Drying

Drying

CHE133

Heat and Mass Transfer Applications

Prepared by:

Rhoda B. Leron, Ph.D.

Drying

The removal of relatively small amounts of water from a material; water is removed as a vapor by air

dry air wet air

Moisture content

Equilibrium moisture content (X*) the moisture content of a solid exposed to air sufficiently long for

equilibrium to be reached (kg of H2O/kg of moisture-free

solid)

Bound water water in the solid that exerts a vapor pressure less than that of liquid water at the same

temperature

* the equilibrium moisture of a given material is

continued to its intersection with the 100% humidity line

Moisture content

Moisture content

Unbound water the excess moisture content of a solid than indicated by intersection with the 100% humidity

line (can still exert a vapor pressure as high as that of

ordinary water)

* Held primarily in the voids of the solid

Free moisture content (X) the moisture in the sample that is above the equilibrium moisture content

* Can be removed by drying under the given %RH

Batch Drying:

Rate of drying curves

In the constant-rate drying period, the surface of the solid is initially very wet and a continuous film of water exists on the drying surface. This water is entirely unbound water and acts as if the solid were not present.

At the critical moisture content, XC, there is insufficient water on the surface to maintain a continuous film of water. The entire surface is no longer wetted, and the wetted area continually decrease until the surface is completely dry.


X = Xt -X * R = -Ls

A

dX

dt


Free moisture, X

Drying Rate, R (kg H2O/h-m

2)

Constant-rate period Falling-rate

period

Xc

Calculations for Constant-Rate

Drying Period

Using experimental drying curves

Drying curve (ex. Fig. 9.5-1a)

Rate-of-drying curve

R = -Ls

A

dX

dt


Drying Period

Over the time interval X1 at t1 = 0 to X2 at t2 = t:

Within the constant-rate period, where X1 and X2 > XC ,

R=constant = RC:

t = dtt1

t2

=Ls

A

dX

RX2

X1

t =Ls

ARC(X1 - X2 )


Drying Period

Using predicted transfer coefficients

where

Ls = kg dry solid used

W = latent heat at TW h = heat transfer coefficient

A = exposed drying area

ky = gas film mass transfer coefficient

Tw = wet bulb temperature

T = dry bulb temperature

Hw = humidity at TW

H = humidity at T

t =LslW (X1 - X2 )

Ah(T -TW )=

Ls(X1 - X2 )

AkyMB(HW -H )


Drying Period

The rate of drying is

For air flowing parallel to the drying surface

At : 45 > T >150C, gas mass velocity, 2450 >G > 29 300 kg/h-m2 or 0.61 > v > 7.6 m/s

RC =h(T -TW )

lW= kyMB(HW -H )

h = 0.0204G0.8

h = 0.0128G0.8

(SI)

(English)


Drying Period

For air flowing parallel to the drying surface

At : 45 > T >150C, gas mass velocity, 2450 >G > 29 300 kg/h-m2 or 0.61 > v > 7.6 m/s.

For air flowing perpendicular to the drying surface

At : 39000 >G > 19 500 kg/h-m2 or 0.90 > v > 4.6 m/s

h = 0.0204G0.8

h = 0.0128G0.8

(SI)

(English)

h =1.17G0.37

h = 0.37G0.37

(SI)

(English)

Calculations for Falling-Rate

Drying Period

t =Ls

A

dX

RX2

X1

Determined

by graphical

integration


Drying Period

t =Ls

A

dX

RX2

X1

If the rate is a linear function of X:

R= aX +b dR= adX

t =Ls X1 - X2( )A R1 -R2( )

lnR1

R2


Drying Period

t =Ls

A

dX

RX2

X1

If the rate is a linear function through the origin:

R= aX dR= adX

t =Ls XC - X

*( )ARC

lnXC - X

*

X2 - X*

Sample Problems

An insoluble wet granular material is dried in a pan

0.457 0.457 m and 25.4 mm deep. The material is

25.4 mm deep in the pan, and the sides and the 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 at 65C and

has a humidity 0.010 kg H2O/kg dry air. Estimate the

rate of drying for the constant-rate period.

Sample Problems

A batch of wet solid whose drying-rate curve

represented by Fig. 9.5-1b is to be dried from a free

moisture content of X1 = 0.38 kg H2O/kg dry solid

to X2 = 0.04 kg H2O/kg dry solid. The weight of

the dry solid is Ls = 399 kg dry solid and A = 18.58

m2 of top drying surface. Calculate the time for

drying.

Continuous drying

Drier

Q

Gas, G

TGs2, H2

Gas, G

TG1, H1

Wet solid

Ls, Ts1, X1

Dried solid

Ls, Ts2, X2

Material balance on moisture:

G (H1 H2) = LS (X1 X2)

Continuous drying

Heat balance: datum = T0oC

Enthalpy of gas: HG = CS (TG T0) + H0

If T0 = 0oC, 0 = 2501 kJ/kg (1075.4 Btu/lbm)

Enthalpy of wet solid: HS = CpS (TS T0) + X CpA (TS T0)

CpS = heat capacity of solid

CpA = heat capacity of moisture

Heat balance on the dryer:

GHG2 + LS HS1 = G HG1 + LS HS2 + Q

Air Recirculation in Driers

H6 = H5 = H2 G1 H1 + G6H2 = (G1 + G6)H4

(G1 + G6)H4 + LSX1 = (G1 + G6)H2 +LSX2

Drying

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Drying