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EG REACTOR DESIGN
As we are using %90 excess of water, reaction will be Pseudo
First Order reaction.
n=1CA=Cao(1-XA)
k=0,23 min-1
Conversion =0,80
Total molar flow rate in = Fti= 7111,5 kmol/h
EO=646,5 kmol/h
Water= 6465,5kmol/h
E (Activation Energy) = 110 kJ/mole
As we know the formula for the volume of the reactor
By integration,
A
AA
A
A
X
XkC
dXFV
0)1(
/0
0
)1ln(
1
/ 00 A
A
A XkC
FV
We need 0AC and 0AF ;
Total Molar Flow Rate In= 0AF = 7111,5 kmol/h
0AC = kmoles of Ethylene Oxide/Volume of the mixture
kmoles of Ethylene Oxide = 646,5 kmol/h
Now we need Volume of the mixture;
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Volume of the mixture = Total weight of the mixture/Density of
the mixture
Total weight of the mixture = 144845,91 kg/h
Now Density of the mixture;
Density of Water ( OH2 ) =]5340.2
)4713.6
4681(1[
01
00
01
02
)4870.2(
3910.4
E
EE
E
By solving the above Equation
Density of Water =
722.1
2487.0
391.4=
091.0
391.4=48.25 kmole/m
3
=48.25kmole/ m3
* 18 kg/kmole
Density of Water ( OH2 ) =868.5 kg/ m3
Density of Ethylene Oxide ( EO) =
])15.469468
1(1[2696.0
26024.0
836.1
By solving the above Equation
= 19775.126024.0
836.1
=
199.0
836.1
= 9.226 kmole/m3
= 9.226 kmole/m3
* 44 kg/kmole
Density of Ethylene Oxide ( EO) = 405.9 kg/ m3
Now
Density of the mixture =
5.868
91,144845
1163709.405
91,144845
91,28475
= (0.09*405.9) + (0.91*868.5)
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= 826,866 kg/ m3
By using Equation for Volume of the mixture, so
Volume of the mixture =
= 175,175 m3
And
0AC == 3.654 kmol/m
3
Now
)1ln(0
0
A
A
X
kC
FAV
By putting the values, we get
Volume of PFR = 20,65 m3
Residence Time:
/ oV v
= 20,65/175,175
= 7,07 min.
Length of Pipe:
Volume of reactor = V = /4 (D2L)
L/D = 200 (P.F.R)
L = 200 * D
V = 50 D3
D = (V/50 )1/3
D = (20,65/50* )1/3
D = 0.508 m
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L = 200 * 0.52 = 101,7 m
Selection of Pipe:
Nominal Pipe Size = 2 Schedule No. 80
Required Flow Area for Plug Flow = /4 (0.508)2
= 0.203 m2
Flow Area of 2, Schedule No. 80
I.D = 1.939 in=0,04925 m
O.D = 2,38 in = 0,065 m
Flow Area of one tube = 1.9048 * 10-3
m2
No. of pipes required/pass = 0.203/1.9048 * 10-3
= 107 pipes
Assume 8 passes:-
Length of each pipe = 107/8
= 13,37 m
Length of pipe = 101,7/8 =12,7 m
Total no. of pipes = 107 * 8
= 856 pipes
Thus Total pipe passes = 8
