CHEMICAL REACTION ENGINEERING HOMEWORK 01 JUAN SEBASTIAN SUAREZ OSPINA GUSTAVO ALBERTO BARRERA SERNA EXERCISE 1 The irreversible gas phase nonelementary reaction:
is to be carried out isothermally in a constant pressure Batch reactor. The feed is at a temperature of 227°C, a pressure of 1013kPa, and its composition is 33.3% A and 66.7% B. Laboratory data taken under identical conditions are as follows: 3
-rA (mol/dm .seg)*10 X
3
0.010 0.0
0.005 0.2
0.002 0.4
0.001 0.6
SOLUTION: Sketch:
Assumptions: The temperature and pressure are constant during the process. The A specie is the limiting reactant. The gas phase is ideal. The inlet molar flow is constant.
The inlet concentration is given by the ideal gas equation:
In concentration terms:
Substituting:
a. Estimate the volume of a plug flow reactor required to achieve 30% conversion of A for an entering 3 volumetric flow rate of 2.0m /min.
The rate law data are: 3
-rA (mol/dm .seg)*10 X
3
0.010 0.0
0.005 0.2
0.002 0.4
0.001 0.6
Now, shall calculate de reciprocal rate data and then plot versus conversion. 3
1/-rA (dm .seg/mol)*10
5
1.0
2.0
5.0
10.0
The plot is shown in the follow figure:
RECIPROCAL RATE 1.20E+06 1.00E+06
) l o m8.00E+05 / g e s . 3
6.00E+05
m d ( A4.00E+05 r / 1
2.00E+05 0.00E+00 0
0.1
0.2
0.3
0.4
0.5
0.6
X
The moles balance for a Plug Flow Reactor is given by:
∫ If the molar flow of the specie A is constant, it can leave the integral sign as follows:
∫ () The volume for a PFR is the area under curve. To realize de operation is useful to use a Simpson´s Rule. In this case for seven points: ∑ ∑ ∫ [ ]
The data to calculate the volume are shown in the follow table: 3
1/-rA (dm .seg/mol)*10 X
5
1.00 0.00
1.27 0.05
1.52 0.10
1.77 0.15
2.00 0.20
2.80 0.25
3.51 0.30
Substituting these values on a Simpson´s Rule:
∫ () ∑ ∑
∫ () ( ) ∫ () The PFR volume is: ∫ () ∫ () Substituting the values:
b. Estimate the volume of a CSTR required to take the effluent from the PFR above and achieve 50% total conversion.
For a CSTR, the moles balance is:
() This volume is the area of the rectangle, this has a height: the reciprocal rate and the conversion difference its base.
()
The plot of the situation is show in the follow figure:
RECIPROCAL RATE 1.20E+06 1.00E+06
) l o m8.00E+05 / g e s . 3
6.00E+05
m d ( A4.00E+05 r / 1
2.00E+05 0.00E+00 0
0.1
0.2
0.3
0.4
0.5
0.6
X
c. What is the total volume of the two reactors? The total volume is the sum of the PFR and CSTR volume:
d. What is the volume of a single plug flow reactor necessary to achieve 60% conversion? 80% conversion?
For a PFR the moles balance is:
∫ () The calculate procedure is similar at the (a) item. In this case is necessary to use more data and then apply the Simpson´s Rule. The data are shown in the follow table: 3
5
1/-rA (dm .seg/mol)*10 X 3 5 1/-rA (dm .seg/mol)*10 X
1.00 0.00 4.25 0.35
1.27 0.05 5.00 0.40
1.52 0.10 6.30 0.45
1.77 0.15 7.50 0.50
2.00 0.20 8.76 0.55
2.80 0.25 10.0 0.60
3.51 0.30 -----
RECIPROCAL RATE 1.20E+06 1.00E+06
) l o m8.00E+05 / g e s . 3
6.00E+05
m d ( A4.00E+05 r / 1
2.00E+05 0.00E+00 0
0.1
0.2
0.3
0.4
0.5
0.6
X
For thirteen points the Simpson´s Rule is:
∫ () ∑ ∑ Replacing the values:
∫ () ( )∑ ∑ ∫ () Then, the volume is: e. What is the volume of a single CSTR necessary to achieve 50% conversion?
For a CSTR:
()
In a CSTR the rate reaction is the exit value. The volume of a CSTR is the area of a rectangle. The plot necessary to obtain the volume is shown in the follow figure. Then:
f. What is the volume of a second CSTR to raise the conversion from 50% to 60%?
The procedure is the same that (e) item. In the plot are both situations 50 and 60% of conversion.
Now, for a 60% conversion in a second CSTR:
() () RECIPROCAL RATE 1.20E+06 1.00E+06
) l o m / 8.00E+05 g e s . 3
6.00E+05
m d ( A4.00E+05 r / 1
2.00E+05 0.00E+00 0
0.1
0.2
0.3 X
0.4
0.5
0.6
g. Plot the rate of reaction and conversion as a function of PFR volume.
The mole balance for a PFR is given by:
∫ ()
The integral sign shall calculate by Simpson’s Rule. For example, when conversion is 0.1:
∫ () ∑ ∑ ( ) ∫ () ∫ ()
The volume is:
RECIPROCAL RATE 1.20E+06 1.00E+06
) l o m / 8.00E+05 g e s . 3
6.00E+05
m d ( A4.00E+05 r / 1
2.00E+05 0.00E+00 0
0.1
0.2
0.3
0.4
0.5
0.6
X
X 5 1/-rA(dm .seg/mol)*10 3 -5 -rA(mol /dm .seg)*10 3 V (dm ) 3
0.0 1.00 1.00 0
0.1 1.52 0.66 28530
0.2 2.00 0.50 75150
0.3 3.51 0.29 147150
0.4 5.00 0.20 258390
0.5 7.50 0.13 422415
0.6 10.00 0.10 653220
The reaction rate and conversion as a function of volume are shown in the follows figures:
RATE OF REACTION 1.00E-05 ) 8.00E-06 g e s . 3
6.00E-06
m d / l o 4.00E-06 m ( A r -
2.00E-06
0.00E+00 0
100000 200000 300000 400000 500000 600000 700000 V (dm3)
CONVERSION 0.6 0.5 0.4 X 0.3
0.2 0.1 0 0
100000
200000
300000
400000
500000
600000
700000
V (dm3)
The plots show us opposites behaviors, the rate of reaction and volume are variables inversely proportional while the conversion and volume are directly proportional. If want to achieve longer conversion the volume reactor increases until a constant value. The reactor size is longer with the decrease of the rate of reaction because the reactants stay more times no reacting. h. Give a critique of the answers to this problem.
The use of Simpson´s Rule is important and useful in the PFR because the polynomial fit isn’t an easy work. This method was applied in the items (a, d, g). For CSTR the principle is calculating the rectangle area resulting of the conversion vs reciprocal rate plot as was made in the (b, e, f) items.
Because the method used the answers are logical, the CSTR volume is longer than a PFR because for it the volume is given by the area under curve while a CSTR is the rectangle area. Find the volume of the PFR necessary to achieve an 80% conversion can be complicated, because don’t have rate of reaction data for a range 60- 80%; besides interpolate can be dangerous because can’t assure that rate of reaction will have the same behavior.
The volume of the CSTR2 is smaller than CSTR1, is logic, because the conversion range for the CSTR1 is 0,5 while for the CSTR2 is 0,1. Analyzing the design equation for a CSTR’s in series the volume is a function of the conversion difference. EXERCISE 2 3 3 Pure A is fed at a volume flow rate of 1000dm /h and at a concentration of 0,005lbmol/ft to an existing a 3 CSTR, which is connected in series to an existing tubular reactor. If the volume of the CSTR is 1200dm and 3 the tubular reactor volume is 600dm , what are the intermediate and final conversions that can be achieved with the existing system? The reciprocal rate is plotted in figure as a function of conversion for the conditions at which the reaction is to be carried out.
SOLUTION:
Moles balance for a CSTR:
()
To obtain the conversion achieved in the CSTR shall develop an iterative procedure. The first step is to suppose conversion values and then calculate F A0/-rA. The second step is read at the same conversion value this ratio and compares these. If the ratios are the same, this value is correct else suppose a new conversion value. The plot given can be better seem in follow figure:
4000
3000
) m d ( A 2000 r / 0 A F
1000
0 0
0.1
0.2
0.3
0.4
0.5
0.6
X
Now:
The table and plot necessaries to find X 1 value are:
X1 supp. 0.2 0.4 0.6 0.5 0.58
FA0/-rA calc. 6000 3000 2000 2400 2069
FA0/-rA read. 1100 1400 2100 1700 2000
Difference 4900 1600 -100 700 69
0.7
0.8
The range of conversion is small, and then can interpolate these values to obtain a difference equal to zero. So the conversion is:
∫ ∫ ()
The PFR volume is given by:
This expression means that the PFR volume is the area under curve. In this case the area is a trapezoid, so the conversion achieved in the device shall be calculated iterating the area and compare with the volume given. The trapezoid area is given by:
Where: B and b are the major and minor bases respectively and h is the height. In this situation b is a constant value and, B and h are variables. Suppose that PFR conversion is 0.7, so:
()
Then, need to read F A0/-rA at this conversion value:
The area is:
The iterative procedure values and plot are shown to follow: X2 supp. 0.7 0.76 0.78 0.79
h 0.112 0.172 0.192 0.202
3
B (dm ) 3000 3600 3800 3900
Then, the conversion achieved in the PFR is:
The plot is shown in the next page.
Atrapezoid 282.24 485.04 560.64 599.94
difference 317.76 114.96 39.36 0.06
EXERCISE 3 According to you, what are the most important equations that were shown in chapters one and two? Describe the restrictions and conditions of their use. SOLUTION:
The most important equations are the continuous flow: CSTR and PFR design equations, because those reactors achieve higher conversion than Batch reactors. PFR The design equation is given by:
∫ ∫ () - This reactor operates to steady state. - The conversion depends of the position. CSTR The design equation is given by:
() - This reactor is a dispositive that operates to steady state. - Shall guarantee the perfect mixed condition, to achieve those characteristics inside the tank can be the same in the outlet. - A good agitation to do the temperature and concentration constants, then rate of reaction also. - It’s commonly used to liquid phase reaction.
