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 The situation

1.1 Interesting questions to ask

2 Process of solution

2.1 Available data and comments

2.2 The mass balances

2.3 Heat balances for all effects

2.4 An analytical procedeure to solve the heat balance equations

2.5 Introducing a spreadsheet

2.6 The heating surfaces

2.7 The heat to the condenser

2.8 The condenser water requirement

2.9 The economy of the evaporator

2.10 The steam consumption

1 The situation

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$50,000\,lb_m/hr$ $12\,psig$ $26\, in \, Hg\, man$ $30\, in\, Hg$ .

$85\,F$ as coolant. For this situation consider the following assumptions,

any BPR (boilint point raise) is negligible,
$c_{pF}=1\, Btu/lb_m\cdot F$ ,
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.

$U_1=600\,Btu/hr\cdot ft^2\cdot F$ $U_2=250\,Btu/hr\cdot ft^2\cdot F$ $U_3=125\,Btu/hr\cdot ft^2\cdot F$ are considered as reasonable for the present operation.

1.1 Interesting questions to ask

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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

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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.

$q$ 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

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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.

$L_1$ $L_2$ $S_1$ $S_2$ $S_3$ the same difficulty appears. How do we find the missing data?

$A_1=A_2=A_3$ : 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)

$p_{W1}$ $p_{L3}$ $p_{L1}$ $p_{L2}$ is by splitting the wholes pressure difference as follows,

$\dfrac{p_{W1}-p_{L3}}{3}=\dfrac{26.7-1.96}{3}$ Eq. (02)

$8.25\,psia$ . And from here, it follows that,

$p_{L1}=18.45\,psia$

$p_{L2}=10.2\,psia$

$L1$ $L2$ 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

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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)

$S_3=10,000\,lb_m/hr$ . 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)

$L_1$ $L_2$ $L_3$ can not be determined directly. At least other two equations are needed.

2.3 Heat balances for all effects

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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)

$W_1$ $L_1$ $L_3$ , 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

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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)

$W_1$ $L_1$ $L_2$ $L_3$ can be found from Eqs. (15-17). It is left to the reader the demonstration of Eqs. (18-20).

2.5 Introducing a spreadsheet

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$W_1$ 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

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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)

$A_1$ $A_2$ $A_3$ 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.

$\approx 1,545\,ft^2$ . 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

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$L_3$ . 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

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$W_{cond}$ $L_3$ $85\,F$ . Then,

$\Delta T_{cond}=125.2-85=40.2\,F$

The stream $W_{cond}$ will be then,

$W_{cond}=\dfrac{q_3}{\Delta T_{cond}}=365,532.49\,lb/hr$ Eq. (25)

$W_{cond}$ $85\,F$ to make this conversion.

2.9 The economy of the evaporator

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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,

$Economy=2.1$

which is reasonable since we expected an economy larger than one.

2.10 The steam consumption

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The reader is referred to the post: Economy in an evaporator for more details on this topic.

$W_1$ . Therefore,

$Steam consumption=19,042.25\,lb_m/hr$

This is the end of the post. I hope, you find it very useful.

Any question? Write in the comments and I shall try to help.

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Algebraic solution for a forward feed evaporator