[1]
A detailed overview of the steps mentioned
Step 1
Convert all the flow rates from kg/hr to kmol/hr to establish the temperature of the varying conversion rates.
Step 2
Derivation of the heat balance around the reactor based on adiabatic operation where
Q = Heat of Reaction + Sensible Heat Term, and since the reaction is adiabatic Q equals zero
[2]
[3]
[4]
[5]
Where
T = Temperature of reactor product
T0 = initial temperature of reactor
FA0 = flow rate of EB into reactor
ΔHR = Heat of reaction
XA = Conversion of EB
ΣFi = total flow rate of substance
Cpi = Heat capacity of reaction
Step 3
To calculate the rate constant for each temperature, by using the expression from
Wenner and Dybal (1948)
Log10k = [6]
Where is in Kelvin
However k is in Imperial units of (lbmol styrene h-1 atm-1 lbcatalyst-1) and to convert to SI units use the following conversion expression
kSIUnit =2.74×10-6 k (kmol s-1 kPa-1 kgcat-1)
Step 4
To determine the equilibrium constant of each XA we use the data given in the table and then plot a graph to determine an equation.
Step 5
To obtain the relationship between partial pressures and conversion rates of the reaction occurring which has a 1:1:1 ratio.
C6H5CH2CH3 C6H5CH=CH2 + H2
The ratio is 1:1:1
Step 6
To obtain the global rate the following equation is used by substituting the variables calculated in the steps above.
[7]
Step 7
To obtain (-rA)bed for each XA value the following equation is used
[8]
With the following data that was given
is 0.45 and is 1440 kgm-3
Step 8
To calculate the bed volume and height by using Simpson’s rule as the area under the graph represents the Vbed.
Where FA0 is the inlet molar flow rate of ethyl benzene of one reactor so since there are five reactors it is required to divide the flow rate of ethyl benzene by 5 as seen below
The graph plotted was (FA0/-rAbed) against XA which is shown below.
Area under the curve is obtained by Simpson’s rule or trapezoidal rule.
s = width of each strip = 0.01
First + Last = sum of the first and last ordinates
4(even) = 4(the sum of the even-numbered ordinates)
2(remainder) = 2(the sum of the remaining odd-numbered ordinates)
Thus the area under the curve is calculated to be ≈ 6.92 m2
The height of each bed is the bed volume divided by the area, where area is . I’ve worked out the area from the given diameter which is 1.5 m. The area is 1.8m2.
Therefore the bed height is 6.92/1.77= 3.91 m.
