Your location：1.1 Dust removal efficiency
According to the design requirements:
The dust concentration of the gas is 1g/m3, while the dust concentration required to be discharged is less than 100mg/m3
Then the dedusting efficiency required to be achieved by the deduster is:
η= （1-0.1） × 100%/1=90%
1.2 Selection of dust remover
The physical and chemical characteristics of dust is an important basis for determining the dust removal scheme. In order to correctly select, install, apply and maintain the dust removal equipment, it is necessary to understand the physical and chemical characteristics of the dust to be treated. In the physical characteristics of dust, the particle size is the most critical characteristic data. The applicable relationship between general dust remover and dust particle size is shown in the following table:
| Type of dust remover | Particle size/um | |
| Gravity settling chamber | >50 | |
| Inertial precipitator | 20-50 | |
| Cyclone dust remover | General specifications | 20-200 |
| Advanced specifications | 5-30 | |
| Wet dust remover | Water bath dust remover | 1-10 |
| Venturi dust remover | 0.5-10 | |
| Bag filter | 0.2-10 | |
| Electrostatic precipitator | 0.2-5 | |
1.3 Advantages and disadvantages of gravity settling chamber and selection of cyclone dust collector
1 Advantages and disadvantages of gravity settling chamber
Advantages: simple structure, less investment, easy maintenance and management, small pressure loss (50-130Pa).
Disadvantages: large floor area, low dedusting efficiency (only used as the pre dedusting device of high-efficiency deduster to remove large and heavy particles)
2. Type selection of cyclone dust removal
General experience method
1) Calculation of required dedusting efficiency
2) . Select the type of dust remover
It shall be determined according to smoke characteristics such as dust concentration, particle size distribution and density, and factors such as dust removal requirements, allowable resistance and manufacturing conditions
3) Determine the inlet gas velocity according to the allowable pressure drop, or take it as 10~25 m/s, which can be obtained from the relevant calculation of industrial ventilation and instrument selection
4)、Calculate the inlet area A, inlet width b and height h of the precipitator according to the treated gas volume and inlet wind speed
5)、Determine the geometric dimensions of each part
The scale of cyclone dust removal is shown in the following table:
Section II Calculation of exhaust air volume of ventilation and dust removal system in spraying workshop
According to the design requirements, it is proposed to adopt the local dedusting upper suction gas collecting hood
Control wind speed Vx=0.4m/s
Size of the operating table: 0.8m long and 0.5m wide
The distance from the operating table to the hood is H=0.4m
The size of the hood: long side A=0.8+0.4 × zero point four × 2=1.12m
Short side B=0.5+0.4 × zero point four × 2=0.82m
Perimeter of hood: P=2 × （1.12+0.82）=3.88m
Then the exhaust air volume required for a single dust producing point is:
L=KPHVx=1.4 × three point eight eight × zero point four × 0.4=0.869 m3/s
As there are 12 operation consoles in the spraying workshop, there are 12 dust producing points, so the total exhaust air volume required by the system is:
L total=12 L=12 × 0.869=10.428m3/s
Considering the air leakage of the precipitator and air duct, the calculated air volume of pipe sections 10 and 11, that is, the actual total air volume required by the system is:
L total, real=1.05 × 10.428=10.949 m3/s
Section III Length Calculation of Gravity Settlement Chamber
When t=20 ℃: air density u=1.79 × 10-5Pa.s， ρ g=1.205kg/m3,
Dust particle size d p=100um, dust density ρ p=2700kg/m3,
The width of gravity settlement chamber is set as B=5m, and the velocity in the chamber is v0=1.12m/s
Then the height of the settlement chamber is: H=LTotal/B v0=10.43/(1.12 × 5)= 1.86m
Considering the actual situation, the height H of the settlement chamber is taken as 2m.
The Reynolds coefficient of the settling chamber is:
Rep= v0 ρ g d p/u=1.12 × one point two zero five × one hundred × 10-6/1.79 × 10-5=7.54
Settlement speed:
vs=0.25（ ρ pg）0.67d p（u ρ p）-0.33
=0.25 × （2700 × 9.8）0.67 × one hundred × 10-6 × （1.79 × 10-5 × 2700）-0.33=0.62 m/s
Then the settling chamber length:
A=H v0/ vs=2 × 1.12/0.62=3.6m
The length of settlement chamber is 5m.
Section IV Hydraulic Calculation of Ventilation Duct
As the design requires that 12 operating consoles should be arranged symmetrically every 3 in a row, so one group in every 2 rows should be arranged symmetrically on both sides of the precipitator, so only one group of hydraulic calculation is needed.
The following are the hydraulic calculation steps for ventilation ducts:
(1) Draw the system diagram of ventilation and dust removal system of the spraying workshop according to the design requirements.
(2) Number each pipe segment, and mark the length of each pipe segment and the exhaust air volume of each exhaust point.
(3) Select the most unfavorable loop of the system, which is 1-2-3-7-8-gravity settling chamber deduster cyclone deduster (ordinary type) - 10-fan-11.
(4) According to the air volume of each pipe section and the selected flow rate, determine the section size and unit length friction resistance of each pipe section on the most unfavorable loop.
(5) According to Table 6-4 above, when conveying air containing steel dust, the minimum wind speed in the air duct is 13m/s for vertical air duct and 15m/s for horizontal air duct.
For Segment 1
According to L1=0.869 m3/s, v1=15m/s, the pipe diameter and friction resistance per unit length are found in Appendix 9. The selected pipe diameter shall conform to the uniform standard of ventilation pipe in Appendix 11 as far as possible.
Then D1=250mm Rm1=13.5Pa/m
Similarly, the pipe diameter and specific friction of other pipe sections can be found. See the water conservancy calculation table for details.
(6) Check Appendix 9 to determine the local resistance coefficient of each pipe section.
1) Pipe section 1
Rectangular umbrella cover: a=40: ξ＝ zero point one three
90 ° elbow (R/D=1.5): 1, ξ= zero point one seven
Then:
ξ 1=0.13+0.17=0.30
Z1= ξ 1v12 ρ g /2=0.30 × one hundred and fifty-two × 1.205/2=40.67 Pa
2) Spool 2
Rectangular umbrella cover: a=40 °: ξ＝ zero point one three
DC tee (1-2)
F1/ F2=（250/360）2 =0.48
L1/ L2=0.869/1.738=0.5
It is found that: ξ=- zero point three
Then:
ξ 2=0.13+（-0.3）= -0.17
Z2= ξ 2v22 ρ g /2=-0.17 × one hundred and fifty-two × 1.205/2=-19.60 Pa
3) Spool 3
Rectangular umbrella cover: a=40 °: ξ＝ zero point one three
DC tee (2-3)
F2/ F3=（360/400）2 =0.81
L2/ L3=1.738/2.607=0.67
It is found that: ξ＝ zero point zero six
90 ° elbow (R/D=1.5): 1 ξ= zero point one seven
Then:
ξ 3=0.13+0.06+0.17=0.36
Z3= ξ 3v32 ρ g /2=-0.36 × one hundred and fifty-two × 1.205/2=41.51 Pa
4) Due to symmetry, relevant calculations of pipe section 4 and pipe section 1, pipe section 2 and pipe section 5 are the same
5) Spool 6
Rectangular umbrella cover: a=40 °: ξ＝ zero point one three
DC tee (2-3)
F2/ F3=（360/400）2 =0.81
L2/ L3=1.738/2.607=0.67
Find out ξ＝ zero point zero six
Then:
ξ 6=0.13+0.06 =0.19
Z3= ξ 3v32 ρ g /2=-0.19 × one hundred and fifty-two × 1.205/2=21.91 Pa
6) Spool 7
DC Tee (3-7)
F6/ F7=（400/450）2 =0.80
L6/ L7=2.607/5.214=0.5
Find out ξ＝ zero point two one
Then:
ξ 7=0.21
Z7= ξ 7v72 ρ g /2=0.21 × one hundred and fifty-two × 1.205/2=23.06 Pa
7) Spool 8
Circular tee (confluence) (7-8)
L7/ L8=5.214/10.428=0.5
Find out ξ＝ zero point zero three
90 ° elbow (R/D=1.5): 1 ξ= zero point one seven
Reducer pipe (reducer) at the inlet of dust remover of gravity settling chamber
The inlet size of the precipitator is 300mm × 800mm, reducing pipe length is 500mm
tana=（1000-800）/（2 × 500）=0.2
Then a=11.3 °
Then ξ＝ zero point one zero
Then:
ξ 8=0.03+0.10+0.17=0.30
Z8= ξ 8v82 ρ g /2=0.30 × one hundred and thirty-two × 1.205/2=30.55Pa
8) Spool 9
Reducer at the outlet of gravity precipitator (reducer)
The outlet size of the precipitator is 300mm × 800mm, reducing pipe length 400mm
tana=（950-800）/（2 × 400）=0.19
Then a=10.62 °
Then ξ＝ zero point one zero
Inlet reducer (reducer) of cyclone dust remover (common type)
The inlet size of the precipitator is 300mm × 800mm, reducing pipe length is 500mm
tana=（950-800）/（2 × 500）=0.15
Then a=8.5 °
Then ξ＝ zero point one zero
Then:
ξ 9=0.1 × 2=0.2
Z8= ξ 9v92 ρ g /2=0.2 × one hundred and fifty-two × 1.205/2=23.06 Pa
9) Spool 10
Cyclone dust remover (ordinary type) outlet reducer (reducer)
The outlet size of the precipitator is 300mm × 800mm, reducing pipe length 400mm
tana=（1000-800）/（2 × 400）=0.25
Then a=14.04 °
Then ξ＝ zero point one zero
90 ° elbow (R/D=1.5): 2 ξ= two × 0.17=0.34
Fan inlet reducer
First, it is proposed to select a fan. The inlet diameter of the fan D1=500mm, and the length of the reducer l=300mm
tana=（1000-500）/（2 × 300）=0.83
Then a=39.7 °
Then ξ＝ zero point one zero
Then:
ξ 10=0.1 × 2+0.34=0.54
Z10= ξ 10v102 ρ g /2=0.54 × one hundred and thirty-two × 1.205/2=54.98 Pa
10) Spool 11
Fan outlet reducer
Fan outlet size 420mm × 315mm, diameter D2=420mm
tana=（1000-420）/（2 × 315）=0.92
Then a=42.6 °
Then ξ＝ zero point one zero
Umbrella hood with diffuser (h/D=0.5) ξ＝ zero point six zero
Then:
ξ11=0.1+0.60=0.70
Z10=ξ11v112ρg /2=0.70×132×1.205/2=71.28 Pa
Pipeline hydraulic calculation table:
| Pipe No |
flow （m3/s） |
lengthL(m) |
pipe diameter D(mm) |
Current Speed V(m/s) |
Local resistance coefficient ∑ξ |
Local resistance Z(Pa) |
Friction resistance per unit length Rm(Pa/m) |
Frictional resistance Rml(Pa) |
Pipe resistance Rml+Z (Pa) |
remarks |
| 1 | 0.869 | 5 | 250 | 15 | 0.17 | 40.67 | 13.50 | 67.50 | 108.17 | |
| 2 | 1.738 | 2 | 360 | 15 | -0.17 | -19.60 | 9.00 | 18.00 | -1.60 | |
| 3 | 2.607 | 6 | 400 | 15 | 0.36 | 41.51 | 8.00 | 48.00 | 89.51 | |
| 7 | 5.214 | 4 | 450 | 15 | 0.21 | 23.06 | 6.50 | 26.00 | 49.06 | |
| 8 | 10.428 | 5 | 1000 | 13 | 0.30 | 30.55 | 1.50 | 7.50 | 38.05 | |
| 9 | 10.428 | 2 | 950 | 15 | 0.2 | 23.06 | 2.80 | 5.60 | 28.66 | |
| 10 | 10.949 | 4 | 1000 | 13 | 0.54 | 54.98 | 1.30 | 5.20 | 57.18 | |
| 11 | 10.949 | 8 | 1000 | 13 | 0.70 | 71.28 | 1.30 | 5.20 | 76.48 | |
| 4 | 0.869 | 5 | 250 | 15 | 0.17 | 40.67 | 13.50 | 67.50 | 108.17 | |
| 5 | 1.738 | 2 | 360 | 15 | -0.17 | -19.60 | 9.00 | 18.00 | -1.6 | |
| 6 | 2.607 | 2 | 400 | 15 | 0.19 | 21.91 | 6.50 | 13.00 | 34.91 | Resistance imbalance |
| 6 | 2.607 | 280 | 16.8 | 38.85 | Resistance imbalance | |||||
| duster1 | 1200 | |||||||||
| duster2 | 1200 |
It can be seen from the symmetry that the junction points A, B and D are in equilibrium. Therefore, resistance balancing is not required. The convergence point C is analyzed.
Convergence point C
△P3=89.51pa △P6=34.91pa
Then:
（△P3-△P1）/△P3 =(89.51-34.91)/89.51=61%>10%
In order to balance the resistance of pipeline 3 and 6, it is necessary to change the diameter of pipeline 6 to increase its resistance
Then:
D6’=D3(△P6/△P6’)0.225=400 × (34.91/89.51)0.225=323.6mm
According to the uniform specification of the ventilation pipe, D6,=320mm
At this time, the corresponding resistance is:
△P6，，=34.91（400/320)0.225=36.71pa
（△P3—△P6，，）/△P3=14.4%>10%
At this time, it is still not in the balance state, so continue to reduce the pipe diameter of pipe 6 to 280mm. After calculation, it is in the balance state. During operation, it is supplemented with valve adjustment to eliminate the imbalance.