Environmental Engineering Reference
In-Depth Information
GC, an efficient packed column should have several thousand theoretical plates, and
an efficient capillary column should have more than 10,000 theoretical plates.
Since the ratio in Eq. 10.6 compares the retention time with the peak width, one
can see that N is also a measure of the band broadening per unit time. The narrower
the peak (w) the greater the number of plates (N). The plate number can then be used
to calculate the height (H) of a theoretical plate using column length (L).
H ¼ L
=
N
ð10
:
7Þ
A good column will have a large N and small H. Note that the theoretical plate
model, adapted from distillation, has no physical significance, but is universally used
to explain the separation in the column. One final equation relates resolution (R)to
three dependent variables, that is, number of plates (N), separation factor (a), and the
retention factor (k):
k
B
1þk
ðwhere k ¼ðk
A
þk
B
Þ=
N
p
1
1
4
1
a
R ¼
2Þ
ð10
:
8Þ
From Eq. 10.8, it is important to realize that the resolution can be optimized with
combined values of a, N, and k. The separation factor (a) has the largest effect. If a
separation needs to be improved, it is well worth the effort of increasing a although it
is impossible to give a general proposal on how to do this. If the plate number (N)is
increased, the effect is only by the factor
p
. This means that if the column length is
doubled, the resolution will improve only
p
¼ 1
4. Increasing the retention factor
(k) only has a notable influence on resolution if k is small to start with.
:
EXAMPLE 10.1.
Two chemicals (A and B) are separated on a GC capillary column with
retention times of 14.6 and 18.3 min, and peak widths (at base) 0.80 and 0.93 min for A and
B, respectively. An unretained air peak occurs at 1.1 min. Calculate: (a) retention factor, (b)
separation factor, (c) resolution, (d) average plate number.
SOLUTION:
By substituting the given values of t
m
(1.1 min),
t
rA
(14.6 min), and t
rB
(18.3 min) into Eq. 10.3 to 10.6, we have the following:
t
rA
t
m
t
m
¼
14
:
6
1
:
1
1:1
¼ 12:3;
t
rB
t
m
t
m
¼
18
:
3
1
:
1
1:1
¼ 15:6
ðaÞ k
A
¼
and
k
B
¼
k
B
k
A
¼
t
rB
t
m
t
rA
t
m
¼
15
:
6
12:3
¼
18
:
3
1
:
1
14:61:1
¼ 1:3
ðbÞ a ¼
t
rB
t
rA
ðw
A
þw
B
Þ=2
¼
18:314:6
ð0:80þ0:93Þ=2
¼ 4:35
ðcÞ R ¼
2
¼ 16
2
2
¼ 16
2
t
rA
w
A
14
:
6
0:8
t
rB
w
B
18
:
3
0:93
ðdÞ N
A
¼ 16
¼ 5329;
and N
B
¼ 16
¼ 6195
NðaverageÞ¼ð5329þ6195Þ=2 ¼ 5762:
Recall that a good range of k is in the range of 2-10. The k values for both compounds in this
example exceed this range, implying improvement is needed by making both chemicals less
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