## Lesson 34. PROBLEMS ON DRYING

Module 3. Food dehydration

Lesson 34

PROBLEMS ON DRYING

Example 34.1

Find out the evaporation rate in a drum dryer for given data: steam temperature = 150 °C , vapourisation temperature of milk 103 °C, overall heat transfer coefficient 1.3 kw / m2 -K, drum diameter = 60 cm, length of the drum = 100 cm, latent heat of vapourisation = 2261 kJ / kg. The product is scraped at ¾ of a revolution of the drum. Assume that there are no heat losses to the surroundings.

Solution

heat transfer area,

A = 1 x 0.6 x π x ¾ = 1.413 m2

dw / dØ = 1.3 x 1.413 (150-103)/ 2261

= 0.038 kg/s

= 137.5 kg/h

Example 34.2

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 1.74 kw / m2 -K. 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

To obtain 50 kg of product, F = 50 x 8 = 400 kg/hr

Vapour removed = 50 x 7 = 350 kg/hr

The effective surface area needed for drying is 1.94 m2. Assuming that about ¼ of the surface area would not be used . Then the total surface area required would be : 1.94 x 4/3 = 2.586 m2 . If the length of the drum is taken as 1.37 m, then the diameter of the drum d will be:

π d x 1.37 = 2.586

Thus, d = 0.6 m

Example 34.3

M ilk of 18.5 % TS is dried to 3.8 % moisture 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

Example 34.4

500 kg of milk containing 40 % TS is dried in a spray dryer to a powder containing 3 % moisture. Air temperatures at the entry and exit of air heater are 30 °C and 150 °C respectively. Feed and product temperatures at the entry and the exit of spray dryer are 20 °C and 54°C respectively. Air temperatures at the entry and the exit of spray dryer are 150 °C and 90 °C respectively. 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) the overall thermal efficiency if the radiation loss in the air heater is 10 %,

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 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 2282 kJ/kg

Heat supplied to dry 500 kg milk = 500 (54 – 20) x 3.894 + (293.5 x 2282)

= 735963.5 kJ

The specific heat of air can be taken as 1.005 kJ/ kg K and the mass density of air at the mean temperature i.e. 120 °C is 0.9 kg/m3

(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.

But the actual amount of heat which is used to dry 500 kg milk = 735963.5 kJ

The thermal efficiency of air drying = 735963.5 / 1177541.6 = 62.5 %

(c)

The amount of air required in kg = 12205kg

Heat taken by air in the air heating section = 12205 x 1.005 (150 – 30)

= 1471923 kJ

At 7.2 kg/ cm2 absolute pressure, the latent heat of steam about 2065.5 kcal/kg

The amount of steam required at 80 % heating efficiency =

1471923/(2265.5 x 0.8) = 890.7kg

(d)

The amount of steam required per kg of water evaporated = 896.7 / 293.5 = 3.05 kg

(e)

Where,

R = % radiation heat loss

Thi = Air temperature at the inlet of air heater

Tho = Air temperature at the outlet of air heater

Th0 = Tai

= (1- 0.1)(150 – 90)/ (150 - 30) = 0.45

Example 34.5

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 104675 kJ/h. Find out the air flow rate required.

Solution:

Based on enthalpy and moisture balance:

ma = Mass flow rate of air (kg dry air/h)

ms = Mass flow rate of solid (kg dry solid/h)

Ta1 = temperature of air at the inlet = 190 °C

Ta2 = temperature of air at the exit = 80 °C

Tp1 = temperature of feed at the inlet = 30 °C

Tp2 = temperature of product at the exit = 50 °C

Ha1 , Ha2 are enthalpy content of air at the inlet and exit ( kJ/ kg dry air)

Hp1 , Hp2 are enthalpy content of feed and product ( kJ/ kg dry air)

W1 , W2 are moisture content in air at the entry and exit (kg moisture/ kg dry air)

w1 , w2 = moisture content in feed and product (kg moisture/ kg dry solid)

Cpp = specific heat of dry solids

Cpw = specific heat of water

From psyc. Chart:

Air contains 0.011 kg water/kg of dry air = W1

Assuming dry milk solids have specific heat of 2.32 kJ / kg

Feed rate = 1000 x 0.96 x 100 / 45 = 2133 kg / hr

ms = 1000 x 0.96 = 960 kg solid/h

w1 = 53745 moisture/ solid concentration of milk

= 1.22 kg of water /kg of dry solids

w2 = 4/96 = 0.042 kg of water / kg of dry solids.

Enthalpy:

Ha1 = 1.005 Ta1 + W1 (2500.5 + 1.884 Ta1)

= 1.005 x 190 + 0.011 (2500.5+1.884 x 190)

= 222.39 kJ / kg air

Ha2 = 1.005 Ta2 + W2 (2500.5 + 1.884 Ta2)

= 1.005 x 80 + W2 (2500.5+1.884 x 80)

= 80.4 + 2651.22 W2

= 222.39 kJ / kg air

Hp1 = Cpp Tp1 + w1 Cpw Tp1

= 2.32 x 30 + 1.22 x 4.187 x 30

= 222.84 kJ/ kg dry solid

Hp2 = Cpp Tp2 + w2 Cpw Tp2

= 2.32 x 50 + 0.042 x 4.187 x 50

= 124.8 kJ/ kg dry solid

Enthalpy balance:

Enthalpy in system = Enthalpy out system

ma Ha1 + ms Hp1 = ma Ha2 + ms Hp2 + Ql

ma x 222.39 + 960 x 222.84 = ma ( 80.4 + 2651.22 W2) + 960 x 124.8 + 104675

142 ma – 2651.22 ma W2 = 224483 ------------ eq. (1)

Moisture balance:

Moisture lost by product = moisture gained by air

ms(w1 – w2) = ma (W2 – W1)

960 (1.22 - 0.042)= ma (W2 – 0.011)

ma(W2 - 0.011) = 1130.88 -------------------- eq. (2)

By solving both equations:

ma = 26663 kg/hr

W1 = 0.011

From eq.(2)

W2 = (1130.88/ ma) + 0.011

W2 = 0.0534

Humidity of air exhaust is 0.0534 kg/kg dry air

Example 34.6

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

TS in feed = 42%

TS in product = 97%

Powder production rate 2000 kg/h

Specific humidity of outdoor 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

Spray dryer inlet air temp = 200 °C

Efficiency of steam air heater = 80 %

Steam pressure at the air heater = 21 kg/cm2

Condensate is discharged at saturation temp.

Specific heat of air Cpa =1.005 kJ/kg

Solution :

w1 = 58 / 42 = 1.380 kg of water/ kg of dry solid

w2 = 3 / 97 = 0.0309 kg of water/ kg of dry solid

Cp = specific heat of air = 0.24 kcal/ kg of dry air

tho = outlet air temperature of air at the air heater = 200 °C = ta1

thi = inlet air temperature of air at the air heater = 32 °C

Hs = enthalpy of steam at inlet of heater = 2800 kJ / kg

Hc = enthalpy of condensate at the outlet = 895.6 kJ / kg

ή = efficiency = 80%

moisture balance:

moisture lost by the product = moisture gained by the air

thus, 2(970)(1.38 – 0.0309) = ma (0.052 – 0.0119)

thus, ma = 65268.18 kg/h

steam consumption (kg/h) = ma . Cpa . (thi – tho)/ [(Hs – Hc). ή

= 65268.18 x 1.005 (200-32) x 0.8/ (2800-895.6)

= 7233 kg/h

= 7233 kg/h (by putting values)

Total water evaporated in dryer / h = total quantity of concentrated milk – total powder

= ((2000 x 0.97 ) / 0.58)- 2000

= 3344.83-2000

=1344.83 kg/h

Specific steam consumption = kg of steam / kg of H2O evaporated

= 7233 /1344.83

= 5.378 kg steam / kg water evaporation

Example 36.7

From the following data given for a spray dryer determine hot air flow rate and humidity of air at the exit of the dryer and quantity of fuel oil requirement for the air heater.

TS in feed = 41%

Moisture content in product = 3.5 %

Temp. of feed = 45 °C

Temp. of product at the exit of dryer = 60 °C

Feed rate = 50000 kg / hr

Specific humidity of inlet air (33°C, 45 % RH) = 0.14 kg water/ kg of dry air

Inlet and exit air temperatures are 210 °C and 85 °C resp.

Specific heat of feed = 2.32 kJ/kg

Heat losses from the dryer is 2 x 105 kJ/h

The calorific value of oil =42000 kJ/kg,

Heat transfer efficiency in the air heater = 85%

Solution:

w1 = 59 / 41 = 1.439 kg of water/ kg of dry solid

w2 = 3.5 / 96.5 = 0.03626 kg of water/ kg of dry solid

Cpp = 2.32 kJ/kg

Ta1 = 210 °C , Ta2 = 85 °C

Tp1=45 °C, Tp2 = 60 °C

Ql = 2 x 105 kJ/h

Quantity of feed ms = (5000 x 0.965) = 4825 kg/h

Ha1 = 1.005 Ta1 + W1 (2500.5 + 1.884 Ta1)

= 1.005 x 210 + 0.014 (2500.5+1.884 x 210)

= 251.6 kJ / kg air

Ha2 = 1.005 Ta2 + W2 (2500.5 + 1.884 Ta2)

= 1.005 x 85 + W2 (2500.5+1.884 x 85)

= 85.42 + 2660.64 W2

Hp1 = Cpp Tp1 + w1 Cpw Tp1

= 2.32 x 45 + 1.22 x 4.187 x 45

= 375.53 kJ/ kg dry solid

Hp2 = Cpp Tp2 + w2 Cpw Tp2

= 2.32 x 60 + 0.03626 x 4.187 x 60

= 148.31 kJ/ kg dry solid

ma Ha1 + ms Hp1 = ma Ha2 + ms Hp2 + Ql

ma x 256.1 + 4825 x 375.53 = ma ( 85.42 + 2600.64 W2) + 4825 x 148.31 + 2 x105

or 170.68 ma – 2660.64 ma W2

= 915595.75 ----------- eq (1)

ms(w1 – w2) = ma (W2 – W1)

or 4825(1.439 – 0.03626) = ma (W2 – 0.014)

or ma (W2 – 0.014) = 6768.22 --------------------- eq(2)

by solving eq (1) and (2)

ma= 141821.56 kg/h, W2 = 0.0617 kg/kg dry air

thus total energy requirement / h = ma x Cpp (210- 33)

= 25227928 kJ/h

Or 7007.75 kW

Quantity of oil requirement = (42000 x 0.85)/35700 kg/h = 706.66 kg/h

Example 34. 8

Find out the steam consumption of roller dryer for the given data:

Water evaporation rate = 25 kg H2O / hr . m2

Initial feed concentration = 30 % TS, Field temperature 45°C

Final product TS = 96 %

Drum diameter = 1.5 m, drum length = 2.5 m

Solution:

Total surface area = 2πrl = 11.78 m2

Total water evaporated /hr = 25 x 11.78 = 294. 52 kg H2O

For 100 kg of feed, feed with TS = 30% and final TS = 96 %;

Total water evaporated = Feed - product

= 100 – (100 x 0.3)/0.93 = 68.75 kg/hr

For 294.52 kg water evaporated, kg of powder = 31.25 x 294.52/ 68.75

= 133.87 kg/hr --- (1)

Feed rate, F= kg water evouparation kg product = 428.39 kg/hr

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 Quantity of vapour)

Ms x 2204.04 = 428.39 x 2.32 x (100 – 45) + 294.52 x 2261

Ms = 326.94 kg/hr

Steam consumption = 326.94 kg steam / hr

(it is assumed that condensate is coming out at saturation temperature)

Thus, specific steam consumption

= (kg of steam / hr )/(kg of water evaporated/hr)

= 326.93/ 294.52 = 1.11kg of steam/ kg of water evaporated

Questions:

Solve the following examples

Example-1

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

Example-2

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.

Example-3

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.

Last modified: Thursday, 27 September 2012, 7:17 AM