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The Single Effect Evaporator
Vacuum Evaporation

Heat Transfer in Evaporators

Frequently in the food industry a raw material or a potential foodstuff contains more water than is required in the final product. When the foodstuff is a liquid, the easiest method of removing the water, in general, is to apply heat to evaporate it. Evaporation is thus a process that is often used by the food technologist.

The basic factors that affect the rate of evaporation are the:

rate at which heat can be transferred to the liquid,
quantity of heat required to evaporate each kg of water,
maximum allowable temperature of the liquid,
pressure at which the evaporation takes place,
changes that may occur in the foodstuff during the course of the      evaporation process.

Considered as a piece of process plant, the evaporator has two principal functions, to exchange heat and to separate the vapour that is formed from the liquid.

Important practical considerations in evaporators are the:

maximum allowable temperature, which may be substantially below 100°C.
promotion of circulation of the liquid across the heat transfer surfaces, to attain reasonably high      heat transfer coefficients and to prevent any local overheating,
viscosity of the fluid which will often increase substantially as the concentration of the dissolved materials      increases,
tendency to foam which makes separation of liquid and vapour difficult.


The typical evaporator is made up of three functional sections: the heat exchanger, the evaporating section, where the liquid boils and evaporates, and the separator in which the vapour leaves the liquid and passes off to the condenser or to other equipment. In many evaporators, all three sections are contained in a single vertical cylinder. In the centre of the cylinder there is a steam heating section, with pipes passing through it in which the evaporating liquors rise. At the top of the cylinder, there are baffles, which allow the vapours to escape but check liquid droplets that may accompany the vapours from the liquid surface. A diagram of this type of evaporator, which may be called the conventional evaporator, is given in Fig. 8.1.

FIG. 8.1 Evaporator
Figure 8.1 Evaporator

In the heat exchanger section, called a calandria in this type of evaporator, steam condenses in the outer jacket and the liquid being evaporated boils on the inside of the tubes and in the space above the upper tube plate. The resistance to heat flow is imposed by the steam and liquid film coefficients and by the material of the tube walls. The circulation of the liquid greatly affects evaporation rates, but circulation rates and patterns are very difficult to predict in any detail. Values of overall heat transfer coefficients that have been reported for evaporators are of the order of 1800-5000 J m-2 s-1 °C-1 for the evaporation of distilled water in a vertical-tube evaporator with heat supplied by condensing steam. However, with dissolved solids in increasing quantities as evaporation proceeds leading to increased viscosity and poorer circulation, heat transfer coefficients in practice may be much lower than this.

As evaporation proceeds, the remaining liquors become more concentrated and because of this the boiling temperatures rise. The rise in the temperature of boiling reduces the available temperature drop, assuming no change in the heat source. And so the total rate of heat transfer will drop accordingly. Also, with increasing solute concentration, the viscosity of the liquid will increase, often quite substantially, and this affects circulation and the heat transfer coefficients leading again to lower rates of boiling. Yet another complication is that measured, overall, heat transfer coefficients have been found to vary with the actual temperature drop, so that the design of an evaporator on theoretical grounds is inevitably subject to wide margins of uncertainty.

Perhaps because of this uncertainty, many evaporator designs have tended to follow traditional patterns of which the calandria type of Fig. 8.1 is a typical example.

Vacuum Evaporation

For the evaporation of liquids that are adversely affected by high temperatures, it may be necessary to reduce the temperature of boiling by operating under reduced pressure. The relationship between vapour pressure and boiling temperature, for water, is shown in Fig. 7.2. When the vapour pressure of the liquid reaches the pressure of its surroundings, the liquid boils. The reduced pressures required to boil the liquor at lower temperatures are obtained by mechanical or steam jet ejector vacuum pumps, combined generally with condensers for the vapours from the evaporator. Mechanical vacuum pumps are generally cheaper in running costs but more expensive in terms of capital than are steam jet ejectors. The condensed liquid can either be pumped from the system or discharged through a tall barometric column in which a static column of liquid balances the atmospheric pressure. Vacuum pumps are then left to deal with the non-condensibles, which of course are much less in volume but still have to be discharged to the atmosphere.

Heat Transfer in Evaporators

Heat transfer in evaporators is governed by the equations for heat transfer to boiling liquids and by the convection and conduction equations. The heat must be provided from a source at a suitable temperature and this is condensing steam in most cases. The steam comes either directly from a boiler or from a previous stage of evaporation in another evaporator. Major objections to other forms of heating, such as direct firing or electric resistance heaters, arise because of the need to avoid local high temperatures and because of the high costs in the case of electricity. In some cases the temperatures of condensing steam may be too high for the product and hot water may be used. Low-pressure steam can also be used but the large volumes create design problems.

Calculations on evaporators can be carried out combining mass and energy balances with the principles of heat transfer.

EXAMPLE 8.1. Single effect evaporator: steam usage and heat transfer surface
A single effect evaporator is required to concentrate a solution from 10% solids to 30% solids at the rate of 250 kg of feed per hour. If the pressure in the evaporator is 77 kPa absolute, and if steam is available at 200 kPa gauge, calculate the quantity of steam required per hour and the area of heat transfer surface if the overall heat transfer coefficient is 1700 J m-2 s-1 °C-1.
Assume that the temperature of the feed is 18°C and that the boiling point of the solution under the pressure of 77 kPa absolute is 91°C. Assume, also, that the specific heat of the solution is the same as for water, that is 4.186 x 103 J kg-1°C-1, and the latent heat of vaporization of the solution is the same as that for water under the same conditions.

From steam tables (Appendix 8), the condensing temperature of steam at 200 kPa (gauge)[300 kPa absolute] is 134°C and latent heat 2164 kJ kg-1; the condensing temperature at 77 kPa (abs.) is 91°C and latent heat is 2281 kJ kg-1.

Mass balance (kg h-1)

Heat balance
Heat available per kg of steam
                                       = latent heat + sensible heat in cooling to 91°C
                                       = 2.164 x 106 + 4.186 x 103(134 - 91)
                                       = 2.164 x 106 + 1.8 x 105
                                       = 2.34 x 106 J

Heat required by the solution
                                       = latent heat + sensible heat in heating from 18°C to 91°C
                                       = 2281 x 103 x 167 + 250 x 4.186 x 103 x (91 - 18)
                                       = 3.81 x 108 + 7.6 x 107
                                       = 4.57 x 108 J h-1

Now, heat from steam       =  heat required by the solution,
Therefore quantity of steam required per hour = (4.57 x 108)/(2.34 x 106)
                                       = 195 kg h-1

Quantity of steam/kg of water evaporated = 195/167
                                       = 1.17 kg steam/kg water.

Heat-transfer area
Temperature of condensing steam = 134°C.
Temperature difference across the evaporator = (134 - 91) = 43°C.
Writing the heat transfer equation for q in joules/sec,
                                     q = UA

              (4.57 x 108)/3600 = 1700 x A x 43
                                     A = 1.74 m2

Area of heat transfer surface = 1.74 m2

(It has been assumed that the sensible heat in the condensed (cooling from 134°C to 91°C) steam is recovered, and this might in practice be done in a feed heater. If it is not recovered usefully, then the sensible heat component, about 8%, should be omitted from the heat available, and the remainder of the working adjusted accordingly).


In evaporators that are working under reduced pressure, a condenser, to remove the bulk of the volume of the vapours by condensing them to a liquid, often precedes the vacuum pump. Condensers for the vapour may be either surface or jet condensers. Surface condensers provide sufficient heat transfer surface, pipes for example, through which the condensing vapour transfers latent heat of vaporization to cooling water circulating through the pipes. In a jet condenser, the vapours are mixed with a stream of condenser water sufficient in quantity to transfer latent heat from the vapours.

EXAMPLE 8.2. Water required in a jet condenser for an evaporator
How much water would be required in a jet condenser to condense the vapours from an evaporator evaporating 5000 kg h-1 of water under a pressure of 15 cm of mercury? The condensing water is available at 18°C and the highest allowable temperature for water discharged from the condenser is 35°C.

Heat balance

The pressure in the evaporator is 15 cm mercury = Zrg = 0.15 x 13.6 x 1000 x 9.81 = 20 kPa.
From Steam Tables, the condensing temperature of water under pressure of 20 kPa is 60°C and the corresponding latent heat of vaporization is 2358 kJ kg-1.

Heat removed from condensate
                                                         = 2358 x 103 + (60 - 35) x 4.186 x 103
                                                         = 2.46 x 106 J kg-1
Heat taken by cooling water
                                                         = (35 - 18) x 4.186 x 103
                                                         = 7.1 x 104 J kg-1

Quantity of heat removed from condensate per hour
                                                         = 5000 x 2.46 x 106 J
Therefore quantity of cooling water per hour
                                                         = (5000 x 2.46 x 106)/7.1 x 104
                                                         = 1.7 x 105 kg

EXAMPLE 8.3. Heat exchange area for a surface condenser for an evaporator
What heat exchange area would be required for a surface condenser working under the same conditions as the jet condenser in Example 8.2, assuming a U value of 2270 J m-2 s-1 °C-1, and disregarding any sub-cooling of the liquid.

The temperature differences are small so that the arithmetic mean temperature can be used for the heat exchanger (condenser).
Mean temperature difference
                                                           = (60 - 18)/2 + (60 - 35)/2
                                                           = 33.5°C.

The data are available from the previous Example, and remembering to put time in hours.
Quantity of heat required by condensate = UA
                                5000 x 2.46 x 106 = 2270 x A x 33.5 x 3600
and so                                              A = 45 m2
Heat transfer area required                    = 45 m2
This would be a large surface condenser so that a jet condenser is often preferred.


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Unit Operations in Food Processing. Copyright © 1983, R. L. Earle. :: Published by NZIFST (Inc.)
NZIFST - The New Zealand Institute of Food Science & Technology