2
Now in order to find the total area we need to calculate the number of peas. As first we calculate the volume of the bed which is given by
volume of bed = (width)(length)(depth) = (0.75) (4) (0.1) = 0.3 m3
since there is a voidage of 0.4 cm this means the only the 60% of the volume above is occupied by peas and thus
volume of peas in the bed = (0.6) (0.3) = 0.18 m3
In other words, the effective length of the bed of peas is 60% of the belt length and thus:
volume of peas in the bed = (0.75) [(0.6) (4)] (0.1) = 0.18 m3
Now we calculate the average volume of 1 pea
average volume of 1 pea = 4/3 pr3 = (4/3) (3.142)(0.003)3 = 113 (10-9)
Therefore the total number of peas will be:
volume of peas in the bed/volume of 1 pea = 0.18/113 (10-9) = 1592920
and the total area is
total area = (number of peas) (area of 1 pea) = (1592920) (113 10-6) = 180 m2
Finally, we can calculate the drying rate using the formula
dWqh
A (Tdry - Twet)
--- =
- ----- = -------------------
dtll
from tables of steam latent heat is 2296 kJ/kg = 2296 103 J/Kg):
drying rate = (23.3) (180) (43)/ 2296 (103) = 0.078 kg s-1
In order to calculate the drying time, we need to calculate the kg of water removed which can be accomplished by mass balance. The mass of peas can be calculated using the bulk density (610 kg/m3) and the total volume of bed (0.3 m3)
mass = (610)(0.3) = 183 Kg
The initial moisture content is 78% which means that solids content is 22% therefore:
solids = (0.22)(mass of peas) = (0.22)(183) = 40.26 kg solidskg water = 183 - 40.26 = 142.74
Since the critical moisture content (i.e., the moisture at the end of constant-rate period) is 300% we have that
Kgwater/Kgdry solids = Kgwater/40.26 = 3Kgwater = (40.26) (3) = 120.78
Thus the mass of water lost is
142.74 - 120.78 = 21.96 Kg water lost
From the drying rate we can now calculate the drying time:
drying rate = 0.078 = 21.96/tt = 21.96/0.078 = 281 s = 4.7 min
To calculate the energy consumption we need to know the latent heat of evaporation at 85°C which can be read off from tables of steam (2296 kJ/kg = 2296 103 j/Kg):
energy = (drying rate) (latent heat) = (0.078 kg s-1) (2296 103) = 179088 W = 179 kW
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2. A food with 77% moisture content (wet basis) has a critical moisture content of 30% (wet basis). Consider that the food has a cube shape with 5-cm sides, a density of 950 kg/m3 and that drying take places from the entire surface of food. By knowig that the constant drying rate is 0.1 kg water/m2s, calculate the time required to reach the begin of the falling-rate period when one cube of food is dryed.
The time needed to reach the falling-rate period is calculated considering that the constant drying rate is given by:
rate = (Water removed)/m2s = 0.1 kg water/m2 s
The rate above is for a surface area of 1 m2. Now if we want the rate for one cube we need to know the area of the cube:
A = (0.0052)(6) = 0.015 m2
and therefore
rate = (0.1 kg/m2s) (0.015 m2) = 0.0015 kg/s
Now we can write
rate = (water removed)/s = 0.0015 kg/s
In order to calculate the drying time we need to calculate the amount of water removed from one piece of food. From the given density (950 kg/m3) we can calculate the initial mass of a cube (volume = 0.053):
m = (950)(0.05)3 = 0.11875 kg
Since the initial moisture content is 77%, the solids content is 23% and thus the amount of solids in one cube is:
(0.11875 kg)(0.23) = 0.0273 kg solids
and the amount of water is
0.1187 - 0.0273 = 0.0914 kg water
At the beginning of the falling-rate period the critical moisture is 30% (wet basis) and from definition of wet-basis moisture we have
0.3 = Wf/(Wf + Kg solids)
since the amount of solids is conserved:
0.3 = Wf/(Wf + 0.0273)
from which
Wf = 0.017
and the amount of water removed from one cube is
0.0914 - 0.0117 = O.O79 kg.
Finally
Rate = O.O79/t = 0.0015 kg/st = O.O79/0.0015 = 53 s
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3. A food is to be dried in a cabinet dryer from 68% moisure content (wet basis) to 5.5% moisture content (wet basis). Assuming that
1. air enters the system at 54°C and 10% RHCalculate the the quantity of air required to dry 1 kg of product.2. air leaves the system at 30°C and 70% RH
3. the temperature of the product is constant at 25°C
The mass balance around the dryer can be written as follows:
(kg air in)(Yin) + (kg dry solids in)(Win) = (kg air out)(Yout) + (kg dry solids out)(Wout)
where Y and W indicate the absolute humidity of air and the moisture content on dry basis.
Y can be deduced by the psychrometric chart and it results that
Yin = 0.0094 kg water/kg airYout = 0.0186 kg water/kg air
The moisture content on dry basis can be calculated from the moisture content on wet basis by using the formula previously described
Wdb = Wwb/1-Wwb
where Wdb and Wwb are the moisture content on dry and wet basis respectively. Therefore
Win = 0.68/(1-0.68) = 2.125 kg water/kg solids
Wout = 0.055/(1-0.055) = 0.0582 kg water/kg solids
By substituting these values in the equation of the mass balance (see above), we have
(kg air in)(0.0094) + (kg dry solids in)(2.125) = (kg air out)(0.0186) + (kg dry solids out)(0.0582)
It should be evident that:
kg dry solids in = kg dry solids out (only water is lost from product)kg air in = kg air out (no air is produced or transformed into the system)
and considering 1 kg of product, we have that
(kg air)(0.0094) + (2.125) = (kg air)(0.0186) + (0.0582)2.125 - 0.0582 = (kg air)(0.0186) - (kg air)(0.0094)
2.06 = 0.0092 kg air
kg air = 2.06/0.0092 = 224 kg air
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4. Diced carrots, with 60% wet-basis moisture content, enter a fluidized-bed dryer at 25°C and exit from the dryer with 10% moisture content (wet-basis). The product is dryed by using air (700 kg dry air/hr at 120° C) obtained by heating ambient air (20°C and 60% RH). Calculate the production rate of dryed carrots.
Assume that
product leaves the dryer at the wet-bulb temperature of airTotal Mass balanceair leaves the dryer with a temperature 10°C above the product temperature
the specific heat of solids is 2.0 kJ/kg °C.
the specific heat of air is 1.005 kJ/kg °C.
(Kg air in)(Yin) + (Kg drycarrot in)(Win) = (Kg air out)(Yout) + (Kg dry carrot out)(Wout)
where Y and W indicate the absolute humidity of air and the moisture content on dry basis.
The absolute humidity of entering air can be deduced from humidity chart and it results that
Yin = 0.009 kg water/kg dry air (at 20°C and 60% RH)
Note that the entering air is at 120° C, but it was obtained by heating air at at 20°C and 60% RH and thus these last two values should be used to calculate the absolute humidity.
The moisture content on dry basis can be calculated from the moisture content on wet basis by using the formula previously described
Wdb = Wwb/1-Wwb
where Wdb and Wwb are the moisture content on dry and wet basis respectively. Therefore
Win = 0.6/(1-0.6) = 1.5 kg water/kg solidsWout = 0.1/(1-0.1) = 0.11 kg water/kg solids
Considering that air in = air out and dry carrot in = dry carrot out (indicated as ms), see previous problem, we can write
In = (700 kg dry air)(0.009 kg water/kg dry air) + ms (1.5 kg water/kg solids)Out = (700 kg dry air)(Yout) + ms (0.11 kg water/kg solids)
6.3 kg water + ms (1.5 kg water/kg solids) = (700 kg dry air)(Yout) + ms (0.11 kg water/kg solids)
1.5 ms - 0.11 ms = 700 Yout - 6.3
1.39 ms = 700 Yout - 6.3Yout = (1.39 ms - 6.3)/700
Energy balance
The energy balance is made by taking 0°C as reference temperature. Let's first consider all the entering streams
Input
_Air: 700 kg of dry air at 120°C with specfic heat 1.005 kJ/kg/ °C
Hair = (1.005) (700) (120 - 0) = 84420 kJ
_Water vapor: air (700 kg at 120 °C) contains 0.009 kg water/kg dry air. From steam tables the enthalpy is 2705 kJ/kg and thus
Hsteam = (700)(0.009)(2705) = 17041 kJ
_solids: solids with specific heat 2.0 kJ/kg°C enters at 20°C thus
Hsolids = ms (2.0)(25-0) = (50) (ms)
_moisture: the enthalpy of water at 25°C is (steam table) 104.87 kJ/kg thus
Hmoisture = ms (1.5 kg water/kg solids)(104.87 kJ/kg) = ms (157.3)Input = 84420 kJ + 17041 kJ + (50) (ms) kJ + ms (157.3) kJ
Input = 101461 + (ms)(207.3)
Output
First let's calculate the wet bulb temperature of air. Drying air is at 120°C and was obtained by heating ambient air (25°C and 60% RH). The vapor pressure of water at 25°C is 3.17 kPa (steam table) and for the definition of RH we have
0.6 = Pw/Ps = Pw/3.17
from which Pw = (0.6)(3.17) = 1.90. At 120°C the vapor pressure is 198.5 (steam table) and thus
RH% = (1.90)(100)/(198.5) = 0.96%
From psychrometric chart the wet bul-temperature, for air with 0.95% RH and Tdry = 120°C, results to be 38°C. Therefore according to the given conditions:
Tproduct = 38Tair = 38 + 10 = 48°C
_air: _Air: 700 kg of dry air at 48°C with specfic heat 1.005 kJ/kg/ °C
Hair = (1.005) (700) (48 - 0) = 33768 kJ
_Water vapor: air (700 kg at 120 °C) contains 0.11 kg water/kg dry air. From steam tables the enthalpy is 2705 kJ/kg and thus
Hsteam = (700)(Yout)(2592) = 1814400 (Yout) kJ
_solids: solids with specific heat 2.0 kJ/kg°C exit at 38°C thus
Hsolids = ms (2.0)(38-0) = 76 (ms)
_moisture: the enthalpy of water at 38°C is (steam table) 209.31 kJ/kg thus
Hmoisture = ms (0.11 kg water/kg solids)(209.31 kJ/kg) = ms (23.023)Output = 33768 kJ + 199584 (Yout) kJ + (76) (ms) + ms (23.023)
Output = 33768 kJ + 199584 (Yout) kJ + 99 (ms)
By matching In and Out:
101461 + (ms)(207.3) = 33768 + 1814400 (Yout)+ 99 (ms)(ms)(207.3) - 99 (ms) = 33768 + 1814400 (Yout) - 101461
108 ms = 1814400 (Yout) - 67693
From mass balance we get
Yout = (1.39 ms - 6.3)/700
Thus
108 ms = 1814400 [(1.39 ms - 6.3)/700] - 67693