Problem1 Drying

1
Problem-1 (Drying)

• An insoluble wet granular material is dried in a
  pan 0.457 x 0.457 m 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. The air is at 65.6 oC and
  has a humidity of 0.01 kg of water/kg dry air.
  Estimate
• The rate of drying for the constant-rate period.
  Assume convective heat transfer coefficient for
  the parallel flow can be expressed by the
  following empirical equation h0.0204 G0.8,
  where mass velocity, G air density x air flow
  velocity.
• If the air stream were to flow normal to the
  drying surface (all other conditions remaining
  same) what would be the new drying rate? Assume
  convective heat transfer coefficient for air flow
  normal to the drying surface can be expressed by
  the following empirical equation h1.17 G0.37.
  (Assume the correlation holds for the G values
  used).
• If the air velocity is decreased 50 and its
  temperature is increased 10 oC, what should be
  the new drying rate?

2
Problem-1 (Drying)
Given Velocity of drying air, Vair 6.1
m/s Temperature of drying air, Tair 65.6
oC Absolute humidity of drying air, ? 0.01 kg
water/kg dry air Estimate Drying rate for the
constant-rate period, Rc ? If air were to flow
normal to drying surface, Rc ? If Vair 0.5 x
6.1 m/s, Tair (65.610) oC, Rc ?
3
Problem-1 (Drying)
Solution Part-1 At constant drying-rate period,
moisture will evaporate from the product at wet
bulb temperature, Twb. From humidity chart,
following adiabatic saturation line from Tair
65.6 oC and ? 0.01 kg water/kg dry air, wet
bulb temperature, Twb can be determined as Twb
29 oC and corresponding absolute humidity ?wb
0.026 kg water/kg dry air Humid volume, VH
(22.41/273) Tair (1/28.97 ?/18.02) Tair
(2.83 x 10-3 4.56 x 10-3 ?)
(65.6273)(2.83 x 10-3 4.56 x 10-3 x 0.01)
0.974 m3/kg dry air Density of moist air, ?moist
air 1 kg dry air ? (10.01) kg moist air ?
1.01 kg moist air ?moist air 1.01 / 0.974
1.037 kg/m3 Mass velocity, G ?moist air x Vair
1.037 x 6.1 x 3600 22772.5 kg/m2 hr
4
Problem-1 (Drying)
For parallel flow, convective heat transfer
coefficient, h 0.0204 G0.8 0.0204 x
(22772.5)0.8 62.45 W/m2 K
62.45 J/s m2 K From steam table, latent heat of
vaporization at Twb 29 oC is hfg 2433 kJ/kg
2433000 J/kg For constant rate
period Rc hfg h(Tair Twb) Rc x 2433000
62.45 (65.6 29) Rc 9.394 x10-4 kg/m2 s
3.38 kg/m2 hr For surface area A 0.457 x
0.457 0.208849 m2 Drying rate RcA 3.38 x
0.208849 0.706 kg of water/hour Note If
thickness of layer is increased, no change of Rc
in the constant rate period
5
Problem-1 (Drying)
Part-2 As G 22772.5 kg/m2 hr gt 19,500, a higher
degree of uncertainty exist in estimation of
Rc For flow normal to the drying surface
convective heat transfer coefficient, h 1.17
G0.37 1.17 x (22772.5)0.37
47.91 W/m2 K 47.91 J/s m2 K For
constant rate period Rc hfg h(Tair Twb) Rc x
2433000 47.91 (65.6 29) Rc 7.207x10-4 kg/m2
s 2.594 kg/m2 hr For surface area A
0.457 x 0.457 0.208849 m2 Drying rate RcA
2.594 x 0.208849 0.54 kg of water/hour
6
Problem-1 (Drying)
Solution Part-3 Velocity of drying air, Vair
0.5 x 6.1 m/s Temperature of drying air, Tair
(65.610) oC 75.6 oC Absolute humidity of
drying air, ? 0.01 kg water/kg dry air Again
from humidity chart, following adiabatic
saturation line from Tair 75.6 oC and ? 0.01
kg water/kg dry air, wet bulb temperature Twb can
be determined as Twb 30.5 oC Humid volume, VH
(22.41/273) Tair (1/28.97 ?/18.02) Tair
(2.83 x 10-3 4.56 x 10-3 ?)
(75.6273)(2.83 x 10-3 4.56 x 10-3 x 0.01)
1.0024 m3/kg dry air Density of moist air, ?moist
air 1 kg dry air ? (10.01) kg moist air ?
1.01 kg moist air ?moist air 1.01 / 1.0024
1.0076 kg/m3 Mass velocity, G ?moist air x
Vair 1.00076 x (0.5 x 6.1) x 3600
11063.45kg/m2 hr
7
Problem-1 (Drying)
For parallel flow, convective heat transfer
coefficient, h 0.0204 G0.8 0.0204 x
(11063.45)0.8 35.05 W/m2 K
35.05 J/s m2 K From steam table, latent heat of
vaporization at Twb 30.5 oC is hfg 2429
kJ/kg 2429000 J/kg For constant rate
period Rc hfg h(Tair Twb) Rc x 2429000
35.05 (75.6 30.5) Rc 6.5 x10-4 kg/m2 s
2.34 kg/m2 hr For surface area A 0.457 x
0.457 0.208849 m2 Drying rate RcA 2.34 x
0.208849 0.489 kg of water/hour
8
Problem-2 (Drying)

• The experimental average diffusion coefficient of
  moisture in a given wood is 2.97 x 10-6 m2/hr.
  Large planks of this wood 25.4 mm thick are dried
  from both sides by air having a humidity such
  that the equilibrium moisture content in the wood
  is 0.04 kg water/kg dry wood. The wood is to be
  dried from a total moisture content of 0.29 to
  0.09 kg water/kg dry wood. Calculate
• Drying time.
• If the drying takes place at a higher temperature
  when the effective diffusivity of water is
  increased tenfold, what would be the final
  moisture content of the plank if the drying time
  is still kept the same? (Note that the simple
  diffusion model assumes drying to occur very
  slowly under isothermal conditions)
• How long will it take to dry a ball made of the
  same wood, 2.54 cm in diameter? All other
  conditions are the same as in (a)

9
Problem-2 (Drying)
Given Diffusivity of water in wood DL 2.97 x
10-6 m2/hr Air relative humidity is such that the
equilibrium moisture content Xm 0.04 kg
water/kg dry wood Initial moisture content
Xm,initial 0.29 kg water/kg dry wood Final
moisture content Xm,final 0.09 kg water/kg dry
wood Estimate Drying time, t ?
10
Problem-2 (Drying)
Part-a
Only free moisture can participate in transport
process One dimensional diffusion equation in
Cartesian coordinates Where Xm free
moisture content Xm,total
Xm If Xm,total Xm free moisture Xm 0 and
no more drying for this drying condition
11
Problem-2 (Drying)
Keeping only the first term of the infinite
series (see any book on conduction) and
integrating across thickness to obtain average
moisture content,
(1)
Xm,initial 0.29 0.04 0.25 kg water/kg dry
wood Xm,final 0.09 0.04 0.05 kg
water/kg dry wood
12
Problem-2 (Drying)
DL 2.97 x 10-6 m2/hr H/2 25.4/2 12.7 mm
0.0127 m
Part-b
DL 2.97 x 10-6 x 10 2.97 x 10-5 m2/hr t
30.8 hours Insert in equation (1) above to find
(Other parameters held constant)
Part-c
Solution (approximate) to the one-dimensional
diffusion equation in spherical geometry
R Radius of sphere
13
Problem-2 (Drying)
Note We can use equation (1) to determine DL for
a given solid by properly designing a drying
experiment to ensure one dimensional diffusion.
Once DL is determined, it can be used to estimate
the falling rate period for any other geometry of
the same solid over the same moisture content
range and at the same temperature. DL can vary by
two to three orders-of-magnitude for a given
solid during a single drying run. It is a
sensitive function of both moisture content and
temperature.