CHAPTER 8
EVAPORATION (cont'd)

MULTIPLE EFFECT EVAPORATION

Feeding of Multiple Effect Evaporators
Advantages of Multiple Effect Evaporators

An evaporator is essentially a heat exchanger in which a liquid is boiled to give a vapour, so that it is also, simultaneously, a low pressure steam generator. It may be possible to make use of this, to treat an evaporator as a low pressure boiler, and to make use of the steam thus produced for further heating in another following evaporator called another effect.

Consider two evaporators connected so that the vapour line from one is connected to the steam chest of the other as shown in Fig. 8.2, making up a two effect evaporator.

Figure 8.2 Double effect evaporator – forward feed

If liquid is to be evaporated in each effect, and if the boiling point of this liquid is unaffected by the solute concentration, then writing a heat balance for the first evaporator:

                q₁ = U₁A₁(T_s - T₁) = U₁A₁ DT₁                                                              (8.1)

where q₁ is the rate of heat transfer, U₁ is the overall heat transfer coefficient in evaporator 1, A₁ is the heat-transfer area in evaporator 1, T_s is the temperature of condensing steam from the boiler, T₁ is the boiling temperature of the liquid in evaporator 1 and DT₁ is the temperature difference in evaporator 1, = (T_s - T₁).

Similarly, in the second evaporator, remembering that the "steam" in the second is the vapour from the first evaporator and that this will condense at approximately the same temperature as it boiled, since pressure changes are small,

                q₂ = U₂A₂(T₁ - T₂) = U₂A₂ DT₂

in which the subscripts 2 indicate the conditions in the second evaporator.

If the evaporators are working in balance, then all of the vapours from the first effect are condensing and in their turn evaporating vapours in the second effect. Also assuming that heat losses can be neglected, there is no appreciable boiling-point elevation of the more concentrated solution, and the feed is supplied at its boiling point,

                q₁ = q₂

Further, if the evaporators are so constructed that A₁ = A₂, the foregoing equations can be combined.

                U₂/U₁ = DT₁/DT₂.                                                                                (8.2)

Equation (8.2) states that the temperature differences are inversely proportional to the overall heat transfer coefficients in the two effects. This analysis may be extended to any number of effects operated in series, in the same way.

Feeding of Multiple Effect Evaporators

In a two effect evaporator, the temperature in the steam chest is higher in the first than in the second effect. In order that the steam provided by the evaporation in the first effect will boil off liquid in the second effect, the boiling temperature in the second effect must be lower and so that effect must be under lower pressure.

Consequently, the pressure in the second effect must be reduced below that in the first. In some cases, the first effect may be at a pressure above atmospheric; or the first effect may be at atmospheric pressure and the second and subsequent effects have therefore to be under increasingly lower pressures. Often many of the later effects are under vacuum. Under these conditions, the liquid feed progress is simplest if it passes from effect one to effect two, to effect three, and so on, as in these circumstances the feed will flow without pumping. This is called forward feed. It means that the most concentrated liquids will occur in the last effect. Alternatively, feed may pass in the reverse direction, starting in the last effect and proceeding to the first, but in this case the liquid has to be pumped from one effect to the next against the pressure drops. This is called backward feed and because the concentrated viscous liquids can be handled at the highest temperatures in the first effects it usually offers larger evaporation capacity than forward feed systems, but it may be disadvantageous from the viewpoint of product quality.

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 = U₁A₁DT₁ and in the second effect, q₂ = U₂A₂DT₂.

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

Since the overall conditions are the same, DT_s = DT₁+ DT₂, 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 = U_s, and A₁ = A₂. The problem then becomes to find A_s for the single-effect evaporator that will evaporate the same quantity as the two effects.

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

                    DT₁ = DT₂
               and DT_s= DT₁ + DT₂ = 2DT₁

                   DT₁ = 0.5DT_s

        Now q₁ + q₂ = U₁A₁DT₁ + U₂A₂DT₂
                           = U₁(A₁+ A₂) DT_s/2
          but q₁ + q₂ = q_s

                 and q_s = UA_sDT_s

so that (A₁ + A₂)/2 = 2A₁/2 = A_s

            That is A₁ = A₂ = A_s

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. The analysis can be extended to cover any number of effects and leads to the same conclusions. In multiple effect evaporators, steam economy has to be paid for by increased capital costs of the evaporators. Since the heat transfer areas are generally equal in the various effects and since in a sense what you are buying in an evaporator is suitable heat transfer surface, the n effects will cost approximately n times as much as a single effect.

Comparative costs of the auxiliary equipment do not altogether follow the same pattern. Condenser requirements are less for multiple effect evaporators. The condensation duty is distributed between the steam chests of the effects, except for the first one, and so condenser and cooling water requirements will be less. The optimum design of evaporation plant must then be based on a balance between operating costs which are lower for multiple effects because of their reduced steam consumption, and capital charges which will be lower for fewer evaporators. The comparative operating costs are illustrated by the figures in Table 8.1 based on data from Grosse and Duffield (1954); if the capital costs were available they would reduce the advantages of the multiple effects, but certainly not remove them.

TABLE 8.1
STEAM CONSUMPTION AND RUNNING COSTS OF EVAPORATORS

EXAMPLE 8.4. 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.

		Solids	Liquids	Total
	Feed	50	450	500
	Product	50	117	167
	Evaporation			333

Heat balance
From 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

                   U₁A₁DT₁  = U₂A₂ DT₂ = U₃A₃DT₃
And DT₁ + DT₂ + DT₃ = (134 - 86) = 48°C.

                  Now, if A₁  = A₂ = A₃
                   then DT₂  = U₁DT₁ /U₂ and DT₃ = U₁DT₁ /U₃
          so that DT₁(1 + U₁/U₂ + U₁/U₃) = 48,

DT₁ x [1 + (2270/2000) + (2270/1420)]  = 48
                                            3.73DT₁ = 48
                                                          DT₁          = 12.9°C,

                   DT₂ = DT₁ x (2270/2000) = 14.6°C
             and DT₃ = DT₁ 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 w_s is the quantity of steam condensed per hour in effect 1, then

                            w₁ x 2200 x 10³ = w₂ x 2240 x 10³
                                                    = w₃ x 2294 x 10³
                                                    = w_s 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
                                               w_s = 115 kg h^-1

Steam consumption
It required 115 kg steam (w_s) 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 = A₁ + A₂ + A₃ = 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.

Evaporation > VAPOUR RECOMPRESSION

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