Dairy Process Engineering

Lesson 12. CLASSIFICATION OF DRYERS, SPRAY DRYERS, DRUM DRYERS AND PROBLEMS ON DRYERS

Module 2. Drying

Lesson 12
CLASSIFICATION OF DRYERS, SPRAY DRYERS AND PROBLEMS ON DRYERS

12.1 Introduction

There are different types of dryers available to meet the wide range of requirement of food and dairy industry. Depending on the quantity of moisture present initially and the final moisture content to be retained in the food, the type of drier, air supply temperatures etc are chosen. The dryers can also be classified as per the scale of operation, source of energy, the physical state of the feed.

Fig. 12.1 Classification of dehydration systems

Dehydration systems can be classified as per the scale of operation

1. Domestic Open sun drying

• Solar drying

2. Small scale Roller dryingTunnel drying

• Cabinet/ Tray
• Trough drying
• Microwave drying
• Dielectric

3. Medium and Large scale Spray

• Fluidized bed
• Freeze
• Puff
• Vacuum
• Bin
• Pneumatic
• Rotary
• Tower

12.2 Drum Drying

The drum dryer is an indirect type dryer in which the milk to be dried is maintained in a thin film on a rotating steam heated drum. The milk being dried is spread over the outside surface of the dryer. Clinging to it and drying continues as the hot drum rotates. At the end of a revolution, the drum comes to a ‘doctor blade’ which scrapes the dried film from the drum, when the product has made about three-quarter of a complete rotation on the drum surface. The process is also known as roller drying or film drying.
Drum drying requires less space and is more economical than spray dryers for small volumes. The ratio of steam consumption to water evaporation is from 1.2 to 1.6:1. The major disadvantages are that the product may have a scorched flavour, and solubility is much lower (85%) because of protein denaturation.

12.2.1 Classification of drum dryers

The drum dryers may be classified according to:

1. Number of drums (a) single drum, (b) double drum, or (c) twin drum.
2. Pressure surrounding the product (a) atmospheric, and (b) vacuum.
3. Feeding arrangement: (a) nip feed, (b) splash feed, (c) dip feed or (d) roller feed.
4. Material of construction: (a) alloy steel, (b) stainless steel, or (c) chrome, or nickel plate steel Fig. 12.1; 12.2; 12.3; 12.4; 12.5; 12.6; & 12.7 shows different types of drum dryers and their feed mechanisms.

12.2.2 Construction of drum dryers

The double drum atmospheric dryer is most commonly used in the dairy industry (Fig. 12.1). Liquid is fed at about 70 ºC from a trough or perforated pipe into a pool in the space above and between the two rolls. Heat is transferred by conduction to the liquid, which is partly concentrated in the space between the rolls. Subsequently, all the liquid is vapourized as the drum turn, leaving a thin layer of dried product which is scraped off by doctor blades into conveyor below.

Water vapour above the dryer has a lower density than the air surrounding the unit, and will rise. Vaporized moisture is removed through a vapour hood above the drums. The lower edge of the hood is formed into a trough to drain away moisture which may accumulate because of condensation. Adequate air flow must be over the drum surface to carry away moisture.

The different feeding arrangements are shown in Fig: 12.2 . Dip feed is the most simplest type of feed, suitable for materials with a high rate of sedimentation. Dip feed is used for certain suspensions of solids, usually with recirculation of material in the tray. Roller feed is suitable for glutinous materials such as starch. The product may be placed in its natural form or condensed before it is fed to the dryer. Milk is usually precondensed (not more than 30%) for single drum units, and preheated for double drums.
The doctor blade, a sharp hard flexible knife, scrapes the dried material from the drum. The blades are made of spring steel if the surface of the drum is hard and for soft drum bronze is used. The blades are positioned at an angle of 15 to 30º with the surface. The conveyor for each drum discharges the product into for sizing the dried product.

The inside of drum is heated with steam at 3-6 kg/cm2. Drum temperatures of below 130 ºC is suggested, as the product temperature approaches the temperature of the steam. The drums used are 60 to 120 cm in diameter, and upto 360 cm in length. Drums are carefully machined, inside and outside, otherwise a difference in thickness will change heat transfer rate and drying will not be uniform. The speed of the drums is adjustable from 6 to 24 rpm. The speed is important as it affects the thickness of milk film and the time of solid contact with metal. The speed of the drum depends upon the concentration of the milk and the final moisture content. Both drums turn at the same speed.

The time that the solid is in contact with metal is 3 s or less. The product is removed after 3/4 to 7/8 of a revolution of the drum has taken place. The thickness of the film is quite critical in controlling the operation. It is important that the material be spread evenly over the drum in order to have a uniform product. The surface tension and the rheological properties of the material controls this.

The spacing between the drum affects the film thickness. One drum of a double drum dryer is mounted on a stationary bearing and the other on a movable bearing to adjust the spacing between the drums. The spacing between the drums is about 0.5 to 1.0 mm. If the clearance is below this, there will be product damage.

As the heat is removed from steam, it is condensed. The condensate moves to the bottom of the drum and must be removed by a pump or siphon. Flooding of the inside with condensate reduces the heat transfer rate. The drying capacity is proportional to the active drum area.

Fig. 12.1 Twin drum dryer

Fig. 12.2 Roller dryers with (a) dip feed (b) application by roller and (c)thin layer application with dip feed roller and transfer roller

Fig. 12.3 Twin roller dryer with spray film feed

Fig. 12.4 Single roller dryer with feed rollers fitted on top

Fig. 12.5 Twin roller dryer with reservoir

Fig. 12.6 Steam inlet in to drum for heating and the condensate discharge in roller driers

Fig. 12.7 Roller dryer with fluidized bed

12.2.3 Rate of evaporation in drum dryers

The rate of moisture removal by a drum dryer is essentially a constant rate of water evaporation as the product is continuously fed between the drums and the dried product is removed. The equation governing the rate of evaporation in a drum dryer is:

Where, = ts - ta
latent heat of vaporization.
the mean temperature difference between the roller surface and the product.

The overall coefficient is from 1000 to 1800 kcal/hr m2 c under optimum condition, although it may be only 1/10th of these values when conditions are adversed. Since the thickness of drum wall is small compared to the diameter of the drum, the area A can be regarded simply as the outer surface area of the drum. The U depends on hs , k and hp and other coefficients.

Where, hs : equivalent film coefficient of steam, Kcal/ hr m2c

x: thickness of metal , m
k : thermal conductivity of metal, kcal/ m hr c
hp : equivalent film coefficient of product
hc : convection coefficient
hr: radiation coefficient
he : evaporation film coefficient

The factor having the greatest effect on U is the condition of the liquid film and the drum speed. Drying rates for drum dryers can be extremely high when thin film of low viscosity is evaporated, and it is thus permissible to use high temperature. In addition to assuring the adequate heat transfer the drying system must provide for removal of water vapour. If the speed of a particular drum is measured, then the U value and the moisture content of the product will be increased, if the conditions are unchanged. The overall thermal efficiency of drum dryer is 35 – 80 %.
The moisture content x of a milk product containing mw amount of water and ms amount of dry matter including fat is given by:

The proportion of dry matter (TS) is given by :

During the evaporation of a product from the original moisture content x0 containing amount of water mwo to a final moisture content x containing an amount of water mw, the following amount is removed.

Calculated as a fraction of the water originally present, the following is obtained:

12.3 Spray Drying

Spray drying is the transformation of feed from a fluid state into a dried particle form by spraying the feed into a hot drying medium. Spray drying is considered to be one of the best methods for drying of food materials. This method is applied to fluids of high moisture content and high viscosity or of a slightly paste like character. The major advantages of spray drying are:

(i) High production rate
(ii) Gentle drying
(iii) Short drying period
(iv) Superior flavour, appearance, and solubility of product
(v) Continuous single step operation
(vi) Uniform product
(vii) Plant can be easily automated
(viii) No product contamination
(ix) High thermal efficiency

A conventional spray dryer consists of the following main components (Fig. 12.8)

(1) Drying chamber
(2) Hot air system and air distribution
(3) Feed system
(4) Atomizing device
(5) Powder separation system
(6) Pneumatic conveying and cooling system
(7) Fluid bed after-drying/cooling
(8) Instrumentation and automation

In a spray dryer, the milk is pumped to a nozzle or rotary valve disc atomizer which sprays the feed in fine droplets into a drying chamber. The droplets are subject to a stream of hot air flowing either counter-currently or cocurrently in relation to the falling droplets. Thereby, the droplets of milk are dried so the dry matter remains as powder particles, which fall down towards the bottom of the chamber from where it is removed more or less continuously, as shown in the Fig. 12.8. On account of the large liquid surface created by atomization, the evaporation takes place very quickly, usually at a very low temperature, irrespective of whether drying air of a very high temperature is used. The hot air applies heat to the droplets and carry away the vapour evolved. The temperature of the milk droplets is kept down to the wet bulb temperature and does not exceed the calculated value of 49-54 ºC. The (Fig. 12.9) shows heat recuperate type air-liquid-air, while Fig-12.10 shows single stage spray dryer with heat recuperator type air-liquid-air., ( shows animation of condensing and drying plant.

Fig-12.10 Single stage spray dryer with heat recuperator type air-liquid-air.

The spray dryer consists of mainly four components:

1. Heating of the drying air: air heaters with accompanying fans, air filters dampers and ducts.
2. Atomization of feed into a spray: atomizer with feed supply system of pumps, tanks and feed pretreatment equipment
3. Contacting of air and sprays and drying of sprays: drying chamber with air disperser, product and exhaust air outlets.
4. Recovery of dried products and final air cleaning: complete product recovery with product discharge, transport and packing, air exhaust system with fans, wet scrubbers, damper and duct.

12.4 Classification of Spray Dryers

Spray dryers are mainly classified according to:

1. Method of atomization:(a) pressure atomization (b) centrifugal atomization, (c) pneumatic atomization
2. Method of heating air (a) steam (b) furnace oil, (c) electricity
3. Position of drying chamber (a) vertical (b) horizontal
4. Direction of air flow in relation to product flow (a) counter current (b) co current (c) mixed current
5. Pressure in dryer (a) atmospheric (b) vacuum
6. Shape of the bottom of the chamber (a) flat bottom (b) conical bottom

Problems on Dryers

Drum Drying

The rate of moisture removal by a drum dryer is essentially a constant rate of water evaporation as the product is continuously feed between the drums and the dried product is removed. The equation governing the rate of evaporation in a drum dryer is:

Where, = ts - ta
latent heat of vaporization.
the mean temperature difference between the roller surface and the product.
The overall coefficient is from 1000 to 1800 kcal/hr m2 c under optimum condition, although it may be only 1/10th of these values when conditions are adversed. Since the thickness of drum wall is small compared to the diameter of the drum, the area A can be regarded simply as the outer surface area of the drum. The U depends on hs , k and hp and other coefficients.

Where, hs : equivalent film coefficient of steam, Kcal/ hr m2c = 4000
x: thickness of metal , m
k : thermal conductivity of metal, kcal/ m hr c
hp : equivalent film coefficient of product = 4000
hc : convection coefficient = 5.0
hr: radiation coefficient = 7.5
he : evaporation film coefficient = 25.0

The factor having the greatest effect on U is the condition of the liquid film and the drum speed. Drying rates for drum dryers can be extremely high when thin film of low viscosity is evaporated, and it is thus permissible to use high temperature. In addition to assuring the adequate heat transfer the drying system must provide for removal of water vapour. If the speed of a particular drum is measured, then the U value and the moisture content of the product will be increased, if the conditions are unchanged. The overall thermal efficiency of drum dryer is 35 – 80 %.
the moisture content x of a milk product containing mw amount of water and ms amount of dry matter including fat is given by:

The proportion of dry matter (TS) is given by :

During the evaporation of a product from the original moisture content x0 containing amount of water mw0 to a final moisture content x containing an amount of water mw, the following amount is removed.

Calculated as a fraction of the water originally present, the following is obtained:

Mechanism of drying solids

Drying process can be divided into two period:

(a) constant rate period
(b) falling rate period

In a constant drying rate period, a material or mass of material contain so much water that liquid surface exists will dry in a manner comparable to an open faced body of water.diffuion of moisture from within the droplet maintains saturated surface conditions and as long as these lasts, evaporation takes place at constant rate. When a solid is dried under constant drying conditions, the moisture content x¬t typically fals as shown in the graph A in the figure.

The graph is linear at first, then curves and eventually levels off. Graph B shows the drying rate : it is horizontal at first indicating that the drying rate is constant, than it curves downward and eventually falls. When the material has reached its equilibrium moisture content, reaches zero.

Constant rate drying period will proceed until free moisture appears from the surface, the moisture removal rate will then become progressively less. The moisture content at which the drying rate ceases to be constant is known as the critical moisture content. During the constant rate period, the moisture from interior migrates to the surface by various means and is vapourised. As the moisture content is lowered, the rate of migration to the surface is lowered. If drying occurs at too high temperatures, the surface forms the layer of closely packed shrunken cells which are sealed together. This presents a barrier to moisture migration and tends to keep the moisture sealed within. This condition is known as ‘case hardening’.

The constant rate period is characterized by a rate of drying independent of moisture content. During this period, the solid is so wet that a continous film of water exists over the entire drying surface, and this water acts as if solids where not there. As long as the rate of surface evaporation continues, the rate of drying is governed by equation:

Where,

w = kg of moisture
Ø = time
h = heat transfer coefficient between air and moisture kcal/ kg hr °C
ta = dry bulb temperature of air, °C
ts = surface temperature, °C
A= area, m2
ΔHv = heat of vapourisation at ts ,Kcal/ kg
Ka = mass transfer coefficient (kg/ hr m)
H¬s = humidity of saturated air at the surface temperature.
H¬a = humidity of air

The constant rate period ends when the migration rate of water from the interior of the surface becomes less than the rate of evaporation from yhe surface. The period subsequent to the critical point is called ‘the falling rate period’. Beyond this point ,the surface temperature rises, and the drying rate falls off rapidly. The falling rate period take a far longer time than the constant rate period, even though the moisture removal may be much less. The drying rate approaches zero at some equilibrium moisture content.

Drying in falling rate period involves two processes:

a) movement of moisture within the material to the surface
b) removal of the moisture from the surface.

The method used to estimate drying rates and drying times in the falling rate period depends on wether the solid is porous or non porous. In a non porous material, once there is no superficial moisture, further drying can occur only at a rate governed by diffusion of internal moisture to the surface. In a porous material other mechanism appears, and drying may even takes place inside the solid instead of at the surface.
Problem: skim milk of x0 = 0.91 is evaporated to x = 0.55. find out the % of water evaporated based on the original water content.
Solution:

= 1 – [0.55 (1 – 0.91) / 0.91 (1 – 0.55)] = 0.88
i.e. 88 % of the water, based on the original water content, has been removed by evaporation.

Problem

Find out the evaporation rate in a drum dryer for given data: steam temp = 150 °C , vapourisation temperature of milk 103 °C, overall heat transfer coefficient 1200 kcal / h m2 °C, drum diameter = 60 cm, length of the drum = 100 cm, latent heat of vapourisation = 540 kcal / kg. the product is scraped at ¾ of a revolution of the drum.

Solution

Heat transfer area, A = 1 x 0.6 x Π x ¾ = 1.413 m2
dw / dØ = 1200 x 1.413 x (150 – 103)/ 540 = 147.5 kg/ hr

Problem: A drum dryer is designed for drying a product from an initial TS of 12 % and a final moisture content of 4 %. An average temperature difference between the roller surface and the product of 65 °C will be used and the overall heat transfer coefficient is 1500 kcal/ hr m2 °C. Determine the surface area of the roller required to provide a production rate of 50 kg product / hr.
Solution: Basis :1 kg of product P = F – V
F : amount of feed / kg of product
V : amount of vapour / kg of product.
Solid balance:
0.96 x 1 = 0.12 F
F = 8 kg ; V = 7 kg

From 50 kg product, F = 50 x 8 = 400 kg/hr
Vapour removed = 50 x 7 = 350 kg/hr

Using equation:
350 = 1500 x A x 65 / 540
Thus, A = 1.93 m2

The effective surface area needed for drying is 1.93 m2. Assuming that about ¼ of the surface area would not be used . Then the total surface area required would be : 1.93 x 4/3 = 2.583 m2 . If the length of the drum is taken as 1.37 m, then the diameter of the drum d will be:
Πd x 1.37 = 2.58
Thus, d = 0.6 m

Problem

Milk of 18.5 % TS is dried to 3.8 % on a drum dryer at a rate of 10.5 kg of dried product / hr. The diameter of the drum is 60 cm and length 90 cm and the product is scrapped at ½ of a revolution. The steam temperature is 160 °C and vaporization point of the moisture is 104 °C. Find out the overall heat transfer coefficient.

Solution

basis: 1 kg of milk
V + P = 1 (material balance)
Where, V = amount of water evaporated
P = amount of the product
0.185 + 1 = 0.962 (material balance)
Thus, V = 0.808 kg and P = 0.192 kg / kg wt.
Where, P = 10.5 kg, then,
dw/dØ = 10.5 x 0.808 / 0.192 = 44.18 kg / hr
The heat transfer area,A = 0.9 x 0.6 x Π /2 = 0.84
Using the equation:

44.18 = U x 0.847 (160 – 104) / 540
Thus, U = 0.5029 kcal / hr m2 °C

SPRAY DRYING

1. Where, W = mass flow rate of air, kg/hr
C = specific heat of air, kcal / kg °C
t1 = inlet air temperature of dryer, °C
t2 = outlet air temperature of dryer, °C
q = total rate of heat transfer kcal / hr
Ua = volumetric heat transfer coefficient (150 – 270 kcal/ hr m2 °C)
= mean temperature difference throughout the dryer, °C
Vd = volume of dryer , m3
2. Dryer efficiency:

Where, R = radiation loss, % of total temperature drop in dryer
t0 = atmospheric air temperature, °C
t1 = temperature to which air is heated, °C
t2 = temperature of air leaving dryer, °C
overall efficiency of dryer ranges between 50 – 60 %

Problem

500 kg of milk containing 60 % water is dried to a moisture content of 3 %, at an initial temperature of 20 °C in a spray dryer. An inlet temperature to heater is 30 °C and air is heated to 150 °Cto dry the product. The temperature of air leaving the dryer is 90 °C and that of milk particles are 54 °C. calculate (a) the amount of air required (m3) to dry the milk. (b) thermal efficiency of air drying (c) the amount of steam required at 7.2 kg/cm2 pressure to heat the air if the efficiency of the air heater is 80 % (d) the amount of steam required/ kg of water evaporated, and (e) if the radiation loss is 10 %, find out the overall thermal efficiency.

Solution

Basis: 1 kg of milk
V + P = 1 (material balance)
Where, V = amount of water evaporated
P = amount of the product
0.4x 1 = 0.97 P (material balance)
Thus, P = 0.413 kg and V = 0.587 kg / kg wt.
The amount of dried product from 500 kg milk = 500x 0.413 = 206.5 kg
Total amount of water removed = 500 x 0.587 = 293.5 kg

Heat supplied to milk = sensible heat to raise the raw milk.
Temperature from 20 °C to 54 °C + latent heat of vapourisation.
The latent heat of vapourisation corresponding to a saturation temperature of 90 °C is about 545 kcal/kg
Heat supplied to dry 500 kg milk = (500 (54 – 20) x 0.93) + (293.5 x 545)
= 175767 kcal.
The specific heat of air can be taken as 0.24 kcal/ kg °C and the mass density of air at the mean temperature i.e. 120 °C is 0.9 kg/m3
Using eq.:

175767 = W x 0.24 (150 – 90 )
Thus, W = 13562 m3
The amount of air required to dry the milk = 13562 m3 or 12205 kg.
(B)
The temperature of hot air should have been brought down to 54 °C i.e. equal to the temperature of dried milk , but in actual the air outlet temperature is 90 °C.
The maximum amount of heat which can be given by the hot air
= 13532 x 0.9 x0.24 x (150 – 54) = 281221 kcal
But the actual amount of heat which is used to dry 500 kg milk = 175767 kcal
The thermal efficiency of air drying = 175767/ 281221 = 0.62
(c)
The amount of air required in kg = 13562x0.9 = 12205kg
Heat taken by air in the air heating section = 12205 x 0.24 (150 – 30)
= 351504 kcal
At 1.2 kg/ cm2 absolute pressure, the latent heat of steam about 490 kcal/kg
The amount of steam required at 80 °C heating efficiency =
351504/(490 x 0.8) = 896.7kg
(d)
The amount of steam required per kg of water evaporated = 896.7 / 293.5 = 3.05 kg
(e)
Using eq.:
= (1- 0.1)(150 – 90)/ (150 - 30) = 0.45

Problem

Skim milk of 89% water content is evaporated to 54% moisture content. Find ot the % water evaporated based on original water content.

Problem

Find out the evaporation rate in a drum dryer when steam temperature is 145 °C, vapourisation temperature of milk is 104 °C, overall heat transfer coefficient 1300 kcal/hr m2 °C, drum diameter = 50 cm, drum length 90 cm, and latent heat of vapourisation = 539 kcal/kg. The product is scraped at 7/8 of a revolution of the drum.

Problem

A drum dryer is designed for drying a product from initial total solid content of 16% and a final moisture content of 4%. An average temperature difference between the roller surface and the product of 70 °C will be used, and overall heat transfer coefficient is 1400 kcal/ m2 hr °C. Determine the surface area of the roller required to provide a production rate of 40 kg product / hr.

Problem

Milk of 17 % solids is dried to 3 % on a drum dryer at a rate of 15 kg of dried product per hr . the drum dryer diameter is 45 cm and length is 75 cm and the product is scrapped at ¾ revolution. The steam temperature is 150 °C and vapourisation point of the moisture is 103 °C. find out the overall heat transfer coefficient.

Problem

1000 kg of milk containing 52 % water is dried to a moisture content of 2.5 % at an initial temperature of 30 °C in a spray drier. An inlet to heater is 35 °C and air is heated to 160 °C to dry the product. The temperature of air leaving the dryer is 85 °C and that of milk particles 52 °C. Calculate (a) amount of air required in m3 to dry the milk (b) thermal efficiency of air drying (c) amount of steam required at 7 kg/cm2 pressure to heat the air if the efficiency of the air heater is 75 %. (d) amount of steam required / kg of water evaporated (e) if the radiation loss is 8 %, find out the overall thermal efficiency.

Problem

1000 kg/hr of dried product (4 %moisture) is produced in a spray dryer. The atmospheric air at 30 °C and 40 %RH is heated to 190 °C and it is exhausted at 80 °C . the concentrated milk having 45 % TS by weight. Feed temperature of concentrated milk is 30 °C. dried product comes out at 50 °C. heat losses from the dryer is 25000 kcal/hr. Find out the air rate required.

Solution

Based on enthalpy and moisture balance:
Qa = quantity of air
Ta1 = temperature of air feed
Ha1 = enthalpy of concentrate feed, kcal/ kg of dry air
W1 = moisture, kg of water/ kg of dry air
Q1 = heat loss / hr
W2 = moisture after drying, kg of water/ kg of dry air
Ms = solid in feed = solid in final product (assuming no loss)
Hs1 , Hs2 = kcal/ kg of solid
Ws = kg of water/ kg of solids
From psyc. Chart:
Air contains 0.011 kg water/kg of dry air = w1
Assuming dry milk solids have specific heat of 0.534 kcal / kg of ----
Feed rate = 1000 x 0.96 x 100 / 45 = 2133 kg / hr
Ws1 = 53745 moisture/ solid concentration of milk
= 1.22 kg of water /kg of dry solids
Ws2 = 4/96 = 0.042 kg of water / kg of dry solids.
Enthalpy:
Ha1 = 0.24 (dbt) +W1(597.2 + 0.45 dbt)
= 0.24 (190) +W1(597.2 + 0.45 x 190)
= 53.11
Ha2 = 0.24 (80) +W2(597.2 + 0.45 x 80)
= 19.2 + 693.2 W2
Hs1 = feed is hot air
= enthalpy of dry solids + enthalpy of moisture
= (specific heat x t x 1 ) + ( Ws1 x 1 x t)
= (0.554 x 30 x 1) + (1.22 x 1 x 30)
= 53.22 kcal / kg of dry solids
Hs2 = enthalpy of dry solid + enthalpy of moisture
= (0.554 x 50 x 1) + (0.042 x 1 x 50)
= 29.8 kcal / kg of dry solids
Enthalpy balance:
Enthalpy in system = enthalpy out system
Ha1 + Hs1 = Ha2 + Hs2 +Q1 per unit
Thus, (Qa x Ha1) + (Hs1 x 2133) = Ha2 + Hs2
Thus, 53.11 Qa + (960 x 53.22) = 25000 + Qa (922 + 633.2 W2)
So, 33.91 Qa = 633.2 Qa W2 = 2517 --------------------- eq. (1)

Moisture balance

Moisture lost by product = moisture gained by air
960 (Ws1 – Ws2) = Qa (W2 – W1)
Thus, Qa (W2 – 0.011) = 1130.88 -------------------- eq. (2)
By solving both equations:
QaW2 = 1130.88 + 0.11 Qa
Qa = 26663 kg/hr
W1 = 0.011
W2 = 0.0534
Humidity of air exhaust = total moisture / quantity of air

Problem

From the following data given for the spray dryer, determine air flow rate and steam consumption.

Specific humidity of inlet air (32°C, 40 % RH) = 0.0119 kg water/ kg of dry air
Specific humidity of outlet air (90°C) = 0.052 kg water/ kg of dry air
Ws1 = 58 / 42 = 1.380 kg of water/ kg of dry solid
Ws2 = 3 / 97 = 0.0309 kg of water/ kg of dry solid
Cp = specific heat of air = 0.24 kcal/ kg of dry air
to = outlet air temperature
ti = inlet air temperature
Hs = enthalpy of steam at inlet of heater = 668.7 kcal / kg
Hc = enthalpy of condensate outlet = 84.95 kcal / kg
n = efficiency

Moisture balance

moisture lost by the product = moisture gained by the air
thus, 2(970)(1.38 – 0.0309) = Qa (0.052 – 0.0119)
thus, Qa = 63835.463 kg/hr

steam consumption (kg/hr) = La . Cp . (t0 – t1)/ [(Hs – Hc).h]
= 3038.89 kg/hr (by putting values)
Total water evaporated in dryer / hr = total concentration of milk – total powder
Thus, 2000 – [2000 x 42 x 1.03 / 100]
= 1134.8 kg/hr
Specific steam consumption = kg of steam / kg of H2O evaporated
= 2.677 kg steam / kg H2O evaporated

Problem

Data

Feed rate = 50000 kg / hr
W1, Specific humidity of inlet air (33°C, 45 % RH) = 0.14 kg water/ kg of dry air
W2, Specific humidity of outlet air (85°C) = ? kg water/ kg of dry air
Ws1 = 59 / 41 = 1.439 kg of water/ kg of dry solid
Ws2 = 3.5 / 96.5 = 0.03626 kg of water/ kg of dry solid
specific heat of feed = 0.554 kcal/ kg of dry air
ti = inlet air temperature (dbt) = 210 °C
calculate the quantity of feed to be taken for the system.

Solution: Ha1 = 0.24 dbt + W1 (597.2 + (0.45 x 210))
= 60.08 kcal/ kg of dry air
Ha2 = 0.24 dbt + W2 (597.2 + (0.45 x dbt))
= 20.4+ 635.45 W2

Problem

Find out the steam consumption of roller dryer for the given data:
Water evaporation rate = 25 kg H2O / hr
Initial feed concentration = 30 % TS
Final product TS = 96 %
Drum diameter = 1.5 m, drum length = 2.5 m

Solution

Total surface area = 2Πrl = 11.775 m2
Total water evaporated /hr = 25 x 11.75 = 294.25 kg H2O

For 100 kg of feed, feed with TS = 30% and final TS = 96 %;
Total water evaporated = concentration x (milk / hr – powder / hr)
= 1.04(100 – 30) = 68.8 kg/hr
For 294.25 kg water evaporated, kg of powder = 31.2 x 294.25/ 68.8
= 133.438 kg/hr --- (1)
Heat supplied by steam = heat gained by the product
Thus, steam quantity x latent heat of steam at 2 kg/cm2
= m.s.dt + (heat of vapourisation x vapour)
V= F (1- Xf/Xl) = 294
(X).(525.9) = (427.65)(0.554)(100.45) + 294 x 540
X = 326.67 kg/hr

Steam consumption = 326.67 kg steam / hr
(it is assumed that condensate is at 98 °C)
Thus, specific steam consumption
= (kg of steam / hr )/(kg of water evaporated/hr)
= 326.67 / 294.25 = 1.11kg of steam/ kg of water evaporated
(losses are neglected)
Hs1 = enthalpy of dry solid + enthalpy of moisture
= (sp. Heat x t x 1) + (Ws1 x t x 1)
{where, t = temperature of feed = 45 °C and let sp. Heat of feed = 0.554}
Thus, Hs1 = (0.554 x 45 x 1) + (1.439 x 45 x 1) = 84.68 kcal/ kg of dry solids

Hs2 = enthalpy of dry solid + enthalpy of moisture of product
Thus, Hs2 = (0.554 x 60 x 1) + (0.03626 x 60 x 1)
= 35.4156 kcal/ kg of dry solids
Enthalpy balance:
Enthalpy in system = enthalpy out of the system
Ha1 + Hs1 = Ha2 + Hs2 + Q1 (where = heat loss, kcal/ hr)
(Qa x Ha1) + (Hs1 + 4825) = Q1 + Qa (Ha2) + (Hs2 x 4825)
211828.62 = 80.48 Qa + 635.45 W2. Qa -------------------------- (1)

Moisture balance:
Moisture lost by the product = moisture gained by the air
4825 (1.439 – 0.036267) = Qa (W2 – 0.014)
6768.22 = Qa W2 – 0.014 Qa --------------------------- (2)

By solving eq. 1 & 2
Qa = 45750.949 kg/hr (air flow)
Thus, total kcal required/ hr = Qa x sp. Heat x temperature difference
= 45750.949 x 0.24 x (210 - 33)
= 1943499.9 kcal / hr
Calorific value of oil = 9985 kcal/kg
But heat transferred = 85% (efficiency)

Thus, actually we get = 9985 x 0.85 = 8487.25 kcal/kg

Thus quantity of oil required = total Kcal required/ (kcal / kg of oil)
= 1943499.9 / 85 = 228.99 kg/hrs