Algebraic solution for a forward feed evaporator
- Approximation based on constant $c_{pF}$ -
Physics and solution procedure of multiple effect evaporator can be more challenging single stage ones. In this post, a situation with a triple effect evaporators is used to reworked to introduce spreadsheets as a time saving tool. The price to be paid is that some algebraic work has to be done.
This example was adapted from the old companion Process Heat Transfer by Donald Q. Kern. The idea of an analytical approach and the introduction of spreadsheets would not have come up without Kern presenting his process of solution.
The situation, presented originally by Kern, is ideal since from the data provided the solution process leading to nearly $A_1=A_2=A_3$ is straight so that no iterative procedure is required. The reader should be aware of this.
CONTENTS
1.1 Interesting questions to ask
2.1 Available data and comments
2.3 Heat balances for all effects
2.4 An analytical procedeure to solve the heat balance equations
2.8 The condenser water requirement
1 The situation
Consider the selection of a triple effect evaporator with feedforward mode operation (comparison with backward feed and mixed feed is omitted). The evaporator is to concentrate a dilute chemical solution from 10% to 50% by weight. The feed to the equipment shall be $50,000\,lb_m/hr$. Also, the heating medium shall be steam at $12\,psig$ with the third effect operating in vaccum at $26\, in \, Hg\, man$ with reference to average barometric pressure of $30\, in\, Hg$.
After the evaporator a barometric condenser (open to the atmosphere) uses water at $85\,F$ as coolant. For this situation consider the following assumptions,
- any BPR (boilint point raise) is negligible,
- the solution across all effects has an average heat capacity $c_{pF}=1\, Btu/lb_m\cdot F$,
- the condensate from each effect drains at saturation conditions,
- heat losses across all effects are very small.
As part of the selection process the following questions need to be answered,
- the steam consumption,
- the heating surface required for each effect,
- the heat going into the condenser and
- the mass flow rate of cooling water required in the barometric condenser after the evaporator.
 |
| Fig. 01 Triple effect evaporator sketch. |
Because the rate of evaporation at each effect is different so will be the over-all heat transfer coefficients since conditions are different from one unit to the other. Therefore, $U_1=600\,Btu/hr\cdot ft^2\cdot F$, $U_2=250\,Btu/hr\cdot ft^2\cdot F$ and $U_3=125\,Btu/hr\cdot ft^2\cdot F$ are considered as reasonable for the present operation.
1.1 Interesting questions to ask
One theoretical assertion on multiple effect evaporators is that the heating surfaces should be the same, because of technical reasons, while the overall heat transfer coefficient changes across the effects. But, will the heating surfaces really be the same?
Let us consider the case in which the heating surfaces are not equal, how do you force the other variables to satisfy this condition? Is there a method to ensure that $A_1=A_2=A_3$?
For short, the heating surfaces are not the same but nearly and that is enough for most of the situations. In othe words, the real physical evaporator has equal heating surfaces but not in the paper, in the theoretical estimations.
2 Process of solution
The process of solution is not too difficult but some math is needed. With this at hand the solution is easily automated in a spreadsheet. One way of having a better perspective of the process of solution is,
- by looking at both the mass and heat balances across all effects,
- finding the most data available from tables or other sources,
- make a comparison of available data and equations and
- look for methods of solution of the system of equations.
The last item is the one at which math skills are needed along with spreadsheet logical.
The reader is referred to the post: Mass and heat balance in a multiple effect evaporator - forward feed, for more details on the equations to be used in this siituation.
In order to estimate $q$ for the whole evaporator mass flow rates and steam data are needed mass and heat balances are part of the solution. As you may have imagined already the mass balance outputs mass flow rates to be used in the heat balance so that there is no scope of the tedious calculations.
2.1 Available data and comments
From the unknown data, the most easy to find is that related to the steam, say: temperature, pressure, enthalpy of vapor or liquid or evaporation. In order to look at these variables, please see carefully the sketch in Fig. 01.
For this purpose, a table of found data is presented below.
| Stream $W_1$ | Stream $L_3$ |
|---|---|
| $p_{W1}=12\, psig$ $p_{W1}=26.7\,psia$ |
$p_{L3}=-26\,in\,Hg\,man$ $p_{L3}=1.96\,psia$ |
| $T_{W1}=243.52\,F$ | $T_{L3}=125.20\,F$ |
| $H_{W1}=1,167.72\,Btu/lb_m$ | $H_{L3}=1,115.45\,Btu/lb_m$ |
| $\lambda_{W1}=949.67\,Btu/lb_m$ | $\lambda_{L3}=1,022.26\,Btu/lb_m$ |
The reader is referred to posts: Example #1, Example #2 and to Enthalpies in the steam tables, to see how steam tables are used to find data as those presented in the table above.
As you can see, for the present situation there is lack of information for steam streams $L_1$ and $L_2$. For liquor streams $S_1$, $S_2$ and $S_3$ the same difficulty appears. How do we find the missing data?
Kern comes up with an experienced based (or rule of thumb) helpful insight about multiple effect evaporators working in forward feed mode and $A_1=A_2=A_3$: the difference in the pressures between effects will be nearly equal. Since there is no better idea at hand we are constrained to take:
$p_{W1}-p_{L1}=p_{L1}-p_{L2}=p_{L2}-p_{L3}$ Eq. (01)
Since all pressure differences are equal so that starting with $p_{W1}$ we end with $p_{L3}$ the way of finding $p_{L1}$ and $p_{L2}$ is by splitting the wholes pressure difference as follows,
$\dfrac{p_{W1}-p_{L3}}{3}=\dfrac{26.7-1.96}{3}$ Eq. (02)
so that every pressure difference in Eq. (02) should be $8.25\,psia$. And from here, it follows that,
Based on the approximated values of pressure in streams $L1$ and $L2$ steam data can now be easily find. This data is presented in the table below,
| Stream $L1$ | Stream $L_2$ |
|---|---|
| $p_{L1}=18.45\,psia$ | $p_{L2}=10.2\,psia$ |
| $T_{L1}=223.61\,F$ | $T_{L2}=194.03\,F$ |
| $H_{L1}=1,154.65\,Btu/lb_m$ | $H_{L2}=1,143.43\,Btu/lb_m$ |
| $\lambda_{W1}=962.96\,Btu/lb_m$ | $\lambda_{L2}=981.33\,Btu/lb_m$ |
Notice that this approach can not be entirely true but for real life applications this can work as an approximation to be corrected later from operation data.
Also, due to the lack of information, the process of solution lies on several assumptions with reasonable strong physics. However, for multiple effect evaporators more data is needed and so deeper thoughts on how such an equipment works become present.
As the number of effects increases, the calculations become messy and numerical methods are required. Here, we will take a dive into an approach of such cases.
2.2 Mass balances
The only stream available is that for $F$. However, $S_3$ is required for the heat balance so that a mass balance for the solute needs to be done,
$F_1\,x_{F1}=S_\,x_{S3}$ Eq. (03)
and after substitution of available data the following is obtained: $S_3=10,000\,lb_m/hr$. Once $S_3$ is known the whole water evaporated from the solution can be estimated from a global mass balance,
$F_1=S_3+L_1+L_2+L_3$ Eq. (04)
which gives,
$L_1+L_2+L_3=40,000\,lb_m/hr$ Eq. (05)
You must agree that the individal streams leaving each effect $L_1$, $L_2$ and $L_3$ can not be determined directly. At least other two equations are needed.
2.3 Heat balances for all effects
For this task 3 equations arise: one for each effect. These can be expressed as follows,
$W_1\, \lambda_{W1}=L_1\, \lambda_{L1} + F_1\,c_{pF1}\left( T_{S1}-T_{F1} \right)$ Eq. (06)
$L_1\, \lambda_{L1}=L_2\, \lambda_{L2} + S_1\,c_{pS1}\left( T_{S2}-T_{S1} \right)$ Eq. (07)
$L_2\, \lambda_{L2}=L_3\, \lambda_{L3} + S_2\,c_{pS2}\left( T_{S3}-T_{S2} \right)$ Eq. (08)
Looking at Eqs. (06-08) there are some data missing: all $c_p$'s and all $S$'s. So that there is an important number of unknowns. This is partially true since,
$c_{pF1}=c_{pS1}=c_{pS2}=1\,Btu/lb_m\cdot F$ Eq. (09)
and mass flow rates $S_1$ trhough $S_3$ in terms of the evaporated water at each effect as,
$S_1=F_1-L_1$ Eq. (10)
$S_2=S_1-L_2=F_1-L_1-L_2$ Eq. (11)
In this way, Eqs. (06-08) reduce to,
$W_1\, \lambda_{W1}=L_1\, \lambda_{L1} + F_1\,\left( T_{S1}-T_{F1} \right)$ Eq. (12)
$L_1\, \lambda_{L1}=L_2\, \lambda_{L2} + \left(F_1-L_1\right)\,\left( T_{S2}-T_{S1} \right)$ Eq. (13)
$L_2\, \lambda_{L2}=L_3\, \lambda_{L3} + \left(F_1-L_1-L_2\right)\,\left( T_{S3}-T_{S2} \right)$ Eq. (14)
In Eqs. (12-13) appear a total of 4 unknowns: $W_1$, $L_1$ to $L_3$, so that one equation is missing. In fact, this equation is already known: Eq. (05). In this way, the heat balance reduces to solve simultaneously Eqs. (05,12-14).
2.4 An analytical procedeure to solve the heat balance equations
If you look carefully to Eqs. (05,12-14) there is an analytical solution so that no numerical method is required. For this situation, such solution is in an recursive way,
$L_1=\dfrac{W_1\,\lambda_{W1}-F_1\,\left( T_{S1}-T_{F1} \right)}{\lambda_{L1}}$ Eq. (15)
$L_2=\dfrac{L_1\,\lambda_{L1}-\left(F_1-L_1\right)\,\left( T_{S2}-T_{S1} \right)}{\lambda_{L2}}$ Eq. (16)
$L_3=\dfrac{L_2\,\lambda_{L2}-\left(F_1-L_1-L_2\right)\,\left( T_{S3}-T_{S2} \right)}{\lambda_{L3}}$ Eq. (17)
and finally,
$W_1=\dfrac{40,000-A\,F_1}{\lambda_{B\, W1}}$ Eq. (18)
where,
$A=-\dfrac{\left( T_{S1}-T_{F1} \right)}{\lambda_{L1}} - \dfrac{\left( T_{S1}-T_{F1} \right)}{\lambda_{L2}} - \dfrac{\left( T_{S2}-T_{S1} \right)}{\lambda_{L2}} - \dfrac{\left( T_{S1}-T_{F1} \right)\left( T_{S2}-T_{S1} \right)}{\lambda_{L1}\,\lambda_{L2}}$
$-\dfrac{T_{S1}-T_{F1}}{\lambda_{L3}} - \dfrac{T_{S2}-T_{S1}}{\lambda_{L3}} - \dfrac{\left( T_{S1}-T_{F1} \right)\left(T_{S2}-T_{S1}\right)}{\lambda_{L1}\,\lambda_{L3}} - \dfrac{T_{S3}-T_{S2}}{\lambda_{L3}}$
$-\dfrac{\left( T_{S1}-T_{F1} \right)\left(T_{S3}-T_{S2}\right)}{\lambda_{L1}\,\lambda_{L3}} - \dfrac{\left( T_{S1}-T_{F1} \right)\left(T_{S3}-T_{S2}\right)}{\lambda_{L2}\,\lambda_{L3}} - \dfrac{\left( T_{S2}-T_{S1} \right)\left(T_{S3}-T_{S2}\right)}{\lambda_{L2}\,\lambda_{L3}}$
$-\dfrac{\left( T_{S1}-T_{F1} \right)\left( T_{S2}-T_{S1} \right)\left(T_{S3}-T_{S2}\right)}{\lambda_{L1}\,\lambda_{L2}\,\lambda_{L3}}$ Eq. (19)
$B=\dfrac{1}{\lambda_{L1}} + \dfrac{1}{\lambda_{L2}} + \dfrac{ T_{S2}-T_{S1}}{\lambda_{L1}\, \lambda_{L2}} + \dfrac{1}{\lambda_{L3}} + \dfrac{ T_{S2}-T_{S1}}{\lambda_{L1}\, \lambda_{L3}} + \dfrac{ T_{S3}-T_{S2}}{\lambda_{L1}\, \lambda_{L3}}$
$ + \dfrac{ T_{S3}-T_{S2}}{\lambda_{L2}\, \lambda_{L3}} + \dfrac{ \left(T_{S2}-T_{S1}\right)\left(T_{S3}-T_{S2}\right)}{\lambda_{L1}\,\lambda_{L2}\, \lambda_{L3}}$ Eq. (20)
where as you can see, once estimated $W_1$ from Eq. (18) streams $L_1$, $L_2$ and $L_3$ can be found from Eqs. (15-17). It is left to the reader the demonstration of Eqs. (18-20).
2.5 Introducing a spreadsheet
It is clear, by simple inspection, that evaluation of Eqs. (19-20) to obtain $W_1$ from Eq. (18) and from here to evaluate Eqs. (15-17), to find streams $L_1$, $L_2$ and $L_3$, with a handheld calculator could be not only tedious but time consumming and subject to possible errors anywhere in the process. Thus, the best tool at this time is a spreadsheet.
Following this strategy, a spreadsheet was build in the same logic as mentioned before. Next, this spreadsheet is shown.
 |
| Fig. 02 A screenshot of the spreadsheet built to evaluate Eqs. (15-20). See results in cells in green. |
The greatest advantage of this spreadsheet is that by changing the the input data, all vapor streams are calculated readily. Also, you can use this very same spreadsheet to include further calculations for heating surfaces.
2.6 The heating surfaces
Heating surfaces can be easily calculated from the formulas,
$A_1=\dfrac{q_1}{U_1\left( T_{W1}-T_{S1} \right)}$ Eq. (21)
$A_2=\dfrac{q_2}{U_1\left( T_{L1}-T_{S2} \right)}$ Eq. (22)
$A_3=\dfrac{q_3}{U_1\left( T_{L2}-T_{S3} \right)}$ Eq. (23)
from which it can be easily seen that all data are known. Using again the already built spreadsheet the temperature differences and heat coming into each effect can be calculated first so that $A_1$, $A_2$ and $A_3$ can be quickly estimated. This is shown next.
 |
| Fig. 03 Calculation of heating surfaces (see cells in green). |
Notice, that these results are in agreement with those presented by Kern in his book.
As side comment, the evaporator has an average heating area of $\approx 1,545\,ft^2$. This is good result since the second effect, having the largest separation of the average, has a reasonable maximum 4.15% of difference.
2.7 The heat to the condenser
The heat sent from the evaporator to the condenser is just that carried by stream $L_3$. Thus, this is estimated as,
$q_3=L_3\,\lambda_{L3}=14,694,405.95\,Btu/hr$ Eq. (24)
You should accept that the idea of the heat going into the condenser is the same used for the evaporator.
2.8 The condenser water requirement
In fact this parameter refers to the mass flow rate of water to be used as cooling media, say $W_{cond}$. For this calculation a temperature gradient is needed: that driving the heat transfer inside the condenser. This is given between the vapor stream $L_3$ and the cooling water at $85\,F$. Then,
The stream $W_{cond}$ will be then,
Notice, that Eq. (25) comes from the idea of the familiar equation of heat. Also, pay attention to the units since confusion on this matter may occur. Because $W_{cond}$ is liquid water it could be better expressed as volumetric flow rate. You can next use the density of the cooling water at $85\,F$ to make this conversion.
2.9 The economy of the evaporator
The reader is referred to the post: Economy in an evaporator for more details on this topic.
In this case, the economy of the evaporator is just,
$Economy=\dfrac{L_1+L_2+L_3}{W_1}$ Eq. (26)
Substitution of known data produces,
which is reasonable since we expected an economy larger than one.
2.10 The steam consumption
The reader is referred to the post: Economy in an evaporator for more details on this topic.
This calculation is omitted since the steam consumption is directly $W_1$. Therefore,
This is the end of the post. I hope, you find it very useful.
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Ildebrando.
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