HPLC HINTS & TIPS for Chromatographers

Tip# 108: HPLC to UHPLC Conversion Notes:

The use of ultra-high performance liquid chromatography (UHPLC) columns to reduce analysis times and sometimes improve detection limits is a hot topic. UHPLC presents a number of new issues. The incorporation of smaller 1.9 to 3.0 micron particles and smaller frits will raise backpressures and increase system wear and tear. Smaller diameter lines are often used (I.D. of 0.12mm or less) which can increase blockages and clogs if you do not filter your mobile phase and samples through a 0.45 or 0.2 micron filters. Piston seals and valve rotors can wear out early due to the very high pressures, heating and stress imposed on them. You should monitor your HPLC system carefully over time and consider increasing the frequency of preventative maintenance and inspection services as well.

I must answer twenty or so questions each week in the area of UHPLC.The most common questions deal with selection of an UHPLC column and making adjustments for the changes in: (1) Column Dimensions; (2) Flow Rate and (3) Injection Volume. The good news is that all of these questions can be answered with some basic math.

(1)      COLUMN DIMENSIONS: Let's start by making things as simple and brief as possible (this is supposed to be a "hint & tip", not a thirty page article). When initially converting from a conventional HPLC column (5 micron) to an UHPLC column (1.9 to 3 micron particle) select a column with the same I.D. and length. This way only the particle size changes. *I like to change one variable at a time. If you would like to change the column length to take advantage of some of the increased efficiency (and decreased pressure!) which results from smaller particles, then please refer to the following equation.

• 'Lc1' = Length of Column #1 in mm; 'Lc2' = Length of Column #2 in mm; 'p1' = particle size of Column #1 in microns; 'p2' = particle size of Column #2 in microns.

EQUATION A:    'Lc2' = ('Lc1' * 'p2') / 'p1'

Example: Column # 1 is a standard HPLC column;  4.6 mm x 250 mm (5u) Column. You want to find out the length of an equivalent column which uses 1.9 micron particles instead of the 5 micron particles.

'Lc2' = (250 * 1.9) / 5 ; Answer is: 'Lc2' = 95 mm. *So a 10 cm long column would be a good choice here.

(2)    FLOW RATE: Flow rate is directly proportional to column diameter and as we saw above in Equation A, the particle size can also affect it too. If you keep the column length and internal diameter the same, then the linear flow will be unchanged with the same particle size. A change to the particle size alone will change the flow rate as follows: 'Fc2' = 'Fc1' x ('p1'/'p2').

A change to a smaller diameter column to compensate for the improved efficiency will require a change to the original flow rate to preserve the linear velocity. Please refer to the following equation.

• 'Fc1' = Flow Rate of Column #1 in ml/min; 'Fc2' = Flow Rate of Column #2 in ml/min; 'd1' = Column #1 Diameter in mm; 'd2' = Column #2 Diameter in mm.

EQUATION B:     'Fc2' = ('d2' / 'd1')^2 * 'Fc1'

Example: Column # 1 is a standard 4.6 mm ID Column. You want to find out what the linear flow rate should be if you use a smaller diameter column (2.1mm in this example).

'Fc2' = (2.1/4.6)^2 * 1.000 ; Answer is: 'Fc2' = 0.208 ml/min. *A flow rate of 200 ul/min would be fine.

(3) INJECTION VOLUME: A change in the column dimension may require a change to the injection volume. The smaller the internal volume of the column, the smaller the injection volume. To calculate the linear change in volume, please refer to the following equation.

• 'V1' = Injection Volume #1 in ul; 'V2' = Injection Volume #2 in ul; 'L1' = Column #1 Length in mm; 'L2' = Column #2 Length in mm; 'd1' = Column #1 Diameter in mm; 'd2' = Column #2 Diameter in mm.

EQUATION C:     'V2' = 'V1' * {('d2'^2 * 'L2') / ('d1'^2 * 'L1')}

Example: Current injection volume is 10 ul. Column # 1 is a standard 4.6 mm ID x 250 mm Column. You want to find out what the equivalent injection volume should be for a 2.1 mm ID x 150 mm column.

'V2' = 10 * (2.1^2 * 150) / (4.6^2 * 250) ; Answer is: 'V2' = 1.25 ul.

> Bill Letter, 11/22/10