Mass and heat balance in a single stage evaporator

Process mass and energy balance in an evaporator are key for the estimation of:

• operation conditions in the equipment,
• efficiency of the evaporation process,
• design of equipment,
• prediction of production scenarios,

among other reasons. However, this is task involving a mix of physics and math  along with a good dose of common sense and engineering pragmatism.

 CONTENTS

 1 How does an evaporator work?

 1.1 Some futher considerations on evaporator calculations

 2 The mass balance in a single stage evaporator

 3 The heat balance in a single stage evaporator

 3.1 What if you were interested in the condensate flow rate

 4 What you know about mass and heat balance in an evaporator

1 How does an evaporator work?

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This may sound redundant but it is better (just in case) to present a brief explanation of how a regular evaporator works and about the evaporation mechanism.

Please, read the post Evaporation - General comments for more details on the evaporation mechanism.

An evaporator will usually have two inlets:

• one for the process fluid feed,
• one for steam (if water vapor is the heating medium);

and two outlets:

• one for the vapor of solvent (separated from the original solution),
• one for, water, condensate,
• and one for the concentrated process fluid (which is also the valuable product).

Fig. 01 Mass and enthalpy variables for the different streams in an evaporator

The sketch above presents all relevant variables for evaporator streams. Notice that $H$ was used for enthalpy of vapor streams and $h$ for the enthalpy of liquid streams.

Now, making use of the variables in Fig. 01 the evaporator working principle is depicted. First, a stream $F$ consisting of the concentrated solution is fed into the evaporator. The concentrated solution is formed of two components: a solute and a solvent. For simplicity, only the solute is concentration is made referenced to as the fraction $x$ since that for the solvent would be $1-x$. Therefore, in stream $F$ there is a fraction $x_F$ while at stream of the product, or thick liquor, the corresponding fraction would be $x_S$. As the stream $F$ enters the tubular space design for its indirect heating in the equipment a part of it evaporates and leaves through an exit at the top which is represented as the stream $E$.

Also, at the same time $F$ enters water vapor (steam) also enters, continuously, in the form of stream $W$, the evaporator through an circuit external to the compartiment flooded by the liquid fluid in $F$. For short, as you may have already imagined, an evaporator is just a heat exchanger designed to evaporate something. As the heat is indirectly transferred into the process fluid in $F$ the steam goes into a change of phase so that it becomes liquid. The action of gravity helps to drain this liquid water continuously from the equipment in the form of stream $C$. The thermal energy, $H_W$, available in $W$ is used to evaporate the solvent so that part of this energy is in the leaving streams $C$, $S$ and $E$.

More complex configurations than that presented in Fig. 01 are found in practice. For example:

• the stream $F$ can be pre-heated by the condensate in a form of avoiding waste of energy and sudden heating of the process fluid,
• the vapor in stream $E$ may be valuable so that further calculations for a condenser would be required,
• for chemical process applications multiple effects (single stage evaporators in series) may be needed.

1.1 Some futher considerations on evaporator calculations

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Operation of process equipment is not ideal  and design of evaporators, and its operation consequently, may become tricky. Therefore, in engineering some licenses are usually taken to go around a number of difficulties. Here are some of these, in the context of evaporators:

• the process fluid entering the evaporator $F$, inside it and during the process should be well mixed so that it has a uniform concentration. This is true for large, horizontally, compartiments but not for evaporators 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 evaporators made of long vertical tubes,
• the fluid leaving the boiling chamber in stream $E$ is only the solvent, which is usually water. Also, its temperature and pressure are exactly those of the boiling solution,
• once the steam has transferred its available energy to the process fluid it condensates and leaves the evaporator as stream $C$ at the water vapor pressure. 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 evaporator into the surroundings. This not completely true but if proper thermal insulation is installed the effect of theses losses is negligible. Also, for large evaporators the heat transfer surface is so large that heat losses through the shell (external metal surface working as case) are little.

2 The mass balance in a single stage evaporator

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The mass balance is made, as usually, with reference to an amount of time (1 hour commonly) and this can be made by two approaches:

• one global mass balance and
• one balance for the solute.

The global mass balance across the evaporator can be written as follows,

$F+W=S+C+E$        Eq. (01)

which of course reduces to,

$F=S+E$        Eq. (02)

since $W=C$. Also, $F$, $W$, $S$, $C$ and $E$ are commonly presented as mass flow rate using 1 hour as base. $lb/h$ and $kg/h$ are usual units.

The mass balance for the solute can be expressed as,

$F\,x_F=S\,x_S$        Eq. (03)

where $x_F$ and $x_S$ are the fractions of the solute in the $F$ and $S$ streams.

3 The heat balance in a single stage evaporator

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This is also called enthalpic balance. The energy balance for the evaporator in Fig. 01 follows the same ideas as for the mass balance. The energy entering the system through all streams ($F$, $W$) should be the same that in all leaving streams ($C$, $E$, $S$) plus, if we are too worried, possible heat losses (like radiation).

        Eq. (04)

For this energy balance, enthalpies are used. However, since every stream going in and out from the evaporator is different in mass flow rate and composition every contribution in Eq. (04) should include this feature. Thus, Eq. (04) may be written as follows,

$F\,h_F + W\,H_W = S\,h_S + E\,H_E + C\,h_C$        Eq. (05)

where the heat loss by radiation term was omitted because theses losses are usually small (as explained above). $H$ and $h$ were used to represent the liquids and vapors enthalpies. Notice that the enthalpy of the steam $H_W$ should be exactly the enthalpy of the water vapor, say $H_{vap}$, which is the summation of the enthapy of the liquid, say $H_{liq}$, plus the evaporation enthalpy, say $H_{evap}$. In other words, by $H_W$ we mean all the energy in the steam fluid. On the other hand, the condensate will carry only the energy in $H_W$ corresponding to the liquid so that: $h_C=H_{liq}$. This last statement is not always true but on steady conditions and under the assumption that the condensate remains at its boiling temperature.

The explanation presented in last paragraph is key to understand why in the enthalpic balances for evaporators the energy carried by the condensate is not taken into account nor the whole energy $H_{vap}$ of the steam but its evaporation enthalpy $H_{evap}$, which is the available energy to be transferred to the solution. Well, the stream of condensate is not usually of interest and it is not measured. As you may have already noticed, in mass balance the condensate mass flow rate was no considered. In fact, in most textbooks the condensate flow rate is not even supplied as part of the calculation data and if provided, not used at all.

Futhermore, the enthalpy of evaporation $H_{evap}$ is defined as the energy required for the change of phase of liquid water at its boiling temperature into vapor (evaporation). In other words: $H_{evap}=H_{vap}-H_{liq}$, which as you may have already noticed is also the definition of the latent heat $\lambda_W$. 

Finally, since you do not want any unnecessary complexities in your calculations, you would wish to delete the term corresponding to the thermal energy taken out by the condensate (because it is worthless). Fortunately, only a small correction in the thermal energy supplied by the steam term has to be done: $H_W$ must be exchanged by $\lambda_W$.

Therefore, and with the above explanation in mind, the Eq. (05) is reduced to,

$F\,h_F + W\,\lambda_W = S\,h_S + E\,H_E$        Eq. (06)

The parameters $\lambda_W$ and $H_E$ can be obtained from the steam tables for water and, $h_F$ and $h_S$ from a plot of enthalpy vs concentration for the solution at hand. The enthalpy vs concentration plots are usually made for a pressure of $1\, atm$ so that if pressure on the fluid is important some corrections would be needed.

In engineering calculations enthalpy is commonly used with units of: energy per unit mass like $kJ/kg$ or $Btu/lb_m$, for example.

For many cases a plot of enthalpy vs concentration may not exist so that neither $h_F$ nor $h_S$ can be found. For such situations a rough approximation is used based on an engineering approximation from thermodynamics. In thermodynamics, the enthalpy change for a process, at constant pressure, is defined as,

$\Delta H=\int_{T_R}^{T_N}\,c_p\, dT$        Eq. (07)

where subscripts $R$ and $N$ are for reference and new. This means that if you need the enthalpy at a new temperature $T_N$ then a reference enthalpy at a reference temperature must be known first. Notice that in Eq. (07) $c_p$ is a function of temperature. From Eq. (07), the enthalpy changes in Eq. (06) for the solution can be roughly estimated as,

$S\,h_S-F\,h_F=F\,\int_{T_F}^{T_S}\, c_p\, dT$        Eq. (08)

I accept that the right hand side of Eq. (08) is forced (perhaps too much) but I did not have a better idea. The right hand side of Eq. (08) do not consider the effect of different mass flow rates: $S$ and $F$. Equation (08) may create other kind of problems since $c_p$ may not be known as a function of temperature but just as a constant. Also, $c_p$ depends on the solute concentration in the streams!

Therefore, since, may be, you do not have $c_p$ neither as a function of temperature nor as a function of the concentration very large errors may be introduced. If $c_p$ is constant, the integral in Eq. (08) can be easily performed. Then, the heat transferred between the stream $F$ and the thick liquor $S$, as the energy required to get $S\,h_S$ from $F\,h_F$ can be approximated as follows,

$S\,h_S-F\,h_F=F\,c_{pF}\left( T_S-T_F \right)$        Eq. (09)

where $T_F$ and $T_S$ are the temperatures of the streams $F$ and $S$, respectively, and $c_{pF}$ is the heat capacity of the fluid in stream $F$.

Just to be clear, even if it is too repetitive, this is just a rough approximation with the following weakesses,

• the heat capacity $c_{pF}$ is not a constant. In fact, it will depend, by definition, on the temperature and in this case, for sure, on the concentration in $F$ and $S$ as well,
• if the difference in temperature $T_S-T_F$ is too much, then larger estimation errors should be expected. How small should $T_S-TF$ need to be? It will depend on the nonlinear dependence of $c_p$ on the temperature and the concentration,
• large concentration operations would introduce estimation errors of unkwnown magnitud,
• the mass is not the same at both temperatures $T_S$ and $T_F$.

Despite, the above drawbacks of using Eq. (09), $c_{pF}$ can be approximated, to include the effect of the concentration, as follows,

$c_{pF}=1-\left(1-c_{pR}\right)\dfrac{x}{x_R}$        Eq. (09)

where $c_{pR}$ and $x_R$ are a heat capacity and a solute fraction of reference. You should be aware that Eq. (10) works if the heat capacity changes linearly, or almost, with the concentration of the solute.

Substitution of Eq. (09) into Eq. (06) produces,

$W\, \lambda_W=E\,H_E+F\,c_{pF}\left( T_S-T_F \right)$        Eq. (10)

As you may have already notice, the last term in Eq. (09) resembles the idea of enthalpy of vaporization for stream since only the change in energy from $F$ to $S$ is being considered. This means, that Eq. (10) needs a correction before being used. This is as follows.

The left hand side and the last term in the right hand side of Eq. (10) only consider the change in enthalpy but the first term $E\,H_E$ takes into account the energy from the fusion to the vapor phase so that this term needs to corrected according to the other ones. $E\,H_E$ must be exchange by $E\,\lambda_E$ where $\lambda_E$ corresponds to the latent heat or enthalpy of vaporization of the steam $E$. Therefore, Eq. (10) is corrected as,

$W\, \lambda_W=E\, \lambda_E+F\,c_{pF}\left( T_S-T_F \right)$        Eq. (11)

Equation (11) is the same energy balance as that presented in Eq. (06). Equation (06) should always be preferred if data is available while Eq. (11) should be used with care considering the drawbacks mentioned before.

3.1 What if you were interested in the condensate flow rate

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Estimation of the condensate flow rate may become of interest for sizing of peripherial equipment or accesories to the evaporator which is at the heart of our process.

At this point two possible ideas can be considered for the estimation of the condensate flow rate. In other words, how much water would condensate from the incomming steam.

• one way is to proceed from the mass balance were it can appears directly and
• another way can be through Eq. (05) were it also appears. You may first proceed with Eq. (06) to solve for some unknowns and then go back to use Eq. (05)

4 What you know about mass and heat balance in an evaporator

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Please, follow the link below to answer a quizz and check what you have learned or understood about mass and heat balance in an evaporator.

What you know about mass and heat balance in an evaporator

This is the end of the post and I hope it would be really helpful.

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

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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
• A list of process variables
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• The lines of Dühring

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Ildebrando.