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A simple and rapid method for calculating the positive pressure air volume in clean room

Popularity:18 Release time:2021-06-01

A simple and rapid method for calculating the positive pressure air volume in clean room

In clean room design, the first thing designers face when planning drawings is the calculation of the positive and negative pressure air supply in clean room. However, in various design manuals about clean room, the calculation of the positive pressure in clean room lists the complex calculation process. Is there a faster and easier way for engineers who build clean rooms?

Calculation method of differential pressure in clean room

Simple and fast calculation method of positive pressure air volume

A good sealed clean room, in the use of the process, the main air leakage way has the following four:

(1) air leakage between doors and Windows; (2) air leakage when opening the door; (3) air leakage in the air shower room and the transfer window; (4) indoor process exhaust air.

1, gap air leakage calculation

Calculation method 1:

V = 1.29 (delta P) 1/2

Delta V = S * n

△P: Pressure difference between clean room and clean room (Pa)

V: Wind speed through the gap (m/s)

S: Gap area (m2)

V: Leakage air volume through the gap (m3/h)

Example: Assumption conditions: room positive pressure of 20Pa, door gap length of 3.6m, window gap length of 40m, assumed gap width of 0.01m

Door gap area S1=0.01*3.6=0.072m2, window gap area S2=0.002*40=0.08m2

Leakage air volume V = s = (S1 + S2) * V * 3600 * 1.29 * (delta P) = 1/2 (0.072 + 0.08) * 3600 * 1.29 * 1/2 = 3157 m3 / hr (20)

Calculation method two:

Pressure difference method: L=0.827×A× (ΔP) 1/2×1.25=1.03375×A× (Hg P) 1/2

In the formula, L -- positive pressure air leakage volume (m3/s); 0.827 -- air leakage coefficient; A -- total effective air leakage area (m2); ΔP - pressure difference (Pa); 1.25 -- Additional coefficient of imprecision

2. The leakage air volume of the door

Assumptions: room positive pressure ΔP=20Pa, door area S3=0.9*2.00=1.8m2, wind speed V =1.29(△P)1/2=5.77m/s, opening times N =1 /hr, opening time T =5s

Leakage air volume Q=S3* V * T * N =1.8*5.77*5*1*= 51.93m3 /h

Open the door once an hour, open 5 seconds, leakage air volume 51.93m3 /h

3, the air leakage of the air shower and the transfer window

Assumption: Air Shower volume is 15m3, closed and seamless

Analysis: Take the air shower room as an example

(1) The pressure in A/S is normal 101325Pa at opening time

(1) The pressure in A/S is 101325Pa at normal pressure when closing, which remains unchanged

(2) When opening, the pressure in A/S is 101325Pa at normal pressure, which remains the same, but after stabilizing, the pressure becomes 101325+ 20Pa at the same pressure in the room

(2) The pressure in A/S is 101325+20Pa at normal pressure at the time of closing, which remains unchanged

Conclusion: Therefore, the air volume that needs to be supplemented is 15m3, and the air volume of the confined space when the space pressure changes to 20Pa

According to the ideal gas equation pV =mRT, (P pressure in Pa; V volume, unit m3; M gas mass unit kg; R gas constant is equal to 287; T gas Kelvin temperature, unit K, assuming room temperature of 25℃, Kelvin temperature of 298K)

Be air quality with m = (delta P * V)/(R * T) = (20 * 15)/(287 * 298) = 0.0035 kg = 3 l

Relatively speaking, the leakage of the air shower and the transfer box is small (under the condition of good airtight) and can be ignored.

4, process exhaust air leakage air volume

The process exhaust leakage air volume is the process exhaust air volume. The sum of the above four kinds of leakage air volume is the air supply volume needed to maintain the positive pressure of the room.

Why do we sometimes fail to achieve the ideal positive pressure?

1. Insufficient fresh air supply: recalculate fresh air demand

2, leakage is too large: find out the cause of leakage, find out the solution

3. The exhaust air volume is too large: mainly the exhaust air volume of the process exceeds the standard, and the exhaust air volume is recalculated

4. The air conditioning return air volume is too large: readjust the air volume

Application example of simple calculation method - in clean room construction for investors to save energy, reasonable positive pressure value is very important for clean room indicators to meet the standard and reduce operating costs. The engineering equipment of semiconductor plant consumes about 49% of the total power consumption, the cold engine consumes 35% of the total power consumption, and the clean room air handling unit consumes 23% of the total power consumption. The greater the positive pressure set value of the clean room, the greater the air volume of the air handling unit, the higher the load of the cold motor, and the lower the power consumption, that is, to reduce 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

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