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:
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, Ts 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, = (Ts - 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,
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,
Further, if the evaporators are so constructed that A1 = A2, the foregoing equations can be combined.
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.
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.
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.
shall now consider a single-effect evaporator, working under the same
pressure as the first effect
Since the overall conditions are the same, DTs = 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 = 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),
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.
Equating the heat transfer in each effect:
q1 = q2 = q3
= U2A2 DT2
Now, if A1 = A2 = A3
x [1 + (2270/2000) + (2270/1420)] = 48
x (2270/2000) = 14.6°C
And so the evaporating
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
w1 x 2200 x 103 = w2 x 2240
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
(113 x 2200 x 1000)/3600
= 2270 x A1 x 12.9
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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