Column designing Numerical

1. PROCESS EQUIPMENT DESIGNING
TERM PAPER
Name: Shreem Rasaniya
Roll No. : 19BCH059
College: Institute of Technology, Nirma University

2. Q: A fluidized bed is used in the production of aniline by the hydrogenation of nitrobenzene. Single-
stage cyclones, followed by candle filters, are used to remove fines from the gases leaving the
fluidized bed. The reactor operates at a temperature 270oC and a pressure of 2.5 bar. The reactor
diameter is 10 m. Hydrogen is used in large excess in the reaction, and for the purposes of this
exercise the properties of the gas may be taken as those of hydrogen at the reactor conditions. The
density of the catalyst particles is 1800 kg/m3.
• The estimated particle size distribution of the fines is:
Particle size, m 50 40 30 20 10 5 2
Percentage by 100 70 40 20 10 5 2
weight less than
• A 70 per cent recovery of the solids is required in the cyclones.
• For a gas flow rate of 100,000 m3/h, at the reactor conditions, determine how many cyclones
operating in parallel are need and design a suitable cyclone. Estimate the size distribution of the
particles entering the filters.

3. •Data:
• Temperature : 270oC = 543k
• Flowrate of the gas : 1,00,000 m3/h = 1,00,000/3600 = 27.778 m3/s
• ∆𝜌1 = 2000 𝑘𝑔/𝑚3
• ∆𝜌2 = 1800 𝑘𝑔/𝑚3 (𝜌2 (particle) = 1800 kg/m3 and
𝜌2 (gas) =
𝑃𝑀
𝑅𝑇
=
2
8.314472∗10−5 ∗
2.5
543
= 0.11074 kg/m3)
• 𝜇1 = 0.018𝑐𝑝 𝜇2 = 0.0135𝑐𝑝 (it is found from graph)
• We use High throughput cyclone

4. https://images.app.goo.gl/Pje5AgEC6ZR4NoT58
0.0135

5. • Area of inlet at 15m/s = Flowrate/ velocity = 27.778/15 = 1.851 m2
• Now Calculate Column Diameter:
For high throughput cyclone:
0.375Dc*0.75Dc = Area = 1.851
Now, Dc =
1.851
0.375∗0.75
= 2.5654 𝑚

6. • Now calculate the scaling factor:
𝑑2
𝑑1
=
𝐷𝑐2
𝐷𝑐1
3
∗
𝑄1
𝑄2
∗
∆𝜌1
∆𝜌2
∗
𝜇2
𝜇1
0.5 (put all the values)
𝑑2
𝑑1
=
2.5654
0.203
3
∗
669
100000
∗
2000
1800
∗
0.0135
0.018
0.5
= 3.3542

7. • Performance of the cyclone:
• We got 64.8 % efficiency which is not near to 70, so Now we have to increase the velocity and make
calculation at 20 m/s.
min max %range mean dia mean dia/ scale factor efficincy at scaled dia
total collected
particulate
50 more 0 50 14.906 85 0
40 50 30 45 13.41 82 24.6
30 40 30 35 10.43 72 21.6
20 30 20 25 7.45 60 12
10 20 10 15 4.47 48 4.8
5 10 5 7.5 2.235 30 1.5
2 5 3 3.5 1.0430 10 0.3
0 2 2 1 0.298 0 0
64.8

8. • Now we calculate at 20 m/s.
• Area of inlet at 20 m/s = Flowrate/ velocity = 27.778/25 = 1.3889 m2
• Now Calculate Column Diameter:
For high throughput cyclone:
0.375Dc*0.75Dc = Area = 1.3889
Now, Dc =
1.3889
0.375∗0.75
= 2.222 𝑚

9. • Now calculate the scaling factor:
𝑑2
𝑑1
=
𝐷𝑐2
𝐷𝑐1
3
∗
𝑄1
𝑄2
∗
∆𝜌1
∆𝜌2
∗
𝜇2
𝜇1
0.5 (put all the values)
𝑑2
𝑑1
=
2.222
0.203
3
∗
669
100000
∗
2000
1800
∗
0.0135
0.018
0.5
= 2.7043

10. • Performance of the cyclone:
• In this case we got 70.65% which is nearer to our requirement.
min max %range mean dia mean dia/ scale factor efficincy at scaled dia
total collected
particulate
50 more 0 50 18.48887116 90 0
40 50 30 45 16.63998404 85 25.5
30 40 30 35 12.94220981 80 24
20 30 20 25 9.244435579 70 14
10 20 10 15 5.546661347 50 5
5 10 5 7.5 2.773330674 35 1.75
2 5 3 3.5 1.294220981 10 0.3
0 2 2 1 0.369777423 5 0.1
70.65

11. • Now we calculate pressure drop:
Equation for Pressure drop:
∆𝑃 =
𝜌𝑓
203
𝑢1
2
∗ 1 + 2∅2 2𝑟𝑡
𝑟𝑒
− 1 + 2𝑢2
2
…1
Here u2 is exit velocity

12. Area at exit(Ae) = 3.14*(0.75Dc)2 /4 = 3.14*(0.75*2.222)2 /4 = 2.1805
m2
Exit velocity(u2) = Flowrate/ Ae = 27.778/2.1805 = 12.73885 m/s
rt =
𝐷𝑐
2
+
0.375𝐷𝑐
2
=
2.222
2
+
0.375∗2.222
2
= 1.52778 m
re =
0.75𝐷𝑐
2
=
0.75∗2.222
2
= 0.8333 m

13. As = 3.14*Dc*L = 4*3.14*Dc2 (here L= 4Dc)
So that, As = 3.14 *4*(2.222)2 = 62.05615 m2
Now ,
𝜑 = 𝑓𝑐
𝐴𝑠
𝐴1
= 0.005
62.05615
1.3889
= 0.2234
rt/re = 1.52778/0.8333 = 1.8333
So, that from this graph ∅ = 1.3
1.3

14. • Put all the values in the equation 1
• ∆𝑃 =
0.11074
203
20 2 ∗ 1 + 2(1.3)2 2 ∗ 1.8333 − 1 +

15. THANK YOU