Evaporation

Dr. Mahesh kumar

Dehydration, Concentration, EvaporationLecture Notes

Dehydration

Drying – End product is solid. e.g. milk powder, dried juices, grains, starch etc.

Concentration – End product is liquid, e.g. concentrated milk, juices, soups etc.

Evaporation:

Evaporation is the partial removal of water from liquid food by boiling. Separation is achieved by

exploiting the difference in volatility between water and solute.

What is meant by concentration?

Concentration means removal of moisture so that the solid content of a liquid food increases.

Concentration is measured using Brix. e.g. Juice squeezed from mature oranges has

concentration of 10-12 Brix. After concentration, juice changes to 60-65 Brix.

Why Concentrate?

1. Increase shelf life: Solid contents are increased, which reduces water activity. e.g. jams

and molasses.

2. Pre-concentrate liquids (juice, milk) before drying, freezing, canning etc. to reduce weight

and volume. Saves energy during subsequent unit operations.

3. Preserve nutritional value

4. Reduce packaging, storage and shipping costs.

5. Improve sensory attributes. Carmelized syrups. Caramelized means thermal degradation

of sugars leading to the formation of volatiles and brown-colored products. The process is

catalyzed by acids or bases and generally requires temperatures > 120oC and pH between 3

and 9.

6. Convenience foods (mainly dried or concentrated). Have high shelf life, to be diluted by

consumer. Examples: fruit juices, soups, tomato products etc.

7. Convenient ingredients for secondary food manufacture – liquid pectin, fruit concentrates

for ice cream, baking etc.

8. Encapsulate flavors

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Methods for Removing Water

a) Mechanical – Centrifugation.

b) Chemical – Reverse Osmosis. Apply pressure on the product side to expel water.

c) Freeze concentration – Freeze the food and separate the ice by draining the juice using a semi-

permeable membrane. Does not cause heat damage. Solids may be lost along with ice. Capital

and operating costs are higher.

Above two methods are used for heat sensitive foods such as apple juice, wine, vinegar, beer,

grape juice etc.

d) Heat Application – to cause evaporation of water. Evaporation is the most popular method for

dehydrating liquid foods. Its disadvantages are that thermal degradation may occur.

Questions: Which of these techniques is the best?Answer: None, it depends upon the product you want to dry. Whether a product is sensitive to heat, properties of the raw liquid and end product, capital cost, environmental factors etc.For orange juice and milk evaporation is used. Freeze drying is used for delicate products such as wine, vinegar, beer, grape juice, apple juice. Reverse osmosis is more popular for producing drinking water than for juices. Centrifugation is more commonly used to separate fat from whey.

Losses during Evaporation1. Flavor Loss2. Color Loss3. Heat Damage

At atmospheric pressure higher temperature is required to boil a liquid. Lower temp of evaporation by applying vacuum so that losses are minimized.

A brief historyIn the mid 90s evaporators operated at low temperature in the range of 18-50 C. Concentrates prepared using these evaporators were not pasteurized and thus were prone to microbial spoilage. By 1955 heating units were added before evaporation to pasteurize the foods. In the early 1960s evaporators employing high temperature short time treatment were designed. In these designs pasteurization was performed along with evaporation. HTST also leads to inactivation of pectinesterase (an enzyme, which is bound to the cell walls with electrostatic forces. It hydrolyzes the pectin to form pectic acids. Pectic acids lead to reduction in viscosity of juices, which is a loss in juice quality. Pasteurization inactivates this enzyme). HTST evaporators generally have two shapes – tubular and plate type. Popular name for tubular designs are TASTE, which stands for Thermally Accelerated Short Time Evaporator. The highest temperature in HTST is in the range of 95-105 C.

Evaporator Construction Industrial evaporators normally consists of

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a) A heat exchanger to supply sensible heat and latent heat of evaporation to the feed. This part is usually called calanderia. In the food industry normally saturated steam is used as heating medium.

b) A separator in which the vapour is separated from the concentrated liquid phase.c) A condenser to effect condensation of vapour and its removal from the system

In many cases, the first two sections are contained in a single vertical cylinder

Most evaporators are heated by steam condensing on outside of metal tubes and material to

be evaporated (concentrated) flows inside the tubes. Usually steam is at a low pressure below 3

atms abs. and often boiling liquid is under moderate vacuum at pressure down to about 0.05 atm

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abs. Reducing the boiling temperature of the liquid increases the temperature difference between

steam and the boiling liquid and thus increases the heat transfer rate in the evaporator.

The capacity of an evaporator is determined by the amount of heat transferred to the fluid by the

heat exchanger.

If q is the amount of heat transferred then heat balance in the calandria gives

q = mf cp (Tb- Ti) + mv (hg – hf)

Where mf is mass flow rate of feeding liquidcp specific heat of the concentrated productmv mass flow rate of vapourhg enthalpy of vapourhf enthalpy of water component of feed that is converted into vapourTb temperature of liquid in the evaporatorTi temperature of incoming feed.

Mass Balance

Feed (mf) = Concentrated product (mp ) + Vapour produced (mv)

Solid in = Solid out

mf x xf = mp x xp

Where xf is the fraction of solids in the feedxp is the fraction of solids in the concentrated product

mv = mf - mf ( xf / xp ) = mf ( 1 - xf / xp )

The overall rate of heat transfer Q from the heating medium to the boiling liquid across the intervening wall and surface films is often known as heat load and is given as

Q = U A ΔT

The various factors influencing U are:

1) The condensing film heat transfer coefficient on the steam side of heat exchanger.

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2) The boiling liquid film coefficient on liquid side of heat exchanger3) The scale or fouling factor on both inner or outer walls bounding the heat transfer

surface4) Thermal resistance of the wall material

Requirements for good evaporation

1. Rapid heat transfer – thin films formed at high flow velocities, large surface area. 2. Higher viscosity limits heat transfer. 3. Lower evaporation temperature by vacuum.4. Efficient energy use and recovery – multiple stages – co-current (steam and product) or

countercurrent. Vapors from product coming off of first stage are used to heat feed to second stage.

5. Equipment design for easy cleaning & good sanitation and operation (process does not pasteurize product).

EXAMPLE 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.

Solution: From steam tables, 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)

  Solids Liquids Total

Feed 25 225 250

Product 25 58 83

Evaporation 167

Heat balance

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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 areaTemperature 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 DT

              (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).

STEAM TABLE - SATURATED STEAM

Temperature Pressure(Absolute) Enthalpy (sat. vap.)

Latent heat Specific volume

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(°C) (kPa) (kJ kg-1) (kJ kg-1) (m3 kg-1)

    Temperature Table    

0 0.611 2501 2501 206

1 0.66 2503 2499 193

2 0.71 2505 2497 180

4 0.81 2509 2492 157

6 0.93 2512 2487 138

8 1.07 2516 2483 121

10 1.23 2520 2478 106

12 1.40 2523 2473 93.9

14 1.60 2527 2468 82.8

16 1.82 2531 2464 73.3

18 2.06 2534 2459 65.0

20 2.34 2538 2454 57.8

22 2.65 2542 2449 51.4

24 2.99 2545 2445 45.9

26 3.36 2549 2440 40.0

28 3.78 2553 2435 36.6

30 4.25 2556 2431 32.9

40 7.38 2574 2407 19.5

50 12.3 2592 2383 12.0

60 19.9 2610 2359 7.67

70 31.2 2627 2334 5.04

80 47.4 2644 2309 3.41

90 70.1 2660 2283 2.36

100 101.4 2676 2257 1.67

105 120.8 2684 2244 1.42

110 143.3 2692 2230 1.21

115 169.1 2699 2217 1.04

120 198.5 2706 2203 0.892

125 232.1 2714 2189 0.771

130 270.1 2721 2174 0.669

135 313.0 2727 2160 0.582

140 361.3 2734 2145 0.509

150 475.8 2747 2114 0.393

160 617.8 2758 2083 0.307

180 1002 2778 2015 0.194

200 1554 2793 1941 0.127

    Pressure Table    

7.0 1.0 2514 2485 129

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9.7 1.2 2519 2479 109

12.0 1.4 2523 2473 93.9

14.0 1.6 2527 2468 82.8

15.8 1.8 2531 2464 74.0

17.5 2.0 2534 2460 67.0

21.1 2.5 2540 2452 54.3

24.1 3.0 2546 2445 45.7

29.0 4.0 2554 2433 34.8

32.9 5.0 2562 2424 28.2

40.3 7.5 2575 2406 19.2

45.8 10.0 2585 2393 14.7

60.1 20.0 2610 2358 7.65

75.9 40.0 2637 2319 3.99

93.5 80.0 2666 2274 2.09

99.6 100 2676 2258 1.69

102.3 119 2680 2251 1.55

104.8 120 2684 2244 1.43

107.1 130 2687 2238 1.33

109.3 140 2690 2232 1.24

111.4 150 2694 2227 1.16

113.3 160 2696 2221 1.09

115.2 170 2699 2216 1.03

116.9 180 2702 2211 0.978

118.6 190 2704 2207 0.929

120.2 200 2707 2202 0.886

127.4 250 2717 2182 0.719

133.6 300 2725 2164 0.606

138.9 350 2732 2148 0.524

143.6 400 2739 2134 0.463

147.9 450 2744 2121 0.414

151.6 500 2749 2109 0.375

167.8 750 2766 2057 0.256

179.9 1000 2778 2015 0.194

* Reproduced with permission from J. H. Keenan et al., Steam Tables - lnternational Edition in Metric Units, John Wiley, New York, 1969.

Types of Evaporators

The more common types of evaporators include:• Batch pan• Forced circulation• Natural circulation• Wiped film/ Agitated film

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• Rising film tubular• Plate equivalents of tubular evaporators• Falling film tubular• Rising/falling film tubular

Batch Pan

Next to natural solar evaporation, the batch pan (Figure 1) is one of the oldest methods of concentration. It is somewhat outdated in today’s technology, but is still used in a few limited applications, such as the concentration of jams and jellies where whole fruit is present and in processing some pharmaceutical products. Up until the early 1960’s, batch pan also enjoyed wide use in the concentration of corn syrups. With a batch pan evaporator, product residence time normally is many hours. Therefore, it is essential to boil at low temperatures and high vacuum when a heat sensitive or thermo degradable product is involved. The batch pan is either jacketed or has internal coils or heaters. Heat transfer areas normally are quite small due to vessel shapes, and heat transfer coefficients (HTC’s) tend to be low under natural convection conditions. Low surface areas together with low HTC’s generally limit the evaporation capacity of such a system. Heat transfer is improved by agitation within the vessel. In many cases, large temperature differences cannot be used for fear of rapid fouling of the heat transfer surface. Relatively low evaporation capacities, therefore, limit its use.

Natural Circulation

Evaporation by natural circulation is achieved through the use of a short tube bundle within the batch pan or by having an external shell and tube heater outside of the main vessel (Figure 2). The external heater has the advantage that its size is not dependent upon the size or shapeof the vessel itself. As a result, larger evaporation capacities may be obtained. The most common application for this type of unit is as a reboiler at the base of a distillation column.

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Rising Film TubularConsidered to be the first ‘modern’ evaporator used in the industry, the rising film unit dates back to the early 1900’s. The rising film principle was developed commercially by using a vertical tube with steam condensing on its outside surface (Figure 3). Liquid on the inside of the tube is brought to a boil, with the vapor generated forming a core in the center of the tube. As the fluid moves up the tube, more vapor is formed resulting in a higher central core velocity that forces the remaining liquid to the tube wall. Higher vapor velocities, in turn, result in thinner and more rapidly moving liquid film. This provides higher HTC’s and shorter product residence time.The development of the rising film principle was a giant step forward in the evaporation field, particularly in product quality. In addition, higher HTC’s resulted in reduced heat transfer area requirements and consequently, in a lower initial capital investment.

Falling Film Tubular

Following development of the rising film principle, it took almost half a century for a falling film evaporation technique to be perfected (Figure 4). Falling film distribution generally is based around use of a perforated plate positioned above the top tube plate of the calandria. The falling film evaporator does have the advantage that the film is ‘going with gravity’ instead of against it. This results in a thinner, faster moving film and gives rise to an even shorter product contact time and a further improvement in the value of HTC. To establish a well-developed film, the rising film

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unit requires a driving film force, typically a temperature difference of at least 25°F (14°C) across the heating surface. In contrast, the falling film evaporator does not have a driving force limitation—permitting a greater number of evaporator effects to be used within the same overall operating limits. For example, if steam is available at 220°F (104°C), then the last effect boiling temperature is 120°F (49°C); the total available ΔT is equal to 100°F (55°C). In this scenario a rising film evaporator would be limited to four effects, each with a ΔT of 25°F (14°C). However, using the falling film technique, it is feasible to have as many as 10 or more effects.

Forced Circulation

The forced circulation evaporator (Figure 6) was developed for processing liquors which are susceptible to scaling or crystallizing. Liquid is circulated at a high rate through the heat exchanger, boiling being prevented within the unit by virtue of a hydrostatic head maintained above the top tube plate. As the liquid enters the separator where the absolute pressure is slightly less than in the tube bundle, the liquid flashes to form a vapor. The main applications for a forced circulation evaporator are in the concentration of inversely soluble materials, crystallizing duties, and in the concentration of thermally degradable materials which result in the deposition of solids. In all cases, the temperature rise across the tube bundle is kept as low as possible, often as low as 3-5°F (2-3°C). This results in a recirculation ratio as high as 220 to 330 lbs (100 to 150 Kg) of liquor per pound (kilogram) of water evaporated. These high recirculation rates result in high liquor velocities through the tube which help to minimize the build up of deposits or crystals along the heating surface. Forced circulation evaporators normally are more expensive than film evaporators because of the need for large bore circulating pipe work and large recirculating pumps. Operating costs of such a unit also are considerably higher.

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Wiped Film/ Agitated film

The wiped or agitated thin film evaporator has limited applications due to the high cost and is

confined mainly to the concentration of very viscous materials and the stripping of solvents down

to very low levels. Feed is introduced at the top of the evaporator and is spread by wiper blades on

to the vertical cylindrical surface inside the unit. Evaporation of the solvent takes place as the thin

film moves down the evaporator wall. The heating medium normally is high pressure steam or oil.

A high temperature heating medium generally is necessary to obtain a reasonable evaporation rate

since the heat transfer surface available is relatively small as a direct result of its cylindrical

configuration. The wiped film evaporator is satisfactory for its limited applications. However, in

addition to its small surface area, it also has the disadvantage of requiring moving parts such as the

wiper blades which, together with the bearings of the rotating shaft, need periodic maintenance.

Capital costs in terms of dollars per pound of solvent evaporated also are very high.

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Performance of Tubular Evaporators

The performance of tubular evaporators is expressed in terms of its CAPACITY and ECONOMY

Capacity is defined as the amount of water vaporized per unit time (kg of vapor/hour)Economy is the quantity of water vaporized by one kg of steam fed to the unit.

Evaporator capacity is influenced by

Rate of heat transfer (Q = UAΔT) which is dependent for a given evaporator upon ΔT (ie temperature of condensing steam and temperature of boiling liquid inside the evaporator. Higher is this temp. difference, higher will be the capacity of the evaporating unit.

ΔT is affected by boiling point elevation of concentrating liquid, liquid head (depth of liquid in the tubes of evaporator) and friction of flow of heated vaporized water in contact with tube surface as well as heat transfer coefficients.

Evaporator economy

The chief factor influencing the economy of an evaporator is number of effects. Multiple effect evaporator consist of a series of vacuum boiling cell arranged in series in such a way that each succeeding evaporator has lower pressure therefore boils at lower temperature. Thus steam fed in the calandria of first body produces vapour from the liquid feed which has sufficient energy to boil the liquid feed in the next unit. This way depending upon the number of effects the steam used once produces vapour in the evaporators in the series.

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Advantages of Multiple Effect Evaporators

At first sight, it may seem that the multiple effect evaporator has all the advantages, the heat is used over and over again and we appear to be getting the evaporation in the second and subsequent effects for nothing in terms of energy costs. Closer examination shows, however, that there is a price to be paid for the heat economy.

In the first effect, q1 = U1A1ΔT1 and in the second effect, q2 = U2A2ΔT2.

We shall now consider a single-effect evaporator, working under the same pressure as the first effect                           qs = UsAsΔTs, where subscript s indicates the single-effect evaporator.

Since the overall conditions are the same, ΔTs = ΔT1+ ΔT2, as the overall temperature drop is between the steam-condensing temperature in the first effect and the evaporating temperature in the second effect. Each successive steam chest in the multiple-effect evaporator condenses at the same temperature as that at which the previous effect is evaporating.

Now, consider the case in which U1 = U2 = Us, and A1 = A2. The problem then becomes to find As for the single-effect evaporator that will evaporate the same quantity as the two effects.

From the given conditions and from eqn. (8.2),

                    ΔT1 = ΔT2

               and ΔTs= ΔT1 + ΔT2 = 2ΔT1

                  ΔT1 = 0.5ΔTs

        Now q1 + q2 = U1A1ΔT1 + U2A2ΔT2

                           = U1(A1+ A2) ΔTs/2           but q1 + q2 = qs

                 and qs = UAsΔTs

so that (A1 + A2)/2 = 2A1/2 = As

            That is A1 = A2 = As

The analysis shows that if the same total quantity is to be evaporated, then the heat transfer surface of each of the two effects must be the same as that for a single effect evaporator working between the same overall conditions.

TABLE 8.1STEAM CONSUMPTION AND RUNNING COSTS OF EVAPORATORS

Number of effects

Steam consumption(kg steam/kg water

evaporated)

Total running cost (relative to a single- effect

evaporator)

One 1.1 1

Two 0.57 0.52

Three 0.40 0.37

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EXAMPLE . Triple effect evaporators: steam usage and heat transfer surface Estimate the requirements of steam and heat transfer surface, and the evaporating temperatures in each effect, for a triple effect evaporator evaporating 500 kg h-1 of a 10% solution up to a 30% solution. Steam is available at 200 kPa gauge and the pressure in the evaporation space in the final effect is 60 kPa absolute. Assume that the overall heat transfer coefficients are 2270, 2000 and 1420 J m-2 s-1 °C-1 in the first, second and third effects respectively. Neglect sensible heat effects and assume no boiling-point elevation, and assume equal heat transfer in each effect.

Mass balance (kg h-1)Solids Liquids Total

Feed 50 450 500

Product 50 117 167

Evaporation     333

Heat balanceFrom steam tables, the condensing temperature of steam at 200 kPa (g) is 134°C and the latent heat is 2164 kJ kg -1. Evaporating temperature in final effect under pressure of 60 kPa (abs.) is 86°C, as there is no boiling-point rise and latent heat is 2294 kJ kg-1.

Equating the heat transfer in each effect:

                             q1 = q2 = q3

                   U1A1ΔT1  = U2A2 ΔT2 = U3A3ΔT3 And ΔT1 + ΔT2 + ΔT3 = (134 - 86) = 48°C.

                  Now, if A1  = A2 = A3

                   then ΔT2  = U1ΔT1 /U2 and ΔT3 = U1ΔT1 /U3

          so that ΔT1(1 + U1/U2 + U1/U3) = 48,

ΔT1 x [1 + (2270/2000) + (2270/1420)]  = 48                                            3.73ΔT1 = 48                                                          ΔT1          = 12.9°C,

                   ΔT2 = ΔT1 x (2270/2000) = 14.6°C             and ΔT3 = ΔT1 x (2270/1420) = 20.6°C

And so the evaporating temperature:in first effect is          (134 - 12.9) = 121°C; latent heat (from Steam Tables) 2200 kJ kg-1. in second effect is     (121 - 14.6) = 106.5°C; latent heat 2240 kJ kg-1

in the third effect is (106.5 - 20.6) = 86°C, latent heat 2294 kJ kg-1

Equating the quantities evaporated in each effect and neglecting the sensible heat changes, if w1, w2, w3 are the respective quantities evaporated in effects 1,2 and 3, and ws is the quantity of steam condensed per hour in effect 1, then

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                            w1 x 2200 x 103 = w2 x 2240 x 103

                                                    = w3 x 2294 x 103

                                                    = ws x 2164 x 103

The sum of the quantities evaporated in each effect must equal the total evaporated in all three effects so that:

                                w1 + w2 + w3 = 333 and solving as above,

                                               w1 = 113 kg h-1    w2 = 111kg h-1     w3 = 108kg h-1

                                               ws = 115 kg h-1

Steam consumptionIt required 115 kg steam (ws) to evaporate a total of 333 kg water, that is                                         0.35kg steam/kg water evaporated.

Heat exchanger surface.Writing a heat balance on the first effect:

(113 x 2200 x 1000)/3600 = 2270 x A1 x 12.9                        A1 = 2.4 m2 = A2 = A3

                       total area = A 1 + A 2 + A 3 = 7.2 m2.

Note that the conditions of this example are considerably simplified, in that sensible heat and feed heating effects are neglected, and no boiling-point rise occurs. The general method remains the same in the more complicated cases, but it is often easier to solve the heat balance equations by trial and error rather than by analytical methods, refining the approximations as far as necessary.

Thermo Vapor Recompression (TVR)

When steam is available at pressures in excess of 45 psig (3 barg) and preferably over 100 psig (7 barg), it will often be possible to use thermo vapor recompression. In this operation, a portion of the steam evaporated from the product is recompressed by a steam jet venturi and returned to the steam chest of the evaporator. A system of this type can provide a 2 to 1 economy or higher depending on the product the steam pressure and the number of effects over which TVR is applied. TVR is a relatively inexpensive technique for improving the economy of evaporation. TVR can also be used in conjunction with multi-effect to provide even larger economies (Figure 18). Shown in (Figure 19) are the economies that can be achieved. Thermocompressors are somewhat inflexible and do not operate well outside the design conditions. Therefore if the product is known to foul severely, so that the heat transfer coefficient is significantly reduced, it is best not to use TVR. The number of degrees of compression is too small for materials that have high boiling point elevation.

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Mechanical Vapor Recompression (MVR)

Thermodynamically, the most efficient technique to evaporate water is to use mechanical vapor thermorecompression. This process takes the vapor that has been evaporated from the product, compresses the vapor mechanically and then uses the higher pressure vapor in the steam chest The vapor compression is carried out by a radial type fan or a compressor. The fan provides a relatively low compression ratio of 1:30 which results in high heat transfer surface area but an extremely energy efficient system. Although higher compression ratios can be achieved with a centrifugal compressor, the fan has become the standard for this type of equipment due to its high reliability, low maintenance cost and generally lower RPM operationThis technique requires only enough energy to compress the vapor because the latent heat energy is always re-used. Therefore, an MVR evaporator is equivalent to an evaporator of over 100 effects. In practice, due to inefficiencies in the compression process, the equivalent number of effects is in the range 30 to 55 depending on the compression ratio.The energy supplied to the compressor can be derived from an electrical motor, steam turbine, gas turbine and internal combustion engine. In any of the cases the operating economics are extremely good. Since the costs of the compressors are high, the capital cost of the equipment will be significantly higher than with multi-effect. However in most cases, for medium size to large evaporators, the pay back time for the addition capital will only be 1 to 2 years. Like the one TVR, the two MVR system is not appropriate for high fouling duties or where boiling point elevation is high.

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