DRYING - Universiti Teknologi MalaysiaDRYING • freeze-dried for biological & pharmaceutical...
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DRYING • freeze-dried for biological & pharmaceutical materials • as a preservative technique esp. food • removal of relatively small amount of water or organic liquids • final processing step before packaging 2 methods of drying: 1. batch - material put into dryer & drying proceeds for a given period of time 2. continuous - material continously added to dryer & continously removed 3 catagories of drying: 1. Direct contact with heated air at atmospheric pressure 2. Vacuum drying - heated indirectly either by contact with a metal wall or by radiation 3. freeze drying - water is sublimed from the frozen material
Transcript of DRYING - Universiti Teknologi MalaysiaDRYING • freeze-dried for biological & pharmaceutical...
DRYING
• freeze-dried for biological & pharmaceutical materials
• as a preservative technique esp. food
• removal of relatively small amount of water or organic liquids
• final processing step before packaging
2 methods of drying:
1. batch - material put into dryer & drying proceeds for a given period of time
2. continuous - material continously added to dryer & continously removed
3 catagories of drying:
1. Direct contact with heated air at atmospheric pressure
2. Vacuum drying - heated indirectly either by contact with a metal wall or by
radiation
3. freeze drying - water is sublimed from the frozen material
EQUILIBRIUM MOISTURE CONTENT, X*
• varies greatly with type of material for
any given % relative humidity
• depends on structure of solid, temperature
& moisture content of gas
• lowest moisture content obtainable at equilibrium
• on dry basis (kg of water/ kg of moisture-free solid)
• decreases with increase in temperature
• assumed constant for moderate temperature
ranges
EQUILIBRIUM MOISTURE CONTENT, X*
• Unbound water = excess water held primarily in the voids of the solid
• free moisture content, X - moisture above the equilibrium moisture content
- can be removed by drying
X = Xt - X*
where
Xt = total moisture content
X* = equilibrium-moisture content – cannot be predicted
• bound water - the minimum moisture a material can carry
- intersection of 100% humidity line in equillibrium water
content vs relative humidity
Moisture amount that
can be release/free
during drying
Drying Curve
Drying Curve
RATE OF DRYING CURVES
• experimental determination
data : WS = weight of dry solid
W = total weight of wet solid vs time t
To obtain as free moisture content X vs time t:
free moisture content, X = Xt - X*
• batch drying
total moisture content , Xt =
WWS
WS
Measure sample every
minutes
RATE OF DRYING CURVES
• experimental determination
• batch drying
To obtain as rate of drying, R :
Get slopes of tangents at different values of t :
where
A = exposed surface area for drying
RWs
AdXdt
RATE OF DRYING CURVES
X = (0.35 + 0.325)/2 = 0.338
R = -21.5 (-0.07) = 1.493
dX
dtX
t
(0.35 - 0.325)
(1.68 - 2.04) 0.07
RATE OF DRYING CURVES
Point A’ : hot solid
Point B-C: constant-rate drying period in which 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. Surface temperature
TW
Point AB : Warming up (unsteady) period where the
solid surface conditions come into equilibrium with the
drying air.
Point C : critical free moisture content, XC , where
the drying rate starts falling and surface
temperature rises. Insufficient water on surface
RATE OF DRYING CURVES
Point C-D : first falling-rate drying period which
surface is drying out. Rate of water to surface is less
that rate of evaporation from surface
Point D : surface completely dry
Point D-E : second falling-rate period in which
evaporation is from inside of solid.
Point E : equilibrium moisture content, X*,
where no further drying occur
CONSTANT RATE OF DRYING PERIOD
1. Experimental drying curve
2. Predicted mass-and-heat coefficients
Experimental drying curve:
1. Drying curve X vs. t
2. Rate-of-drying curve R vs. X
To determine the time required for drying from X1 to X2:
Under similar conditions to actual process
where
RC = constant rate of drying
tW
S
ARC
(X1X
2)
WS = kg of dry solid used
A = exposed surface area for drying
FALLING-RATE OF DRYING PERIOD
1. Graphical integration
To determine the time required for drying from X1 to X2:
t W
S
A
dX
RX
2
X1
WS
AX
1
R
av
Most accurate
FALLING-RATE OF DRYING PERIOD
2. Special cases
To determine the time required for drying from X1 to X2:
a) Rate is linear function of X
tW
S(X
1X
2)
A(R1R
2)
lnR1
R2
FALLING-RATE OF DRYING PERIOD
2. Special cases
b) Rate is a linear function thru’ origin (a straight line from C to E at the
origin)
To determine the time required for drying from X1 to X2:
or
and
tW
SXC
ARC
lnRC
R2
tW
SXC
ARC
lnXC
X2
RRC
X
XC
EXAMPLE 9.7-1
EXAMPLE 9.7-2
Repeat Example 9.7-1, but as an approximation assume a straight line for
the rate R versus X through the origin from point Xc to X = 0 for the
falling-rate period.
Drying Curve
HUMIDITY & HUMIDITY CHART
• Dew point - temp. at which a mixture of air-water would be saturated
• Humidity, H - kg of water vapour in 1 kg of dry air
H18.0228.97
pA
PpA
• Saturation humidity, HS
HS18.0228.97
pAS
PpAS
• Percentage humidity, HP
Hp100 H
HS
• Percentage relative humidity, HR ( HR HP)
HR100 pA
pAS
where pA = partial pressure of water vapour in air
H = humidity of air
pAS = saturated partial pressure of water vapour in air
HS = humidity of saturated air
HUMIDITY CHART
HUMIDITY & HUMIDITY CHART
• Humid heat, cS - amount of heat required to raise the temp. of 1 kg dry air plus
water vapour present by 1K
cS (kJ/kg dry air.K) = 1.005 + 1.88H
• Humid volume, H - total volume of 1 kg dry air plus water vapour present at 1
atm & given gas temperature
H (m3/kg dry air) = (2.83 x 10-3 + 4.56 x10-3H)T
T is in Kelvin
• H –Humidity found from chart
•Total enthalpy of 1 kg of air plus its water vapour, Hy
Hy (kJ/kg dry air) = (1.005 + 1.88H)(ToC- 0) + 2501.4H
DRY & WET BULB TEMPERATURE
Dry bulb temperature:
the ordinary temperature you
measure with a thermometer
wet cloth/wick
Air flow
Wet-bulb temperature : decreases in
temperature below the dry-bulb
temperature until the rate of heat transfer
from the warmer air to the wick is just
equal to the rate of heat transfer needed to
provide for the evaporation of water from
the wick into the air stream.
HUMIDITY CHART H
um
idity
0.0115
Wet bulb temp. =20oC, dry bulb temp. = 30oC,humidity = ?
HUMIDITY CHART H
um
idity
Importance of pyschrometric analysis for drying
Air: T1, H1 Air: T2, H2
T1 T2
H1
H2
Assumption: evaporation surface
is a liquid film
Constant wet-bulb temperature
process
CONSTANT RATE OF DRYING PERIOD
Predicted mass-and-heat coefficients:
Mass transfer of water vapour
Steady-state : rate of mass transfer = rate of heat transfer
To determine the time required for drying from X1 to X2:
Heat transfer furnishes the latent heat of evaporation
Assumptions:
1. Only convective heat transfer to solid surface from hot gas to
surface
2. Mass transfer is from surface to hot gas
CONSTANT RATE OF DRYING PERIOD
where
A = exposed drying area (m2)
W = latent heat at TW (J/kg)
Rate of drying, RC:
T, TW = temp. of gas & surface of solid, respectively (oC)
MA,MB = molecular weight of water & air, respectively
h = heat-transfer coefficient (W/m2.K)
RC q
AW
h(TT
W)
W
kyM
B(H
WH)
CONSTANT RATE OF DRYING PERIOD
Air flowing parallel to the drying surface
(T = 45-150oC, G = 2450 -29300 kg/h.m2, = 0.61-7.6 m/s)
h = 0.0204G0.8
Air flowing perpendicular to the drying surface
(G = 3900 -19500 kg/h.m2, = 0.9-4.6 m/s)
h = 1.17G0.37
where G = mass velocity =
To determine the time required for drying from X1 to X2:
tW
SW(X
1X
2)
Ah(T-TW)
W
S(X
1X
2)
AkyM
B(H
W-H )
Example 3 • A granular insoluble solid material wet with water is
being dried in the constant-rate period in a pan 0.61m x 0.61m and the depth of material is 25.4 mm. The sides and bottom are insulated. Air flows parallel to the top drying surface at a velocity of 3.05 m/s and has a dry bulb temperature of 60oC and a wet bulb temperature of 29.4oC. The pan contains 11.34 kg of dry solid having a free moisture content of 0.35 kg H2O/kg dry solid, and the material is to be dried in the constant-rate period to 0.22 kg H2O/kg dry solid.
a) Predict the drying rate period and the time in hours needed.
b) Predict the time needed if the depth of material is increased to 44.5 mm
Example 4
• An insoluble wet granular material is dried in a pan 0.457 x 0.457 m (1.5 x 1.5 ft) and 25.4 mm deep. The material is 25.4 mm deep in the pan, and the sides and bottom can be considered to be insulated. Heat transfer is by convection from an air stream flowing parallel to the surface at a velocity of 6.1 m/s (20 ft/s). The air is at 65.6oC (150oF) and has a humidity of 0.010 kg H2O/kg dry air. Estimate the rate of drying for the constant-rate period using SI and English units.
Industrial Training
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