Over the past 40 years the falling film evaporator 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 evaporation takes place is increased. Fig. 2 shows a diagram of a falling film evaporator.

The liquid to be evaporated is evenly distributed on the inner surface of a tube (see page 39). 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. See Fig. 3. 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 Fig. 3a. 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 vapour 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 vapour. The separated concentrate is discharged (usually by 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.

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    (1)

See Fig. 4, 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.

e = A/C = C-Concentrate/C-Feed  (2)

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	B= A×e-1/e	(3)
	C	C= A×1/e	(4)
Evaporated quantity B	A	A= B×e/e-1	(5)
	C	C= b×1/e-1	(6)
Concentrate quantity C	A	A= C×e	(7)
	B	B= C×(e-1)	(8)

Where	A:	feed in kg/h
	B:	evaporation in kg/h
	C:	concentrate in kg/h
	e:	evaporation ratio See                         &n bsp;            &nbs p;                           &n bsp;         formula (2)

Since milk, due to the protein content, is a heat-sensitive product, evaporation (i.e. boiling) at 100ºC 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 main-tained by condensing the vapour by means of cooling water. A vacuum pump or simi-lar is used to evacuate incondensable gases from the milk.

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

Boiling temperature	ºC	100	60
Heating	Kcal/kg	94	54
Evaporation	Kcal/kg	539	564
Vacuum	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 vapour, see Fig. 4, from the evaporated milk contains almost all the applied energy, it is obvious to utilize this to evaporate more water by condensing the vapour. This is done by adding another calandria to the evaporator. This new calandria - the second effect - where the boiling temperature is lower, now works as condenser for the vapours from the first effect, and the energy in the vapour is thus utilized as it condenses.

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

Boiling point ºC	Vacuum m WG	corresp. to mm Hg abs	≈m above  sea level	Volume of water vapour
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 vapour 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 20-30ºC) condensing the vapour 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 vapours, and crystallization of lactose determine the practical limit being about 45ºC.

From Fig. 5 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.