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Chapter 8
8-1. To use CRW, you must first specify a mixture name – in this case “Homework 6 2018” and
then searching for the chemicals in the database and adding them to the mixture. There can
be some confusion on what exact chemicals to use from the database. In this case, the
Once the mixture has been populated, we then click the Compatibility Chart button at the
top of the program and our compatibility chart is displayed. The result is shown below.
We have 10 possible binary combinations with these 5 chemicals. Five combinations are
incompatible and 4 have cautions. Only one combination is compatible.
2
PREDICTED HAZARDS REPORT
--------------------------------------------------
Chemicals and Reactive Groups in this Mixture:
--------------------------------------------------
ACETIC ANHYDRIDE mixed with itself -
INTRINSIC REACTIVE HAZARDS:
CUMENE HYDROPEROXIDE mixed with ACETIC ANHYDRIDE -
PREDICTED HAZARDS:
CUMENE HYDROPEROXIDE mixed with itself -
INTRINSIC REACTIVE HAZARDS:
METHANOL mixed with ACETIC ANHYDRIDE -
PREDICTED HAZARDS:
METHANOL mixed with CUMENE HYDROPEROXIDE -
PREDICTED HAZARDS:
3
METHANOL mixed with itself -
INTRINSIC REACTIVE HAZARDS:
SULFURIC ACID mixed with ACETIC ANHYDRIDE -
PREDICTED HAZARDS:
POTENTIAL GASES:
SULFURIC ACID mixed with CUMENE HYDROPEROXIDE -
PREDICTED HAZARDS:
SULFURIC ACID mixed with METHANOL -
PREDICTED HAZARDS:
Nitrogen Oxides
SULFURIC ACID mixed with itself -
INTRINSIC REACTIVE HAZARDS:
WATER mixed with ACETIC ANHYDRIDE -
PREDICTED HAZARDS:
WATER mixed with CUMENE HYDROPEROXIDE -
PREDICTED HAZARDS:
--- END OF HAZARDS FOR THIS MIXTURE PAIR ---
WATER mixed with METHANOL -
PREDICTED HAZARDS:
WATER mixed with SULFURIC ACID -
PREDICTED HAZARDS:
WATER mixed with itself -
INTRINSIC REACTIVE HAZARDS:
1. For the binary combinations that are not compatible, special precautions must be done
to ensure these chemicals are not stored together nor accidently mixed. These
combinations are:
2. For the cautions, we need to be aware of these potential reaction hazards. Note that
several of the cautions are predicted to result in an exothermic reaction:
8-2. It can be easily shown that:
0
/ 1 /
Bx Bx T T T
. Thus, the solution here follows
the solution of Example 8-4 and Figure 8-12 in the textbook.
0
6
0 0
0 0
0
5 1
8 1
exp /
exp
exp /
4.54 10 sec
exp 30
4.85 10 sec
a g
a g
k T A E R T
k T k T
AE R T
e. Equation 8-21 provides the location of the maximum rate:
12 4 1
0.8211
The actual temperature is computed using Equation 8-14:
0
0
0.8211
432 K
m
F
T T
xT T
T
8-3. a. The total energy in the liquid is:
0
P
Q mC T T
dQ dT
7
8-4. Start with Equation 8-20:
n
dx
d
Differentiate with respect to
:
1
n n
d dx
Set equal to 0 and solve for x. Divide first by
exp
Bx
:
1
n n
8-5. a. Substituting the numbers provided into Equations 8-17 and 8-18:
612 K 350 K 0.75
350 K
29.1 kJ/mol
10.00
0.008314 kJ/mol K 350 K
F o
o
a
g o
T T
BT
E
R T
b. From Equation 8-21:
8
12 4 1
2
1
12 10 10 4 1.75
2 1 0.75
0.667 1.038
0.693
m
m
x n n B
nB
x
This can be converted to a temperature using Equation 8-14:
350 K
0.693
612 K 350 K
531 K
o
m
F o
T T T
xT T
T
The maximum temperature rate is provided by Equation 8-19:
1
1 exp 1
10 0.75 0.693
n
dx Bx
x
d Bx
3 1
9.08 10 sec
Substituting into Equation 8-26:
o 3 1
22.3 K/s
F o o
dT dx
T T k T
dt d
c. We can use Figure 8-2 to determine the time. For
10
and B = 0.75 we get
0.205
m
. We can convert to actual time using Equation 8-25 with n = 1.
3 1
0.205
9.08 10 sec
22.6 seconds
m
o
tk T
So after the reaction is detected at the onset temperature, there are only 22.6 seconds
until the maximum rate is reached! This is not very much time.
9
8-6. The data are shown below. Since the data points are not equally spaced, we need to use
the Lagrange derivative method to calculate the slope at each point.
Given the data points:
and we wish to compute the derivative at any arbitrary point x where
0 2
x x x
, then
the derivative at x is given by:
1 2 0 2 0 1
0 1 2
0 1 0 2 1 0 1 2 2 0 2 1
2 2 2
x x x x x x x x x
a y b y c y
x x x x x x x x x x x x
Then
dy
dx
To calculate the derivative at the middle point then
1
x x
and the formulas are
simplified somewhat.
1 2 1 0 2 1 0
0 1 2
0 1 0 2 1 0 1 2 2 0 2 1
2
x x x x x x x
a y b y c y
x x x x x x x x x x x x
To calculate the derivative at the very first point, then
o
x x
and we get the following
equations:
1 2 0 2 1
0 1 0 2 1 0 1 2 2 0 2 1
2
o o
x x x x x x x
To calculate the derivative at the very last point, then
2
x x
and we get the following
equations:
2 1 2 0 2 1
0 1 0 2 1 0 1 2 2 0 2 1
2
x x x x x x x
The raw data table is shown below:
RSST Data
Time Temp. Pressure Time Temp.
********** Lagrange Derivative **********
(min) (deg. C) (psia) (s) (K) a b c dT/dt -1000/T -1000/RgT (dT/dt)/(Tf-T) ln(col L)
0 12.0 309.6 0 285.15 -1.402 1.9022 -0.4885 0.01122 -3.507 -0.4218 7.017E-05 -9.565E+00
101.2 171.9 332.3 6072 445.05 72.963 -92.8646 17.2973 -2.604295 -2.247 -0.2703 2.140E+02 5.366E+00
11
Plot of temperature rate (K/s) vs. -1000/T to get onset and final temperatures
Results from above plot:
Straight line begins at: -3.273 = 305.5301 K = 32.38009 =Onset temperature
Max. rate occurs at: -2.367 = 422.4757 K = 149.3257
Final temeprature occurs at: -2.247 = 445.0378 K = 171.8878 =Final temperature
0.001
0.01
0.1
1
10
-3.500 -3.300 -3.100 -2.900 -2.700 -2.500 -2.300 -2.100
12
Once we have these results, we can use the theoretical model to compare the theoretical results
with the experiment results. This is always important!
y = 68.005x + 16.729
R² = 0.9986
-13
-11
-9
-7
-5
-3
RESULTS: OTHER RESULTS:
13
Calculating temperature from theoretical model
Time step: 0.0003
Tau x dx/dtau Time (s) T (K)
0.009 0.009468 1.111523 3848.9 306.851
0.0093 0.009802 1.115648 3855.8 306.898
0.0096 0.010136 1.119802 3862.7 306.944
0.0099 0.010472 1.123985 3869.6 306.991
0.0102 0.01081 1.128198 3876.5 307.038
14
8-7. From Equation 8-20:
n
dx
d
Differentiate with respect to
:
21
2
1
1 1 exp
1 exp 1 exp
0
n
n
n n
d x n x Bx
d
n x Bx x B Bx
The term
exp
Bx
can be dropped out of each term.
Divide by
1
n
x
:
1
1 0
1
1
1
n x B
n
Bx
n
xB
n
x
B
Which is the desired result.
Comparison of Experimental Data with Theoretical model - See calculations below
250
270
290
310
330
350
370
390
410
430
450
0 1000 2000 3000 4000 5000 6000
15
8-8. Start with Equation 8-21:
Equation 8-18
o
a
g o
E
R T
Substitute these into Equation 8-21 above:
max
2
o
g o g o a o
n R T R T E T
But the square root term in the bottom equation can be simplified to:
a
E
Then,
4
1
g F
a a
nR T
E E
T T T
16
8-9. From Equation 8-34,
S C
s P c P
S
s P
m C m C
m C
Here:
3.60 gm
2.27 J/gm K
8.773 gm
0.502 J/gm K
S
S
P
c
c
P
m
C
m
C
Substituting into Equation 8-34,
3.60 gm 2.27 J/gm K 8.773 gm 0.502 J/gm K
3.60 gm 2.27 J/gm K
1.53
This is a typical phi-factor for an ARC.
8-10. a. The maximum energy generation will occur at the maximum self-heat rate. From
Example 8-4, the max. self-heat rate is 104 K/min = 1.73 K/s. The heat release rate is
given by:
P
dQ dT
mC
dt dt
Substituting the numbers provided:
4 5
4.0 10 J/s 1.44 10 kJ/hour
dQ
Other considerations:
b. Heat capacity of water:
Not much difference.
5
P
Q mC T
17
8-11. Reactor volume = 10 m3
Specific gravity of liquid mixture = 0.97
Heat of reaction = 67.8 kJ/mol
a. For the fastest reaction, operate at the maximum self-heat rate temperature. For a 2:1
b. Molecular weights:
Methanol, CH3OH: 32
46.7 moles
x
Thus,
c. Maximum cooling duty = 30 MJ/min = 5
5.00 10 J/s
Let
m
= molar addition rate of acetic anhydride
45.1 kg/min
d. Total addition time: 4763 kg acetic anhydride
105.5 min 1.75 hours
45.1 kg/min
