Forward feed operation mode
In this operation mode, the solvent vapor and the liquor, whose concentration increases with every effect, travel in the same direction. From the energy point of view, this mode of operation is the less efficient in the usage of the energy available in the steam.
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| Fig. 01 Variables in a triple-effect chemical evaporator in forward feed mode operation. |
However, this arrangement of the effects seems to be natural and so, the easiest to understand.
CONTENTS
2.2 Mass balance on the solute
4 Heating surface requirements
5 What you know about mass and heat balance in a multiple effect evaporator
1 Some important assumptions
The mass and heat balance shall be based on observations of the typical functioning of the evaporator so that the balances and future calculations were easy to conduct. These are as follows,
- the process fluid entering the first effect $F_1$ and subsequents $(S_1, S_2)$ should be well mixed, so that these have uniform concentration. This is true for large, horizontally, compartiments but not for effects made of long vertical tubes in which not turbulent conditions may exist,
- the hydraulic head or hydraulic load due to depth of the fluid in the heating chamber and pressurization at the top may influence the boiling temperature of the solution. Again, this may cause trouble in effects made of long vertical tubes,
- the fluid leaving the boiling chamber in streams $(L_1,L_2,L_3)$ is only the solvent, which is usually water. Also, its temperature and pressure are exactly those of the boiling solution in the corresponding effect,
- once the steam has transferred its available energy to the process fluid it condensates and leaves the evaporator as streams $(C_1,C_2,C_3)$ at the water vapor pressure of the corresponding effect. One may say that the liquid is in fact subcooled (liquid at a temperature slightly below its boiling temperature) but this effect is neglected because its contribution to evaporation process is small,
- no heat losses occur from the effects into the surroundings. This not true since it is known that multiple effect evaporators loss more heat that single stage designs. However, let us consider that proper thermal insulation is installed so that the effect of theses losses is negligible,
- there are no increments in the boiling point rise BPR with respect to the first effect. This means that the largest BPR will only exist at the first effect and that it will decrease with every subsequent effect,
- the heat transfer area is the same for each effect. There can be cases were the area changes from one effect to another one, but it is not common practice. Besides, if the heat transfer area is the same calculations are greatly reduced.
2 The mass balance
This can be divided into a global mass balance and for the solute in every stream leaving every effect. Since the arrangement of streams is different for all streams, some assertions discussed below may be helpful.
2.1 Global mass balance
This shall be made only on the incomming stream of dilute solution $F$ since the mass flow rate of steam entering in effect 1, as $W$, should be the same leaving as the stream of condensate $C$. Therefore, a global mass balance can be thought as,
Eq. (01)and in terms of the variables used in Fig. 01, Eq. (01) becomes,
$F_1=S_3+L_1+L_2+L_3$ Eq. (02)
As you can see the streams $S_2$ and $S_3$ are not included in this balance since inconsistency with $L_1$ trhough $L_3$ would arise. The summation $L_1+L_2+L_3$ represent the solvent vapor extracted from the solution across the whole process.
Individual mass balances for every effect can also be written. Using the same idea of Eq. (01) the following equations come up,
$F_1=S_1+L_1$ Eq. (03)
$S_1=S_2+L_2$ Eq. (04)
$S_2=S_3+L_3$ Eq. (05)
Of course, if Eqs. (03-05) are added up Eq. (02) is obtained. These equations may look like redundant stuff but perhaps these may be useful for a tricky situation.
2.2 Mass balance on the solute
Again, a gloabal balance can be thought in the same way as for single stage evaporators. Here is the idea,
Eq. (06)which in terms of the variables in use is rewritten as,
$F\,x_F=S_3\,x_{S3}$ Eq. (07)
In Eq. (07) $x_F$ and $x_{S3}$ are the solute fraction in the feeding and tick liquor streams, respectively. Of course, since the liquor thickens from the first effect to the third one: $x_F<x_{S1}<x_{S_2}<x_{S3}$.
As in the previous sectoion, a set of equations for the solute balance in each effect can be written. These are,
$F\,x_F=S_1\,x_{S1}$ Eq. (08)
$S_1\,x_{S1}=S_2\,x_{S2}$ Eq. (09)
$S_2\,x_{S2}=S_3\,x_{S3}$ Eq. (10)
Notice, that combination of Eqs. (08-10) produces the mass balance Eq. (07).
3 The heat balance
The heat balance is performed in terms of the enthaplies. In this way, the same idea of balance for a single stage evaporator holds for a multiple effects unit,
Eq. (11)Please, refer to post: Mass and heat balance in a single stage evaporator for more details on the simplification of the term Heat in the condensate in Eq. (11).
You should be aware of the contributions of heat loss by radiation indicated in Eq. (11) since for multiple effect evaporators it can be significant.
Instead of a global heat balance, individual effects balance shall be considered. Then, by using Eq. (11) for each of the three effects the following equations are obtained.
For the 1st effect,
$w_1\, \lambda_{W1} + F\,h{F1} = L_1\, H_{L1} + S_1\, h_{S1}$ Eq. (12)
For the 2nd effect,
$L_1\, \lambda_{L1} + S_1\, h_{S1} = L_2\, H_{L2} + S_2\, h_{S2}$ Eq. (13)
For the 3rd effect,
$L_2\, \lambda_{L2} + S_2\, h_{S2} = L_3\, H_{L3} + S_3\, h_{S3}$ Eq. (14)
As a side comment. If Eqns. (12-14) were compared with those presented in Kern's Process Heat Transfer you will see a difference since its equations have only three terms in each balance and the specific heat is introduced. It left to the reader the demonstration of going from the equations of this post to those presented by Kern's.
4 Heating surface requirements
This section it shall be considered that the heat transfer area $A$ is the same in each effect. Despite this assumption, the overall heat transfer coefficient $U$ may different for each effect.
As in the previous section, we shall proceed by one effect at the time. Also, as for the case of a single stage evaporator, the heat transferred in an effect may be expressed as,
Eq. (15)in which for subsequent effects to the first one the steam will be in fact the solvent vapor of the preceeding effect. Therefore, using the general heat equation,
$q=UA\Delta T$ Eq. (16)
The following expressions for each effect are obtained. For the 1st effect,
$A_1=\dfrac{W_1\, \lambda_{W1}}{U_1\left( T_{W1}-T_{S1} \right)}$ Eq. (17)
For the 2nd effect we obtain,
$A_2=\dfrac{L_1\, \lambda_{L1}}{U_2\left( T_{L1}-T_{S2} \right)}$ Eq. (18)
and, finally, for the 3rd effect it follows,
$A_3=\dfrac{L_2\, \lambda_{L2}}{U_3\left( T_{L2}-T_{S3} \right)}$ Eq. (19)
One final comment on the heating area subject. From the mathematical point of view, having $A_1=A_2=A_3$ and so on is convenient since it helps to reduce the number of unknowns in design calculations. Perhaps the reason of this license has more to do with the solution procedure than with the construction of a multiple effect evaporator with units of different heating surfaces.
5 What you know about mass and heat balance in a multiple effect evaporator
Please, follow the link below to access a quizz and check what you have learned about this topic
What you know about mass and heat balance in a multiple effect avaporator
This is the end of the post and I hope you will find it useful.
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- The pros and cons of bimetallic thermometers
- Some examples of temperature instruments
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- What is a process variable?
- What are the most important process variables?
- Time dependence of process variables
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- Evaporation - General comments
- The lines of Dühring
- Mass and heat balance in a single stage evaporator
- Example: Mass and and heat balance in an evaporator
- Multiple effect evaporation - General comments
Ildebrando.
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