Planning of Distillation and Absorption Columns

Abstract

First, this chapter presents general information on the planning of a distillation column. It then introduces the prerequisite for the planning of a distillation or absorption plant, particularly the physical data and the vapour–liquid equilibrium of the components involved. The chapter shows how to prepare a mass and energy balance for the desired separation. It covers fixing the separation conditions according to the available heating and cooling media. The general criteria for selecting equipment parts for the given problem definition is provided. The equipment parts are as follows:

• appropriate column internals for the separation problem

• condensers, reboilers and vacuum pumps

• control equipment according to the requirements

• heating with steam or thermal oil

• cooling with cooling water or air

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

1.1 Planning Information

The basis for each plan of a distillation or absorption plant is the determination of the required number of trays and the required reflux ratio for fractionating, the required absorption fluid flow rate for absorption or the stripping gas flow rate.

These calculations are mostly performed with computers. However, care must be taken in the unchecked acceptance of a computer generated result.

There are several measured physical properties and measured vapour–liquid equilibrium data showing comparably deviating results. Small inaccuracies in vapour pressure or different equilibrium data or the choice of the calculation model for the equilibrium can result in substantial deviations in the design.

Figure 1.1 shows how the required number of trays and the reflux ratio change if the equilibrium is better or worse, by the order of 0.2%, than that assumed. Since the number of trays in an existing plant cannot be increased distillation has to be achieved with a higher reflux ratio, for instance, with R = 20 instead of R = 15, if the vapour–liquid equilibrium is worse.

Fig. 1.1

Required theoretical trays as a function of the reflux ratio for different relative volatilities

Figure 1.2 shows the influence of tray efficiency on the required reflux ratio.

Fig. 1.2

Actual trays required for different tray efficiencies

If the tray efficiency is poor the reflux ratio has to be increased, for instance from R = 16 to R = 18.5 for 60 trays, if the tray efficiency is only 75% rather than 80%.

The additional vapour and liquid loadings at a higher reflux ratio, due to poorer vapour–liquid equilibrium or poorer tray efficiency, must be considered while designing column internals. Sufficient reserve capacities must be available in the column.

Reboilers and condensers must be dimensioned for higher heat loads at higher reflux ratios.

Rules for a trouble-free operation

The flow chart in Fig. 1.3 shows the required accessories for a distillation plant. Each individual piece of equipment, each pump, each control valve, each vessel and each component of piping must be accurately designed and its control system must function properly:

Fig. 1.3

Fractionation column with feed and product vessels

• A sufficient feed height must be present for top condensers in order to avoid reflux variations [2].

• Where there is condensation in tubes, for instance in air coolers, the maximal vapour flow capacity of the condenser must be determined with an adiabatic flow factor [1, 2].

• The outlet pipe from the condenser to the accumulator should be self-venting [1].

• Vibrations in the U-tube formed by the column and thermosiphon reboiler can be a problem. The frequency of the vibrations depend on the tube length. These vibrations can be removed by applying a greater pressure drop, for instance by using a larger circulation, or an orifice plate, in the downcomer.

• When drawing boiling liquids as side streams from the column, a sufficient liquid height over the nozzle must be provided in order to avoid flash evaporation in the withdrawal pipe [1].

• Measurement nozzles for temperature, pressure and level must be provided.

• The piping with flow meters and control valves must be accessible in the steel framework for maintenance purposes.

• Depending on the mass and energy balance and the available utilities, for instance 12-bar of steam and cooling water at 25 °C, the reboilers, condensers and heat exchangers used for heating the feed and cooling the distillate and bottoms streams must be appropriately dimensioned [2].

• When using the pumps the required Net Positive Suction Head (NPSH) value must be kept in order to avoid cavitation [1].

• In practice the columns are installed on 3-m foundations in order to provide sufficient liquid height.

• If the boiling point is higher than the temperature of the given heating medium the vacuum distillation has to be adopted.

• In order to decide on the dimensions of the vacuum pump the required suction pressure and the required suction capacity is needed [3]. There should be a minimum pressure drop in the vacuum piping.

• The choice of the column internals should be determined using the allowable pressure loss and the available construction height, for instance in a production hall.

• Demisters should be used in order to minimize the loss of valuable materials and to protect the vacuum pumps and compressors.

• When deciding upon the dimensions of the piping for the product and utility streams, high-pressure losses or cavitation on bottlenecks should be avoided [1].

• The control valves must be designed for a functional working pressure drop [1].

• If sharp reduction occurs in valves which are designed too large, choked flow with cavitation may occur in the valve.

Construction notes

• The columns must be calculated based on the pressure or vacuum and wind loads. The columns should not sway at high wind velocities.

• In construction, consideration must be given to how internals can be installed into the column (manways) and what type of support is necessary in the column.

• The support rings must have tight tolerances in order to allow an even installation of the trays or liquid distributors.

• The nozzles for the evaporator, condenser, feed and side streams or side strippers must be properly arranged.

• Sufficient flexibility must be provided for thermal expansion of hot piping in order to avoid large thermal stresses and protect the equipment and pump nozzles. Leakages must be avoided especially in the vacuum columns.

• In order to avoid heat loss, and to provide surface protection, hot surfaces are appropriately insulated, mostly with mineral wool mats [3].

What basic data are required for the design?

• Flow rates with physical properties and equilibrium data, compositions, temperatures and pressures.

• Heating and cooling medium along with temperatures and pressures: steam, organic heat transfer media, cooling water, cold water, and cooling brine.

• Required materials: steel, stainless steel, monel, plastic, and graphite.

1.2 Mass Balance for the Separation Task

The starting point for every separation calculation is a mass balance with the required component distributions.

Example 1.2.1: Mass balance for a four-component mixture

Component	Feed				Distillate			Bottoms
	M	kg/h	kMol/h	x E	kg/h	kMol/h	x D	kg/h	kMol/h	x B
Light comp.	78	1952	25	0.25	1952	25	0.48	0.0	0.0	0.0
LK	92.1	2303	25	0.25	2081	22.6	0.44	219	2.4	0.05
HK	106.1	2652	25	0.25	228	2.15	0.04	2425	22.85	0.48
Heavy comp.	104.1	2602	25	0.25	239	2.3	0.04	2580	24.77	0.47
		9508	100		4500	52.05		5008	47.95

Conversions:

$$ {\text{kmol/h}} = \frac{\text{kg/h}}{{M}} $$

• M = mole weight

• Average mole weight M _m = ∑ x_i * M _i

• Feed: M _m = 95.075

• Distillate: M _m = 86.37

• Bottoms: M _m = 104.46

• LK = light key component

• HK = heavy key component

• x _E = feed compositions (molfraction)

• x _D = distillate compositions (molfraction)

• x _B = bottoms compositions(molfraction)

Conversion of weight, volume, and mole percentages

$$ \begin{array}{*{20}l} {{\text{Weight}}\%A = \frac{{{\text{Vol}}\%A\,*\,\rho_{\text{A}} }}{{{\text{Vol}}\%A\,*\,\rho_{\text{A}} \,+\, {\text{Vol}}\% B\,*\,\rho_{\text{B}} }}} \hfill &{{\text{Vol}}\% A = \frac{{{\text{Weight}}\% A/\rho_{\text{A}} }}{{{\text{Weight}}\% A/\rho_{\text{A}}\, +\, {\text{Weight}}\% B/\rho_{\text{B}} }}} \hfill \\ {{\text{Weight}}\% A = \frac{{{\text{Mol}}\%A\,*\,M_{\text{A}} }}{{{\text{Mol}}\% A\,*\,M_{\text{A}} \,+ \, {\text{Mol}}\% B\,*\,M_{\text{B}} }}} \hfill &{{\text{Mol}}\% A = \frac{{{\text{Weight}}\% A/M_{\text{A}} }}{{{\text{Weight}}\% A/M_{\text{A}} \,+ \,{\text{Weight}}\%B/M_{\text{B}} }}} \hfill \\ {{\text{Molfraction}}\;A = \frac{{{\text{Mol}}\% }}{100}} \hfill &{{\text{Weightfraction}}\;A = \frac{{{\text{Weight}}\% }}{100}} \hfill \\ \end{array} $$

1.3 Separation Conditions

First, the pressure and temperature in the column have to be fixed for the given problem definition.

• With a steam heated reboiler the boiling temperature in the bottom should be at least 20 °C below the dew point temperature of the available heating steam. A remedial measure for lowering the boiling temperature is vacuum distillation.

• Often the bottom temperature must not exceed a certain value in order to avoid thermal cracking of the product. One supporting measure for lowering the boiling temperature is vacuum distillation.

• A high pressure drop in the column, for instance from bubble cap trays with large slot seals, increases the bottom pressure and therefore increases the boiling temperature. One remedial measure is to have column internals demonstrating low pressure losses, for instance random or structured packing.

• With low top temperatures, for instance those under 30 °C, chilled water must be used to cause condensation. Alternatively, a higher pressure would also increase the dew point.

The decision becomes diffithe energy balance, insignificantcult if in the bottom a high boiling material can only be evaporated under vacuum and the separated low-boiling component on the top can only be condensed under pressure (e.g., stripping of gasoline from gasoil in order to improve its flash point). One remedial measure is steam-stripping distillation.

1.4 Vapour-Liquid Equilibrium [4]

The appropriate equilibrium- and physical property models must be chosen for the given mixture.

1.4.1 The Ideal Equilibrium According to Raoult–Dalton

.

1.4.2 Equations of State

The equation of states being: Benedict–Webb–Rubin for KWST C₁ to C₇ (BWR); Soave–Redlich–Kwong (SRK) for hydrocarbons; Redlich-Kwong (RK); and Peng–Robinson (PR) for small non-polar molecules

1.4.3 Equilibrium Models for Non-polar Materials Considering the Non-ideal Behaviour in the Liquid Phase Based on Pure Component Data

The models include: Chao–Seader (CS) for hydrocarbons from −20 to 260 °C and Grayson–Streed (GS) | CS with corrected data for higher temperatures and pressures, which is better suited to mixtures containing hydrogen.

1.4.4 Equilibrium Models for Polar Components with High Non-ideal Behaviour in the Liquid Phase

Here is a list of models with interaction parameters from measured vapour–liquid equilibria data for mixtures: Margules; van Laar; Wilson; NRTL; Uniquac; and UNIFAC (which is a group contribution method for structural groups).

1.5 Energy and Mass Balance in the Column

These balances are required for the fluid dynamic design of stages or packings and for the dimensioning of reboilers and condensers.

1.5.1 Mass Balance (Fig. 1.4)

In the rectification section—the section of the column above the feed stage—results from the vapour and liquid loading from the distillate flow rate and the reflux ratio. The loadings in kg/h or kMol/h are constant if the molar latent heat does not change and if no side draws exist. If the reflux feedback to the column is subcooled the vapour and liquid loadings increase. In the stripping section—the section of the column under the feed stage—the heat loss of the column must be additionally considered as must the thermal condition of the feed which is characterized by the q-value.

Fig. 1.4

Mass balance of a fractionation column

Due to the fact that the feed rate is mostly fed into the column in a subcooled condition the vapour and liquid flow rates in the stripping section increase accordingly.

It has to be considered that, over the length of the column, the physical properties change as a function of pressure, temperature, and composition. In particular, in vacuum distillation columns, the vapour volume changes greatly as pressure changes due to the pressure loss at the stages.

Example 1.5.1

Here is a calculation of the vapour loading for 2500 kg/h vapour at different pressures.

At column top	10 stages below
P = 6.65 mbar	P = 26.6 mbar
T = 185 °C	T= 220 °C
ϱ D = 0.0473 kg/m3	ϱ D = 0.178 kg/m3
V = 52,854 m3/h	V = 14,045 m3/h

where T = temperature; P = pressure; V = vapour volume (m³/h); and ϱ _D = vapour density (kg/m).

At the top of the column the vapour loading is greater than 10 stages below by a factor of 3.7.

In a 3.5 m diameter column the vapour flow velocity at the top of the column is 1.52 m/s and 10 stages below it is only 0.4 m/s.

Checks must be completed to identify if the separation is hindered by droplet entrainment at high vapour flow velocities or by weeping at low gas flow velocities.

Explanations for the mass and energy balances in Figs. 1.4 and 1.5

Rectification section:

Fig. 1.5

Energy balance for a fractionation column

• D = distillate flow rate (kg/h)

• E = feed flow rate (kg/h)

• G _K = vapour flow rate to the condenser = (RV + 1) * D (kg/h)

• RV = reflux ratio

• G _V = vapour flow rate in the stripping section = G _K + G _R (kg/h)

• G _R = vapour flow rate for heating the reflux to top temperatures

$$ G_{\text{R}} = \frac{{R*c*(t_{\text{K}} - t_{\text{R}} )}}{r}\;\left( {\text{kg/h}} \right) $$

• R = reflux flow rate (kg/h)

• c = specific heat capacity (Wh/kg K)

• t _K = column top temperature (°C)

• t _R = reflux temperature (°C)

• L _V = liquid flow rate in the stripping section = R + G _R = RV * D + G _R (kg/h).

Stripping section:

• B = bottom draw flow rate (kg/h)

• G _A = vapour flow rate in the stripping section = G _V + G _E + G _W (kg/h)

• G _E = vapour flow rate for heating the feed (E) from the feed temperature (t _Z) to the column temperature (t _E) on the feed stage

• E = feed flow rate (kg/h)

• G _W = vapour flow rate for balancing the heat losses (kg/h)

• r = latent heat (Wh/kg)

• L _A = liquid flow rate in the stripping section (kg/h).

Energy:

• Q _C = condensation duty = G _K * (r + c * (t _K − t _R)) (W)

• r = latent heat (Wh/kg)

• c = specific heat capacity (Wh/kgK)

• t _K = top temperature (°C)

• t _R = reflux temperature (°C)

• q _R = heating load for the reflux (R)

• q _R = R * c * (t _K − t _R) (W)

• q _D = heat content of the distillate (D)

• q _D = D* c * t _D (W)

• D = distillate flow rate (kg/h)

• t _D = distillate temperature (°C)

• q _E = heat content of the feed

• q _E = E * c * t _Z (W)

• t _Z = feed temperature (°C)

• t _B = bottom draw temperature (°C)

• q _B = heat content of the bottom draw (B)

• q _B = B * c * t _B (W)

• Q _R = reboiler duty = G _A * r (W) = Q _C + q _D + q _B + q _V − q _E (W)

• G _A = vapour flow rate in the stripping section (kg/h)

• q _V = Q _W = heat loss of the column (W).

When determining the flow loadings in the rectification and stripping sections the thermal condition q of the feed mixture must be considered. This q value also has an influence on the required number of theoretical stages. The calculation of the q value is covered in Chap. 3.

• q = 1 for the liquid feed at bubble point temperature.

• q = 0 for the vapour feed at dew point temperature.

• q = 0.75 for a two-phase mixture with 75% liquid.

• q < 0 for superheated vapour.

• q > 1 for the liquid feed below the bubble point temperature.

Calculation formula for the loading in the column:

$$ \begin{aligned} G_{\text{V}} & = \left( {RV + 1} \right)*D = G_{\text{A}} + \left( {1 - q} \right)*E\;({\text{kg/h}}) \\ L_{\text{V}} & = RV*D\;({\text{kg/h}}) \\ G_{\text{A}} & = G_{\text{V}} - \left( {1 - q} \right)*E = L_{\text{A}} - B\;({\text{kg/h}}) \\ L_{\text{A}} & = L_{\text{V}} + q*E\;({\text{kg/h}}) \\ G_{\text{V}} - G_{\text{A}} & = \left( {1 - q} \right)*E\;({\text{kg/h}}) \\ L_{\text{A}} - L_{\text{V}} & = \left( {1 - q} \right)*E\;({\text{kg/h}}) \\ \end{aligned} $$

.

1.5.2 Energy Balance (Fig. 1.5)

The required heating energy is supplied by the reboiler with the necessary cooling energy by the condenser. Within the column the heating energy is transported with the vapour and the cooling energy with the liquid. For a subcooled reflux or a cold feed more vapour is needed to heat the liquid. The condensed vapour for heating increases the exiting liquid flow rate. The calculation equations listed in Figs. 1.4 and 1.5 are used whilst preparing the energy and mass balances. The use of these equations is shown in the following examples.

Example 1.5.2.: Flow and energy balance of a distillation column

Calculation data:

• Feed flow rate, E = 2467 kg/h

• Distillate flow rate, D = 740 kg/h

• Bottom draw flow rate, B = 1727 kg/h

• Feed temperature, t _Z = 174 °C or 90 °C

• Temperature at the feed stage, t _E = 174 °C

• Top temperature, t _K = 160 °C

• Bottom temperature, t _B = 210 °C

• Latent heat, r = 100 Wh/kg

• Specific heat capacity, c = 0.4 Wh/kgK

• Reflux ratio, RV = 12.5

• Reflux temperature, t _R = 160 °C or 153 °C.

Example 1.5.2.1: Mass and energy balance for liquid feed with a bubble point temperature (q = 1) Without heat loss

$$ t_{\text{R}} = t_{\text{K}} = 1 60\,^\circ {\text{C}} \quad t_{\text{Z}} = t_{\text{E}} = 1 7 4\,^\circ {\text{C}} \quad q = 1 \quad {\text{RV}} = 1 2. 5\quad q_{\text{V}} = Q_{\text{W}} = 0\,{\text{W}} $$

Mass balance in the rectification section:

• Vapour flow rate in the rectification section, G _V = (RV + 1) * D = (12.5 + 1) * 740 = 9990 kg/h

• Liquid flow rate in the rectification section, L _V = RV * D = 12.5 * 740 = 9250 kg/h

Mass balance in the stripping section:

• Vapour flow rate, G _A = G _V − (1 − q) * E = 9990 − (1 − 1) * 2467 = 9990 kg/h

• Liquid flow rate, L _A = L _V + q * E = 9250 + 1 * 2,467 = 11,717 kg/h

• Bottom draw flow rate B = L _A − G_A = 11,717–9990 = 1727 kg/h

• Distillate flow rate D = E − B = 2467 − 1727 = 740 kg/h

Energy balance:

$$ {\text{Condenser}}\,{\text{duty}},\,Q_{\text{C}} = G_{\text{K}} *r = 9990* 100 = 999{,}000\,{\text{W}} $$

$$ {\text{Reboiler}}\,{\text{duty}}\,Q_{\text{R}} = Q_{\text{C}} + q_{\text{D}} + q_{\text{B}} + q_{\text{V}} - q_{\text{E}} $$

• Distillate heat q _D = D * c * t _D = 740 * 0.4 * 160 = 47,369 W

• Bottom draw heat q _B = B * c * t _B = 1727 * 0.4 * 210 = 145,068 W

• Feed heat q _E = E * c * t _E = 2467 * 0.4 * 174 = −171,703 W

$$ Q_{\text{R}} = 999{,}000 + 47{,}369+ 145{,}068+ 0{-}171{,}703= 1{,}019{,}734\,{\text{W}} $$

Cross-check:

$$ G_{A} = \frac{1019734}{100} = 10197\;{\text{kg/h}}\; > \;9990\;{\text{kg/h}} $$

$$ L_{\text{A}} = G_{\text{A}} + B = 10{,}197+ 1727= 11{,}924\,{\text{kg}}/{\text{h}} > 11{,}717\,{\text{kg}}/{\text{h}} $$

For energy balances, insignificant higher vapour and liquid loadings result in the stripping section at q = 1.

Example 1.5.2.2: Mass and energy balance for vapour feed at dew point temperature (q = 0) without heat losses

$$ t_{\text{R}} = t_{\text{K}} = 1 60\,^\circ {\text{C}} \quad t_{\text{Z}} = t_{\text{E}} = 1 7 4\,^\circ {\text{C}} \quad q = 0\quad {\text{RV}} = 1 2. 5\quad q_{\text{V}} = Q_{\text{W}} = 0\,{\text{W}} $$

Mass balance in the rectification section:

• Vapour flow rate, G _V = (RV + 1) * D + (1 − q) * E

• Vapour flow rate G _V = (12.5 + 1) * 740 + (1 − 0) * 2467 = 12,457 kg/h

• Liquid flow rate, L _V = G _V − D = 12,457 − 740 = 11,717 kg/h

• Distillate flow rate, D = G _V − L _V = 12,457 − 11,717 = 740 kg/h

Mass balance in the stripping section:

• Vapour flow rate, G _A = G _V − (1 − q) * E = 9990 − 0(1 − 0) * 2467 = 12,457 − 2467 = 9990 kg/h

• Liquid flow rate, L _A = L _V + q * E = 11,717 + 0 * 2467 = 11,717 kg/h

• Bottom draw flow rate, B = L _A − G _A = 11,717–9990 = 1727 kg/h

Energy balance:

$$ {\text{Condenser}}\,{\text{duty}},\,Q_{\text{C}} = G_{\text{K}} *r = 12{,}457*100 = 1{,}245{,}700\,{\text{W}} $$

$$ {\text{Reboiler}}\,{\text{duty}},\,Q_{\text{R}} = Q_{\text{C}} + q_{\text{D}} + q_{\text{B}} + q_{\text{V}} - q_{\text{E}} $$

• Distillate heat q _D = D * c * t _D = 740 * 0.4 * 160 = 47,360 W

• Bottom draw heat q _B = B * c * t _B = 1727 * 0.4 * 210 = 145,068 W

• Feed heat q _E = E * (c * t _E + r) = 2467 * (0.4 * 174 + 100) = 418,403 W

$$ Q_{\text{R}} = 1{,}245{,}700 + 47{,}360 + 145{,}068+ 0 - \, 418{,}403= 1{,}019{,}725\,{\text{W}} $$

Cross-check:

$$ G_{A} = \frac{1{,}019{,}725}{100} = 10,197\;{\text{kg/h}}\; > \;9990\;{\text{kg/h}} $$

$$ L_{\text{A}} = G_{\text{A}} + B = 10{,}1 9 7+ 1 7 2 7= 1 1{,}9 2 4\,{\text{kg}}/{\text{h}} > 1 1{,}7 1 7\,{\text{kg}}/{\text{h}} $$

With the energy balance, insignificant higher vapour and liquid loadings result in the stripping section.

Example 1.5.2.3: Mass and energy balance for q = 0.75 (25% vapour) and Q _W = 0

$$ t_{\text{R}} = t_{\text{K}} = 1 60\;^\circ {\text{C}}\;t_{\text{Z}} = t_{\text{E}} = 1 7 4\,^\circ {\text{C}}\;q = 0. 7 5\quad {\text{RV}} = 1 2. 5\quad q_{\text{V}} = Q_{\text{W}} = 0\,{\text{W}} $$

Mass balance in the rectification section:

• Vapour flow rate, G _V = (RV + 1) * D + (1 − q) * E = (12.5 + 1) * 740 + (1 − 0.75) * 2467 = 10,607 kg/h

• Liquid flow rate, L _V = G _V − D = 10,607 − 740 = 9867 kg/h

• Distillate flow rate, D = G _V − L _V = 10,607 − 9867 = 740 kg/h

Mass balance in the stripping section:

• G _A = G _V − (1 − q) * E = 10,607 − (1 − 0.75) * 2467 = 10,607 − 617 = 9990 kg/h vapour

• L _A = L _V + q * E = 9867 + 0.75 * 2467 = 11,717 kg/h liquid in the striping section

• Bottom draw, B = L _A − G _A = 11,717–9990 = 1727 kg/h bottom draw flow rate

Energy balance:

$$ {\text{Condenser}}\,{\text{duty}},\,Q_{\text{C}} = G_{\text{K}} *r = 10{,}607* 100 = 1{,}060{,}700\,{\text{W}} $$

$$ {\text{Reboiler}}\,{\text{duty}},\,Q_{\text{R}} = Q_{\text{C}} + q_{\text{D}} + q_{\text{B}} + q_{\text{V}} - q_{\text{E}} $$

• Distillate heatq _D = D * c * t _D = 740 * 0.4 * 160 = 47,360 W

• Bottom draw heatq _B = B * c * t _B = 1727 * 0.4 * 210 = 145,068 W

• Feed heatq _E = E * (c * t _E + (1 − q) * r) = 2467 * (0.4 * 174 + 25) = 233,378 W

$$ Q_{\text{R}} = 1,0 60, 700 + 4 7, 3 60 + 1 4 5,0 6 8+ 0 - 2 3 3, 3 7 8= 1,0 1 9, 7 50\,{\text{W}} $$

Cross-check:

$$ G_{A} = \frac{1019750}{100} = 10197\;{\text{kg/h}}\; > \;9990\;{\text{kg/h}} $$

$$ L_{\text{A}} = G_{\text{A}} + B = 10, 1 9 7+ 1 7 2 7= 1 1 , 9 2 4\,{\text{kg}}/{\text{h}} > 1 1 , 7 1 7\,{\text{kg}}/{\text{h}} $$

With the energy balance, insignificant higher vapour and liquid loadings result in the stripping section.

Example 1.5.2.4: Mass and energy balance for q = 1 considering a heat loss of Q _W = 60 kW in the column

$$ t_{\text{R}} = t_{\text{K}} = 1 60\,^\circ {\text{C}}\quad t_{\text{Z}} = t_{\text{E}} = 1 7 4\,^\circ {\text{C}}\quad q = 1\quad {\text{RV}} = 1 2. 5\quad q_{\text{V}} = Q_{\text{W}} = 60\,{\text{kW}} $$

Mass balance in the rectification section:

• Vapour flow rate, G _V = (RV + 1) * D + (1 − q) * E = (12.5 + 1) * 740 + (1 − 1) * 2467 = 9990 kg/h

• Liquid flow rate, L _V = G _V − D = 9990 − 740 = 9250 kg/h = RV * D = 12.5 * 740

• Distillate flow rate, D = G _V − L _V = 9990–9250 = 740 kg/h

Mass balance in the stripping section with G _W for the heat loss Q _W = q _V = 60 kW

$$ G_{\text{W}} = \frac{{Q_{\text{W}} }}{r} = \frac{60000}{100} = 600\;{\text{kg/h}} $$

• Vapour flow rate, G _A = G _V − (1 − q) * E + G _W = 9990 − (1 − 1) * 2467 + 600 = 10,590 kg/h

• Liquid flow rate, L _A = L _V + q * E + G _W = 9250 + 1 * 2467 + 600 = 12,317 kg/h

• Bottom draw flow rate, B = L _A − G _A = 12,317 − 10,590 = 1727 kg/h

Energy balance:

$$ {\text{Condenser}}\,{\text{duty}},\,Q_{\text{C}} = G_{\text{K}} *r = 9990* 100 = 999{,}000\,{\text{W}} $$

• Distillate heat q _D = D * c * t _R = 740 * 0.4 * 160 = 47,360 W

• Bottom draw heat q _B = B * c * t _B = 1727 * 0.4 * 210 = 145,068 W

• Feed heat q _E = E * c * t _Z = 2467 * 0.4 * 174 = 171,703 W

• Heat loss Q _W = q _V = 60,000 W

$$ {\text{Reboiler}}\,{\text{duty}},\,Q_{\text{R}} = Q_{\text{C}} + q_{\text{D}} + q_{\text{B}} + Q_{\text{W}} - q_{\text{E}} $$

$$ Q_{\text{R}} = 999{,}000 + 47{,}360 + 145{,}068+ 60{,}000{-} 171{,}703= 1{,}079{,}725\,{\text{W}} $$

Required vapour flow rate to bring the required heat from the bottom:

$$ G_{\text{A}} = \frac{1{,}079{,}725}{100} = 10{,}797\;{\text{kg/h}}\; > \;10{,}590\;{\text{kg/h}} $$

$$ L_{\text{A}} = G_{\text{A}} + B = 10{,}797+ 1727= 12{,}524\,{\text{kg}}/{\text{h}} > 12{,}317\,{\text{kg}}/{\text{h}} $$

With this energy balance, rather higher vapour and liquid loadings result in the stripping section due to heat losses.

Example 1.5.2.5: Mass and energy balance for real conditions with heat loss, subcooled reflux, and subcooled feed

$$ t_{\text{Z}} = 90\,^\circ {\text{C}}\quad t_{\text{R}} = 1 5 3\,^\circ {\text{C}}\quad q = 1 3 3 6\quad {\text{heat}}\,{\text{loss}}\,Q_{\text{W}} = q_{\text{V}} = 60\,{\text{kW}} $$

Mass balance in the rectification section considering the subcooled reflux:

$$ {\text{Reflux}}\,{\text{rate}}\,R = {\text{RV}}*D = 1 2. 5* 7 40 = 9 2 50\,{\text{kg}}/{\text{h}} $$

Calculation of the vapour flow rate, G _R , for heating the subcooled reflux to the top temperature:

$$ G_{\text{R}} = \frac{{R*c*(t_{\text{K}} - t_{\text{R}} )}}{r} = \frac{{9250*0.4*\left( {160 - 153} \right)}}{100} = 259\;{\text{kg/h}} $$

• Vapour flow rate, G _V = (RV + 1) * D + G _R = (12.5 + 1) * 740 + 259 = 10,249 kg/h

• Liquid flow rate, L _V = R + G _R = 9250 + 259 = 9509 kg/h

• Distillate flow rate, D = G _V − L _V = 10,249 − 9509 = 740 kg/h.

Mass balance in the stripping section considering the heat loss of the column and the subcooled feed:

$$ {\text{Heat}}\,{\text{loss}}\,{\text{of}}\,{\text{the}}\,{\text{column}},\,Q_{\text{W}} = q_{\text{V}} = 60{,}000\,{\text{W}} $$

Required vapour flow rate, G _W, to balance the heat loss

$$ G_{\text{W}} = \frac{{Q_{\text{W}} }}{r} = \frac{60{,}000}{100} = 600\;{\text{kg/h}} $$

Calculation of the q value for the subcooled feed with 90 °C

$$ q = 1 + \frac{{0.4*\left( {174 - 90} \right)}}{100} = 1,336 $$

Calculation of the vapour flow rate, G _E , for heating the subcooled feed to the bubble point temperature of the feed stage:

$$ G_{\text{E}} = \left( {q - 1} \right)*E = \left( { 1. 3 3 6- 1} \right)* 2. 4 6 7= 8 2 8. 9\,{\text{kg}}/{\text{h}} $$

• Vapour flow rate, G _A = G _V + G _E + G _W = 10,249 + 828.9 + 600 = 11,677.9 kg/h

• Vapour flow rate, G _A = G _V − (1 − q) * E + G _W = 10,249 − (1 – 1.336) * 2467 + 600 = 11,677.9 kg/h

• Liquid flow rate, L _A = L _V + (q * E) + G _W = 9,509 + 1,336 * 2,467 + 600 = 13,404.9 kg/h

• Bottom draw flow rate, B = L _A − G _A = 13,404.9 − 11,677.9 = 1727 kg/h

Energy balance:

$$ {\text{Condenser}}\,{\text{duty}},\,Q_{\text{C}} = G_{\text{K}} *\left( {r + c*\left( {t_{\text{K}} - t_{\text{R}} } \right)} \right) = 9990*[ 100 + 0. 4*\left( { 160 - 153} \right)] = 1{,}026{,}972\,{\text{W}} $$

• Distillate heat q _D = D * c * t _R = 740 * 0.4 * 153 = 45,288 W

• Draw stage heat q _B = B * c * t _B = 1727 * 0.4 * 210 = 145,068 W

• Feed heat q _E = E * c * t _Z = 2467 * 0.4 * 90 = 88,812 W

• Heat loss Q _W = q _V = 60,000 W

$$ {\text{Reboiler}}\,{\text{duty}} \,Q_{\text{R}} = Q_{\text{C}} + q_{\text{D}} + q_{\text{B}} + Q_{\text{W}} - q_{{{\text{E}} }} $$

$$ Q_{\text{R}} = 1026{,}972+ 45{,}288+ 145{,}068+ 60{,}000{-} 88{,}812= 1{,}188{,}516\,{\text{W}} $$

Required vapour flow rate to bring the required heat to the bottom:

$$ G_{\text{A}} = \frac{1{,}188{,}516}{100} = 11{,}885\;{\text{kg/h}}\; > 11{,}677.9\;{\text{kg/h}} $$

$$ L_{\text{A}} = G_{\text{A}} + B = 11{,}885+ 1727= 13{,}612\,{\text{kg}}/{\text{h}} > 13,404.9\,{\text{kg}}/{\text{h}} $$

With this energy balance, slightly higher vapour and liquid loadings result for the stripping section.

1.5.3 Required Column Diameter

The required flow cross section, A, in the column or the column diameter, D, for the vapour flow rates, G _V and G _A, in the rectification and stripping section are determined with the F factor (see Chaps. 9 and 10)

$$ \begin{aligned} A & = \frac{{G({\text{kg/h}})}}{{3600*F*\sqrt {\rho_{\text{V}} } }} = \frac{G}{{3600*w*\rho_{\text{V}} }}({\text{m}}^{ 2} )\quad \quad F = w*\sqrt {\rho_{\text{V}} } \\ D & = \sqrt {\frac{4*A}{\pi }} \;({\text{m}})\quad \quad \\ \end{aligned} $$

where A = flow cross section (m²); D = column diameter (m); w = vapour flow velocity (m/s); and ρ _D = vapour density (kg/m³)

Example 1.5.3.1: Determination of the required column diameter

$$ G_{\text{V}} = 5000\,{\text{kg}}/{\text{h}}\quad F = 1. 2\quad \rho_{\text{V}} = 1. 4 5\,{\text{kg}}/{\text{m}}^{3} $$

$$ \begin{aligned} w & = \frac{F}{{\sqrt {\rho_{\text{V}} } }} = \frac{1.2}{{\sqrt {1.45} }} = 1\;{\text{m/s}} \\ A & = \frac{5000/3600}{{1.2*\sqrt {1.45} }} = 0.96\;{\text{m}}^{ 2} = \frac{5000/3600}{1*1.45} = 0.96\;{\text{m}}^{ 2} \\ D & = \sqrt {\frac{0.96*4}{\pi }} = 1.1\;{\text{m}} \\ \end{aligned} $$

1.6 Selection of Column Internals

In selecting the column internals the following points have to be considered: throughput capacity, pressure loss, number of separation stages, side draw facility, fouling dangers (for instance, by residue or tar).

The internals determine the required height of the column. For instance you may have a 13-m high column for 50 theoretical stages with gauze packing or 25-m high column for 50 theoretical stages with cross flow stages.

Design criteria

Random packed and structured packed columns:

• HTU value = packing height for a transfer unit (m packing)

• HETP value = packing height for a theoretical stage (m packing)

• Flooding factor and pressure drop

• Minimum irrigation rate

• Internals: support plates, liquid distributors, liquid collectors and redistributors, and gas distributors

Design information:

• Random packing: NT = 1.5–2 theoretical stages per metre of packing height

• Sheet packing: NT = 2–3 theoretical stages per metre of packing height

• Gauze packing: NT = 4–5 theoretical stages per metre of packing height

• Ratio of column diameter/packing diameter ≈ 10: 1

• Gas loading factor, $F = w*\sqrt {\rho_{\text{G}} } = 2{-} 2. 5$

• Liquid loading ∼4–80 m³/m²h

• Pressure drop ∼1–4 mbar/m

• A good liquid distribution is very important for to the overall effectiveness of the process.

Tray columns:

• NT = 1.6 theoretical stages per metre of column height for valve, sieve, tunnel, or bubble cap trays

• Allowable vapour velocities in view of entrainment and pressure drop

• Dimensioning of the downcomer area for the liquid

Design information:

• Gas loading factor, $F = w*\sqrt {\rho_{\text{G}} } = 1. 5{-} 2$.

• Gas velocity = 80% of w _max.

• Determination of the column diameter assuming an 80% active area for the vapour and a 20% active area for the liquid downcomer.

• Free hole area ∼8–10% of the cross sectional area.

• Weir height ∼30–60 mm.

• Weir overflow height ∼5–40 mm.

• Tray spacing ∼400–600 mm.

Cross-check calculations for weeping, entrainment, and flooding for all column cross sections with different vapour and liquid loadings should be carried out. In Chaps. 9 and 10 the fluid dynamic dimensioning is covered in detail.

1.7 Condensers [2]

The selection of an adequate condenser and the calculation of dew and bubble points as well as the condensation lines and the determination of the heat transfer, and overall heat transfer, coefficients for a given problem are dealt with in numerous examples found in the “Heat Exchanger Design Guide” [2].

The different condenser construction types are shown in Fig. 1.6:

Fig. 1.6

Condenser construction types

• Condensation horizontal in shells or in the tubes.

• Condensation vertical in shells or in the tubes.

1.8 Reboiler [2]

The different evaporator types are shown in Figs. 1.7 and 1.8. The advantages and disadvantages of the different construction types along with the designs with the calculations of the overall heat transfer coefficients are covered in the “Heat Exchanger Design Guide” [2]. Reboiler/evaporator types are:

Fig. 1.7

Thermosiphon and forced circulation evaporators

Fig. 1.8

Reboiler types for distillation plants

• Thermosiphon circulation reboiler, vertical or horizontal.

• Thermosiphon once-through reboiler, vertical or horizontal.

• Forced circulation and flash evaporator.

• Shell-and-tube and internal evaporators (heating coils).

• Falling film reboiler.

1.9 Vacuum Pumps [3]

First the required suction capacity is determined for a given problem and then a suitable vacuum pump, with a corresponding suction capacity at the required operating pressure, is selected.

In addition to the leak rate from leakages, non-condensable gases from reactions or degassings have also to be considered.

In the “Wärmetausch-Fibel II” [3] it is shown how required suction capacities for different problem definitions is determined in addition to what has to be considered in the selection of the different vacuum pumps, i.e.:

• Required suction capacity for the evacuation.

• Required suction capacity for inert gases and vapours.

• Determination of the leak rate from the equipment.

• Flow velocities and pressure drops in the vacuum lines.

• Unloading of the vacuum pumps by condensation.

• Optimal combination of different vacuum pumps.

Selection of vacuum pumps with advantages and disadvantages:

• Oil lubricated rotary disk pumps (problem: vapor condensation).

• Liquid ring pumps (problem: cavitation and suction capacity).

• Dry running vacuum pumps (problem: high temperature and explosion protection).

• Steam jet pumps (problem: waste water contamination).

1.10 Control Facilities [5]

The required process conditions in a column are determined by the composition of the feed mixture and the required specifications for the top and bottom product. In order to achieve the desired separation the column must operate under equilibrium conditions.

• The evaporator must steadily supply the required heating energy.

• The pressure in the column must be held constant.

• The condensation of the vapour must be correctly controlled.

• The flows must be fed and drawn steadily.

• Instruments for the following problem definitions are required:

• Feed flow control and control of product draws from the column: distillate, side streams and bottom products.

• Level control and heat supply control from steam or hot oil.

• Cooling of the condenser and the after coolers for the products.

• Column pressure control.

Often a special control has to be installed for the special separation problem. In the following text are some examples of the control facilities in distillation columns (Figs. 1.9, 1.10, 1.11, 1.12, 1.13, 1.14, 1.15, 1.16, 1.17, 1.18, 1.19, 1.20, 1.21, 1.22, 1.23, 1.24, 1.25 and 1.26).

Fig. 1.9

Conventional column control according to F.G. Shinskey (5)

Fig. 1.10

Suitable control for small bottom rates according to F.G. Shinskey (5)

Fig. 1.11

Feed control from tank

Fig. 1.12

Level control for bottom product

Fig. 1.13

Distillate control for constant feed

Fig. 1.14

Distillate control for variable feed

Fig. 1.15

Cascade control for feed

Fig. 1.16

Heat supply on level control

Fig. 1.17

Heat supply on flow control

Fig. 1.18

Heat supply on temperature control

Fig. 1.19

Stable heat flow by temperature and flow control

Fig. 1.20

Optimum temperature point for controlling heat supply

Fig. 1.21

Temperature controlled reflux

Fig. 1.22

Differential pressure control for heat supply

Fig. 1.23

Temperature controlled cooling water

Fig. 1.24

Pressure control by venting

Fig. 1.25

Pressure control with cooling water

Fig. 1.26

Pressure control by flooding condenser bundle

1.11 Heating Systems with Steam or Hot Oil [3]

The most important aspects in the selection of a heating system with steam- or liquid-heating fluids are covered in [3] and what has to be considered in the design and control of the heating system is also shown.

1.11.1 Heat Transfer Coefficients

Figure 1.27 shows that the heat transfer coefficients of condensing steam and hot water are substantially better than the heat transfer coefficients of organic heat transfer fluids.

Fig. 1.27

Heat transfer coefficients of condensing steam, hot water, and hot oil

1.11.2 Steam Heating [3]

Steam heating is preferentially used because it has great advantages:

• Extensive isothermal heating over the whole heating area.

• Very good heat transfer coefficients >6000 W/m²K.

• No large circulation rates as in heat transfer fluids.

Figure 1.28 shows the steam feed control, for fast control, whereas in Fig. 1.29 the control of the condensate drain is shown for a very wide working range [3].

Fig. 1.28

Steam feed control for steam heated equipment

Fig. 1.29

Condensate drain control for steam heated equipment

1.11.3 Heating with Hot Oil [3]

An organic heat transfer agent has the advantage that the heating system and the heated equipment only needs to be designed for low pressures even at high temperatures above 200 °C.

Disadvantages:

• Poor heat transfer coefficients in comparison to steam or hot water.

• No isothermal heating.

• Large heat exchange areas are required.

• Large heat transfer fluid rates must be moved by pumping.

The possible heating systems are shown in Figs. 1.30 and 1.31.

Fig. 1.30

Primary heating circuit for hot oil

Fig. 1.31

Hot oil heating with secondary circulations

In the primary heating circulation, according to Fig. 1.30, all heat exchangers are fed with the highest oil temperature exiting the oil pipe still. Therefore, thermal damage of the product can occur.

The heating loads are influenced by the heating demands of other consumers and by the variations in the heater. By throttling the hot oil rate the flow velocity is reduced and hence so is the overall heat transfer coefficient.

Heating with secondary circulations is shown in Fig. 1.31 and is considered much better.

1.12 Cooling Systems [3]

1.12.1 Cooling Water Circulation Systems

Figure 1.32 shows the different possible systems for cooling the water: river water cooling, air cooling, as well as open and closed cooling towers.

Fig. 1.32

Cooling water systems

1.12.2 Comparison Between a Cooling Tower and an Air Cooler

Advantages of the cooling tower

• Lower cooling water temperatures than with air coolers because the wet-bulb temperature and not the dry-air temperature determines the possible cooling.

• Due to the constant wet-bulb temperature there are only low variations in the cooling water flow temperature.

• Lower investment costs.

Disadvantages of the cooling tower

• Treatment and fresh water costs leading to higher operating costs.

• Enrichment of salts and of air in the cooling water leading to corrosion and precipitation.

• Salt precipitation makes this method unsuitable for higher temperatures >50 °C.

• Fog and ice formation at cold temperatures.

Advantages of the air cooler

• No problem with corrosion, salt precipitation, biological fouling, and freezing.

• No additional water and water treatment needed – inhibiting costs.

• No problems with product contamination.

Disadvantages of air coolers

• Strong dependence on the air temperature leading to high water exit temperatures in the summer.

• Overdimensioned design for hot days in summer leading to overcooling in the winter.

• Maldistribution by wind and fouling.

The main problem with air coolers is their dependency on the ambient temperature which can be in the range 35–40 °C for summer daytime. This can be resolved by spraying water which leads to evaporation and thus a reduction in the air temperature. For example, 35 °C hot air with a 40% relative humidity can be cooled to 27 °C using this technique.

The following points must be considered thereby

• The danger of salt or calcium precipitation from salty water exists. It is recommended to use condensate during the hottest summer days.

• The indosed water droplets must vaporize before the heat exchanger in order to affect air cooling. The residence time must therefore be sufficient.

• Salty droplets must not hit and evaporate on the finned tubes of the heat exchanger.

1.12.3 Cooling Water by Evaporating a Refrigerant or Adiabatic Evaporation

If very low water temperatures are required a refrigeration unit must be used. The flow chart in Fig. 1.33 illustrates this mode of operation [3].

Fig. 1.33

Chilled water circulation cooled by a refrigeration unit

Another method is adiabatic water evaporation in the vacuum. In Fig. 1.34 the cooling time at different suction capacities of the vacuum pump is shown.

Fig. 1.34

Cooling times for 1 tonne of water at different suction capacities, S

1.12.4 Direct Condensation in Columns [6]

If the vapour to be condensed contains a high melting point product, for instance Naphthalene with a melting point of 80 °C, these products crystallize out in a water-cooled condenser and create a blockage. In such cases it makes more sense to bring the vapour into direct contact with a suitable lean oil, for instance tar oil, in a packed column for direct condensation.

Another example is the direct condensation of solvents from exhaust air. The heat transfer coefficient in the condensing solvent vapour, containing inert gas, is poor and the undesired fog formation results in heavy cooling.

In such cases a cold wash is adopted. The exhaust air stream is washed with a cold solvent. For instance exhaust air containing methanol is washed with −20 °C cold liquid methanol or on the other hand exhaust air containing gasoline with −30 °C cold liquid gasoline.

The flow sheet of a “cold wash with its own juice” is shown in Fig. 1.35.

Fig. 1.35

Cold wash for solvent-rich exhaust air