Drying Problems

8/13/2019 Drying Problems

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Drying Calculations

The typical drying rate curve

Drying Curves: Drying RatePlotted Against Moisture Content

0

0.4

0.8

1.2

1.6

2

0 100 200 300 400 500 600

X gH 2 O/100g solids

g H 2

O / m

i n . 1

0 0 g s o

l i d s

ConstantRate

FallingRate 1

FallingRate 2

Critical MoistureXc

Or often drying rate is plottedagainst “free moisture” X-X e

0

100

200

300

400

500

600

700

0 200 400 600 800 1000 1200

Time (min)

X

g H

2 O / 1 0 0 g s o

l i d s

Xe

The moistureattained at verylong time

0

0.4

0.8

1.2

1.6

2

0 100 200 300 400 500 600

X- X e gH2O/100g solids

g H 2

O / m

i n . 1

0 0 g

The nature of the curvedepends on the product beingdried, for example . . .


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Drying-Time Prediction! First need to determine the rate of

moisture removal! The rate of moisture removal varies in

different regimes! Constant in constant rate period! May decline linearly or by some other

function in the falling rate periods

In general, the drying rate is given by

t =W

s

A

dX

R X 2

X 1 "

Where W s is the weight of the product solids A is the exposed surface area

For constant rate drying, R=R c isconstant. Calculating drying time is easy

t c

=

W s

ARc

( X 2 " X 1)

Free Moisture

D r y

i n g

R a

t e

For falling rate periods, R varies

Free Moisture

D r y

i n g

R a

t e

We don’t always know themechanism, or the importantproperties for predicting dringin the falling rate periods.

Sometimes, we just fit thecurves

If it has just 1 falling rate, with Rfalling linearly

Free Moisture

D r y

i n g

R a

t e

" dX

dt =

Rc

X c

# X

t F

=

X c

Rc

! n X

c

X

$

% &

'

( )

If it has more than 1 falling rate period

Free Moisture

D r y

i n g

R a

t e

t F

= X

c 1 " X r 1 R

c

#ln X

c 1 " X r 1 X

c 2 " X r 1

$

% &

'

( ) +

X c 1 " X r 1

Rc

# X

c 2 " X r 2 X

r 2 " X r 1#ln

X c 2 " X r 2

X " X r 2

$

% &

'

( )

Xc1

Xc2

Xr1


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Or we can use graphicalintegration

t =W

s

A

dX

R X 2

X 1 "

Sometimes we try to predictdrying rate by equations! Usually break the problem up into

constant rate and falling rate periods

Constant rate period! Getting heat from the air to the surface

is rate limiting. Thus,q = hA(T a " T s) = ṁc# H v

T a = air temperature

T s = product surface temperature ($ T wb )

ṁc = rate of H 2O vapor transfer# H v = heat of vaporization Ta

Ts

heat

watervapor

Based on heat transfer

t c

=

" H v (w o # w c )

hA (T a # T s )

That is, the drying time in the constant rate period(from w o to w c)

And we might be able to estimate h from the air velocity.For example, for forced flow of air past spheres

Nu = 2 + 0.60Re0.5

+ Pr 1/ 3

where

Nu =hd ck

Re = " vd cµ

Pr =µ c p

k

convective heat transferconductive heat transfer

molecular diffusion massmolecular diffusion heat

turbulence

measures thickness ofmass and thermal

boundary layer

Based on mass transfer

t c =0.622 RT A (wo " wc )k m AM w (W s " W a )

w = moisture content

k m = convective mass transfercoefficient (m / s)

M w = molecularweight of water

W a = absolutehumidityof air (kgwater / kgdry air )

W s = absolutehumiditysurface (kgwater / kgdry air )

T A = heatedtemperature (K )


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Falling Rate Periods! Getting water from inside product to

surface is limiting! The model used depends on the

mechanism of moisture transfer

! If the mechanism is only simplediffusion, the resistance tomovement is measured by thediffusion coefficient D

! The particular equation depends onthe product geometry

Infinite plate:

t F =4 d c

2

" 2 D

! n8

" 2

wc # w e

w # w e

$

% &

'

( )

*

+ ,

-

. /

d c = characteristic half-thickness (m)D = diffusion coefficient (m 2/s)

Infinite cylinder :

t F =d

c2

" 2 D! n

4" 2

wc # w e

w # w e

$

% &

'

( )

*

+ ,

-

. /

Sphere :

t F =d

c2

" 2 D

! n6

" 2

wc # w e

w # w e

$

% &

'

( )

*

+ ,

-

. /

! Of course, there may be othermechanisms at play than just simplediffusion

! Also, there may be more than onefalling rate regime


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For spray drying

t =" H v (wo # wc )

4 $ Rd k a (T a # T s)+

R p2

$ 2 D

! n 6$

2

wc # wew # we

%

& '

(

) *

+

, -

.

/ 0

constantrate

fallingrate

Heat and Mass Balances! Mass balance (water that is)

For a countercurrent system

IN OUTṁ aW 2 + ṁ p w1 = ṁ aW 1 + ṁ p w 2

ṁ a = air flow (kg dry air/h)ṁ p = product flow (kg dry solids/h)

W = absolute humidity (kg water/kg dry air)w = product moisture (kg water/kg solids)

Energy balance

ṁ a H a 2 + ṁ p H p1 = ṁ a H a1 + ṁ p H p 2 + q

q = thermal energy loss from system

H a = enthalpy of air (kJ/kg dry air)

H p = enthalpy of product (kJ/kg dry solids)

H a = c s(T a " T o ) + W # H v

H p = c pp (T p " T o ) + wc pw (T p " T o )

Drying Problems

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