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Unformatted text preview: elations that can be used to pin down the unknown fluxes:
n B A (20) k cA ∆ A However, we now have introduced a new unknown cA ∆ . In the next section we show how this information can be used to determine the concentration profile in the gas layer. Analysis of the Diffusion Layer
It is instructive first to consider a limiting case when cA ∆
from (11) to get
AB C cA 0 . First, we use (14) to eliminate B xA
(21) A 1 xA n 1 z Evaluating this expression at the surface of the catalyst gives
AB C xA 1 xA n 1 z k cA ∆ z∆ (22) We would now like to derive an estimate that will give us insight when cA 0
cA ∆ . The way to
proceed is to obtain order of magnitude estimates for the terms in (22). For example
xA cA z C z cA ∆ cA 0 (23) ∆ where (· designates the order of magnitude. Thus we can approximate the LHS of (22) as
)
AB C 1 xA n 1 xA
z z∆ 1 xA ∆ cA ∆ cA 0 AB n1 ∆ (24) 4 IntroReactionDiffusion.nb Substituting (23) into (22) and solving for cA 0 we get
cA 0 cA ∆ 1 (25) where
k∆
1 xA ∆ (26) n1 AB We can now ask the question: When will cA 0
cA ∆ ? It follows from (24) that cA 0
1. If we inspect we see that the requirement for
1 means that cA ∆ when k∆
(27) 1
AB or when
k ∆ diffusional time scale AB Φ2 ∆2 reaction time scale (28) 1 , with Φ The quantity Φ is in a form of a Thiele modulus. Thus if Φ >>1, the reaction time scale ∆/k is much
smaller than the diffusive time scale AB ∆2 . That is, the rate of reaction is much faster than the rate at which mass is transferred by diffusion. This means that the concentration of A at the catalytic surface
will be consumed by the chemical reaction and approach zero:
cA ∆ 0 when Φ2 (29) Sometimes in the literature or reaction engineering textbooks, this condition is phrased as " the reaction
is instantaneous or rapid ", without further qualification. Concentration Profile when Φ
In this section we will solve for the concentration profile when Φ
cA ∆ 0. Thus the boundary value problem is
cA AB z 1 BC1 : cA 0 0 c0
A BC2 : cA ∆ cA
C so that we can assume that z 0 n1 (30) To find the general solution of (29) we integrate once to get
cA AB 1 cA
C K1 (31) z n1 Now this equation can be written as
Log 1
z cA n1
n1 (32) K1 C C AB Integrating a second time gives
Log 1 cA
C n1
n1 K1 z K2
C AB (33) IntroReactionDiffusion.nb 5 It follows from BC1 that
K2 c0
A Log 1 (34) n1 C
and f...
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 Winter '14

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