In a well-closed clean room, the main air leakage routes are as follows:
Air leakage from efficient air supply outlets; air leakage from gaps in interior doors and windows; air leakage during door opening process; air leakage from various purification devices in air showers and transfer windows;
1, the gap leakage calculation
Calculation method one:
v=1.29 (â–³P)1/2
â–³V=S*v
â–³P: clean indoor and outdoor pressure difference (Pa)
v: wind speed (m/s) flowing through the gap
S: gap area (m2)
V: Leakage flow through the gap (m3/h)
Example: Hypothesis: The room has a positive pressure of 20pa, the length of the door seam is 3.6m, the length of the window seam is 40m, and the gap width is assumed to be 0.01m. The door gap area S1=0.01*3.6=0.072m2, and the window gap area S2=0.002*40=0.08m2. Leakage air volume V=s*v=(S1+S2)*3600*1.29*(ΔP)1/2=(0.072+0.08)*3600*1.29*(20)1/2=3157m3/hr.
Calculation method two:
Differential pressure calculation method:
L=0.827×A×(ΔP)1/2×1.25=1.03375×A×(ΔP)1/2
In the formula, L-positive air leakage (m3/s); 0.827—leakage coefficient; A— total effective air leakage area (m2); ΔP—pressure difference (Pa); 1.25—additional coefficient of impreciseness.
2, open the door to leak air volume
Assumptions: room pressure ΔP=20Pa, door area S3=0.9*2.00=1.8m2, wind speed v=1.29(△P)1/2=5.77m/s, opening frequency n=1 times/hr, open time t =5s.
The amount of leakage air Q=S3*v*t*n=1.8*5.77*5*1*=51.93m3/h.
Open the door every hour, open for 5 seconds, leak air volume 51.93m3/h.
3, the air shower room and transmission window air leakage
Assumptions: air shower volume 15m3, closed seamless. The opening and closing sequence is (1) On→ (1) Off→ (2) On→ (2) Off.
Analysis: Take air shower as an example:
(1) At the time of opening, the pressure in the A/S chamber is 101,325 Pa at normal pressure, and the pressure in the A/S chamber at the time of closing is 101,315 Pa at atmospheric pressure, which remains unchanged.
(2) At the time of opening, the pressure in the A/S chamber is 101,325 Pa, which remains unchanged, but after the pressure is stabilized, the pressure becomes 101325+20pa in the same pressure in the room, and the pressure in the A/S chamber is 101325+20Pa at the normal pressure during the off period. change.
Conclusion: Therefore, it is necessary to replenish the volume of air in a confined space where the air volume is 15m3 and the space pressure is changed to 20Pa. According to the ideal gas equation PV = mRT, (P pressure, unit Pa; V volume, unit m3; m gas mass unit kg; R gas constant equal to 287; T gas Kelvin temperature, unit K, assuming room temperature 25 °C, Kelvin The temperature is 298K). Need to add air quality m = (△ P * V) / (R * T) = (20 * 15) / (287 * 298) = 0.0035 kg = 3L.
Relatively speaking, the leakage of the air shower and transmission box is small (in the case of a good seal) and can be ignored.
4, process exhaust air leakage
Process air leakage is the amount of process exhaust. The sum of the above four leak air flows is the amount of air supply required to maintain a positive pressure in the room.
A more rapid calculation method: positive pressure ventilation times (times/hours) during cleanliness, such as Table 1 positive pressure ventilation experience data table, through which you can more quickly calculate the positive pressure air supply, clean room air leakage Cause analysis and clean room air pressure calculation method.
For example, the clean room requires a positive pressure of 20 Pa (the fourth row in the table), a clean room volume of 100 m3, and the clean room is a clean room with an outside window and good airtightness (second column in the table). Then the clean room is maintained. The amount of air supply required for positive pressure V = number of air changes 2.5*100m3=250m3 (Excluding process ventilation)
Why can't we reach the ideal positive pressure sometimes?
1. Insufficient supply of fresh air: Recalculate the demand for fresh air.
2. Excessive leakage: Find out the cause of the leak and find a solution.
3. Exhaust air volume is too high: The process exhaust air volume is excessive and recalculated.
4, air conditioning return air volume is too large: re-adjust the air volume.
Application example of simple calculation method - to save energy for the investor in the construction of clean room, a reasonable positive pressure value is very important for the compliance of various indicators of the clean room and reduce the running cost.
Semiconductor plant engineering equipment consumes about 49% of the total electricity used by the whole plant. Cold motive power accounts for 35% of engineering equipment. Cleanroom air handling units account for 23% of engineering equipment. The larger the positive pressure setting value in the clean room, the greater the air supply volume of the air handling unit, and the higher the load on the cold drive, reducing the power consumption, that is, reducing the consumption of resources and the manufacture of pollution sources. Therefore, based on the above analysis, it can be seen that the positive pressure setting in the clean room is directly related to the operating cost of the plant, the cause of the air leakage in the clean room, and the calculation method of the air pressure in the clean room.
For example, a factory mainly consists of a 150m2 Class 1000 clean room and a 100m2 Class 1 clean room. The positive pressure setting is 50Pa. Now calculate the required fresh air volume with its Class 1 clean room as an example.
Calculation conditions
The area is 100m2, the height is 3m, and the space under the raised floor is 1m. FFU is full of raised floor system. Window 9 (1mx1m), Door 1 (1mx2.1m), Open once per hour, each time for 5 seconds, one air shower, one transfer box (1m3), process air exhaust 1000CMH, airtightness of the entire clean room Good, positive pressure is set at 50Pa.
calculation process:
1, the amount of air leakage leakage
V=(S1+S2)*3600*1.29*(â–³P)1/2
(0.01*1*4*9+0.02*1*2+0.02*2.1*2)*3600*50(1/2)*1.29=4642m3/h
2, open the door to leak air volume
Q=S3*1.29*(â–³P)1/2*t*n
(2.1*1)*(50(1/2)*1.29)*5*1=109m3/h
3, air shower and transmission window is well sealed, assuming no air leakage
4, the process of exhaust capacity of 1000CMH
The amount of fresh air needed to control the positive pressure of 50Pa is 4624+109+1000=5733m3/h. If calculated based on empirical values, the air volume required to control the positive pressure of 50Pa is 5.3*100*4+1000=3120m3/h, whichever is the greater value of 5733m3 /h, the actual residual pressure valve shows a positive pressure of 65 Pa. The owner has requested to reduce the positive pressure to 35 Pa for energy saving considerations.
First, improve the FFU wind speed
way of improvement:
1, through the control computer, adjust the FFU motor speed reduction (original: 1,300rpm).
2. Observe the changes in the pressure value of the clean room (original: 65Pa).
3, continue to reduce the speed until the pressure is 45Pa.
4. Record the final speed of the FFU and observe the cleanroom pressure.
Second, to improve (MAU) wind speed
way of improvement:
1, through the control computer to adjust the MAU motor speed (original: 45Hz).
2. Observe the changes in the pressure value of the clean room (original: 45Pa).
3, continue to reduce the speed until the pressure is 35Pa.
4. Record the final speed of the MAU and observe the cleanroom pressure.
Third, the final energy-saving effect
1, reduce the FFU speed
(1) The FFU speed was reduced from 1,300 rpm to 1,100 rpm.
(2) Total savings in FFU motor power consumption: 62Kw.
2, reduce the MAU speed
(1) The pressure in the clean room is adjusted to 35Pa, and the MAU load is reduced from 45Hz to 41Hz.
(2) The total power consumption of MAU motor can be saved: 4kW.
(3) Reduce the positive pressure in the clean room while reducing the supply of outdoor fresh air (MA) to save energy.
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