Section 3
Theory of Operation
3-5
two cells are identical, then for a given gas concentration, the transmittance
will be the same whether it is computed against zero between the cells or
within either cell.
Now, suppose the reference cell contains CO
2
mole fraction C
r
. If we can
compute what the transmittance would be if C
r
were in the sample cell
against zero in the reference cell, then we will know what the transmittance
actually is in the reference cell when it contains C
r
, and that will tell us how
to correct for the gain change.
The strategy for accomplishing this is straightforward. First, we use C
r
to
solve equation 3-4 for V
r
, giving
V
F
C
T
T
P
P
r
r
o
o
=
−
1
3-5
Then, rearrange equation 3-2b to find
τ
r
given V
r,
τ
r
r
V
K
= −
1
3-6
Equations 3-5 and 3-6 ordinarily apply to the situation where C
r
is in the
sample cell and zero is in the reference cell; but if the cells are identical,
τ
r
also computes the transmittance in the reference cell when C
r
is present. The
gain correction G equals
τ
r
, as shown earlier, so
G
V
K
r
= −
1
3-7
After correcting the differential voltage V for the gain change we can write
GV = V
s
- V
r
. Solving for V
s
and substituting the result into equation 3-4
gives the general expression for computing the gas concentration C in the
sample cell given concentration C
r
in the reference cell,
C
F VG V
P
P
T
T
r
o
o
=
+
(
)
3-8
The differential CO
2
mole fraction
∆
C is simply (C - C
r
).
∆
C
F VG V
P
P
T
T
C
r
o
o
r
=
+
(
)
−
3-9
