Case of a NaOH solution
Here a the case of concentration of NaOH solution in a single stage evaporator is used to show how to make the balances of mass and heat.
This example was adapted from the book of Principles of Unit Operations by Foust A. S. et al.
CONTENTS
1.1 Interesting questions to ask
2.1 Available data and comments
2.2.1 What is the enthalpy $h_F$ of the fed solution?
2.2.2 What is the mass flow rate $F$ of the fed solution?
2.2.3 What is the temperature $T_S$ of the thick liquor?
2.2.4 What is the enthalpy $h_S$ of the thick liquor?
2.2.5 What is the tempetature $T_L$ of the solvent vapor?
2.2.6 What is the enthalpy $H_L$ of the solvent vapor?
2.2.7 What is the mass flow rate $L$ of the solvent vapor?
2.2.8 What is the enthalpy $H_W$ of the steam?
2.2.9 What is the latent heat $\lambda_W$ of the steam?
2.2.10 What is the temperature $T_W$ of the steam?
2.2.11 What is the mass flow rate $W$ of the steam?
2.2.12 What is the temperature $T_C$ of the condensate?
2.2.13 What is the enthalpy $h_C$ of the condensate?
1 The situation
In this situation a single stage evaporator is used to increase the concentration of a NaOH solution from 9% to 53%. As heating medium the evaporator uses steam coming from a boiler in another part of the process plant facilities. The boiler supplies the steam at $54\, psia$ while the evaporator only requires steam at an average pressure of $4.7\, psia$. This would mean that the manometric pressure in the evaporating solution chamber could be even negative which makes sense since the temperature difference in the head compartiment (space occupied by the solvent vapor) could cause pressure variations too, so that the evaporator should be capable of sustain those mechanical stresses.
The NaOH solution at 9% enters the evaporator with a temperature of $100\,F$. At the thick liquor outlet a stream of $10,000\, lb_m/h$ of concentrated solution, at 53%, leave the evaporator.
 |
| Fig. 01 Evaporator sketch with given data |
1.1 Interesting questions to ask
Several questions arised from the situation presented above. Of course, some question are related to the mass and energy subject but some other are related to features of the equipment. Here is a list of questions we would need to answer:
- what is the enthalpy $h_F$ of the fed solution?
- what is the mass flow rate $F$ of the fed solution?
- what is the temperature $T_S$ of the thick liquor?
- what is the enthalpy $h_S$ of the thick liquor?
- what is the tempetature $T_L$ of the solvent vapor?
- what is the enthalpy $H_L$ of the solvent vapor?
- what is the mass flow rate $L$ of the solvent vapor?
- what is the enthalpy $H_W$ of the steam?
- what is the latent heat $\lambda_W$ of the steam?
- what is the temperature $T_W$ of the steam?
- what is the mass flow rate $W$ of the steam?
- what is the temperature $T_C$ of the condensate?
- what is the enthalpy $h_C$ of the condensate?
- what is the mass flow rate $C$ of the condensate?
- how much surface would be needed in order to operate the evaporator under the above conditions?
Further discussion can be made around the estimated parameters.
2 Process of solution
The solution to situations like this one may become tricky because: assumptions need to be made, data has to be found (somewhere) and math skills are needed to deal with unknowns.
Based on the list of questions to be asked around this situation is helpful for our frist steps (if you do not know where to start). Therefore, some questions are just reduced to search for data in reliable sources. For example, information about steam properties can be found over the internet or some technical book (see the ASME. Steam tables. Compact edition as a good source).
Please, take a look to already available steam tables in the following Excel file or Google Sheets file, prepared for this purpose.
Parts of the process of solution shall be conducted by using SI units but in general British units shall be used.
Once the easy to find data has been found (in the sources of your preference) mass and energy balances may be performed to find the remaining missing data.
2.1 Available data and comments
Only a few data is available, and shown in Fig. 01 according to every stream, but the description of the process are good clues to determine the missing information.
As you can see, this is a general situation since nothing is being said about the type of evaporator or about some other physical feature. For example, determination of the required heating area will depend on the configuration of equipment and its heat transfer coefficient.
2.2 Answer to questions
2.2.1 What is the enthalpy $h_F$ of the fed solution?
The enthalpy $h_F$ must be read from a diagram of Enthalpy of the solution vs solution concentration at a given pressure. Therefore, to find this unknown data this diagram has to be found first. Fortunately, experiments on this matter has already been reported by J. W. Bertetti and W. L. McCabe in their paper of 1936 (see https://doi.org/10.1021/ie50315a025) and H. R. Wilson and W. L. McCabe in their paper of 1942 (see https://doi.org/10.1021/ie50389a009).
Just as a cultural note: W. L. McCabe is the same guy of the textbook.
In the work of Bertetti and Mc Cabe report the enthalpy vs NaOH solution concentration data for dilute solutions while Wilson and McCabe the corresponding data for strong solutions.
Fortunately, all this data can be found readily plot gathering all the information so that despites the possible error while reading on the diagram it saves a lot of work (which would be out of the scope of this post).
 |
| Fig. 02 Plot of Enthalpy of solution ($Btu/lb_m$) vs NaOH solultion concentration (%). See red line for text explanation. |
Now, since the solution concentration at the inlet is 10% and at $T_F=100\,F$ all we need to do is to locate the point over curve for $100\,F$ where a vertical line starting at concentration 50% is crossed (following the red line). Next, from this point to the ordinate we read $h_F=60\,Btu/lb_m$.
2.2.2 What is the mass flow rate $F$ of the fed solution?
This unkown seem to be difficult to calculate but you would agree that mass balances usually help in this cases. Now, if you take a look to the solute mass balance equation,
$Fx_F=Lx_L$ Eq. (01)
you will notice that all parameters in Eq. (01) are known but $F$. Thus, since
$x_F=0.1$, $x_L=0.5$ and $L=10,000\,lb_m/h$
a simple isolation of $F$ from Eq. (01) produces,
$F=50,000\,lb_m/h$.
Please, refer to post: Mass and heat balance on a single stage evaporator for more details on this subject.
2.2.3 What is the temperature $T_S$ of the thick liquor?
$T_S$ is, actually, the temperature at which the solution is boiling. All we know about this temperature is that it is larger than that of solvent and that the solution is boiling at its vapor pressure. Then with this in mind, you may use the corresonding plot of Dühring, if you have it, to estimate the temperature $T_S$.
Plesase, refer to the post: The lines of Dühring for more details on this subject.
Let us consider for a moment that no plot of Dühring is available and you need to built your own for this case. This can be easily built from the data for NaOH solutions at 50% weight concentration. This data can be found in the International Critical Tables, from which an extract of this information is presented next
| Concentration [weight %] |
Vapor pressure [mm Hg] | |||||
|---|---|---|---|---|---|---|
| At 20C | At 40C | At 60C | At 80C | At 100C | At 120C | |
| 0 | 17.539 | 55.34 | 149.46 | 355.47 | 760 | 1488.9 |
| 5 | 16.9 | 53.2 | 143.5 | 341.5 | 730 | 1430 |
| 10 | 16 | 50.6 | 137 | 325.5 | 697 | 1365 |
| 20 | 13.9 | 44.2 | 120.5 | 288.5 | 621 | 1225 |
| 30 | 11.3 | 36.6 | 101 | 246 | 537 | 1070 |
| 40 | 8.7 | 28.7 | 81 | 202 | 450 | 920 |
| 50 | 6.3 | 20.7 | 62.5 | 160.5 | 368 | 770 |
From the table above we are only interested in the row for concentration 50%. Next, if the corresponding steam data is searched then, the following table comes up,
| Solution vapor pressure [mm Hg] |
Solution vapor pressure [MPa] |
Water boiling point [C] |
Solution boiling point [C] 50% |
|---|---|---|---|
| 6.3 | 0.0008399286 | 4.432998338 | 20 |
| 20.7 | 0.0027597654 | 22.68551616 | 40 |
| 62.5 | 0.008332625 | 42.2750375 | 60 |
| 160.5 | 0.021398181 | 61.45468555 | 80 |
| 368 | 0.049062496 | 80.83249792 | 100 |
| 770 | 0.10265794 | 100.3249859 | 120 |
(The pressures presented in the table above are absolute)
And since we are only interested in the water boiling temperatures at $4.7\,psia=0.0324054\,MPa\,(abs)$ it is not difficult to find that the corresponding boiling temperature for water and the solution are $69.16\,C=156.49\,F$ and $87.96\,C=190.33\,F$, respectively. This confirms that the boiling temperature of the solution is larger than that of the solvent, by $33.84\,F$ in this case.
Then, the temperature we are looking for is: $T_S=190.33\,F$. Notice that this result is different, along with the boiling temperature rise, to that presented in the book if Foust et al and in my defense all I can say is that they use a graphical method rather than an analytical one (as in here). This suspicion is quite true since the water boiling temperature is very close to that presented in the textbook.
2.2.4 What is the enthalpy $h_S$ of the thick liquor?
 |
| Fig. 03 Plot of Enthalpy of solution ($Btu/lb_m$) vs NaOH solultion concentration (%). See green line for text explanation. |
2.2.5 What is the temperature $T_L$ of the solvent vapor?
In this case we should recall an assumption, based on the physics of what is actually happening in the evaporator, on the solvent vapor: this will have the same temperature of the boiling solution.
In other words, the only way the solvent in the solution splits from it is by boiling it so that the vapor, even if it is water, will have the temperature of the boiling solution. This is: $T_L=87.96\,C=190.33\,F$.
You may argue that water does not form vapor at $81.96\,C$ but remember that this is water at vapor pressure (not as in your home) with certain amount of energy (called enthalpy) allowing it this behavior.
2.2.6 What is the enthalpy $H_L$ of the solvent vapor?
In this case the solvent vapor in stream $L$ is water in vapor phase. The fluid in $L$, as mentioned before at $T_L=87.96\,C=190.33\,F$, should have an enthalpy corresponding to the steam at that temperature. Therefore, if we search for this temperature in a steam table the following entries appear,
 |
| Fig. 04 Fragment of data from a steam table. This data corresponds to $T_L$. |
In Fig. 04 it can be seen that the enthalpy is barely $H_L=2,656\,kJ/kg=1,141.87, Btu/lb_m$. You will notice that this result is very close to that presented in the textbook of Foust.
Please, take a look to already available steam tables in the following Excel file or Google Sheets file, prepared for this purpose.
In Fig. 04, it is obvious that $0.06502\,MPa\,(abs)\neq 4.7\,psia$. Well, the solvent vapor is at $4.7\,psia$ but its temperature is greater than $69.16\,C$. The only way for this occurs is by keeping pressure constant at $4.7\,psia$ but increasing its energy (or enthalpy).
2.2.7 What is the mass flow rate $L$ of solvent vapor?
The best tool at hand for this unknown is a mass balance on the solution. This can be expressed as,
$F=S+L$ Eq. (02)
Since we already know that $F=50,000\,lb_m/h$ and $S=10,000\,lb_m/h$, the stream $L$ is easily isolated from Eq. (02), it follows that,
2.2.8 What is the enthalpy $H_W$ of the steam?
The steam entering the evaporator is at $54\,psia$ as stated at the beginning. Therefore, all we need to do is to search in the steam table for the enthalpy of the vapor phase. Here is an extract of this table,
 |
| Fig. 05 Fragment of data from a steam table. Here, data close to $54\,psia=0.372317\,MPa(abs)$ is shown for linear interpolation purposes. |
Next, if we take the columns for pressure and enthalpy for the vapor, of table in Fig. 05, it is found that $H_W=2,735.23\,kJ/kg=1,175.94\,Btu/lb_m$, approximately.
2.2.9 What is the latent heat $\lambda_W$ of the steam?
The latent heat is exactly the same as the enthalpy of evaporation which is the energy to required to transform liquid into vapor at the same temperature and vapor pressure.
Now, if the columns of pressure and enthalpy of evaporation, indicated as HV in the last column, in the table shown in Fig. 05 and a linear interpolation to for $54\,psia=0.372317\,MPa(abs)$ is performed, $\lambda_W$ is found. Therefore, $\lambda_W=2,141.30\,kJ/kg=920.59\,Btu/lb_m$, approximately.
2.2.10 What is the temperature $T_W$ of the steam?
This unknown is easily found too. For this the same steam data presented in Fig. 05 can be used. In this case the columns for pressure and temperature are used.
A simple linear interpolation will show that $T_W=141.03\,C=285.85\,F$, approximately.
2.2.11 What is the mass flow rate $W$ of the steam?
It would seem that for $W$ the same strategy of a mass balance, for streams $W$ and $C$, can be implemented. However, $C$ is unknown too!. Therefore, a different approach must be used.
Please, read the post: Mass and heat balance in a single stage evaporator for more details on the energy balance for evaporators calculations.
One tool we have not used is the energy balance. In fact, for this situation, we have never spoken about this equation. What does it looks like?
$Fh_F + W\lambda_W = Sh_S + LH_L$ Eq. (03)
A simple inspection of Eq. (03) tells us that all parameters are known but $W$, which is what we are looking for. Therefore, isolation of $W$ from Eq. (03)
and substitution of known data gives
2.2.12 What is the temperature $T_C$ of the condensate?
The condensate is the liquid produced from the steam that entered the evaporator and which has yielded its evaporation enthalpy $H_{Evap}$ to indirectly heat the solution. Therefore, the liquid is at the same pressure that the steam: $54\,psia$ in this case. You may argue that the liquid water temperature may be lower but these losses are negligible during the steady operating conditions of the equipment.
Also, the temperature $T_C$ of the condensate should be the same than that for the steam: $T_W$. Therefore, $T_C=T_W=285.85\,F$.
2.2.13 What is the enthalpy $h_C$ of the condensate?
Based on the arguements already presented for $T_C$, the enthalpy $h_C$ is easily estimated from the steam data of table in Fig. 05. Then, by a linear interpolation between the columns for pressure and liquid enthalpy (5th column) it comes out that: $h_C=593.65\,kJ/kg=255.22\,Btu/lb_m$.
2.2.14 What is the mass flow rate $C$ of the condensate?
This a question of quick answer. If the mass flow rate of steam entering the evaporator is $W$ therefore the same mass flow rate must be expected in $C$. This means that: $C=48,691.38\,lb_m/h$. It does not matter if $F$ is steam and $C$ liquid since the balance is made with respect to the mass (no density is taken ito account).
Another way of estimating $C$ can be through an energy balance as was done for the estimation of $W$. From the fundamentals of the energy balance for an evaporator, another way of writting Eq. (03) is,
$Fh_F + WH_W = Sh_S + LH_L+Ch_C$ Eq. (04)
where as you can see all data is known but $C$. Therefore, a simple isolation of $C$ and further substitution of available data gives the result.
Please, read the post: Mass and heat balance in a single stage evaporator for more details on the energy balance for evaporators calculations.
$C=\dfrac{Fh_F + WH_W-Sh_S - LH_L}{h_C}$
$C=\dfrac{(50,000)(60) + (48,691.38)(1,175.94) - (10,000)(215) - (40,000)(1,141.87)}{255.22}$
$C=48,716.17\,lb_m/h$
As you can see, in this new calculation $F\neq C$!. The explanation of this discrepancy is that the energy balance is not complete. There are energy losses not considered in Eq. (04), as mentioned before, so that it produces a maximum error of 0.05%. This error is small and negligible but must important: I can live with that.
2.2.15 How much surface would be needed in order to operate the evaporator under the above conditions?
Since the evaporator works on pool boiling, an engineer would question: how big should we built the evaporator? (so that it can work as previously calculated).
This is a question of multiple answers since in practice exist different evaporator designs with different efficiencies which were created in response to specific applications. Therefore, in order to give the reader an idea of the answer two different escenarios shall be considered.
The over-all heat transfer in an evaporator is studied from the familiar heat equation,
$q=U_0A_0(\Delta T)$ Eq. (05)
where $q$ is the heat or thermal energy, $U_0$ is the over-all heat transfer coefficient, $A_0$ is the surface on which the heat transferred and $\Delta T$ the temperature gradient driving the heat transfer. The parameter $q$ can also be understood as the energy been used in the process, which for our case van be expressed as,
$q=W\lambda_W$ Eq. (06)
The temperature difference $\Delta T$ comes from the gap between the steam entering the evaporator and the boiling solution.
The over-all heat transfer coefficient $U_0$, given in $[Btu/h\,ft^2F]$ or $[W/m^2K]$, depends on the configuration of the evaporator and some physical properties of the wotking fluid. For this situation an evaporator of low efficiency $U_0=200\,Btu/h\,ft^2F$ and another one of higher efficiency with $U_0=1,500\,Btu/h\,ft^2F$, shall be considered.
After combination of Eqs. (05-06), $A_0$ is isolated,
$A_0=\dfrac{W\lambda_W}{U_0(\Delta T)}$ Eq. (07)
Using the known data it produces for low $U_0$,
$A_0=\dfrac{(48,691.38)(920.59)}{(200)(285.85-190.33)}$
and for highger $U_0$,
$A_0=\dfrac{(48,691.38)(920.59)}{(1,500)(285.85-190.33)}$
Notice the difference in required heating surface for each $U_0$. Ultimately, the calculated surface should be compared with the available models of evaporators or used to estimate the number of tubes for the passage of the steam. The pipe size and their schedule are interesting subjects out of the scope of this post.
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| Fig. 06 Table of results for the concentration of NaOH in a single stage evaporator |
This is the end of the post. I hope you will find it interesting and useful.
Other stuff of interest
- LE01 - AC and DC voltage measurement and continuity test
- The pros and cons of bulb thermometers
- The pros and cons of bimetallic thermometers
- Some examples of temperature instruments
- Minor losses - Formulas
- What is a process variable?
- What are the most important process variables?
- Time dependence of process variables
- A list of process variables
- PID controllers for temperature
- Valves - Gate design
- Evaporation - General comments
- The lines of Dühring
- Mass and heat balance in a single stage evaporator
Ildebrando.
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