SALEEM INDIA BLOG: Dairy Falling Film Evaporators

Monday, September 01, 2008

Dairy Falling Film Evaporators

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Over the last 40 years the falling film evaporators has practically replaced the forced recirculation evaporator used until then. This type of evaporator is desirable from a product point of view, as it offers a short holding time. Further, the amount of product in the evaporator is reduced and the surface from which the evaporator takes place is increased. Figure 1 shows a diagram of a falling film evaporator.

Figure 1 Falling Film Recirculation Evaporator

The liquid to be evaporated is evenly distributed on the inner surface of a tube. The liquid will flow downwards forming a thin film, from which the boiling/evaporation will take place because of the heat applied by the steam. The steam will condense and flow downwards on the outer surface of the tube. A number of tubes are built together side by side. At each end the tubes are fixed to tube plates, and finally the tube bundle is enclosed by a jacket, see Figure 2.

Figure 2 Evaporator Calandria

The steam is introduced through the jacket. The space between the tubes is thus forming the heating section. The inner side of the tubes is called the boiling section. Together they form the so-called calandria. The concentrated liquid and the vapor leave the calandria at the bottom part, from where the main proportion of the concentrated liquid is discharged. The remaining part enters the subsequent separator tangentially together with the vapor. The separated concentrate is discharged (usually be means of the same pump as for the major part of the concentrate from the calandria), and the vapour leaves the separator from the top. The heating steam, which condenses on the outer surface of the tubes, is collected as condensate at the bottom part of the heating section, from where it is discharged by means of a pump.

In order to understand the heat and mass transfer, the basis for the evaporation, it is necessary to define various specific quantities.

Figure 3 One-stage evaporator. Definition of various specific quantities and the corresponding heat-flow diagram.

From a given quantity of feed (A) part of the solvent is evaporated (B) leaving the concentrate or the evaporated product (C). And thus

A=B + C

See Figure 3, showing specific quantities and the corresponding heat flow diagram.

The evaporation ratio (e) is a measure for the evaporation intensity and can be defined either as the ratio between the amount of feed and concentrate or the ratio between the solids percentage in the concentrate and in the feed.

C=A /C=(C -Concentrate)/(C-Feed)

If the concentrations or the evaporation ratio are known the quantities A, B or C can be calculated, if one of them is known.

Given Quantity	to be found	Formula
Quantity to be treated A	B C	B=A X (e-1)/e C=A X 1/e
Evaporated Quantity B	A C	A=B X e/(e-1) C=B X 1/(e-1)
Concentrate Quantity C	A	A=C X e
Where: A: feed in kg/hr B: evaporator in kg/h C: concentrate in kg/h E: evaporation ration

Since milk, due to the protein content, is a heat-sensitive product, evaporation (i.e. boiling) at 100 ° C will result in denaturation of these proteins to such an extent that the final product is considered unfit for consumption. The boiling section is therefore operated under vacuum, which means that the boiling/evaporation takes place at a lower temperature than that corresponding to the normal atmospheric pressure. The vacuum is created by a vacuum pump prior to start-up of the evaporator and is maintained by condensing the vapor by means of cooling water. A vacuum pump or similar is used to evacuate incondensable gases from the milk.

At 100° C the evaporation enthalpy of water is 539 Kcal/kg and at 60°C it is 564 Kcal/kg. As the milk has to be heated from e.g. 6°C to the boiling point, and as energy, approx. 20 Kcal/kg, is required to maintain a vacuum corresponding to a boiling point of 60° C, we get the following energy consumption figures, provided we estimate the heat loss to be 2%:

Boiling temperature	°C	100	60
Heating	Kcal/Kg	94	54
Evaporation	Kcal/kg	539	564
Vaccum	Kcal/kg	_	20
Net Energy Consumption	Kcal/Kg	633	638
Heat loss, approx.	Kcal/kg	15	15
Total Energy Consumption	Kcal/kg	648	653

Corresponding to about 1.1 kg steam/kg evaporated water.

To simplify the following examples we will use 1 kg steam/kg evaporated water.

As vapor, see Figure 3, from the evaporated milk contains almost all the applied energy, it is obvious to utilize this to evaporate more water by condensing the vapor. This is done by adding another calandria to the evaporator. This new calandria - the second effect - now works as condenser for the vapors from the first effect, and the energy in the vapor is thus utilized as it condenses.

In order to obtain a temperature difference in the second effect between the product and vapor coming from the first effect, the boiling section of the second effect is operated at a higher vacuum corresponding to a lower boiling point.

Given Quantity	Vacumm m WG	Corresp. to mm HG abs	= m above sea level	Volume of water vapor
100	0	760	0	1.7 m³/kg
85	4.5	434	5,200	2.8 m³/kg
70	7.2	233	10,000	4.8 m³/kg
60	8.3	149	14,000	7.7 m³/kg
50	9.1	92	18,000	12.0 m³/kg
40	9.6	55	22,000	19.6 m³/kg

A third effect heated by vapor from the second effect, and so forth, can of course be added. The limit is the lowest vacuum obtainable, and that is decided from the amount and temperature of the cooling water (usually 20 - 30°C) condensing the vapor from the last effect, whereby the vacuum is maintained. Using ice-water or direct expansion of freon to bring down the last effect boiling temperature is of course theoretically possible, but other factors such as viscosity of the product, volume of the vapors, and crystallization of lactose determine the practical limit being about 45°C.

From Figure 4 we can see that 1 kg of steam can evaporate 2 kg of water and by applying a third effect 3 kg of water is evaporated using only 1 kg of steam.