1. To calculate the efficiency of spray dryer in term of material and heat balance.

2. To know the operation of a spray dryer. The operation include the performance of the dryer by varying feed concentration, hot water flow rate and cool water flow rate.

Spray drying process is a transformation process where feed in a form of fluid is changed to a dry product by spraying through a drying medium that’s hot. Using Atomization of feed into the spray dryer, which is the process where the fluid is into millions of small particles, which forms a spray, does this. The performance of a spray dryer is critically dependent on the drop size produced by the atomizer in which the gaseous medium mixes with the drops.

The most important thermenologies in drying are :

1. Humidity

- This is defined as the mass of water carried per unit of any air and can be calculated as

Kmol of water vapour = ___ P_w____

Kmol dry air ( P - P_w )

P = Total pressure

P_w = Partial pressure of water vapour

2. Humidity of saturated air Ho

- Defined as the humidity of air when saturated with water vapour. The air is in equilibrium with water at the given temperature and pressure.

3. Percentage humidity

Humidity of air x 100 = H x 100

Humidity of saturated air Ho

4. Percentage of relative humidity

- Is defined as :

Partial pressure of water vapour in air x 100

Vapour pressure of water at the same temperature

5. Humid volume

-Defined as the volume of unit mass of dry air and associated vapour. If atmospheric pressure is used, it can be calculated :-

22.4 ( T ) + 22.4 H ( T ) m³/ kg where T is in Kelvin

29 273 18 273

6. Saturated volume

-Defined as the volume of unit mass of dry air together with the water vapour required saturating it.

7. Humid heat

- Defined as the heat required to raise unit mass of dry air and associated vapour through 1^oK at constant pressure.

8. Dew point

- Defined as the temperature at which condensation will first occur when air is cooled.

9. Wet bulb temperature

- Defined as the temperature at which water is at equilibrium with air in sufficiently small surface.

Rate of heat transfer required from gas to liquid = hA ( O - O_w )

where ;

A = is surface area

O = is temperature of the air stream

O_w = Wet bulb temperature

h = heat transfer coefficient

10. Mass rate of vaporization

h_DAM_W ( P_Wo - P_W )

RT

h_D A M_A ( P - P_W )_mean ( H_W- H )

RT

h_DAr _A

For this drying process, the amount of mass in for the air and feed is the same as the amount of mass out. But the total heat for air and feed that goes in the system is not the same as the total heat that comes out because some heat is excepted through the surrounding. The spray dryer is made to produce the dry product with the highest termal efficient possible. The termal efficiency for the spray dryer is defined as :-

Heat used for vaporization

Input heat

The termal efficiency for this drying effect will increase with the increase of the inert temperature. Unfortunately, in most process, an inert temperature that’s too high will degrade the product. That is why inert temperature of 180^oC is chosen. An inert temperature that is too low will increase the humidity in the final dry product. It is important to control the termal efficient and the use of heat in the dryer to control the lost and quality of the product.

-The drying process for the spray dryer depends a lot of the humidity of the air that leaves the dryer. The humidity of the outlet air is a combination of the humidity in dry air and dry feed. To maximize the drying process, the dry air has to be dehumidified.

The termal efficiency can be improved by :-

1. Using air inert temperature that’s reasonably high

2. Using the outlet temperature that’s reasonably low

3. Using a insulated drying chamber

4. Dehumidifying the drying air.

1) The equipment used is the B191 Mini Spray Dryer. The equipment is already installed and readies to be operate.

2) Preparation is made for the spraying process.

3) The compressed air tab is opened and the green main switch is on.

4) The spray flow is opened by opened the valve until the desired through put is indicated.( manufacturer’s recommendation 600 L/hr )

5) The aspirator is turned on. The desired aspirator capacity is set with the arrow keys

( maximum capacity 35 m³/h, manufacturer’s recommendation 100% ).

6) As soon as the inlet temperature has stabilized, the outlet temperature is adjusted using distilled water. Increasing or decreasing the pump output (to maintain the maximum permitted product temperature, outlet temperature corresponds to maximum product temperature) could control the outlet temperature. The temperature is recorded.

7) A beaker of distilled water (approximately 100 ml ) is placed on the supporting table and the feed hose is immersed in the beaker.

8) The pump is set to small nominal value and it is increased slowly in steps until the desired outlet temperature is reached (maximum output 1800 ml/h) . The pump is switch on.

9) When the desired outlet values are reached, product spraying can be started. The feed hose is removed from the beaker of distilled water and it is immersed immediately in the solution to be sprayed. Depending on the concentration of the product ( water content ), the outlet temperature must now be adjusted to some extent. Stirred the milk continuously while the operation going on.

10) As soon as the product has been sprayed, the feed hose is rinsed immediately by reimmersing it in the vessel of distilled water. As soon as the hose and nozzle have been rinsed, the unit is switched off.

11) These are the steps to take while switched off the unit :

i) The pump should remain switched on for a while to prevent the outlet temperature from rising too high in the case of temperature- sensitre products.

ii) The pump is waited until the inlet temperature falls below 70^o C.

iii) The aspirator is switched off.

iv) The product vessel is removed.

12. The product in the glass parts is to be weighing.

1. Milk concentration = 5%
2. Weight of milk powder = 5.0019g
3. Volume of water added to milk = 95.0ml
4. Weight of milk + water + beaker = 208.5036g
5. Empty beaker weight (100ml) = 110.37g
6. Empty spray dryer beaker = 334.00g
7. Spray dryer beaker + milk powder = 335.29g
8. Weight of milk powder = 1.29g
9. Time = 35 minute
10. Initial inlet temperature = 180 ºC
11. Initial outlet temperature = 120 ºC
12. Initial temperature dry bulb = 29 ºC
13. Final temperature dry bulb = 52 ºC
14. Initial temperature wet bulb = 27 ºC
15. Final temperature wet bulb = 43.5 ºC

1. To calculate the mass balance:

Rate of air = 35 m³

j

Rate of output pum = 1800 ml× 0.05

j

= 90 ml

j

convert it into m³/j

= 90 ml × 1 m³

j 10⁶ ml

= 90 × 10^-6 m³

j

\ Rate of output pum = 90 × 10^-6 m³

j

Then, calculate the density of milk,

Density of milk = mass of milk solution

volume of milk solution

= 98.13 g

100 ml

= 0.9813 g × 10⁶ ml × 1 kg

ml 1 m³ 1000 g

 

= 981.3 kg

m³

 

 

Interpolation for 95% Humidity

 

Humid volume = Volume of dry air + (Volume of saturated wet air – Volume of dry

air) × Relatif humidity

 

Inlet volume of wet air = 0.85 + (0.88 – 0.85) × 0.95

= 0.8785 m³

kg dry air

 

Outlet volume of wet air = 0.91 + (0.99 – 0.91) × 0.6

= 0.9580 m³

kg dry air

 

Mass of water in air = rate of air × absolute humidity

humid volume

 

Mass of water in inlet air = 35 × 0.026

0.8785

= 1.0359 kg water

hour

 

 

Mass of water in outlet air = 35 × 0.058

0.9580

= 2.1190 kg water

hour

 

Mass of water vaporize into air = 2.1190 – 1.0359

= 1.0831 kg water

hour

 

Mass of water vaporize from milk solution

= rate of solution × density of milk solution × percentage of water in the solution

= 90 × 10^-6 × 981.3 × 0.95

= 0.08390 kg water

hour

 

Percentage of efficiency vaporization = 0.08390 × 100%

1.0831

= 7.7464 %

 

Percentage accumulation of milk powder = mass of accumulation dried milk

Initial mass of milk powder

= 1.29 g × 100%

5 g

= 25.8 %

 

2. To calculate the thermal equilibrium

Enthalpy of air = enthalpy dry air [ (enthalpy of saturated vapor – enthalpy of

dry air) × relative humidity ]

Enthalpy of inlet air = 30 + [ (98 – 30) × 0.95 ]

= 94.6 kJ

= 41940 Jmol^-1

= 41940 J × 1 mol

mol 18 g water

= 2328 J

g

= 2328 kJ

kg

Heat needed to vaporize milk solution

= m^oD H_n

= 1.0831 × 2328

= 2521.45 kJ

Heat loss to surrounding = Heat produced – heat needed to vaporize milk solution

= 3362.73 – 2521.45

= 841.28 kJ

hour

Thermal efficiency = Heat needed to vaporize water × 100%

Heat produced

= 2521.45 × 100%

3362.73

= 75.0 %

The result obtained from the experiment show that the efficiency of vaporization of milk (5% concentration) is rather low, that is 7.7464%. The percentage accumulation of milk powder is 25.8% while the thermal efficiency is 75.0%.

From these results, the efficiency of the spray dryer is rather low. This is due to several factors. Although the thermal efficiency is 75%, there is still a large amount of heat loss to the surrounding. The evaporization process of milk (especially low concentration milk) needs a large amount of heat. Loss of heat to surrounding means less milk is being evaporized. The heat is lost through convection, conduction and radiation because the spray dryer is not properly insulated.

Besides that, the inlet temperature could be too high. High inlet temperature would cause a layer of "crust" to form and envelop the milk droplets. This happens when water from the outer layer of the droplets evaporates unevenly leaving a higher density zone at the outer layer. A layer of "crust" will form as more water evaporates. The crust will decrease the rate of evaporization.

The other factor involved is parallax error when taking the thermometer’s reading or because of its position. In addition to that the connection of the thermometer unit and the air outlet hose is loose.

1. The milk solution is constantly stirred
2. The thermometer unit is properly connected
3. Avoid parallax error when taking the temperature reading
4. Set a lower or a more desible inlet temperature so an even evaporization of droplets is obtained.

The spray dryer is suitable to evaporate liquefied solution. The efficiency of the spray can be increase through proper procedure.