Distillation Tutorials

Example 1:  Phase Diagram & Equilibrium Curve

Use Raoult's Law and calculate the vapour and liquid compositions in equilibrium (in mole fractions, y and x) for the benzene-toluene system using vapour pressure data measure at a pressure of 101.32 kPa as shown in Table E-1 below 

Table E-1. Temperature-Vapour Pressure Data for Benzene-Toluene at 101.32 kPa

Next plot the phase diagram (T-x-y diagram) and equilibrium curve for the benzene-toluene system at 101.32 kPa

Solution to Example 1

Write Raoult's Law for a binary mixture of A and B;

pA = Pvp,A xA ; pB = Pvp,B xB = Pvp,B (1 - xA)

The total pressure, PT = pA + pB

Replacing for the partial pressures and re-arrange, we have:


PT= PVP,A * XA + (PVP,B - PVP,B * XA)
PT= (PVP,A - PVP,B) * XA + PVP,B


Re-arrange, we have the expression for XA:  XA= (PT - PVP,B)/(PVP,A - PVP,B)


Since PVP,A * XA = YA * PT; we have the expression for YA: 
YA= (PVP,A/PT) * XA

The VLE data can be calculated at each temperature by substituting for total pressure (101.32 kPa) and the appropriate vapour pressures. For example, at 85.0 ^oC,


XA = (101.32 - 46.0)/(116.9 - 46.0) = 0.7803
YA = (116.9/101.32) (0.7803) = 0.9002

We have the following results:

The constant pressure phase diagram can be plotted using the above T-x-y values, and the equilibrium curve plotted using x-y values.

Example 2: Simple Distillation

A 100-g-mole liquid mixture containing 50 mole% n-heptane and 50 mole% n-octane at 30 ^oC is to be subjected to a differential distillation at atmospheric pressure.

The process is stopped when the mole fraction of MVC in the still reaches 0.33.

Determine the amount of composited distillate collected and its mole fraction, (i.e. y_ave).

Equilibrium data for n-heptane and n-octane is given in Table E-2 below (in mole fractions n-heptane in vapour and liquid):

Solution to Example 2

This is the problem of determining L₂ from known values (either given in the problem statement or can be calculated from the information given) of L₁ , x₁ and x₂ .

In this case, L₁ = 100 g-moles, x₁ = 0.50 (n-heptane, the MVC) and x₂ = 0.33

Plot the curve using the equilibrium data given: (see Note)

The RHS = the area under the curve 1/(y-x) bounded between x₁ and x₂ is 0.916. Thus, the LHS = ln (100/L₂) = 0.916.

Solving for L₂:

L₂ = 40.0 g-mole

Average composition in the vapour distilled:

Note: These data are obtained from equilibrium curve as shown below. The equilibrium curve is plotted using the given VLE data.

Example 3: Flash Distillation

An equimolar mixture of benzene and toluene is subjected to flash distillation at a pressure of 1 bar in the separator. Determine the compositions (in mole fraction benzene) of the liquid and vapour leaving the separator when the feed is 25% vaporized.

Estimate the temperature in the separator.

Equilibrium data for benzene-toluene system at 1 bar is given in Table E-3 below:

What are the concentrations in the vapour and liquid, and the separator temperature the feed is:

· 0% vapourised 
· 50% vapourised
· 75% vapourised
· 100% vapourised ?

Plot the operating lines for each of these cases and discuss how the operating lines change as f changes.

Solution to Example 3

The operating line equation is as follows, with f = 0.25 (25% of feed is vapourised) and xF = 0.50 (equimolar mixture, MVC = benzene)

Y = -{(1-f)/f} X + (XF/f)

Locate the first point ( x = x_F = 0.50, y = x_F = 0.50 ) on the 45^o diagonal.

Locate the second point using the operating line equation.

Plot the operating line by joining the 2 points. Intersection between the operating line and equilibrium curve gives the solution for x_B and y_D.

From graph - this is shown as case (b) - we have:

x_B = 0.44, y_D = 0.66

The separator temperature can be estimate by interpolation or determined from the equilibrium phase diagram. As shown in the Figure, this is approximately 92.4 ^oC.

The same calculations can be repeated for the other scenarios and the results are tabulated in the following Table:

We can see from the results that there is a trade-off between quantity and concentration: the higher the fraction of feed vapourised (i.e. the larger the vapour product), the lower the mole fraction MVC in the vapour.

Example 4: Continuous Distillation

A distillation column operating at 1 atm is to be designed for separating an ethanol-water mixture. The feed is 20 mole% ethanol and the feed flow rate is 1000 kg-mole/hr of saturated liquid. A distillate composition of 80 mole% ethanol and a bottoms composition of not more than 2 mole% ethanol are desired. The reflux ratio is 5/3.

Determine:
(a) the total number of equilibrium stages required 
(b) the optimum feed plate location
(c) the distillate and bottoms flow rates in kg-mole/hr

Equilibrium data for ethanol-water system at 1 atm pressure are in Table E-4 given below:

Analyse the graphical solution. What can you conclude about the extent of separation in the rectifying section and stripping section respectively?

Obtain the mole fractions in liquid and vapour for each tray and plot a concentration profile for your design (vertical-axis: number of trays, top-down, horizontal-axis: mole fractions x and y).

Solution to Tutorial 4

Given:

F = 1000 kg-mole/hr, x_F = 0.20 (ethanol is MVC)
Feed is saturated liquid, thus q = 1.0
x_D = 0.80, x_B = 0.02 (maximum); R = 5/3

First, plot the equilibrium curve using the VLE data given. Note that you need to convert mole% ethanol into mole fraction. Then, apply the McCabe-Thiele method to find the number of theoretical (equilibrium) trays required for the separation.

Step 1:

Since R and xD are known (5/3, and 0.80 respectively), plot the ROL equation - it passes through x_D on the 45^o diagonal, i.e. the point (0.80, 0.80).

The intercept is 0.80 / (5/3+1) = 0.30.

Y = -{R/(R+1)}X + (XD/(R+1))

Step 2:

Plot the feed line using the q-line equation with q = 1.0 and x_F = 0.20 - it passes through x_F on the 45^o diagonal.

Y = -{q/(1-q)}X+(1/(1-q))XF

In this case, with q = 1.0, the q-line is a vertical line at x_F = 0.20. Then locate its intersection with ROL.

Step 3:

Plot the SOL, by joining the point x_B = 0.02 on the 45^o diagonal to the intersection point between the ROL and q-line.

Step 4:

Draw triangles starting from x_D (between ROL and equilibrium curve) down to x_B (between SOL and equilibrium curve). The change from ROL to SOL occurs when x_F is reached.

From the graph plotted, number of triangles = 15. Thus, number of equilibrium stages required = 14 + 1 reboiler. Optimum feed plate location = stage #13.

The distillate and bottoms flowrates (D and B) can be obtained using material balances:

Total: F = D + B ; thus, B = 1000 - D

MVC: F x_F = D x_D + B x_B

Substituting for B gives: ( 1000 ) ( 0.20 ) = ( D ) ( 0.80 ) + ( 1000 - D ) ( 0.02 )

Solving gives: D = 230.77 kg-mole/hr

And B = 769.23 kg-mole/hr

Analysis of Design:

It is noted that the Rectifying Section has more trays (12) than the Stripping Section (3, including the feed tray but excluding the reboiler). This is because the relative volatility of ethanol decreases sharply with increase in ethanol concentration.

The relative volatility is very high initially, resulting in fewer trays being required for separation in the Stripping Section. As concentration of ethanol (mole fraction) increases, the relative volatility drops quickly, and becomes smaller as the concentration moves toward azeotropic concentration (at 0.894 mole fraction ethanol). The distance between the ROL and equilibrium curve becomes smaller, signifying a drop in concentration driving force for mass transfer, resulting in more trays being required in the Rectifying Section.

Example 5: Continuous Distillation

For the same separation of Example 4, i.e. a saturated liquid feed of 20 mole% ethanol to be separated into a distillate of 80 mole% ethanol and a bottoms of not more than 2 mole% ethanol, determine the number of theoretical trays required if the reflux ratio used is 11/3.

What can you conclude regarding the relationship between R and N?

Solution to Example 5