Page 260 - Elements of Chemical Reaction Engineering 3rd Edition

232 Collection and Analysis of Rate Data Chap. 5

"i

0.2

0
t (mid
Figure E5-1.1 Graphical differentiation.

Finite Dgference. Next we calculate dPldt from finite difference formulas (5-8)
through (5-10):

- 3Po + 4Pl- P2 - - 3(7.5) + 4( 10.5) - 12.5 - 1.40
-
-
t=O: E) = 2At 2(2.5)
0
f'z-p, - 12.5 -7.5 - l.o
Calculating t=2.5: E) ==---- 2(2.5)
1
t = 5 : Here we have a change in time increments At, betweal
P, and P, and between P, and P,. Consequently, we
have two choices for evaluating (dPldt),:
p3 - Po - 15.8 - 7.5 = o.83
2(5)
- 3P, 4p3 - P4 - - - 3( 12.5) + 4( 15.8) - 17.9 - - o,78
(b) b) = 2(At) 2(5)
2
17.9 - 12.5 = o,54
t = 10: b] = 2(5)
3
19.4 - 15.8 = o.36
t = 15: @) = 2(5)
I t=20: k) P,f4P4-3P5 - 15.8-4(17.9)+3(19.4) = o.24
4
-
2(5)
2At
=
5