Evaporation

To reduce load times, this material is divided into three files, corresponding to the numbered points below. The present file (evap2.html) contains point 2 only.
  1. Evaporator Concepts
  2. Evaporator Modeling
  3. Evaporator Calculations

Evaporator Modeling

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Evaporator Nomenclature In an evaporator, heat is transferred from a heating medium (usually steam) to a solution by conduction through a solid surface (the tube walls). As the solution boils, mass and heat are simultaneously transferred into the vapor phase. Consequently, an evaporator model consists of

Dynamic Model

Begin with the material balances. A total material balance can be written for the steam side (inside the tubes) and for the process side (liquid plus vapor).

Total Material Balance -- Steam Side
'deriv(V;s*?rho?;s,t)=m;si-m;so
The coil has constant volume and is filled with steam, so we can revise this to
V;s*('deriv(?rho?;s,t))=m;si-m;so
In some cases, you may want to represent the vapor density as a function of pressure.

Total Material Balance -- Process Side
'deriv(W;L+W;V,t)=m;f-m-m;v
If the total mass of the process vapor is much smaller than that of the process liquid, it may be possible to neglect the vapor in the total material balance.

Component (Solute) Balance
'deriv(W;L*x;L+W;V*x;V,t)=m;f*x;f-m;x
A solvent balance could also be written, but would not be a mathematically independent equation. Note that the solute is treated as nonvolatile, so that none leaves with the vapor. This is true for most evaporator applications.

Energy Balance -- Steam Side
'deriv(V;s*?rho?;s*H;s,t)=m;si*H;s-m;so*H;c-q;s-Q;1
This expression includes two heat transfer terms: qs, for heat transferred from the steam to the tube metal, and Q1 for heat losses from the steam.

Energy Balance -- Process Side
'deriv(W;L*H+W;L*H;V,t)=m;f*H;f-m*H-m;v*H;v+q-Q;2
The heat transfer terms are q, heat transferred from the tube metal to the process liquid and Q2, heat losses from the process side.

Energy Balance -- Tube Metal
'deriv(W;m*c;p;m*T;m,t)=h;i*A;i*(T;s-T;m)-h;o*A;o*(T;m-T)-Q;3
No material balance is necessary for the tube metal, but heat is transferred into and out of the tube walls, so an energy balance is may be needed. Q3 represents any heat losses from the tube wall.

Two transport equations are also needed:

Heat Transfer -- Steam to Tubes
q;s=h;c*A;i*(T;s-T;m) The inside heat transfer coefficient is a condensing coefficient.

Heat Transfer -- Tubes to Process Liquid
q=h;o*A;o*(T;m-T)

These six differential and two algebraic equations comprise a fundamental, dynamic model of a single effect evaporator.

Steady State Model

For many evaporator calculations, we are less interested in the behavior of the evaporator as a function of time than in broad behaviors. As we start looking at longer and longer times, fluctuations in behavior become less apparent and the system values approach a single operating point, the steady state. When we consider the steady state system, process variables stop changing with respect to time, so the time derivatives of the accumulation terms become zero. This allows us to replace differential equations with algebraic equations, and the steady state model which results is:

Steam Side Material Balance
m;si=m;so=m;s

Process Side Material Balance
m;f=m+m;v

Component (Solute) Balance
m;f*x;f=m*x

Steam Side Energy Balance
0=m;si*H;s-m;so*H;c-q;s-Q;1
If losses are negligible, then Q;1=0 . Substituting the result of the mass balance gives
0=m;s*(H;s-H;c)-q;s=m;s*?lambda?;s-q;s
where ?lambda?;s is the latent heat of the steam. This equation is analogous to Eq. 16.2 in McCabe, Smith, and Harriott.

Process Side Energy Balance
0=m;f*H;f-m*H-m;v*H;v+q-Q;2
Substituting the results of the material balance and neglecting the losses, we can rearrange this to
(m;f-m)*H;v-m;f*H;f+m*H=q
which is equivalent to McCabe et al. Eq. 16.3. The liquor enthalpies depend on the composition and concentration of the solution.

Tube Metal Energy Balance
Neglecting wall capacitance and losses, we obtain
q;s=q
and combining with the other energy balances gives
m;s*?lambda?;s=(m;f-m)*H;v-m;f*H;f+m*H
which is equivalent to the textbook's Eq. 16.4.

Heat Transfer

Several heat transfer effects must be considered when modeling an evaporator:

Transfer from condensing steam vapor to the interior of the coil wall can be represented by a steam film heat transfer coefficient, hc. This is usually high; however, if a large quantity of noncondensibles are present in the steam they can reduce heat transfer. Such noncondensibles should be vented to keep the coefficient up.

Transfer through any scaling or fouling on the tube interior wall is usually not a concern in an evaporator. Industrial boiler feed water is typically treated to keep scaling to a minimum.

Transfer through the tube wall can be represented by a conventional tube wall resistance (wall thickness/thermal conductivity, 'quot(?Delta?x,k)). This is usually not significant unless the evaporator is the agitated film type.

Transfer through fouling on the exterior tube wall can be significant, depending on what material is being evaporated.

Transfer to the boiling liquid depends on the velocity of the liquid. For dilute aqueous solutions, values on the order of 1500- 3000 W/m^2*C are reasonable. The text provides a figure for estimating this value.

All of these factors should be accounted for when calculating the overall heat transfer. A typical expression, in this case treating interior scaling as negligible, is given by

q;o='quot(T;s- T,'quot(1,h;i*A;i)+'quot(?Delta?x,k*A;m)+R;f+'quot(1,h;o*A;o))

It is often more convenient to express this in terms of an overall heat transfer coefficient, U,

q;o=U*A*?Delta?T

Question: Should a log mean temperature difference used?

The overall heat transfer coefficient, U, depends on the properties of the solution, the heating medium, and the surface geometry and type, including smoothness, cleanliness, composition, and thickness of metal. As with heat exchangers, U can be expressed as the inverse of the sum of the heat transfer resistances:

U='quot(1,'Sum(resistance;i,i,1,n))

Heat of Dilution

For some compounds, such as H2SO4, NaOH, and CaCl2, it is necessary to provide for the solution heat of dilution. The most convenient way to accomplish this is to use enthalpy-composition diagrams to obtain enthalpy data.

When it is legitimate to neglect the heat of dilution, the heat transfer is equal to the sum of the sensible and latent heat of the solution, or

m;f*c;p*(T-T;f)+(m;f-m)*?lambda?;v
If it is also acceptable to neglect the boiling point elevation, then the latent heat of the vapor can be approximated by the latent heat of steam.

When the temperature of the feed, Tf is greater that the temperature in the evaporator, flash evaporation occurs and sensible heat is removed instead of being added. When the feed is subcooled, and Tf < T, an extra load is placed on the evaporator.


R.M. Price
Original: 1/7/97
Modified: 3/16/98; 3/6/2003

Copyright 1997,1998, 2003 by R.M. Price -- All Rights Reserved