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Chapters 8, 9 and 18
Submitted by:Chua, PaulineClimaco, Mary AnnZablan, Bianca
Problem 8.3Problem 8.21Problem 18.5Problem 18.23
Problem 8.3
Pure A (CAo=100) is fed to a mixed flow reactor, R and S are formed, and the following outlet concentrations are rerecorded. Find a kinetic scheme to fit this data.Run CA CR CS
1 75 15 102 25 45 30
Problem 8.3
Assuming A→R→S. In Figure 8.14
Results don’t fit the kinetics
Problem 8.3
Assuming A→R; A→S -rR=k1CA -rS=k2CA
For mixed flow: τ1=(CR-CRo)/k1CA
τ2=(CR-CRo)/k2CA
Dividing, CR/Cs = k1/k2
Therefore the assumed scheme is correct with k1=1.5k2
Run CA CR CS CR/CS
1 75 15 10 1.5
2 25 45 30 1.5
Problem 8.21
Chemical A reacts to form R (k1=6 hr-
1) and R reacts away to form S (k2=3 hr-1). In addition, R slowly decomposes to form T (k3= 1 hr-1). If a solution containing 1.0 mol/L of A is introduced into a batch reactor, how long would it take to reach CR
max, and what would be CR max?
Problem 8.21
Given:
k1=6 hr-1
k2=3 hr-1
k3=1 hr-1
From the units of k, it can be concluded that all reactions are 1st order
CAo=1.0 mol/L
Problem 8.21
1. rA=-k1CA
2. rR=k1CA-k2CR-k3CR
3. rS=k2CR
4. rT=k3CR
By equation 2
rR=k1CA-(k2+k3)CR = k1CA-k23CR =dCR/dt and CA=CAoe-k1t
Problem 8.21
dCR/dt = k1CAoe-k1t-k23CR
dCR/dt + k23CR = k1CAoe-k1t
dCR/dt + P(t)y = Q(t)
Problem 8.21
To get the maximized CR
Problem 8.21
t=12.16395 min (time to reach CRmax )
Problem 8.21
To get CRmax
CRmax=0.4444 mol/L
Problem 18.5
A gas containing A (2 mol/m3) is fed (1 m3/hr) to a plug flow reactor with a recycle loop (0.02 m3 loop volume, 3 kg of catalyst), and the output composition from the reactor system is measured (0.5 mol A/m3). Find the rate equation for the decomposition of A for the following cases. Be sure to give the units of –rA’, CA, and k’ in your final expressions:
Very large recycle A→3R, n=1, 50% A-50% inerts in feed
Problem 18.5
A→3R, n=1 (1st order) ε=[1+3-(1+1)]/(1+1)=1
Mixed Flow Reactor
Thus,
Problem 18.23
The first-order decomposition of A is run in an experimental mixed flow reactor. Find the role played by pore diffusion in these runs; in effect, determine whether the runs were made under diffusion free, strong resistance, or intermediate conditions.
dP W CAo ν XA
3 1 100 9 0.412 4 300 8 0.6
Problem 18.23
Run
dp
W CAo
ν XA
CA k’
1 3 1 100 9 0.4
60 6
2 12 4 300 8 0.6
120
3
For no diffusion resistance: k2’/k1’ = 1
For strong resistance k2’/k1’ = dp1/dp2 = 3/12 =
1/4 From the data
k2’/k1’ =3/6 = ½ Thus, the runs were
made in intermediate conditions between strong resistance and no diffusion resistance.
For mixed flow