Week 5 - Exercise

Week 5 - Exercise#

Problem Statement#

A protein mixture has been separated through an ion-exchange chromatography process. The residence time of the column in non-adsorbing conditons is \(t_0=1\) min. The chromatogram resulting from the separation is reported in the following:

Drawing

Determine:#

  • How many components in the mixture can you identify from the chromatogram?

  • What is the selectivity of this column for each couple of components?

  • Assuming a linear adsorption isotherm holds in all cases, what conclusions can you draw about adsorption thermodynamics of each component in the column?

  • What is the number of theoretical plates associated to the elution of each component you have identified? If the coulmn is 10 cm long, what is the height equivalent of a theoretical plate for each component of the mixture?

  • Compute the resolution relative to each couple of components in the chromatogram. Does it depend only on the adsorption thermodynamics?

  • Comment on the separation performances of the column. Does it allow for a sharp resolution of all components?

Solution#

  1. The chromatogram clearly indicates three components, let us indicate them with A, B, and C. The retention times for these components are:

  • \(t_{r,A}=7\)

  • \(t_{r,B}=10\)

  • \(t_{r,C}=15\)

  1. The selectivity can be computed for every pair of components as:

\[ S_{i,j}=\frac{t_{r,i}-t_0}{t_{r,j}-t_0} \]
import numpy as np

tr=np.array([7,10,15]); 
t0=1
SEL=np.zeros((np.size(tr),np.size(tr)));

for i in range(0,np.size(tr)):
    for j in range(0,np.size(tr)):
        SEL[i,j]=(tr[i]-t0)/(tr[j]-t0)

print('Selectivity i,j\n',SEL)

Selectivity i,j
 [[1.         0.66666667 0.42857143]
 [1.5        1.         0.64285714]
 [2.33333333 1.55555556 1.        ]]

  1. Since \(S_{i,j}=H_i/H_j\) we can determine the relative affinity of components A, B and C for the startionary phase. Component C shows the highest affinity for the stationary phase, followed by B and then C.

4/5. The number of theoretical plates associated to the elution of components A, B, and C can be obtained applying the relationship \(N=16*(tr/tw)^2\). The resolution for every couple of components is computed as: \(R_{i,j}=\frac{t_{r,i}-t_{r,j}}{\frac{1}{2}(t_{w,i}+t_{w,j})}\)

import numpy as np

tr=np.array([7,10,15]); 
tw=np.array([1,2,1.5]); 
L=10 #cm

N=(16*(tr/tw)**2)

H=L/N

R=np.zeros((np.size(tr),np.size(tr)));

for i in range(0,np.size(tr)):
    for j in range(0,np.size(tr)):
        R[i,j]=abs((tr[i]-tr[j])/(2*(tw[i]+tw[j])))


print('\nHeight of a theoretical plates ',H,' [-]')
print('Number of theoretical plates ',H,' cm')
print('Resolution i,j\n',R)

Height of a theoretical plates  [0.0127551 0.025     0.00625  ]  [-]
Number of theoretical plates  [0.0127551 0.025     0.00625  ]  cm
Resolution i,j
 [[0.         0.5        1.6       ]
 [0.5        0.         0.71428571]
 [1.6        0.71428571 0.        ]]
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