MODELING CHAR OXIDATION AS A FUNCTION OF PRESSURE ...

2
r obs ′ − [kD Pog + k d + kD kd / ka ] r obs ′ ′ + ( kd )(kDP og ) = 0 (5.38)
2
or r obs ′ − P r obs ′ + Q = 0 (5.39)
where P = (k DP og + k d +k Dk d/k a) and Q = ( k d)(k DP og). The solution of this quadratic
equation is
r obs ′ = (1/2)P[1 − (1− 4Q / P 2 ) 1/2 ] (5.40)
In order to obtain a simple form, Essenhigh went a step further and approximated the
above equation by
r obs ′ = Q/ P (5.41)
Substituting Q and P back into Eq. (5.41) gives
1
′ ′
r obs
=
1
k a P og
+ 1
k d
+ 1
k D P og
This is the "Extended Resistance Equation" (ERE) developed by Essenhigh (1988).
Comparing Eq. 5.41 with Eq. 5.40 leads to
61
(5.42)
(1-4Q/P 2 ) 1/2 = (1- 2Q/P 2 ) (5.43)
This is to say, the above relation was assumed by Essenhigh in simplifying Eq. 5.40 into
Eq. 5.41. Eq. 5.43 can be re-written as:
(1-4Q/P 2 ) = (1- 2Q/P 2 ) 2
Let x = 2Q/P 2 , and the above equation becomes
(5.44)
(1-2x) = (1- 2x + x 2 ) (5.45)
The above equation (and hence the ERE) is valid only if
2Q/P 2 = x