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Example #03: single stage chemical evaporator

Heating surface of a chemical evaporator

- Case of NaOH solution -


Here, the situation presented in Example 01 is reworked for very little concentration by evaporation. The aim of this new example is to use the rough approximation for heat balance based on constant heat capacity.

This example is adapted from the familiar text Transport processes and unit operations by Christie J. Geankoplis. Notice, that some interpretations of the physics of the heat balance are little different.

 CONTENTS

 1 The situation

 1.1 Interesting questions to ask

 2 Process of solution

 2.1 Available data and comments

 2.2 The mass flow rate of the concentrated solution $S$

 2.3 The mass flow rate of water evaporated from the dilute solution $L$

 2.4 A heat balance for $W$

 2.5 The heat $q$ transferred to the process fluid

 2.6 The temperature gradient $\Delta T$

 2.7 The heating area $A$

 2.8 The capacity of the evaporator

 2.9 The economy of the evaporator

 2.10 The steam consumption

1 The situation

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The selection of a single stage chemical evaporator is being work out and some thermal and mechanical features are needed for comparison purposes so that the following is needed,

  • the heating surface $A$ of the evaporator,
  • the capacity of the evaporator,
  • the consumption of steam,
  • the economy of the evaporator,
  • the mass flow rate of concentrated solution and
  • the mass flow rate of water evaporated from the solution.

The evaporator is to concentrate a NaOH solution from 1.0% to 1.5% by weight. For short, the available data is presented in the sketch below.

Fig. 01 A sketch of the single stage evaporator with known data and unknown variables.


Hint:

For the present case you are to consider that the solution is so dilute that it has a boiling temperature close to that of water (on the same conditions) and that since there is no enthalpy data available for such concentrations a good approximation based on the heat capacity of the solution as $c_p=4.14\,kJ/kg\cdot K$ can be used.

1.1 Interesting questions

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Perhaps, the quickest question  could be if the unknowns can really be estimated with little data. Looking at the range of concentrations from 1% to 1.5% you may  wonder if an evaporator would certainly be needed or if there is an application for such an scenario.

The present situation relies on the fact that since the concentration range is little the heat capacity should be almost constant. Then, the reader is challenge to rework this problem with data for enthalpies and check how good the approximation is.

2 Process of solution

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If you pay a little of attention to the list of unknowns some disorder is noticed. Perhaps one may proceed with the calculation in the following order,

  • the mass flow rate of concentrated solution - estimated from a solute mass balance,
  • the mass flow rate of water evaporated from the solution - estimated from a global mass balance,
  • the heating surface $A$ of the evaporator - estimated from the energy carried by $W$,
and using the formulas in the post of Economy of an evaporator, the following can be estimated as well,
  • the capacity of the evaporator,
  • the economy of the evaporator and
  • the consumption of steam.

Hopefully, this plan will work.

2.1 Available data and comments

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Again, if you are working with steam, a good set of steam tables are key. Starting with the pressure of the steam other physical features such as enthalpies and temperatures can be estimated very easily. The same holds for the solution at hand.

First, knowing that the steam $W$ is at $p_W=143.3\,kPa\,abs$, from steam tables the following data are obtained,


$H_W=1,156.87\,Btu/lb_m=748.26\,W/kg$

$H_{W-vap}=958.71\,Btu/lb_m=620.09\,W/kg$

$T_W=229.81\,F=109.9\,C$

Fig. 02 Sample data from steam table for $p_W$.
.

Also, the generated steam in the evaporation chamber is at $p_L=1\,atm\,abs$ so that the physical properties of $L$ can be extracted from a steam table. These are,


$H_L=\,Btu/lb_m=743.99\,W/kg$

$H_{L-vap}=970.23\,Btu/lb_m=627.54\,W/kg$

$T_L=211.85\,F=99.92\,C$

Fig. 03 Sample data from steam table for $p_L$.


Please, read the following post on Enthalpies in the steam tables for more physical insight on this subject.

From the situation statement other data for the solution can be estimated but these are left for furhter sections of this post.

2.2 The mass flow rate of the concentrated solution $S$

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Since $F$ is given, $S$ can be easily calculated from a solute mass balance. This is as follows,

$F\,x_F=S\,x_S$        Eq. (01)

$S=\dfrac{\left(9,072\,kg/hr\right)\left( 0.01\right)}{0.015}$

$S=6,048\,kg/hr$


where $x$ is used to indicate the solute fraction solely.

2.3 The mass flow rate of water evaporated from the dilute solution $L$

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Now that $S$ is known $L$ can be estimated from a global mass balance. This is expressed as,

$F=L+S$        Eq. (02)

and after isolation of $L$ and further substitution ok known data,

$L=3,024\,kg/hr$

2.4 A heat balance for $W$

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In the near future, $W$ shall be needed to estimate $q$ so that because we alredy used all tricks available a heat balance must be attempted. Since the approximation based on the heat capacity $c_{pf}$ is suggested, the following formula applies,

$W\,\lambda_W=L\,\lambda_L+F\,c_{pf}\left( T_S-T_F \right)$        Eq. (03)

direct substitution of known data and further isolation of $W$ produces,

$W=\dfrac{\left( 3,024\,kg/hr \right)\left( 627.54\,W/kg \right) + \left(9,072\,kg/hr\right)\left( 1.15\,W/kg\cdot K \right)\left( 99.92 - 37.8 \right)}{620.09\,W/kg}$

$W=4,105.48\,kg/hr$

In the preious evaluation the temperature difference does not need to be converted to $K$ since what matters is the temperature difference as is.

2.5 The heat $q$ transferred to the process fluid

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This is a component of the formula to be used in the estimation of the heating surface,

$A=\dfrac{q}{U\cdot \Delta T}$        Eq. (04)

Since $q$ is transferred from the steam and this corresponds to the energy available in it, then,

$q=W\,H_{W-vap}$

$q=\left( 4,105.48\,kg/hr \right)\left( 620.09\,W/kg \right)$

$q=2,545,767.09\,W/hr$

2.6 The temperature gradient $\Delta T$

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Again, because the solution is dilute it would boil as if it were water. In this way, the thicken solution would be at the same pressure of the vapor in $L$. Therefore: $T_S=T_L$. In this way,

$\Delta T=T_W-T_S$        Eq. (05)

$\Delta T=109.9\,C-99.92\,C$

$\Delta T=9.98\,C$ or $9.98\, K$

Notice that if the solution were not dilute, the calculation would have taken another direction.

2.7 The heating area $A$

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Since, all variables in Eq. (04) are now known, a simple substitution leads to the answer. This is,

$A=\dfrac{2,545,767.09\,W/hr}{\left( 1,704\,W\cdot m^2 \cdot K \right)\left( 9.98\,K \right)}$

$A=149.70\,m^2$

2.8 The capacity of the evaporator

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The capacity of the evaporator is defined as the amount of solvent it can evaporate from the solution. Therefore, it is just $L$.

$Evaporator\,\,capacity = L=3,024\,kg/hr$

2.9 The economy of the evaporator

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This can be calculated from $L$ and $W$ as follows,

$Economy=\dfrac{3,024}{4,105.48}$

$Economy=0.74$

which seems to be acceptable for a single stage evaporator.

For more details on this parameter read the post: Economy in an evaporator.

2.10 The steam consumption

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As you may have imagine this is just $W$. Therefore,

$Steam \,\,consumption=W=4,105.48\,kg/hr$

This is the end of the solution based on an approximation for the enthalpy for the solution. Remember that the fact that the solution is dilute is taken a little far. Perhaps, the error, if it is noticeable, could be reflected as $S$ not matching the prescribed $x_S$.

I hope, you find this post very useful.

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

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