Separation Processes 1

Evaporators

Evaporators are a common process whenever large amounts of water or another solvent are to be separated from relatively non-volatile components.
They function by simply boiling the volatile component away.
They are often seen when large amounts of fresh water are required. For example, drinking water on ships or for steam ship boilers ( top product).
Evaporators are also used to concentrate aqueous solutions such as fruit juice or orange juice (bottom product).
They are common when the process fluid contains precipitates that need to be crystallised.

We study evaporators first in this course as they are essentially an ideal flash-distillation step (there is a perfect separation of the components). No need to worry about the phase equilibria part of thermodynamics (yet).
Later on, we will have to calculate how much of the two (or more) species is present in the top and bottom product using vapour-liquid equilibrium data, or solubility data.
However, in this course, the defining characteristic for evaporators is that we will assume all of the involatile components (e.g., salt, solids) are recovered in the bottoms product.

There are many designs of evaporator, but there are some fundamental operational classifications.
Evaporators may be single-pass, where the fluid is heated only once, or recirculating, with a long residence time of the treated fluid.
Evaporators may also operate in convective or boiling regimes depending on the state of the fluid at the heat source.
The flow around the heat source may be natural or forced.
We'll look at some common designs now to get a feel for the evaporator process equipment. While looking, try to see why the vapour and liquid outlets might be considered "well-mixed" and in equilibrium.

There are many designs of evaporator:

There is the open-kettle or pan which is a simple evaporator.
It is usually heated through a jacket or coil, and may be directly fired.
They are inexpensive and simple to operate, but the efficiency is poor.
They are common for small scale or old/traditional process designs.
For example, old salt pans used in salt mines or traditional maple syrup refiners (see right).
No longer used except for "traditional" (i.e., inefficient) production.

Pan evaporator used in small-scale maple syrup production. Note the efficient covered evaporator stages to the right, and the final "finishing" stage which is open to be hand operated on the left.

There are many designs of evaporator:

A common and simple design is the horizontal-tube natural circulation evaporator.
The heating steam is condensing on the tube side and the tube bundle is completely submerged in the liquid.
They are simple designs which are relatively easy to clean.
These evaporators are good for low-viscosity fluids with high heat transfer coefficients, as it relies on natural convection.

There are many designs of evaporator:

There is the vertical-tube natural circulation evaporator.
In this evaporator, the boiling liquid is on the tube side, but also flows around the tube bundle in a downcomer (multiple pass).
The heating of the fluid in the tubes causes natural convection to occur.
This pumps the liquid up the tubes increasing the liquid velocity and the heat transfer coefficient.
As such, this design cannot be used with viscous liquids, but it is more efficient than the kettle type evaporator.
The lower portion of the evaporator is known as the calandria.

Two designs of vertical-tube natural circulation evaporator.

There are many designs of evaporator:

There is the long-tube vertical-type evaporator.
The tubes can be 3 to 10m high and the liquid may only pass through the evaporator once ( single-pass).
Boiling causes vapour bubbles to form which pump the liquid at high velocities through the tubes.
Contact times can be low, which is useful in the food industry (no burning of the product).

There are many designs of evaporator:

A variation on the long-tube vertical-type evaporator is the vertically fed falling film type evaporator.
Gravity and expansion of the fluid greatly increase the fluid velocities.
These can be designed with very short contact times.
This is an advantage as damage to sensitive heat-treated fluids is a function of temperature and contact time.

Falling film type evaporator.

There are many designs of evaporator:

There are also forced-flow evaporators.
These may exploit the pressure of a liquid head to prevent boiling in the exchanger.
They may also have a circulating pump to force the flow, or they may rely on the thermosyphon effect.
This is an analogue of the chimney effect, where a strong form of natural convection is used to pump a fluid.
Despite the “natural” origins of the thermosyphon flow, the fluid is being strongly forced through the heat exchanger.

There are many designs of evaporator:

Finally, for viscous fluids, there are evaporators which mechanically coat the heated surface with the fluid (agitated-film).
This ensures good heat transfer regardless of the viscosity and can be used for the drying of very thick pastes.
Deciding which evaporator to use strongly depends on the cost of materials, energy and fluid properties.
Luckily, there are many case studies in the literature which have been collated into tables to aid you in the selection of your evaporator…

Evaporator selection table, taken from C&R vol. 2, pg. 806.

A common feature of all types of evaporators are deentraining devices such as baffles.
These reduce or prevent entrained droplets of liquid exiting in the top stream.
Evaporators are also typically operated as total-condensers on the steam side (only condensate/water is removed from the steam side). This greatly simplifies the heat balance on the steam side, as its the latent heat of steam vapourisation/condensation.

horizontal-tube type evaporator.

When designing your evaporator, you start with a feed stream, and you have to achieve an output specification.
For example, you have a feed stream of 1000 kg/hr of orange juice with a solids concentration of 7.08% w/w.
You need to concentrate it to 50% w/w solids content before it is economical to ship it overseas.
How big should our exchanger be to achieve this? What is the heat transfer area required? What utilities (steam) will be used? What is the overall duty of the evaporator?

Single-stage evaporators are, essentially, a heat exchanger and we roughly size heat exchanges using their area.
To find the size of the evaporator, we need to solve the heat transfer equation for the heat transfer area, A : \begin{align*} Q = U  A  \left(T_{steam} - T_{fluid}\right) \end{align*}
We need to find the duty, Q, the heat transfer coefficient U, the boiling temperature of the evaporator T_steam, and the steam temperature.
The steam temperature is set by the available utilities, and the heat transfer coefficient/area can be calculated together from our Heat, Mass, and Momentum knowledge once the duty is known.
So we must first compute the duty, Q, and we need a simple model of our evaporator to do this…

Our simple model for the evaporator is that it has one input stream, and two output streams.
All of the solids entering in the feed stream F, is recovered in the liquid stream, L.
No heat is lost or gained during this process, except by that removed or supplied by the duty, Q.

With the initial design data, we can see that there are two unknown flowrates of streams we need to solve, L and V.
We need the fundamental equation of process engineering, the balance equation : \begin{align*} \textrm{INPUT}-\textrm{OUTPUT}+\textrm{GENERATION}=\textrm{ACCUMULATION} \end{align*}
Taking a total mass balance and a mass balance in the solids content we have \begin{align*} F &= L + V & x_{s,F} F &= x_{s,L} L \end{align*}
These equations are our general balance equations for an evaporator.

Taking the solids balance, we have \begin{align*} x_{s,F} F &= x_{s,L} L\\ 0.0708\times1000 &= 0.50 \times L\\ L &= 0.0708\times1000 / 0.5 = 141.6 \textrm{kg/hr} \end{align*}
Using the total balance, we have \begin{align*} F &= L + V\\ 1000 &= 141.6 + V\\ V &= 1000 - 141.6 = 858.4 \textrm{kg/hr} \end{align*}
We'll now update our diagram…

To calculate the duty, Q, we need to perform an energy balance to obtain… \begin{align*} \textrm{Energy In} + \textrm{Duty} &= \textrm{Energy Out}\\ F h_F + Q &= L h_L + V h_V \end{align*} where $h_F$ is the enthalpy per mass for the feed stream, $F$.
The enthalpies, $h$, for each stream depend on temperature AND composition!
The most difficult part of evaporator design may be tracking down accurate values for the stream enthalpies.
The difficulty is that there is an infinite selection of mixtures and a very limited amount of experimental data.

\begin{align*} \textrm{Energy In} + \textrm{Duty} &= \textrm{Energy Out}\\ F h_F + Q &= L h_L + V h_V \end{align*}

We will look at enthalpy and boiling point data in a later lecture, and will make some simplifying assumptions for now...
Let's assume that the inlet stream is at its boiling point ($100^{\circ}C$ @ 1atm) and the solids concentration does not affect the boiling temperature.
We also assume that there is no enthalpy of mixing ( concentration does not affect the enthalpy, $h_F=h_L$).
Then the only enthalpy difference will come about due to the boiling of the water content.

\begin{align*} F h_F + Q &= L h_L + V h_V \end{align*}

Using $h_F=h_L$ and $F-L=V$, the energy balance yields: \begin{align*} Q &= V \left(h_V - h_L\right) \end{align*}
We have from the steam tables $h_V-h_L=2260$ kJ/kg at 100 ${}^\circ$ C, therefore \begin{align*} Q &= 858.4\times2260\approx1.9\times10^9\textrm{J}\,\textrm{hr}^{-1}\approx540\textrm{kW} \end{align*}

\begin{align*} Q &= U  A  \left(T_{steam} - T_{fluid}\right) \end{align*}

To calculate the evaporator size, we need the overall heat transfer coefficient, and the temperature difference.
The overall heat transfer coefficient depends on the evaporator type, operating method and flow velocities (see Heat, Mass, and Momentum Transfer.).
You will probably have to iterate to a solution to $U$ in a detailed design (just like your design project).
But there are typical values available in the literature…

Type	Overall U (W/m ${}^2 $ K)
Short-tube vertical, natural	1100–2800
Horizontal-tube, natural	1100–2800
long-tube vertical, natural	1100–2800
long-tube vertical, forced	2300–11000
Agitated film	680–2300

Typical values for the evaporator heat transfer coefficient, $U$, taken from “Transport Processes and Unit Operations,” by C. J. Geankoplis.

\begin{align*} Q &= U  A  \left(T_{steam} - T_{fluid}\right) \end{align*}

Finally, we need the temperatures of the evaporator.
The temperature of the evaporator fluid depends on the solvent boiling point and an effect called boiling point rise.
For simplicity, we can assume at low concentrations the boiling point is the same as for the pure solvent. \begin{align*} T_{fluid} = T_{sat.,solvent} = 100^\circ\textrm{C} \end{align*}

\begin{align*} Q &= U  A  \left(T_{steam} - T_{fluid}\right) \end{align*}

The temperature of the steam depends on its saturation temperature. \begin{align*} T_{steam} = T_{sat.,steam} \end{align*}
This is because the latent heat of condensation of the steam is used to heat the evaporator (the condenser is a total steam condenser).
The saturation temperature depends on the pressure of the available steam, and can be looked up in your steam tables.

To calculate how much steam, $S$, the evaporator will use, we will need another heat balance: \begin{align*} Q &= S \left(h_{steam,vapour} - h_{steam,liquid}\right) \end{align*}
However, if we assume that steam has a latent heat of vaporisation with only a weak temperature and concentration dependence, the amount of steam condensed will be the same amount of water evaporated! \begin{align*} S \approx V \end{align*}
This concludes our roughest design of an evaporator, we will improve this calculation in the coming lectures.