249326256-Drying.pdf

CHEMICAL ENGINEERING SERIES

DRYING Compilation of Lectures and Solved Problems

CHEMICAL ENGINEERING SERIES 2 DRYING

DRYING -

is the removal of relatively small amounts of solvent, at temperatures below its boiling point, by circulating air or some other gas over the material in order to carry away the solvent vapor.

-

This is an adiabatic (constant enthalpy) drying process in which heat required for the vaporization of solvent comes solely from the sensible heat of the frying medium

-

In the usual drying or dehumidification process, water is the solvent and air is the drying medium. The drying process cools the air adiabatically at a constant wet bulb. The dry bulb temperature approaches the wet bulb temperature and could reach it at the saturation point.

Moisture Content, wet basis, Expressed as kg moisture per kg wet solid or kg moisture per combined kg of dry solid and moisture. Moisture Content, dry basis, Expressed as kg moisture per kg dry solid Bound Moisture Is the moisture content of a substance which exerts an equilibrium vapor pressure less than that of the pure liquid at the same temperature; it is the moisture difficult to remove, but which can be removed only under special conditions Unbound Moisture Refers to the moisture content of a substance which exerts an equilibrium vapor pressure equal to that of the pure liquid at the same temperature. Equilibrium Moisture Content, Is the limiting moisture to which a given material can be dried under specific conditions of air temperature and humidity; corresponds to bound moisture Free Moisture Content,

−

Moisture content of a substance in excess of the equilibrium moisture; only free moisture can be evaporated, and the free moisture content of a solid depends upon the vapor concentration in the gas

CHEMICAL ENGINEERING SERIES 3 DRYING

Critical Moisture Content, The average moisture content at the end of constant rate drying period or at the start of the falling rate period Constant Rate Drying Period The drying period during which the rate of water removal per unit of drying surface is constant Falling Rate Drying Period The drying period during which the instantaneous drying rate continually decreases

METHODSOF DRYING OPERATIONS 1. Batch Drying – actually a semi-batch process wherein a quantity of the substance to be dried is exposed to a continuously flowing stream of air into which the moisture evaporates; batch or semi-batch equipment is operated intermittently or cyclically under steady state conditions: the dryer is charged with the substance, which remains in the equipment until dry, whereupon the dryer is emptied and recharged with a fresh batch 2. Continuous Drying – the substance to be dried as well as the gas passes continually through the equipment; no typically stagewise methods are ordinarily used, and all operations involve continuous contact of the gas and the drying substance; continuous dryers are usually operated in steady state fashion METHODS OF SUPPLYING THE HEAT NECESSARY FOR EVAPORATION OF MOISTURE 1. Direct Dryers – heat is supplied entirely by direct contact of the substance with the hot gas into which the evaporation takes place 2. Indirect Dryers – heat is supplied quite independently of the gas used to carry away the vaporized moisture NATURE OF THE SUBSTANCE TO BE DRIED 1. 2. 3. 4.

Rigid Solid – wood or fibreboard Flexible material – cloth or paper Granular solid – mass of crystals Thick paste or thin slurry or a solution

CHEMICAL ENGINEERING SERIES 4 DRYING

HEAT TRANSFER IN DRYERS CALCULATION OF HEAT DUTY: Heat must be applied to a dryer to accomplish the following: 1. Heat the feed (solids and liquid) to the vaporization temperature ( − )+ ( − ) = 2. Vaporize the liquid =

−

3. Heat the product (solids and liquid) to their final temperature = − + − 4. Heat the vapour to its final temperature = −

−

5. Heat the air or other added gas to final temperature The total heat transferred per unit mass of dry bone solid is:

̅ (

=

)+

− +

̇

=

(

− ̅

− −

+

+

−

= + + ) + −

+ +

̅

−

+

−

− (

)+

−

−

+

−

−

In an adiabatic dryer, the heat transferred to the solids, liquid and vapour, comes from the cooling of the gas =

̇

−

HEAT TRANSFER COEFFICIENTS = =

CHEMICAL ENGINEERING SERIES 5 DRYING

EVALUATION OF HEAT TRANSFER COEFFICIENT 1. For air flowing parallel to the drying surface =

.

= .

.

2. For flow of gas perpendicular to the surface, (air velocities between 0.90 and 4.5 m/s) . = . (

∙

;

∙°

∙

)

Constant drying rate is simply, ̇

=

( −

=

)

HEAT TRANSFER UNITS Some adiabatic dryers, especially rotary dryers, are conveniently rated in terms of the number of heat transfer units they contain. =

− −

=

When the initial liquid content of the solids is high and most of the heat transferred is for vaporization, may be taken as the logarithmic mean difference between the dry bulb and wet bulb temperatures =

=

(

−

)− ( − −

− )

= =

=

− − ( − −

=

− ( − −

)

(

−

= )

=

−

)

CHEMICAL ENGINEERING SERIES 6 DRYING

where: = total heat transferred per unit mass of dry bone solid = rate of heat transfer in a section of the dryer = specific heat of dry bone solid = specific heat of liquid = specific heat of vapour = humid heat of gas at inlet humidity = temperature of feed = final solids temperature = vaporization temperature = final vapour temperature = heat of vaporization ̇ = mass rate of dry bone solid ̇ = mass rate of dry gas = over-all heat transfer coefficient = heat transfer area = average temperature difference (not necessarily the logarithmic mean) = dryer volume = volumetric heat transfer coefficient = Nusselt number ℎ = heat transfer coefficient between gas and surface of slab = equivalent diameter = thermal conductivity BATCH DRYING: CALCULATIONS OF DRYING TIME 



 

Drying in batches is relatively an expensive operation and is consequently limited to small-scale operations, to pilot plant and development work and to drying valuable materials whose total cost will be little influenced by added expense in the drying operations Examples of batch dryers o Tray, Cabinet or Shelf Dryers – used for drying solids which must be supported on trays Unless stated otherwise, moisture contents of solids are on the wet basis and should be converted to dry basis before solving any problem Batch dryers operate under constant drying conditions

CHEMICAL ENGINEERING SERIES 7 DRYING

TIME OF DRYING UNDER CONSTANT DRYING CONDITIONS Falling Rate

Constant Rate

R

Xe

Xf

Xc

Xi

Where: = rate of drying, lb H2O/ft2·h or kg H2O/m2·h = rate of drying at falling rate = rate of drying at constant rate = equilibrium moisture content (dry basis) = critical moisture content (dry basis) = final moisture content (dry basis) = initial moisture content (dry basis) = drying time = weight of dry bone solid, kg or lb = total drying area, ft2 or m2 At Constant Drying Conditions (CDC) = −

1. Constant Rate Period – as long as the liquid covers the entire surface of the solid, the rate of drying is constant. During this period, water diffuses through the solid at a rate sufficient to keep the entire surface wet = − = = −

=−

(

−

)

CHEMICAL ENGINEERING SERIES 8 DRYING (

=

)

−

2. Falling Rate Period – when part of the solid surface is no longer wetted by the liquid, the drying rate decreases. Most of the water escapes by vaporizing at the surface of the solid ∝( − ) = ( − ) = − ( −

)= −

= −

− − −

= −

= (

=−

=

(

− )

−

−

− −

)

− −

3. Total Drying Time (Constant Rate + Falling Rate) = + ( − ) ( − = + =

−

+(

−

)

)

− − − −

CHEMICAL ENGINEERING SERIES 9 DRYING

DRYING EQUIPMENT 1. Dryers for Solids and Pastes a. Tray Dryers  Consists of a rectangular chamber of sheet metal containing two trucks that supports racks; each rack carries a number of shallow trays that are loaded with the material to be dried  Heated is circulated at 2 – 5 m/s between trays by fan and motor and passes over heaters; air is distributed uniformly over the stack of trays through baffles  Useful on small production rate; they find most frequent application for valuable products like dyes and pharmaceuticals b. Screen Conveyor Dryers  A layer (25 mm to 150 mm) thick of material to be dried is slowly carried on a travelling metal screen through a long drying chamber or tunnel  The chamber consists of series of separate sections, each with its own fan and air heater. At the inlet end of the dryer, the air usually passes upward through the screen and the solids; near the discharge end, where the material is dry and may be dusty, air is passed downward through the screen. The air temperature and humidity may differ in the various sections to give optimum conditions for drying at each point  Typically 2 m wide and 4 – 50 m long, giving drying times of 5 – 120 minutes; the minimum screen size is about 30 mesh  Handles variety of solids continuously and with a very gentle action; particularly applicable when the drying conditions must be appreciably changed as the moisture content of the solid is reduced c. Tower dryers  Contains a series of circular trays mounted one above the other on a central rotating shaft  Solid feed is dropped on the topmost tray is exposed to a stream of hot air or gas that passes across the tray. The solid is then scrapped off and dropped to the tray below. The flow of solids and gas may be either parallel or counter-current d. Rotary Dryers  Consists of a revolving cylindrical shell, horizontal or slightly inclined toward the outlet  Wet feed enters one end of the cylinder; dry material discharges from the other  Rotary dryers are heated by direct contact of gas with the solids, by hot gas passing through an external jacket, or by steam condensing in a set of longitudinal tubes mounted on the inner surface of the shell

CHEMICAL ENGINEERING SERIES 10 DRYING 

 

The allowable mass velocity of the gas in a direct dryer depends on the dusting characteristics of the solid being dried and ranges from 2,000 to 25,000 kg/m2·h for coarse particles; inlet gas temperatures are typically 120 – 175°C for steam heated air and 550 - 800°C for flue gas from a furnace. Dryer diameters range from 1 – 3 m; the peripheral speed of the shell is commonly 20 – 25 m/min. Direct contact rotary dryers are designed on the basis of heat transfer =

.

. .

∆ = . =

.

∆

.

Where: = rate of heat transfer, BTU/h = dryer volume, ft3 = dryer length, ft ∆ = average temperature difference, taken as logarithmic mean of wet-blub depressions at inlet and outlet of the dryers = mass velovity, lb/ft2·h = dryer diameter, ft = volumetric heat transfer coefficient, BTU/ft3·h·°F e. Screw Conveyor Dryers  A continuous indirect-heat dryer, consisting essentially of a horizontal screw conveyor (or paddle conveyor) enclosed in a cylindrical jacketed shell  Solid fed in one end is conveyed slowly through the heated zone and discharges from the other end. The vapour evolved is withdrawn through pipes set in the roof of the shell  Handles solids that are too fine and too sticky for rotary dryers; they are completely enclosed and permit recovery of solvent vapors with little or no dilution by air. f.

Fluid Bed Dryers  Solid particles are fluidized by air or gas in a boiling-bed unit; mixing and heat transfer are very rapid; wet feed is admitted to the top of the bed; dry product is taken out from the side, near the bottom .

/

= . + .

Where: ℎ = heat transfer coefficient between and gas and solid, BTU/ft2·h·°F = particle diameter, ft = thermal conductivity at mean film temperature, BTU/ft·h·°F

CHEMICAL ENGINEERING SERIES 11 DRYING

g. Flash Dryers  Wet pulverized solid is transported for a few seconds in a hot gas stream  The rate of heat transfer from the gas to the suspended solid particles is high and drying is rapid so that no more than 3 or 4 s is required to evaporate substantially all the moisture from the solid  Flash drying may be applied to sensitive materials that in other dryers would have to be dried indirectly by a much cooler heating medium 2. Dryers for Solutions and Slurries a. Spray Dryers  A slurry or liquid solution is dispersed into a stream of hot gas in the form of a mist of fine droplets. Moisture is rapidly vaporized from the droplets, leaving residual particles of dry solid, which are then separated from the gas stream. The flow of liquid and gas may be co-current, counter current or a combination of both in the same unit  Droplets are formed inside a cylindrical drying chamber by pressure nozzles, two-fluid nozzles, or, in large dryers, high speed spray disks  An equation for the volume-surface mean diameter of the drops from a disk atomizer is: .

.

.

= .

Where: = average drop diameter, m or ft = disk radius, m or ft = spray mass rate per unit length of disk periphery, kg/m·s or lb/ft·s = surface tension of liquid, kg/m3 or lb/ft3 = disk speed, r/s = viscosity of liquid, Pa·s or lb/ft·s = disk periphery, 2Πr, m or ft b. Thin Film Dryers  Competitive with spray dryers but relatively expensive c. Drum Dryers  Consist of one or more heated metal rolls on the outside of which a thin layer of liquid is evaporated to dryness. Dried solid is scraped off the rolls as they slowly revolve

CHEMICAL ENGINEERING SERIES 12 DRYING

PROBLEM # 01. A hot air dryer handles 1,000 kg/h of wet feed with a moisture content of 20% wet basis to reduce the moisture content to 12.5% dry basis. Atmospheric air at 23.9°C with a relative humidity of 60% is preheated to a dry bulb temperature of 82.2°C. The exhaust air leaves the dryer at 60°C. Calculate: (a) the volume of the atmospheric air handled by the preheater, and (b) the duty of the pre-heater in kcal/h. Source: CHE Board Exam Problem (November 1989) SOLUTION:

Air 23.9 C 60% RH HEATER

60 C

82.2 C DRYER

F = 1,000 kg/h xi=0.20

Xf=0.125

1. Final moisture content wet basis =

1+ 0.125 =

= 0.1111

(1 + 0.125)

2. Dry product balance: (1 − ) = 1 − (1 − 0.20) 1,000 =

(1 − 0.1111)

ℎ

= 899.9888

ℎ

CHEMICAL ENGINEERING SERIES 13 DRYING

3. Water evaporated = +ℇ ℇ = 1,000

ℎ

− 899.9888

ℎ

= 100.0112

ℎ

4. Humidity of air entering the pre-heater At 23.9°C, = 22.243 =

100

(60)(22.243) = 13.3458 100 (18)(13.3458) = = 0.0112 (28.84)(760 − 13.3458)

=

5. Humidity of air entering the dryer =

= 0.0112

6. Humidity of air leaving the dryer Assume adiabatic drying conditions, using the psychrometric chart at dry bulb temperature of 60°C = 0.019

7. Amount of dry air entering the dryer ℇ= ( − ) 100.0112

=

ℎ

= 12,821.9487

(0.019 − 0.0112)

8. Specific volume of dry atmospheric air From the psychrometric chart at 23.9°C 1 = 13.4

= 0.8365 16.0185

ℎ

CHEMICAL ENGINEERING SERIES 14 DRYING

9. Humid volume of atmospheric air 0.0112 = 0.8365

0.08205

+ (1

∙ ∙

) 18

= 0.8516

10. Volume of entering air = 12,821.9487 =

,

ℎ

0.8516

.

11. Humid heat of air entering the dryer From the psychrometric chart at = 0.0112 = 0.244

∙°

12. Heat requirement ( − ) = = 12,821.9487 =

,

.

ℎ

0.244

∙°

(82.2 − 23.9)°

(296.9 )

CHEMICAL ENGINEERING SERIES 15 DRYING

PROBLEM # 02. A Proctor & Swartz coconut meat dryer processes hour. The following are other data: Specific heat of coconut meat Density of coconut meat from the shearing and washing section Initial moisture content of coconut meat Final moisture content of coconut meat Coconut meat inlet temperature Coconut meat outlet temperature Drying condition Drying condition RH Barometer reading Compute the heat input to the dryer.

1,133.8 kg of desiccated coconut per -

0.754 kJ/kg·K 1.520 kg/m3

-

20% 1.5% 21°C 38°C 71°C db 20% 89.6 kPa

SOLUTION: Air RH = 20% tF = 71 C

DRYER

F = 1,133.8 kg/h xi = 0.20 tsi = 21 C

xf = 0.015 t sf = 38 C

1. Moisture content (dry basis) 0.20 = = = 0.25 1− 1 − 0.20 0.015 = = = 0.0152 1− 1 − 0.015 2. Mass of bone-dry solid ̇ = 1,133.8

ℎ

80 100

= 907.04

ℎ

3. Mass of water in the feed ̇

= 1,133.8

ℎ

20 100

= 226.76

ℎ

CHEMICAL ENGINEERING SERIES 16 DRYING

4. Mass of product after drying Consider solid balance (1 − ) = 1 − (1 − 0.20) 1,133.8 =

ℎ

(1 − 0.015)

= 920.8528

ℎ

5. Mass of water evaporated Consider over-all material balance = + = 1,133.8

− 920.8528

= 212.9472

ℎ

6. Mass water in the product ̇

= 920.8528

1.5 100

ℎ

= 13.8128

ℎ

7. Vaporization temperature From steam table at 89.6 kPa drying operation = 96.48 ° 8. Sensible heat to raise temperature of solid from initial (21°C) to vaporization temperature (96.48°C) = ̇ − ) , ( = 907.04

ℎ

= 51,621.3879

0.754

(96.48 − 21)

∙

ℎ

9. Sensible heat to raise temperature if water from initial (21°C) to vaporization temperature (96.48°C) = ̇ − ) , ( = 226.76

ℎ

= 71,660.6190

4.1868 ℎ

∙

(96.48 − 21)

CHEMICAL ENGINEERING SERIES 17 DRYING

10. Latent heat to vaporize the moisture = From steam table at 89.6 kPa, = 2,261.55 = 212.9472

2,261.55

ℎ

= 481,591.7526

ℎ

11. Sensible heat to raise temperature of the bone dry solid from vaporization temperature (96.48°C) to final product temperature (38°C) = ̇ − , = 907.04

0.754

ℎ

= −39,994.9492

(38 − 96.48)

∙

ℎ

12. Sensible heat to raise temperature of water remaining in the material from vaporization temperature (96.48°C) to final product temperature (38°C) = ̇ − , = 13.8128

4.1868

ℎ

= −3,381.9821

(38 − 96.48)

∙

ℎ

13. Total heat requirement of the system = + + + + = 51,621.3879

ℎ

+ 71,660.6190

+ −3,381.9821 = 561,496.8282 =

.

ℎ

1ℎ 3,600

ℎ 1 1

ℎ

+ 481,591.7526

ℎ

+ −39,994.9492

ℎ

CHEMICAL ENGINEERING SERIES 18 DRYING

PROBLEM # 03. Ipil-ipil leaves will be dried in a moving train of tray dryers. The wet leaves containing 75% water (wet basis) is to be dried to 15% (wet basis) in trays measuring 1 m x 1.5 m. The wet leaves are spread out in the tray to a uniform thickness of 8 cm. Calculate: a) The number of trays needed to produce 1 metric ton of the dried leaves b) The amount of water removed/MT of product. The density of the wet leaves is 0.75 g/cc c) If dry hot air at 20% RH and a dry bulb temperature of 110°F is blown into the dryer and moist air leaves at 105°F dry bulb and 86°F wet bulb temperatures, how many ft3 of dry hot air will be needed per MT of product? Source: CHE Board Exam Problem SOLUTION: Air 110 F 20% RH

db = 105 F wb = 86 F DRYER

xi=0.75 P = 1 MT xf=0.15

1. Weight of feed (1 − ) = 1 − (1 − 0.15)(1,000 = (1 − 0.75)

)

= 3,400

2. Required number of trays =1

1.5

#

=

#

=

8

1 100

3,400

750 0.12

#

=

.

= 0.12

~

CHEMICAL ENGINEERING SERIES 19 DRYING

3. kg water removed = +ℇ ℇ = 3,400 − 1,000 ℇ= ,

= 2,400

4. humidity of entering air For the entering air, at 20% RH and 110°F: = 1.2763 ° =

100

20 = 0.25526 100 (18)(0.25526) = = 0.0110 (28.84)(14.7 − 0.25526)

= (1.2763

)

5. Humidity of air leaving the dryer For the exit air at Tdb = 105°F and Twb 86°F = 0.023

6. Amount of dry air entering the dryer ℇ= ( − ) 2,400

=

= 200,000

(0.023 − 0.011)

7. Specific volume of dry air From the psychrometric chart at 110°F = 14.3

13. Humid volume of atmospheric air = 14.3

. .

0.0110 +

.

0.7302

(1

= 14.5544

14. Volume of entering air = 200,000 = .

1 0.454

14.5544

) 18

∙ ∙°

(570° )

CHEMICAL ENGINEERING SERIES 20 DRYING

PROBLEM # 04. A wet material from a process plant containing 100% (dry basis) moisture has to be dried to produce a product with 10% moisture. Heated air at 100°C and 10% relative humidity is being supplied to the dryer and leaves at 60°C and a dew point of 52.5°C. Part of the outlet air is re-circulated and mixed with ambient air at 30°C and 70% relative humidity. Neglecting heat losses due to radiation to the surroundings and pre-heating of the solid materials and its receptacle, calculate: (a) volume of ambient air, m3/min, and (b) percentage of the outlet air re-circulated and mixed with ambient air when producing 500 kg/h of product. Source: CHE Board Exam Problem (May 1988) SOLUTION: Air 30 C 70% RH db = 60 C Dew pt = 52.5 C 4

R 2

3 DRYER

Xi=1.00

1. Feed rate Consider dry material balance (1 − ) = 1 − 1 = = = 0.50 1+ 1+1 0.10 = = = 0.0909 1+ 1 + 0.10 (1 − 0.0909) 500 =

ℎ

(1 − 0.50)

= 909.0909

ℎ

2. Water evaporated/removed from the material Consider over-all material balance = + ℇ ℇ = 909.0909

ℎ

− 500

ℎ

1

= 409.0909

ℎ

100 C 10% RH

CHEMICAL ENGINEERING SERIES 21 DRYING

3. Humidity of ambient air For air at 30°C, 70%RH, from table 2-5 CHE HB 8th edition = 31.824 = = =

100 (31.824

)(70) 100

( −

)

=

= 22.2768

(18)(22.2768) = 0.0188 (28.84)(760 − 22.2768)

4. Humidity of air entering the dryer For air at 100°C, 10%RH, from table 2-5 CHE HB 8th edition = 760 = = =

100 (760

)(10) 100

( −

)

=

= 76

(18)(76) = 0.0693 (28.84)(760 − 76)

5. Humidity of leaving the dryer For air at 60°C and dew point of 52.5°C, from the pychrometric chart = 0.0950

6. Amount of dry air required in the dryer ℇ= ( − ) =

409.0909

ℎ

= 15,917.9338

(0.095 − 0.0693)

7. Dry air balance at the entrance of dryer = + = 15,917.9338 − 1 Consider water balance: = + . 15,917.9338 0.0693 ℎ = 0.095 = 11,611.7138 − 0.1979 2 Equate 1 and 2 15,917.9338 − = 11,611.7138 − 0.1979 = 5,368.6822

ℎ

ℎ

−(

) 0.0188

CHEMICAL ENGINEERING SERIES 22 DRYING

8. Specific volume of ambient air From figure 19.2 Unit Operations of Chemical Engineers, 7th edition, McCabe, et.al. . 1 . . . = 13.75 = 0.8584 . . . 16.0185 . 9. Humid volume of ambient air 0.0188

. + .

= 0.8584

= 0.8844

0.08205 (1

∙ ∙

) 18

.

10. Volume of ambient air = 5,368.6822 =

.

1ℎ 60

ℎ

0.8844

.

.

11. Weight of wet air recycled, from equation 1 .

= 15,917.9338 − 5,368.6822 = 10,549.2516 = 10,549.2516 = 11,551.4305

. ℎ

+ 10,549.2516

11,551.4305 17,430.1375

%

=

.

0.0950

ℎ

13. % recycled air =

ℎ

.

12. Weight of wet air leaving the dryer = + . = 15,917.9338 + 15,917.9338 ℎ . = 17,430.1375 ℎ

%

ℎ .

%

. ℎ . ℎ

100

. ℎ

0.0950

(303 )

CHEMICAL ENGINEERING SERIES 23 DRYING

PROBLEM # 05. An adiabatic tunnel dryer reduces the moisture content of pineapple stumps used as a fuel in the boiler of a pineapple cannery. Ambient air is heated and blown through the tunnel dryer countercurrent to the flow of pineapple stumps. The operating conditions are: Pineapple Stumps:

Feed rate Moisture content, feed Moisture content, product Temperature Relative Humidity Inlet temperature Outlet relative humidity

Ambient Air: Hot Air:

100 MT/day 100% Dry basis 30% dry basis 29.4°C 80% 76.9°C 100%

Calculate: (a) the quantity of moisture removed from the pineapple stumps in MT/day; (b) the humidity of the inlet air; (c) the temperature of the air leaving the dryer; (d) the volume of ambient air needed for drying in m3/h CHE Board Exam Problem (May 1993) SOLUTION: Ambient Air T1 = 29.4 C 80% RH T2 = 76.9 C

100% RH

HEATER DRYER

F = 100 MT/day Xi = 1.0

Xf = 0.30

1. Dried pineapple stumps produced (1 − ) = 1 − 1 = = = 0.50 1+ 1+1 0.30 = = = 0.2308 1+ 1 + 0.30 (1 − 0.50) 100 =

(1 − 0.2308)

= 65.0026

CHEMICAL ENGINEERING SERIES 24 DRYING

2. Moisture removed = +ℇ ℇ = 100 − 65.0026 ℇ=

.

3. Humidity of ambient air From table 2-5 CHE HB 8th edition at 29.4°C = 30.745 = =

100

(30.745

)(80) 100

=

( −

)

=

= 24.596

(18)(24.596 ) = 0.0209 (28.84)(760 − 24.596)

4. Humidity of inlet air to dryer Since ambient air undergoes only sensible heating, therefore, = = .

5. Wet bulb temperature of inlet air to the dryer From figure 19.2 (McCabe, et al), for air at 76.9°C and H = 0.0209 = 97° = 36.11° 6. Wet bulb temperature of outlet air from the dryer For adiabatic drying, = = 97° = 36.11° Temperature of the outlet air, outlet air humidity and amount of air needed are all interconnected, thus the remaining questions can be solved only by trial and error. a. Assume value of outlet air dry bulb temperature b. Using the psychrometric chart (figure 19.2) with wet bulb temperature of 97°F, determine the air outlet humidity c. To check if assumption is correct, solve for humidity considering 100% RH, wherein = =

( − ) d. If humidity from (b) is approximately the same as that from (c), then assumption is correct; if otherwise, make new assumptions.

CHEMICAL ENGINEERING SERIES 25 DRYING

Trial

Assume T3

H (from b)

pA3

H (from c)

01 02

97°F 97.5°F

0.040 0.040

0.8698 psi 0.8832 psi

0.0392 0.0399

7. Temperature of outlet air = . ° = . ° 8. Amount of dry air needed ℇ= ( − ) 34.9974 =

= 1,841.9684 (0.0399 − 0.0209)

9. Specific volume of ambient air at 29.4°C From figure 19.2 Unit Operations of Chemical Engineers, 7th edition, McCabe, et.al. . 1 . . . = 13.70 = 0.8553 . . . 16.0185 . 10. Humid volume of ambient air . + .

= 0.8553

= 0.8841

0.0209

0.08205 (1

) 18

.

11. Volume of ambient air = 1,841.9684 =

,

.

1,000

1 24 ℎ

0.8841

∙ ∙

(302.4 )

CHEMICAL ENGINEERING SERIES 26 DRYING

PROBLEM # 06. 10 short tons/h of crushed coal with 15.1% moisture (wet basis) is to be dried to 5% moisture (wet basis) in a counter-current continuous rotary dryer using hot air entering the dryer at 180°F, 10% relative humidity and leaves at 40% RH. How much hot air, in ft3/min will be needed for the operation? Assume adiabatic operation. Source: CHE Board Exam Problem SOLUTION: Hot Air T1 = 180 F 10% RH

40% RH DRYER

F = 10 short ton/h xi = 0.151

xf = 0.05

1. Weight of dried crushed coal (1 − ) = 1 − (1 − 0.151) 10 =

(1 − 0.05)

ℎ

= 9.8368

ℎ

2. Moisture removed = +ℇ ℇ = 10 − 9.8368 = 1.0632

ℎ

3. Humidity of inlet air From appendix 7 (McCabe, et.al) at 180°F = 7.515 =

100

(7.515 )(10) = 0.7515 100 (18)(0.7515) = = = 0.0336 ( − ) (28.84)(14.7 − 0.7515)

=

CHEMICAL ENGINEERING SERIES 27 DRYING

4. Wet bulb temperature of inlet air From figure 19.2 (McCabe, et al) = 106° 5. Wet bulb temperature of outlet air For adiabatic process, = = 106° 6. Humidity of outlet air From figure 12-1 (CHE HB, 8th edition), at Twet bulb of 106°F and 40% RH = 0.045

7. Air requirement ℇ= ( − ) =

1.0632

2,000 ℎ

= 186,526.3158

(0.045 − 0.0336)

ℎ

8. Specific volume of dry air From figure 19.2 (McCabe, et al) = 16.1

9. Humid volume of hot inlet air 0.0336 = 16.1

0.7302

+ (1

) 18

= 16.9723

10. Volume of air required = 186,526.3158 =

,

.

ℎ

1ℎ 60

16.9723

∙ ∙°

(640° )

CHEMICAL ENGINEERING SERIES 28 DRYING

PROBLEM # 07. Adiabatic tunnel dryer handles 100 kg of banana chips per batch reducing the moisture content from 50% to 12%, all on wet basis. Drying takes 10 hours to complete by blowing air at 82°C and 5% RH. The air leaves the dryer saturated. Assuming that the drying rate is constant, calculate: (a) exit temperature of the air; (b) volume of air blown, in m3/h Source: CHE Board Exam Problem (May 1990) SOLUTION: Air T1 = 82 C 5% RH

100% sat’d DRYER

F = 100 kg xi = 0.50

xf = 0.12

1. Amount of dried banana chips (1 − ) = 1 − (1 − 0.50)(100 ) = = 56.8182 (1 − 0.12) 2. Moisture removed = +ℇ ℇ = 100 − 56.8182 = 43.1818 3. Humidity of inlet air From figure 12-3 (CHE HB, 8th edition), at 82°C and 5% RH = 0.017

4. Humidity of outlet air (adiabatic conditions) From figure 12-3 (CHE HB, 8th edition), at 100% saturated = 0.0375

5. Temperature of outlet air = ° = °

CHEMICAL ENGINEERING SERIES 29 DRYING

6. Dry air required ℇ= ( − ) 43.1818 = (0.0375 − 0.017)

= 2,106.4293

7. Specific volume of inlet dry air From figure 12-4 (CHE HB, 8th edition) at 82°C = 16.15

8. Humid volume of inlet air 0.017 = 16.15

0.7302

+ (1 1

= 16.5914

∙ ∙°

) 18

. .

16.0185

. .

= 1.0358

9. Volume of air required = (2,106.4293

) 1.0358

10. Air flow rate Since each batch takes 10 hours of drying time 1 = 2,181.8394 10 ℎ =

.

= 2,181.8394

(640° )

CHEMICAL ENGINEERING SERIES 30 DRYING

PROBLEM # 08. A dryer is to deliver 1,000 kg/h of palay with a final moisture content of 10%. The initial moisture content in the feed is 15% at atmospheric conditions with 32°C dry bulb and 21°C wet bulb. The dryer is maintained at 45°C while the relative humidity of the hot humid air from the dryer is 80%. If the steam pressure supplied to the heater is 2 MPa, determine: (a) palay supplied to the dryer in kg/h; (b) temperature of the hot humid air from the dryer; (c) air supplied to the heater in m3/h; (d) heat supplied by the heater in kW; (e) steam supplied to the heater in kg/h Source: ME Board Exam Problem (October 1985) SOLUTION: Air TDB = 32 C TWB = 21 C T2 = 45 C

80% RH

HEATER DRYER

F = 1,000 kg/h xi = 0.15

xf = 0.10

1. Palay fed to the dryer (1 − ) = 1 − (1 − 0.10) 1,000 = = ,

ℎ

(1 − 0.15) .

2. Moisture removed = +ℇ ℇ = 1,058.8235 − 1,000 = 58.8235

ℎ

3. Humidity of air entering the heater From figure 12-3 (CHE HB, 8th edition), at 32°C dry bulb and 21°C wet bulb = 0.0115

CHEMICAL ENGINEERING SERIES 31 DRYING

4. Humidity of air entering the dryer Since only sensible heating is involved in the heater =

= 0.0115

5. Wet bulb temperature of the air inside the dryer From figure 12-3 (CHE HB, 8th edition), at 45°C dry bulb and 0.0115 humidity = 74.5° 6. Wet bulb temperature of air leaving the dryer Assume adiabatic operation, = = 74.5° 7. Temperature of the air leaving the dryer From figure 12-3 (CHE HB 8th edition), at 80% RH and 74.5°F wet bulb = ° = . ° 8. Humidity of air leaving the dryer From figure 12-3 (CHE HB 8th edition), at 80% RH and 74.5°F wet bulb = 0.0172

9. Air supplied to the dryer ℇ= ( − ) 58.8235

=

ℎ

= 10,319.9123

(0.0172 − 0.0115)

ℎ

10. Specific volume of dry air entering the heater From figure 19-2 (McCabe, et. al), at 32°C = 13.8

11. Humid volume of air entering the dryer 0.0115 = 13.8

0.7302

+ (1 1

= 14.0564 16.0185

) 18

. . . .

= 0.8775

∙ ∙°

(549.6° )

CHEMICAL ENGINEERING SERIES 32 DRYING

12. Air supplied to the dryer = 10,319.9123 = ,

0.8775

ℎ

.

13. Humid heat of air entering the heater From figure 19-2 (McCabe, et. al), at humidity of 0.0115 = 0.244

∙°

14. Heat transfer by the air in the heater ( − ) = = 10,319.9123

ℎ

0.244

∙°

(45 − 32)° = 32,734.7618

15. Heat input to the heater = =

= 32,734.7618 .

16. Steam requirement = From steam table at 2 MPa, = 451.6329 =

32,734.7618 451.6329

=

.

ℎ

ℎ

1ℎ 60

1 14.34

ℎ

CHEMICAL ENGINEERING SERIES 33 DRYING

PROBLEM # 09. A rotary dryer fired with bunker oil of 10,000 kcal/kg heating value (HHV) is to produce 20 MT/h of dried sand with 0.5% moisture from a wet feed containing 7% moisture. Specific heat of sand is 0.21 BTU/lb·°R. Temperature of wet sand is 30°C and temperature of dried product is 115°C. Determine: (a) weight of wet feed; (b) weight of water to be removed in kg/h; (c) heat required; (d) liters of bunker oil per hour if specific gravity of bunker oil is 0.90 and efficiency is 60%. Source: ME Board Exam Problem (November 1983) SOLUTION: Assume water starts to evaporate at 100°C 1. Weight of wet feed (1 − ) = 1 − (1 − 0.005) 20,000 =

ℎ

(1 − 0.07)

=

,

.

2. Water removed = +ℇ ℇ = 21,397.8495 − 20,000 ℇ= ,

.

3. Weight of bone dry sand = (1 − 0.005) 20,000

̇ = 1−

ℎ

= 19,900

ℎ

4. Weight of water in the feed ̇

=

= (0.07) 21,397.8495

ℎ

= 1,497.8495

ℎ

5. Sensible heat to raise temperature of sand from 30 to 100°C = ̇ − ) , ( = 19,900

ℎ

0.21

(100 − 30)

∙

= 292,530

ℎ

6. Sensible heat to raise temperature of water from 30 to 100°C = ̇ − ) , ( = 1,497.8495

ℎ

1.0

∙

(100 − 30) = 104,849.465

ℎ

CHEMICAL ENGINEERING SERIES 34 DRYING

7. Latent heat to vaporize the moisture =ℇ From steam table at 100°C = 539.0599 = 1,397.8495

539.0599

ℎ

= 753,524.567

ℎ

8. Sensible heat to raise temperature of sand from 100 to 115°C = ̇ − , = 19,900

0.21

ℎ

∙

(115 − 100)

= 62,685

ℎ

9. Sensible heat to raise temperature of water in the product from 100 to 115°C = ̇ − , = (1,497.8495 − 1,397.8495)

ℎ

1.0

∙

(115 − 100)

= 1,500

ℎ

10. Total heat requirement of the system = + + + + = 292,530

+ 104,849.465

ℎ

+ 1,500 = ,

,

ℎ

+ 753,524.567

ℎ

.

11. Weight of bunker fuel oil =

1,215,089.032

ℎ

10,000

1 = 202.5148 0.60 ℎ

12. Volume of bunker fuel oil = 202.5148 =

.

ℎ

1 900

1,000

ℎ

+ 62,685

ℎ

CHEMICAL ENGINEERING SERIES 35 DRYING

PROBLEM # 10. A wet material having a critical moisture content of 15% (db) and an equilibrium moisture content of 3% (db) took 6 hours to dry from 45% (db) to 5.5% (db). How long will it take to dry to 15% moisture (db)? Source: CHE Board Exam Problem (January 1974) SOLUTION:

R

Xe 0.03

=

+(

−

−

)

Xf 0.055

Xc 0.15

− −

6ℎ

=

0.45 − 0.15 + (0.15 − 0.03) ln

0.15 − 0.03 0.055 − 0.03

= 12.2892

1. Time to dry from 45% to 15% (CRP) = − = = − =−

(

−

)

= −(12.892)(0.15 − 0.45) = .

Xi 0.45

CHEMICAL ENGINEERING SERIES 36 DRYING

PROBLEM # 11. Wet solids is to be dried from 36% to 8% moisture in 5 hours under constant drying conditions, Xc = 14%, Xe = 4%. Calculate how much longer it would take under the same drying conditions to dry the same materials from 8% to 5.5% moisture. All moisture contents are on dry basis. Source: CHE Board Exam Problem (October 1979) SOLUTION:

R

Xe Xf’ Xf Xc 0.04 0.055 0.08 0.14

1. Consider crying from 36% to 8% - (CRP + FRP) − = − +( − ) − 5 = 0.14 − 0.04 0.36 − 0.14 + (0.14 − 0.04) ln 0.08 − 0.04 = 16.0447

2. Time to dry from 8% to 5.5% (FRP) ( − ) − = − 0.08 − 0.04 = (16.0447)(0.14 − 0.04) ln 0.055 − 0.04 = .

Xi 0.36

CHEMICAL ENGINEERING SERIES 37 DRYING

PROBLEM # 12. Glazed banana chips are dried to its equilibrium moisture content of 12.28% by wt (wet basis) in a tunnel dryer under constant air temperature and humidity in a batch process. Control tests give the following data on the process: Time, minutes Moisture content, % wet basis

0 23.08

30 20.00

60 16.67

90 14.72

120 13.21

150 13.01

From these data, (a) determine the critical moisture content of the banana chips, % wt dry basis; (b) formulate the falling rate equation for the drying operations. Source: CHE Board Exam Problem (November 1988) SOLUTION: Basis: 1 m2 drying area x X* 0.2308 0.3001 0.2000

ΔX

Δθ, min

R

0.2751

0.0501

30

0.00167

0.2250

0.0500

30

0.00167

0.1863

0.0274

30

0.00091

0.1624

0.0204

30

0.00068

0.1509

0.0026

30

0.00009

0.2500

0.1667

0.2000

0.1472

0.1726

0.1321

0.1522

0.1301

∗

Ave X

0.1496

=

1− Plot ΔX vs R

;

=

∆ 1 ∙ ∆

Critical moisture content From the chart, = 0.225 % = 22.5 % ANSWER

0.0018 0.0016

Drying Rate

0.0014 0.0012 0.001 0.0008

Falling rate equation = . − .

0.0006 0.0004 0.0002 0 0.15

0.17

0.19

0.21

0.23

0.25

Moisture Content, % dry basis

0.27

0.29

CHEMICAL ENGINEERING SERIES 38 DRYING

PROBLEM # 13. A batch of wet solid was dried on a tray dryer using constant drying conditions and a thickness of material on the tray of 25.4 mm. Only the top surface was exposed. The drying rate during the constant rate period was R = 2.05 kg H2O/h·m2. The ratio of LS/A used was 24.4 kg dry solid/m2 exposed surface. The initial free moisture was X1 = 0.55 and the critical moisture content XC = 0.22 kg free moisture/kg dry solid. Calculate the time to dry a batch of this material from X1 = 0.45 to X2 = 0.30 using the same drying conditions but a thickness of 50.8 mm, with the drying from the top and bottom surfaces. Source: Transport Process and Unit Operations, by Geankoplis, et al SOLUTION: Assume 1 m2 cross-sectional area of dry solid 1. Volume of dry solid at condition 1 )(0.0254 ) = 0.0254 = = (1 2. Mass of dry solid at condition 1 (1

= 24.4

) = 24.4

3. Density of dry solid 24.4 = = 960.6299 0.0254 4. Volume of dry solid at condition 2 )(0.0508 ) = 0.0508 = = (1 5. Mass of dry solid at condition 2 = 960.6299

(0.0508

) = 48.8000

6. Time to dry solid from 45% to 30% = − = = − =− =−

− (48.8000 (2 1

= .

) 2.05

) ∙ℎ

(0.30 − 0.45)

CHEMICAL ENGINEERING SERIES 39 DRYING

PROBLEM # 14. In commercial practice, rayon-yarn skeins are dried after centrifuging, and the drying occurs in the falling rate period. Experimental data for the drying of a certain type of yarn under constant air-drying conditions have been correlated by the following equation: −

where:

−

= 0.00302

.

(

− )

=

Rate of drying, lb H2O evaporated / lb of dry yarn ·hour

= = = =

Wet air mass velocity, lb wet air/h·ft2 Saturation humidity at wet-bulb temperature of the air Humidity of air Free moisture content of yarn, lb water/lb dry yarn

A compartment dryer at 1 atm is to be used to dry the yarn from 0.80 to 0.01 lb free water per lb dry yarn. The operation will be conducted under conditions approximating constant drying conditions. The average conditions of the air passing over the yarn are to be: Dry bulb temperature = 150°F % Relative Humidity = 10 Air velocity = 600 fpm The equilibrium moisture content of the yarn for the above conditions is 0.036 lb water per lb dry yarn. (a) determine the lbs of water evaporated per 100 lb of dry yarn; (b) final moisture content of the yarn; (c) determine the time required for drying. SOLUTION: 1. Water evaporated ℇ = (100

)(0.80 − 0.01)

ℇ=

2. Final moisture content of the yarn = + = 0.01 + 0.036 = .

3. Air humidity From figure 12-3 (CHE HB 8th edition), for air at 150°F and 10% RH = 0.016

CHEMICAL ENGINEERING SERIES 40 DRYING

4. Air wet bulb temperature From figure 12-3 (CHE HB 8th edition), for air at 150°F and 10% RH = 89.1° 5. Air saturation humidity at wet bulb temperature of 89.1°F From figure 19.2 (Unit Operations of CHE, 7th edition, McCabe, et. al) = 0.031

6. Specific volume of dry air From figure 19.2 (Unit Operations of CHE, 7th edition, McCabe, et. al) at 150°F = 15.3

7. Humid volume of air 0.016 = 15.3

0.7302

+ (1

∙ ∙°

(610° )

) 18

= 15.6959

8. Mass velocity of air =

60

600

(1 + 0.016)

ℎ

15.6959

9. Drying time .

(

− )

−

= 0.00302

−

= 0.00302 (2,330.2855)

−

= 4.0382 = − =

−

= 4.0382 −1 = 4.0382

. .

0.01 = −0.2476 ln 0.8 = .

.

(0.031 − 0.016)

= 2,330.2855

∙ℎ

CHEMICAL ENGINEERING SERIES 41 DRYING

PROBLEM # 15. In a pilot process, mashed potatoes are dried in pans ½ inch deep which are insulated on the bottom. Drying air is at 180°F with a 10% relative humidity. Drying from an initial moisture content of 0.15 lb/lb dry solid requires 6 hours. All drying is in the falling rate period and is diffusion controlled. The equilibrium moisture content is 0.1 lb/lb dry solid when in contact with air at this temperature and humidity. Bulk density of the product is 0.75 g/cc. In the plant process, the potatoes are to be dried from the same initial moisture content as in the pilot process to a final moisture content of 0.25 lb/lb dry solid. The process will take place in a pan dryer with pans 2-in deep but with perforated metal bottoms so that drying occurs from both faces. If the drying air conditions are controlled to duplicate those in the pilot dryer, what drying time will be required? Source: Principles of Unit Operations by Foust, et.al SOLUTION: For the pilot process: assume pan cross sectional area of 1 ft2 1. Volume of dry solid =

= (1

) 0.5

12

= 0.0417

2. Mass of dry solid 62.43 =

(0.0417

0.75

) = 1.9509

3. Constant drying rate for falling rate period (FRP) ( − ) − = − (1.9509

)(0.60 − 0.10)

= = 0.3743

(1 ∙ℎ

)(6 ℎ )

ln

0.60 − 0.10 0.15 − 0.10

CHEMICAL ENGINEERING SERIES 42 DRYING

For the plant process: assume same pan cross-sectional area of 1 ft2 4. Volume of dry solid =

= (1

) 2

12

= 0.1667

5. Mass of dry solid 62.43 =

(0.1667

0.75

) = 7.8038

6. Required drying time for falling rate period (FRP) ( − ) − = − Since drying will occur on both sides, therefore, A must be multiplied by 2 (7.8038

)(0.60 − 0.10)

=

ln (2 1

= .

) 0.3743

∙ℎ

0.60 − 0.10 0.25 − 0.10

CHEMICAL ENGINEERING SERIES 43 DRYING

PROBLEM # 16. Under a constant drying condition a dryer produces 1,080 kilos of 8% wet from a 50% wet material in 6 hours. The critical moisture content of this material is 16%. In the market, this product could be sold under 2 classes as follows: Grade

Moisture Content

Price/kg

I II

9% 5%

20 cents 23 cents

The plant must operate on a continuous 24 hours basis and spends PhP 80 daily for heat, power, labor, fixed charges, etc. if the raw material usually arrives at an average of 50% moisture content and costs PhP 0.12 per kilo received, which grade will you produce? (% are in dry basis). Do not assume any change in the construction and operating conditions, except the rate of feed. SOLUTION: 1. Constant drying rate Since final moisture content is lower than critical moisture and initial moisture is higher than critical moisture, therefore drying is combined constant rate and falling rate = + For CRP: (

=−

)

−

For FRP: =

(

−

)

− −

Assume Xe = 0 = =−

ln (

−

)+

ln

=

−

+

ln

=

−

+

ln

= 1,080

8 (100 + 8)

100 8

= 1,000

CHEMICAL ENGINEERING SERIES 44 DRYING =

1,000 6ℎ

0.50 − 0.16 + 0.16 ln

= 75.1506

0.16 0.08

ℎ

For Grade I product 2. Over-all drying time (

=− = =

)+

− −

+

ln ln

1,000

0.50 − 0.16 + 0.16 ln

75.1506

0.16 0.09

= 5.7492 ℎ

ℎ

3. Wet product daily production (100 + 9) = 1,000 ℎ 100

1 ℎ 5.7492 ℎ

24 ℎ

4. Total Sales ℎ 0.20

= 4,550.1983

ℎ 910.04

=

5. Raw materials required daily = (1 − ) =

1+

= 1−

=

1+

= (1,000 = 1,500

1 1+

)(1 + 0.50) ℎ

1 ℎ 5.7492 ℎ

24 ℎ

= 6,261.7408

6. Cost of raw material ℎ 0.12

= 6,261.7408

=

ℎ 751.41

7. Revenue = =

−( ℎ 910.04

= ℎ 78.63

−

+ ℎ 751.41

+

) ℎ 80

= 4,550.1983

CHEMICAL ENGINEERING SERIES 45 DRYING

For Grade II product 8. Over-all drying time (

=− = =

)+

− −

+

ln ln

1,000

0.50 − 0.16 + 0.16 ln

75.1506

0.16 0.05

= 7.0001 ℎ

ℎ

9. Wet product daily production (100 + 5) = 1,000 ℎ 100

1 ℎ 7.0001 ℎ

24 ℎ

= 3,599.6583

10. Total Sales ℎ 0.23

= 3,599.6583

ℎ 827.92

=

11. Raw materials required daily = (1 − ) =

1+

= 1−

=

1+

= (1,000 = 1,500

1 1+

)(1 + 0.50) ℎ

1 ℎ 7.0001 ℎ

24 ℎ

= 5,142.7837

12. Cost of raw material ℎ 0.12

= 5,142.7837

=

ℎ 617.13

13. Revenue = =

−( ℎ 827.92

−

+ ℎ 617.13

+

) ℎ 80

= ℎ 130.79

THEREFORE, IT IS MORE ECONOMICAL TO PRODUCE THE GRADE II PRODUCT

CHEMICAL ENGINEERING SERIES 46 DRYING

PROBLEM # 17. A 25% solution of a certain soap is fed at 175°F to a double drum dryer (2 ft diameter by 4 ft long) being heated by steam at 40 psig. The drums are rotated at 5 rpm. In 30 minutes, 20 lb of product (2% moisture) was obtained. Calculate: (a) capacity of dryer, in lb water evaporated per sq ft drum area per hour; (b) if the specific gravity of the product was 1.05, how thick was the soap flakes produced; (c) estimate the steam consumption per hour; (d) calculate the over-all heat transfer coefficient for this dryer. SOLUTION: 1. Lb of product produced per hour 20 = = 40 0.50 ℎ ℎ 2. Lb dry solid produced per hour 2 98 = 40 ℎ 100 2

= 39.2

3. Lb feed required per hour (1 − ) = 1 − (1 − 0.02) 40 =

ℎ

0.25

= 156.8

ℎ

4. Water removed per hour = + ℇ ℇ = 156.8 − 40 = 116.8

ℎ

5. Dryer surface area per revolution =2

= 2 (2

)(4

) = 50.2655

6. Dryer capacity ℇ = =

116.8

ℎ

50.2655

= .

∙

ℎ

CHEMICAL ENGINEERING SERIES 47 DRYING

7. Volume of soap flakes produced per batch 40

=

ℎ

= 0.6105

1.05 62.4

ℎ

8. Thickness of soap flakes =

0.6105

=

ℎ

5

50.2655

60 ℎ

= .

9. Steam requirement Assume that the steam requirement is the amount needed to evaporate water from soap and water is removed at 212°F (

=

)+

−

212° = 970.3 = 116.8

1

ℎ

= 117,652.64

∙°

(212 − 175)° + 970.3

ℎ

For the steam = 40 =

= 920.2063

117,652.64

ℎ

920.2063 =

.

10. Over-all heat transfer coefficient ( − ) = 40 = 286.61° = =

117,652.64 (50.2655 .

ℎ )(286.61 − 212)° ∙

∙°

CHEMICAL ENGINEERING SERIES 48 DRYING

PROBLEM # 18. Flourspar (CaF2) is to be dried from 6 to 0.4 percent moisture (dry basis) in a countercurrent adiabatic rotary dryer at a rate of 18,000 lb/h of bone-dry solids. The heating air enters at 1,000°F with a humidity of 0.03 and a wet-bulb temperature of 150°F. The solids have specific heat of 0.48 BTU/lb·°F; they enter the dryer at 70°F and leave at 200°F. The maximum allowable mass velocity of the air is 2,000 lb/ft2·h. (a) what would be the diameter and length of the dryer if Nt = 2.2? Is this a reasonable design? (c) repeat part a if Nt = 1.8. Source: Unit Operations of Chemical Engineering, 7th edition, McCabe, et.al SOLUTION: Assume water is vaporized at 212°F 1. Mass water in feed ,

= 18,000

ℎ

6 100

= 1,080

0.4 100

= 72

ℎ

2. Mass water in product ,

= 18,000

ℎ

ℎ

3. Mass water vaporized/removed ℇ = 1,080 − 72 = 1,008

ℎ

4. Temperature of outlet air Using equation 24.8 (McCabe, et al) − = ln − 1,000° − 150° = + 150° = 244.1827° . 5. Specific heat of air at H = 0.03 Using figure 12-3 CH HB 8th edition = 0.254

∙°

6. Logarithmic temperature difference Using equation 24.7 (McCabe, et al) ( )−( ) − − ∆ = − ln −

CHEMICAL ENGINEERING SERIES 49 DRYING ∆ =

(1,000 − 150) − (244.1827 − 150) = 343.5533° 1,000 − 150 ln 244.1827 − 150

7. Heat the solids from 70°F to 212°F ( − ) = = 18,000

0.48

ℎ

(212 − 70)° = 1,226,880

∙°

ℎ

8. Heat water from feed from 70°F to 212°F ( − ) = = 1,080

1.0

ℎ

(212 − 70)° = 153,360

∙°

ℎ

9. Heat to vaporize water at 212°F = 970.3 = ℇ = 1,008

970.3

ℎ

= 978,062.40

10. Heat the solids from 212°F to 200°F = − = 18,000

ℎ

0.48

∙°

(200 − 212)° = −103,680

ℎ

11. Heat water from product from 212°F to 200°F = − = 72

ℎ

1.0

(200 − 212)° = −864

∙°

ℎ

12. Heat duty of the dryer = + + + + = 1,226,880

ℎ

+ 153,360

+ −864 = 2,253,758.4

ℎ

ℎ

13. Air mass flow rate For adiabatic condition = ( = − ,

)

ℎ

+ 978,062.40

+ −103,680

ℎ

CHEMICAL ENGINEERING SERIES 50 DRYING 2,253,758.4

= 0.254

ℎ

= 11,739.695

(244.1827 − 1,000)°

∙°

14. Cross-sectional area of dryer = =

11,739.695 2,000

ℎ = 5.8698

∙ℎ

15. Dryer diameter = =

4 4(5.8698

)

= .

16. Length of dryer Using equation 24.22 (McCabe, et al) . = 0.125 ∆ 2,253,758.4 = 0.125 (2.7338)(2,000) . (343.5533) = . THIS IS LONGER THAN USUAL ROTARY DRYER DESIGN For Nt = 1.8 1. Temperature of outlet air Using equation 24.8 (McCabe, et al) − = ln − 1,000° − 150° = + 150° = 290.5041° . 2. Logarithmic temperature difference Using equation 24.7 (McCabe, et al) ( )−( ) − − ∆ = − ln − (1,000 − 150) − (290.5041 − 150) ∆ = = 394.1644° 1,000 − 150 ln 290.5041 − 150

ℎ

CHEMICAL ENGINEERING SERIES 51 DRYING

3. Air mass flow rate For adiabatic condition =− ( =− − ,

)

−2,253,758.4

= 0.254

ℎ

(290.5041 − 1,000)°

∙°

4. Cross-sectional area of dryer = =

12,506.1534 2,000

ℎ = 6.2531

∙ℎ

5. Dryer diameter = =

4 4(6.2531

)

= .

6. Length of dryer Using equation 24.22 (McCabe, et al) . = 0.125 ∆ 2,253,758.4 = 0.125 (2.8216)(2,000) . (394.1644) = .

= 12,506.1534

ℎ

CHEMICAL ENGINEERING SERIES 52 DRYING

PROBLEM # 19. Calculate the mean droplet diameter for a spray dryer with a 6-in rotary atomizer turning at 10,000 rpm. The feed rate is 30 lb/min at 120°F. Assume the slurry density is 70 lb/ft3 and the surface tension is that of water. In scaling up to a larger dryer with a 12-in atomizer, what rotation should be used to obtain the same droplet size? Source: Unit Operations of Chemical Engineering, 7th edition (McCabe, et al) SOLUTION: Using equation 24.24 (McCabe, et.al) = 0.4

Γ

.

Γ

1. Disk radius 6 = 2 12

.

.

Γ

= 0.25

2. Disk periphery =2 = 2 (0.25 ) = 1.5708 3. Spray mass rate per unit length of disk periphery Γ= Γ=

1 60

30 1.5708

= 0.3183

∙

4. Disk speed = 10,000

1 60

= 166.6667

5. Density of water at 120°F = 61.71

6. Viscosity of water at 120°F 6.7197 = 0.559

10

∙

= 3.7563

10

∙

CHEMICAL ENGINEERING SERIES 53 DRYING

7. Surface tension of water at 120°F = 4.67

10

8. Droplet size .

Γ

= 0.4

.

.

Γ

Γ .

0.3183

.

Γ

∙ 166.6667

= 61.71

= 0.0104 (0.25

)

.

3.7563

.

10

∙

=

Γ

0.3183

Γ

0.25

∙

⎡ 4.67 10 ⎢ =⎢ ⎢ ⎣

.

= 0.2596

61.71

(1.5708

0.3183

=

∙

12

10

2.54

1 1 10

.

9. Rotation rate for a 12-in atomizer, with the same droplet size = . . . Γ = 0.4 Γ Γ . . . . . Γ Γ 0.4 = 0.4 Γ Γ Γ For a change in diameter of disk atomizer, , , , will remain constant . Γ . 1 . Γ . 1 = Γ Γ Γ .

. .

⎤ ⎥ ⎥ ⎥ ⎦

= 0.4(0.0104)(0.2596)(1.6435)

= 4.4372

Γ

.

∙ ∙

) 32.174

.

=

Γ

.

. .

.

.

.

Γ

.

Γ

= 1.6435

CHEMICAL ENGINEERING SERIES 54 DRYING .

⎡ ⎢ ⎢ ⎣

.

.

.

⎤ ⎡ ⎥ =⎢ ⎥ ⎢ ⎦ ⎣

.

1 .

.

.

.

1 (2

.

(

.

)

.

⎤ ⎥ ⎥ ⎦

1

=

.

1

.

.

.

=

.

.

.

.

.

1 (2

)

.

1

=

.

) =(

.

) .

= = (10,000 = ,

)

6 12

.

.

PROBLEM # 20. If the rate of drying under constant drying conditions with air at 140°F and absolute humidity of 0.03 lb water per lb dry air of unknown mass of solid is given by: = 0.5 −

What is the equilibrium moisture content? Source: CHE Board Exam Problem (May 1985) SOLUTION: 1. Equilibrium moisture content, where R = 0 = 0.5 − 0 = 0.5 − = 0.0116 = . %

ℎ∙